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**Unformatted text preview: **Algorithms
Department of Computer Science
University of Illinois at Urbana-Champaign
Instructor: Jeff Erickson Teaching Assistants:
• Spring 1999: Mitch Harris and Shripad Thite
• Summer 1999 (IMCS): Mitch Harris
• Summer 2000 (IMCS): Mitch Harris
• Fall 2000: Chris Neihengen, Ekta Manaktala, and Nick Hurlburt
• Spring 2001: Brian Ensink, Chris Neihengen, and Nick Hurlburt
• Summer 2001 (I2CS): Asha Seetharam and Dan Bullok
• Fall 2002: Erin Wolf, Gio Kao, Kevin Small, Michael Bond, Rishi Talreja, Rob McCann, and
Yasutaka Furakawa
• Spring 2004: Dan Cranston, Johnathon Fischer, Kevin Milans, and Lan Chen
• Fall 2005: Erin Chambers, Igor Gammer, and Aditya Ramani
• Fall 2006: Dan Cranston, Nitish Korula, and Kevin Milans
• Spring 2007: Kevin Milans
• Fall 2008: Reza Zamani-Nasab
• Spring 2009: Alina Ene, Ben Moseley, and Amir Nayyeri c Copyright 1999–2009 Jeff Erickson. Last update August 16, 2009. This work may be freely copied and distributed, either electronically or on paper.
It may not be sold for more than the actual cost of reproduction, storage, or transmittal.
This work is licensed under a Creative Commons Attribution-NonCommercial-Share Alike 3.0 United States License.
For license details, see http://creativecommons.org/licenses/by-nc-sa/3.0/us/.
For the most recent edition of this work, see http://www.uiuc.edu/~jeffe/teaching/algorithms/. Shall I tell you, my friend, how you will come to understand it?
Go and write a book on it.
— Henry Home, Lord Kames (1696–1782), to Sir Gilbert Elliot
You know, I could write a book.
And this book would be thick enough to stun an ox.
— Laurie Anderson, “Let X=X”, Big Science (1982)
I’m writing a book. I’ve got the page numbers done,
so now I just have to ﬁll in the rest.
— Stephen Wright About These Notes
This course packet includes lecture notes, homework questions, and exam questions from algorithms
courses I taught at the University of Illinois at Urbana-Champaign in Spring 1999, Fall 2000, Spring
2001, Fall 2002, Spring 2004, Fall 2005, Fall 2006, Spring 2007, Fall 2008, and Spring 2009. These
lecture notes and my videotaped lectures were also offered over the web in Summer 1999, Summer
2000, Summer 2001, Fall 2002, and Fall 2005 as part of the UIUC computer science department’s online
master’s program. Lecture notes were posted to the course web site a few days (on average) after each
lecture. Homeworks, exams, and solutions were also distributed over the web.
I wrote a signiﬁcant fraction of these lecture notes in Spring 1999; I revise them and add a few new
notes every time I teach the course. The recurrences ‘pre-lecture’ is very loosely based on a handout
written by Ari Trachtenberg, based on a paper by George Lueker, from an earlier semester (Fall 1998?)
taught by Ed Reingold, but I have essentially rewritten it from scratch.
Most (but not all) of the exercises at the end of each lecture note have been used at least once in
a homework assignment, discussion section, or exam. You can also ﬁnd a near-complete collection of
homeworks and exams from past semesters of my class online at http://www.uiuc.edu/~jeffe/teaching/
algorithms/. A large fraction of these exercises were contributed by some amazing teaching assistants:
Aditya Ramani, Alina Ene, Amir Nayyeri, Asha Seetharam, Ben Moseley, Brian Ensink,
Chris Neihengen, Dan Bullok, Dan Cranston, Johnathon Fischer, Ekta Manaktala, Erin Wolf
Chambers, Igor Gammer, Gio Kao, Kevin Milans, Kevin Small, Lan Chen, Michael Bond,
Mitch Harris, Nick Hurlburt, Nitish Korula, Reza Zamani-Nasab, Rishi Talreja, Rob McCann,
Shripad Thite, and Yasu Furakawa.
Stars indicate more challenging problems; many of these appeared on qualifying exams for the
algorithms PhD stduents at UIUC. A small number of really hard problems are marked with a larger
star; one or two open problems are indicated by enormous stars.
Please do not ask me for solutions to the exercises. If you’re a student, seeing the solution will rob
you of the experience of solving the problem yourself, which is the only way to learn the material. If
you’re an instructor, you shouldn’t assign problems that you can’t solve yourself! (I do not always follow
my own advice; some of these problems have serious bugs.) Prerequisites
For the most part, these notes assume that the reader has mastered the material covered in the
ﬁrst two years of a typical undergraduate computer science curriculum. (Mastery is not the same
thing as ‘exposure’ or ‘a good grade’; this is why I start every semester with Homework Zero.) Speciﬁc
prerequisites include:
• Proof techniques: direct proof, indirect proof, proof by contradiction, combinatorial proof, and
induction (including its “strong”, “structural”, and “recursive” forms). Lecture 0 requires induction,
and whenever Lecture n − 1 requires induction, so does Lecture n.
• Discrete mathematics: Boolean algebra, predicate logic, sets, functions, relations, recursive
deﬁnitions, trees (as abstract objects, not just data structures), graphs
• Elementary discrete probability: expectation, linearity of expectation, independence
• Iterative programming concepts: variables, conditionals, iteration, subroutines, indirection (addresses/pointers/references), recursion. Programming experience in any language that supports
pointers and recursion is a plus.
• Fundamental data structures: arrays/vectors, linked lists, search trees, heaps
• Fundamental abstract data types: dictionaries, stacks, queues, priority queues; the difference
between this list and the previous list.
• Fundamental algorithms: searching, sorting (selection, insertion, merge, heap, quick, anything
but bubble), pre-/post-/inorder tree traversal, depth/breadth ﬁrst search
• Basic algorithm analysis: Asymptotic notation (o, O, Θ, Ω, ω), translating loops into sums and
recursive calls into recurrences, evaluating simple sums and recurrences.
• Mathematical maturity: facility with abstraction, formal (especially recursive) deﬁnitions, and
(especially inductive) proofs; following mathematical arguments; recognizing syntactic, semantic,
and/or logical nonsense; writing the former rather than the latter
Some of this material is covered brieﬂy, but more as a reminder than a good introduction. In the
future, I hope to write review notes that cover more of these topics. Alas, the future has not yet arrived. Acknowledgments
The lecture notes and exercises draw heavily on the creativity, wisdom, and experience of thousands of
algorithms students, teachers, and researchers. In particular, I am immensely grateful to the almost 1200
Illinois students who have used these notes as a primary reference, offered useful (if sometimes painful)
criticism, and suffered through some truly awful ﬁrst drafts. I’m also grateful for the contributions and
feedback from teaching assistants, all listed above.
Naturally, these notes owe a great deal to the people who taught me this algorithms stuff in the ﬁrst
place: Bob Bixby and Michael Perlman at Rice; David Eppstein, Dan Hirshberg, and George Lueker at UC
Irvine; and Abhiram Ranade, Dick Karp, Manuel Blum, Mike Luby, and Raimund Seidel at UC Berkeley.
I’ve also been helped tremendously by many discussions with faculty colleagues at UIUC—Chandra
Chekuri, Edgar Ramos, Herbert Edelsbrunner, Jason Zych, Lenny Pitt, Mahesh Viswanathan, Margaret
Fleck, Shang-Hua Teng, Steve LaValle, and especially Ed Reingold and Sariel Har-Peled. I stole the ﬁrst
iteration of the overall course structure, and the idea to write up my own lecture notes, from Herbert
Edelsbrunner. “Johnny’s” multi-colored crayon homework was found under the TA ofﬁce door among the other Fall
¯
2000 Homework 1 submissions. The square Kuﬁ rendition of the name “al-Khwarizm¯ on the back of
ı”
the cover page is mine.
The following sources have been invaluable references. (This list is incomplete!)
• Alfred V Aho, John E. Hopcroft, and Jeffrey D. Ullman. The Design and Analysis of Computer
.
Algorithms. Addison-Wesley, 1974. (I used this textbook as an undergrad at Rice, and again as a
masters student at UC Irvine.)
• Sara Baase and Allen Van Gelder. Computer Algorithms: Introduction to Design and Analysis.
Addison-Wesley, 2000.
• Mark de Berg, Marc van Kreveld, Mark Overmars, and Otfried Schwarzkopf. Computational
Geometry: Algorithms and Applications. Springer-Verlag, 1997.
• Thomas Cormen, Charles Leiserson, Ron Rivest, and Cliff Stein. Introduction to Algorithms, second
edition. MIT Press/McGraw-Hill, 2000. (This was my recommended textbook until 2005. I also
used the ﬁrst edition as a teaching assistant at Berkeley.)
• Sanjoy Dasgupta, Christos H. Papadimitriou, and Umesh V Vazirani. Algorithms. McGraw-Hill,
.
2006. (This is the current recommended textbook for the undergraduate section of my class.)
• Michael R. Garey and David S. Johnson. Computers and Intractability: A Guide to the Theory of
NP-Completeness. W. H. Freeman, 1979.
• Michael T. Goodrich and Roberto Tamassia. Algorithm Design: Foundations, Analysis, and Internet
Examples. John Wiley & Sons, 2002.
• Dan Gusﬁeld. Algorithms on Strings, Trees, and Sequences: Computer Science and Molecular Biology.
Cambridge University Press, 1997.
• Jon Kleinberg and Éva Tardos. Algorithm Design. Addison-Wesley, 2005. (This is the current
recommended textbook for the graduate section of my class.)
• Donald Knuth. The Art of Computer Programming, volumes 1–3. Addison-Wesley, 1997. (My parents
gave me these for Christmas when I was 14. I actually read it much later.)
• Udi Manber. Introduction to Algorithms: A Creative Approach. Addison-Wesley, 1989. (I used this
textbook as a teaching assistant at Berkeley.)
• Rajeev Motwani and Prabhakar Raghavan. Randomized Algorithms. Cambridge University Press,
1995.
• Ian Parberry. Problems on Algorithms. Prentice-Hall, 1995. (This book is out of print, but it can be
downloaded for ‘free’ from http://www.eng.unt.edu/ian/books/free/license.html .)
• Alexander Schrijver. Combinatorial Optimization: Polyhedra and Efﬁciency. Springer, 2003.
• Robert Sedgewick. Algorithms. Addison-Wesley, 1988. (This book and its sequels have by far the
best algorithm illustrations I’ve seen anywhere.)
• Robert Endre Tarjan. Data Structures and Network Algorithms. SIAM, 1983.
• Robert J. Vanderbei. Linear Programming: Foundations and Extensions. Springer, 2001.
• Class notes from my own algorithms classes at Berkeley, especially those taught by Dick Karp and
Raimund Seidel.
• The Source of All Knowledge (Google) and The Source of All Lies (Wikipedia). Caveat Lector!
With few exceptions, each of these notes contains far too much material to cover in a single lecture.
In a typical 75-minute lecture, I tend to cover 4 to 5 pages of material—a bit more if I’m lecturing to
graduate students than to undergraduates. Your mileage may vary! (Arguably, that means that as I
continue to add material, the label “lecture notes” becomes less and less accurate.)
Despite several rounds of revision, these notes still contain lots of mistakes, errors, bugs, gaffes,
omissions, snafus, kludges, typos, mathos, grammaros, thinkos, brain farts, nonsense, garbage, cruft,
junk, and outright lies, all of which are entirely Steve Skiena’s fault. I revise and update these notes
every time I teach the course, so please let me know if you ﬁnd a bug. (Steve is unlikely to care.)
Whenever I teach the algorithms class, I award extra credit points to the ﬁrst student to post an
explanation and correction of any error in the lecture notes to the course newsgroup. Obviously, the
number of extra credit points depends on the severity of the error and the quality of the correction.
If I’m not teaching the course, encourage your instructor to set up a similar extra-credit scheme, and
forward the bug reports to Steve me!
Of course, any other feedback is also welcome!
Enjoy!
— Jeff It is traditional for the author to magnanimously accept the blame for whatever
deﬁciencies remain. I don’t. Any errors, deﬁciencies, or problems in this book are
somebody else’s fault, but I would appreciate knowing about them so as to determine
who is to blame.
— Steven S. Skiena, The Algorithm Design Manual (1997) c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. Algorithms Lecture 0: Introduction
We should explain, before proceeding, that it is not our object to consider this program with
reference to the actual arrangement of the data on the Variables of the engine, but simply
as an abstract question of the nature and number of the operations required to be perfomed
during its complete solution.
— Ada Augusta Byron King, Countess of Lovelace, translator’s notes for Luigi F. Menabrea,
“Sketch of the Analytical Engine invented by Charles Babbage, Esq.” (1843)
You are right to demand that an artist engage his work consciously, but you confuse two
different things: solving the problem and correctly posing the question.
— Anton Chekhov, in a letter to A. S. Suvorin (October 27, 1888)
The more we reduce ourselves to machines in the lower things,
the more force we shall set free to use in the higher.
— Anna C. Brackett, The Technique of Rest (1892)
The moment a man begins to talk about technique
that’s proof that he is fresh out of ideas.
— Raymond Chandler 0 Introduction 0.1 What is an algorithm? An algorithm is an explicit, precise, unambiguous, mechanically-executable sequence of elementary
instructions. For example, here is an algorithm for singing that annoying song ‘99 Bottles of Beer on the
Wall’, for arbitrary values of 99:
B OTTLESOFBEER(n):
For i ← n down to 1
Sing “i bottles of beer on the wall, i bottles of beer,”
Sing “Take one down, pass it around, i − 1 bottles of beer on the wall.”
Sing “No bottles of beer on the wall, no bottles of beer,”
Sing “Go to the store, buy some more, n bottles of beer on the wall.” The word ‘algorithm’ does not derive, as algorithmophobic classicists might guess, from the Greek
root algos (αλγoς), meaning ‘pain’. Rather, it is a corruption of the name of the 9th century Persian
¯
¯
¯¯
¯
¯
mathematician Abu ’Abd Allah Muhammad ibn Musa al-Khwarizm¯ 1 Al-Khwarizm¯ is perhaps best
ı.
ı
.
¯
¯
known as the writer of the treatise Al-Kitab al-mukhtasar f¯ ¯ ab al-˘abr wa’l-muqabala2 , from which
ıhıs¯
g
.
the modern word algebra derives. The word algorithm is a corruption of the older word algorism (by
false connection to the Greek arithmos (αριθ µoς), meaning ‘number’, and perhaps the aforementioned
αλγoς), used to describe the modern decimal system for writing and manipulating numbers—in
¯
particular, the use of a small circle or sifr to represent a missing quantity—which al-Khwarizm¯ brought
ı
into Persia from India. Thanks to the efforts of the medieval Italian mathematician Leonardo of Pisa,
better known as Fibonacci, algorism began to replace the abacus as the preferred system of commercial
calculation3 in Europe in the late 12th century, although it was several more centuries before cyphers
became truly ubiquitous. (Counting boards were used by the English and Scottish royal exchequers well
into the 1600s.) Thus, until very recently, the word algorithm referred exclusively to pencil-and-paper
1 ¯
¯
‘Mohammad, father of Adbdulla, son of Moses, the Kwarizmian’. Kwarizm is an ancient city, now called Khiva, in the
Khorezm Province of Uzbekistan.
2
‘The Compendious Book on Calculation by Completion and Balancing’.
3
from the Latin word calculus, meaning literally ‘small rock’, referring to the stones on a counting board, or abacus 1 Algorithms Lecture 0: Introduction methods for numerical calculations. People trained in the reliable execution of these methods were
called—you guessed it—computers. 0.2 A Few Simple Examples Multiplication by compass and straightedge. Although they have only been an object of formal
study for a few decades, algorithms have been with us since the dawn of civilization, for centuries before
¯
Al-Khwarizm¯ and Fibonacci popularized the cypher. Here is an algorithm, popularized (but almost
ı
certainly not discovered) by Euclid about 2500 years ago, for multiplying or dividing numbers using a
ruler and compass. The Greek geometers represented numbers using line segments of the appropriate
length. In the pseudo-code below, CIRCLE( p, q) represents the circle centered at a point p and passing
through another point q. Hopefully the other instructions are obvious.4 Z 〈〈Construct the line perpendicular to and passing through P .〉〉
RIGHTANGLE( , P ):
Choose a point A ∈
A, B ← INTERSECT(CIRCLE( P, A), )
C , D ← INTERSECT(CIRCLE(A, B ), CIRCLE(B , A))
return LINE(C , D)
〈〈Construct a point Z such that |AZ | = |AC ||AD|/|AB |.〉〉
MULTIPLYORDIVIDE(A, B , C , D):
α ← RIGHTANGLE(LINE(A, C ), A)
E ← INTERSECT(CIRCLE(A, B ), α)
F ← INTERSECT(CIRCLE(A, D), α)
β ← RIGHTANGLE(LINE( E , C ), F )
γ ← RIGHTANGLE(β , F )
return INTERSECT(γ, LINE(A, C )) C D B β
α F
A E
γ Multiplying or dividing using a compass and straightedge. This algorithm breaks down the difﬁcult task of multiplication into a series of simple primitive
operations: drawing a line between two points, drawing a circle with a given center and boundary point,
and so on. These primitive steps are quite non-trivial to execute on a modern digital computer, but
this algorithm wasn’t designed for a digital computer; it was designed for the Platonic Ideal Classical
Greek Mathematician, wielding the Platonic Ideal Compass and the Platonic Ideal Straightedge. In
this example, Euclid ﬁrst deﬁnes a new primitive operation, constructing a right angle, by (as modern
programmers would put it) writing a subroutine.
Multiplication by duplation and mediation. Here is an even older algorithm for multiplying large
numbers, sometimes called (Russian) peasant multiplication. A variant of this method was copied into
the Rhind papyrus by the Egyptian scribe Ahmes around 1650 BC, from a document he claimed was
(then) about 350 years old. This was the most common method of calculation by Europeans before
Fibonacci’s introduction of Arabic numerals; it was still being used in Russia, along with the Julian
Euclid and his students almost certainly drew their constructions on an αβαξ, a table covered in sand (or very small
rocks). Over the next several centuries, the Greek abax evolved into the medieval European abacus.
4 2 Algorithms Lecture 0: Introduction calendar, well into the 20th century. This algorithm was also commonly used by early digital computers
that did not implement integer multiplication directly in hardware.
x y 123
61
30
15
7
3
1 PEASANTMULTIPLY( x , y ):
prod ← 0
while x > 0
if x is odd
prod ← prod + y
x ← x /2
y← y+y
return p +456
+912
1824
+3648
+7296
+14592
+29184 prod
0
= 456
= 1368
= 5016
= 12312
= 26904
= 56088 The peasant multiplication algorithm breaks the difﬁcult task of general multiplication into four
simpler operations: (1) determining parity (even or odd), (2) addition, (3) duplation (doubling a
number), and (4) mediation (halving a number, rounding down).5 Of course a full speciﬁcation of this
algorithm requires describing how to perform those four ‘primitive’ operations. Peasant multiplication
requires (a constant factor!) more paperwork to execute by hand, but the necessary operations are easier
(for humans) to remember than the 10 × 10 multiplication table required by the American grade school
algorithm.6
The correctness of peasant multiplication follows from the following recursive identity, which holds
for any non-negative integers x and y : if x = 0
0
x·y= x /2 · ( y + y ) A bad example. if x is even x /2 · ( y + y ) + y if x is odd Consider “Martin’s algorithm”:7
BECOMEAMILLIONAIREANDNEVERPAYTAXES:
Get a million dollars.
Don’t pay taxes.
If you get caught,
Say “I forgot.” Pretty simple, except for that ﬁrst step; it’s a doozy. A group of billionaire CEOs would consider
this an algorithm, since for them the ﬁrst step is both unambiguous and trivial, but for the rest of us
poor slobs, Martin’s procedure is too vague to be considered an algorithm. On the other hand, this is a
perfect example of a reduction—it reduces the problem of being a millionaire and never paying taxes to
the ‘easier’ problem of acquiring a million dollars. We’ll see reductions over and over again in this class.
As hundreds of businessmen and politicians have demonstrated, if you know how to solve the easier
problem, a reduction tells you how to solve the harder one.
5 The version of this algorithm actually used in ancient Egypt does not use mediation or parity, but it does use comparisons.
To avoid halving, the algorithm pre-computes two tables by repeated doubling: one containing all the powers of 2 not
exceeding x , the other containing the same powers of 2 multiplied by y . The powers of 2 that sum to x are then found by
greedy subtraction, and the corresponding entries in the other table are added together to form the product.
6
American school kids learn a variant of the lattice multiplication algorithm developed by Indian mathematicians and
described by Fibonacci in Liber Abaci. The two algorithms are equivalent if the input numbers are represented in binary.
7
S. Martin, “You Can Be A Millionaire”, Saturday Night Live, January 21, 1978. Appears on Comedy Is Not Pretty, Warner
Bros. Records, 1979. 3 Algorithms Lecture 0: Introduction Martin’s algorithm, like many of our previous examples, is not the kind of algorithm that computer
scientists are used to thinking about, because it is phrased in terms of operations that are difﬁcult
for computers to perform. In this class, we’ll focus (almost!) exclusively on algorithms that can be
reasonably implemented on a computer. In other words, each step in the algorithm must be something
that either is directly supported by common programming languages (such as arithmetic, assignments,
loops, or recursion) or is something that you’ve already learned how to do in an earlier class (like sorting,
binary search, or depth ﬁrst search).
Congressional apportionment. Here is another good example of an algorithm that comes from
outside the world of computing. Article I, Section 2 of the US Constitution requires that
Representatives and direct Taxes shall be apportioned among the several States which may
be included within this Union, according to their respective Numbers. . . . The Number of
Representatives shall not exceed one for every thirty Thousand, but each State shall have at
Least one Representative. . . .
Since there are a limited number of seats available in the House of Representatives, exact proportional
representation is impossible without either shared or fractional representatives, neither of which are
legal. As a result, several different apportionment algorithms have been proposed and used to round
the fractional solution fairly. The algorithm actually used today, called the Huntington-Hill method or
the method of equal proportions, was ﬁrst suggested by Census Bureau statistician Joseph Hill in 1911,
reﬁned by Harvard mathematician Edward Huntington in 1920, adopted into Federal law (2 U.S.C. §§2a
and 2b) in 1941, and survived a Supreme Court challenge in 1992.8 The input array P [1 .. n] stores the
populations of the n states, and R is the total number of representatives. Currently, n = 50 and R = 435.
APPORTIONCONGRESS( P [1 .. n], R):
H ← NEWMAXHEAP
for i ← 1 to n
r [i ] ← 1
INSERT H , i , P [i ]/ 2
R←R−n
while R > 0
s ← EXTRACTMAX(H )
r [s ] ← r [s ] + 1
INSERT H , i , P [i ]/
R←R−1 r [i ]( r [i ] + 1) return r [1 .. n] Note that this description assumes that you know how to implement a max-heap and its basic
operations NEWMAXHEAP, INSERT, and EXTRACTMAX. (The actual law doesn’t make those assumptions, of
course.) Moreover, the correctness of the algorithm doesn’t depend at all on how these operations are
implemented. The Census Bureau implements the max-heap as an unsorted array stored in a column of
an Excel spreadsheet; you should have learned a more efﬁcient solution in your undergraduate data
structures class.
8 Overruling an earlier ruling by a federal district court, the Supreme Court unanimously held that any apportionment
method adopted in good faith by Congress is constitutional (United States Department of Commerce v. Montana). The
current congressional apportionment algorithm is described in gruesome detail at the U.S. Census Department web site
http://www.census.gov/population/www/censusdata/apportionment/computing.html. A good history of the apportionment
problem can be found at http://www.thirty-thousand.org/pages/Apportionment.htm. A report by the Congressional Research
Service describing various apportionment methods is available at http://www.rules.house.gov/archives/RL31074.pdf. 4 Algorithms 0.3 Lecture 0: Introduction Writing down algorithms Computer programs are concrete representations of algorithms, but algorithms are not programs; they
should not be described in a particular programming language. The whole point of this course is to
develop computational techniques that can be used in any programming language.9 The idiosyncratic
syntactic details of C, Java, Python, Ruby, Erlang, OcaML, Scheme, Visual Basic, Smalltalk, Javascript,
Forth, TEX, COBOL, Intercal, or Brainfuck10 are of little or no importance in algorithm design, and
focusing on them will only distract you from what’s really going on.11 What we really want is closer to
what you’d write in the comments of a real program than the code itself.
On the other hand, a plain English prose description is usually not a good idea either. Algorithms
have a lot of structure—especially conditionals, loops, and recursion—that are far too easily hidden
by unstructured prose. Like any natural languags, English is full of ambiguities, subtleties, and shades
of meaning, but algorithms must be described as accurately as possible. Finally and more seriously,
many people have a tendency to describe loops informally: “Do this ﬁrst, then do this second, and so
on.” As anyone who has taken one of those ‘What comes next in this sequence?’ tests already knows,
specifying what happens in the ﬁrst few iterations of a loop doesn’t say much about what happens in
later iterations.12 Phrases like ‘and so on’ or ‘repeat this for all n’ are good indicators that the algorithm
should have been described in terms of loops or recursion, and the description should have speciﬁed a
generic iteration of the loop. Similarly, the appearance of the phrase ‘and so on’ in a proof almost always
means the proof should have been done by induction.
The best way to write down an algorithm is using pseudocode. Pseudocode uses the structure of
formal programming languages and mathematics to break the algorithm into one-sentence steps, but
those sentences can be written using mathematics, pure English, or an appropriate mixture of the two.
Exactly how to structure the pseudocode is a personal choice, but the overriding goal should be clarity
and precision. Here are the guidelines I follow:
• Use standard imperative programming keywords (if/then/else, while, for, repeat/until, case,
return) and notation (variable ← value, Array[index], function(argument), bigger > smaller, etc.).
Keywords should be standard English words: write ‘else if’ instead of ‘elif’.
• Use standard mathematical notation for standard mathematical stuff. For example, write x and
a b and π instead of sqrt( x ) and power(a, b) and pi. (One exception: never use ∀ to represent a
for-loop!)
• Avoid mathematical notation if English is clearer. For example, ‘Insert a into X ’ is often preferable
to INSERT(X , a) or X ← X ∪ {a}.
9
See http://www.ionet.net/~timtroyr/funhouse/beer.html for implementations of the B OTTLESOFBEER algorithm in over
200 different programming languages.
10
Brainfuck is the well-deserved name of a programming language invented by Urban Müller in 1993. Brainfuck programs
are written entirely using the punctuation characters <>+-,., each representing a different operation (roughly: shift left,
shift right, increment, decrement, input, output, begin loop, end loop). See http://esolangs.org/wiki/Brainfuck for a complete
language description; links to several interpreters, compilers, and sample programs; and lots of related shit.
11
This is, of course, a matter of religious conviction. Linguists argue incessantly over the Sapir-Whorf hypothesis, which states
(more or less) that people think only in the categories imposed by their languages. According to an extreme formulation of
this principle, some concepts in one language simply cannot be understood by speakers of other languages, not just because
of technological advancement—How would you translate ‘jump the shark’ or ‘blog’ into Aramaic?—but because of inherent
structural differences between languages and cultures. For a more skeptical view, see Steven Pinker’s The Language Instinct.
There is admittedly some strength to this idea when applied to different programming paradigms. (What’s the Y combinator,
again? How do templates work? What’s an Abstract Factory?) Fortunately, those differences are generally too subtle to have
much impact in this class.
12
See http://www.research.att.com/~njas/sequences/. 5 Algorithms Lecture 0: Introduction • Indent everything carefully and consistently; the block structure should be visible from across
the room. This is especially important for nested loops and conditionals. Don’t use unnecessary
syntactic sugar (like braces or begin/end tags).
• Don’t typeset keywords in a different font or style. Changing type style emphasizes the keywords,
making the reader think the syntactic sugar is important. It isn’t!
• Each statement should ﬁt on one line, and each line should contain either exactly one statement
or exactly one structuring element (for, while, if). (I sometimes make an exception for short and
similar statements like i ← i + 1; j ← j − 1; k ← 0.)
• Use short, mnemonic algorithm and variable names. Absolutely never use pronouns!
A good description of an algorithm reveals the internal structure, hides irrelevant details, and can be
implemented easily and correctly by any competent programmer in their favorite programming language,
even if they don’t understand why the algorithm works. Good pseudocode, like good code, makes the
algorithm much easier to understand and analyze; it also makes mistakes much easier to spot. The
algorithm descriptions in these lecture notes are (hopefully) good examples of what we want to see on
your homeworks and exams. 0.4 Analyzing algorithms It’s not enough just to write down an algorithm and say ‘Behold!’ We also need to convince ourselves
(and our graders) that the algorithm does what it’s supposed to do, and that it does it efﬁciently.
Correctness: In some application settings, it is acceptable for programs to behave correctly most of
the time, on all ‘reasonable’ inputs. Not in this class; we require algorithms that are correct for all
possible inputs. Moreover, we must prove that they’re correct; trusting our instincts, or trying a few test
cases, isn’t good enough.13 Sometimes correctness is fairly obvious, especially for algorithms you’ve
seen in earlier courses. On the other hand, ‘obvious’ is all too often a synonym for ‘wrong’. Many of
the algorithms we will discuss in this course will require some extra work to prove. Correctness proofs
almost always involve induction. We like induction. Induction is our friend.14
But before we can formally prove that our algorithm does what we want it to do, we have to formally
state what we want it to do! Usually problems are given to us in real-world terms, not with formal
mathematical descriptions. It’s up to us, the algorithm designers, to restate these problems in terms of
mathematical objects that we can prove things about—numbers, arrays, lists, graphs, trees, and so on.
We also need to determine if the problem statement makes any hidden assumptions, and state those
assumptions explicitly. (For example, in the song “n Bottles of Beer on the Wall”, n is always a positive
integer.) Restating the problem formally is not only required for proofs; it is also one of the best ways
to really understand what the problems is asking for. The hardest part of answering any question is
ﬁguring out the right way to ask it!
It is important to remember the distinction between a problem and an algorithm. A problem is a
task to perform, like “Compute the square root of x ” or “Sort these n numbers” or “Keep n algorithms
students awake for t minutes”. An algorithm is a set of instructions to follow if you want to accomplish
this task. The same problem may have hundreds of different algorithms; the same algorithm may solve
hundreds of different problems.
13
14 I say we take off and nuke the entire site from orbit. It’s the only way to be sure.
If induction is not your friend, you will have a hard time in this course. 6 Algorithms Lecture 0: Introduction Running time: The most common way of ranking different algorithms for the same problem is by
how fast they run. Ideally, we want the fastest possible algorithm for our problem. In many application
settings, it is acceptable for programs to run efﬁciently most of the time, on all ‘reasonable’ inputs. Not
in this class; we require algorithms that always run efﬁciently, even in the worst case.
But how do we measure running time? As a speciﬁc example, how long does it take to sing the song
B OTTLESOFBEER(n)? This is obviously a function of the input value n, but it also depends on how quickly
you can sing. Some singers might take ten seconds to sing a verse; others might take twenty. Technology
widens the possibilities even further. Dictating the song over a telegraph using Morse code might take
a full minute per verse. Downloading an mp3 over the Web might take a tenth of a second per verse.
Duplicating the mp3 in a computer’s main memory might take only a few microseconds per verse.
What’s important here is how the singing time changes as n grows. Singing B OTTLESOFBEER(2n)
takes about twice as long as singing B OTTLESOFBEER(n), no matter what technology is being used. This
is reﬂected in the asymptotic singing time Θ(n). We can measure time by counting how many times the
algorithm executes a certain instruction or reaches a certain milestone in the ‘code’. For example, we
might notice that the word ‘beer’ is sung three times in every verse of B OTTLESOFBEER, so the number of
times you sing ‘beer’ is a good indication of the total singing time. For this question, we can give an
exact answer: B OTTLESOFBEER(n) uses exactly 3n + 3 beers.
There are plenty of other songs that have non-trivial singing time. This one is probably familiar to
most English-speakers:
NDAYSOFCHRISTMAS(gifts[2 .. n]):
for i ← 1 to n
Sing “On the i th day of Christmas, my true love gave to me”
for j ← i down to 2
Sing “ j gifts[ j ]”
if i > 1
Sing “and”
Sing “a partridge in a pear tree.” The input to NDAYSOFCHRISTMAS is a list of n − 1 gifts. It’s quite easy to show that the singing time is
n
Θ(n2 ); in particular, the singer mentions the name of a gift i =1 i = n(n + 1)/2 times (counting the
partridge in the pear tree). It’s also easy to see that during the ﬁrst n days of Christmas, my true love
n
i
gave to me exactly i =1 j =1 j = n(n + 1)(n + 2)/6 = Θ(n3 ) gifts. Other songs that take quadratic time
to sing are “Old MacDonald had a Farm”, “There Was an Old Lady Who Swallowed a Fly”, “The House
that Jack Built”, “Green Grow the Rushes O”, “Eh, Compare!”, “Alouette”, “Echad Mi Yodea”, “Chad
Gadya”, and “Ist das nicht ein Schnitzelbank?”15 For further details, consult your nearest preschooler.
OLDMACDONALD(animals[1 .. n], noise[1 .. n]):
for i ← 1 to n
Sing “Old MacDonald had a farm, E I E I O”
Sing “And on this farm he had some animals[i ], E I E I O”
Sing “With a noise[i ] noise[i ] here, and a noise[i ] noise[i ] there”
Sing “Here a noise[i ], there a noise[i ], everywhere a noise[i ] noise[i ]”
for j ← i − 1 down to 1
Sing “noise[ j ] noise[ j ] here, noise[ j ] noise[ j ] there”
Sing “Here a noise[ j ], there a noise[ j ], everywhere a noise[ j ] noise[ j ]”
Sing “Old MacDonald had a farm, E I E I O.”
15 Wakko: Ist das nicht Otto von Schnitzelpusskrankengescheitmeyer?
Yakko and Dot: Ja, das ist Otto von Schnitzelpusskrankengescheitmeyer!! 7 Algorithms Lecture 0: Introduction
ALOUETTE(lapart[1 .. n]):
Chantez « Alouette, gentille alouette, alouette, je te plumerais. »
pour tout i de 1 á n
Chantez « Je te plumerais lapart[i ]. Je te plumerais lapart[i ]. »
pour tout j de i − 1 á bas á 1
Chantez « Et lapart[ j ] ! Et lapart[ j ] ! »
Chantez « Ooooooo! »
Chantez « Alouette, gentille alluette, alouette, je te plumerais. » For a slightly more complicated example, consider the algorithm APPORTIONCONGRESS. Here the
running time obviously depends on the implementation of the max-heap operations, but we can certainly
bound the running time as O(N + RI + (R − n) E ), where N is the time for a NEWMAXHEAP, I is the time
for an INSERT, and E is the time for an EXTRACTMAX. Under the reasonable assumption that R > 2n (on
average, each state gets at least two representatives), this simpliﬁes to O(N + R( I + E )). The Census
Bureau uses an unsorted array of size n, for which N = I = Θ(1) (since we know a priori how big the
array is), and E = Θ(n), so the overall running time is Θ(Rn). This is ﬁne for the federal government,
but if we want to be more efﬁcient, we can implement the heap as a perfectly balanced n-node binary
tree (or a heap-ordered array). In this case, we have N = Θ(1) and I = R = O(log n), so the overall
running time is Θ(R log n).
Sometimes we are also interested in other computational resources: space, randomness, page faults,
inter-process messages, and so forth. We use the same techniques to analyze those resources as we use
for running time. 0.5 A Longer Example: Stable Marriage Every year, thousands of new doctors must obtain internships at hospitals around the United States.
During the ﬁrst half of the 20th century, competition among hospitals for the best doctors led to earlier
and earlier offers of internships, sometimes as early as the second year of medical school, along with
tighter deadlines for acceptance. In the 1940s, medical schools agreed not to release information until a
common date during their students’ fourth year. In response, hospitals began demanding faster decisions.
By 1950, hospitals would regularly call doctors, offer them an internship, and demand an immediate
response. Interns were forced to gamble if their third-choice hospital called ﬁrst—accept and risk losing
a better opportunity later, or reject and risk having no position at all.
Finally, a central clearinghouse for internship assignments, now called the National Resident Matching
Program, was established in the early 1950s. Each year, doctors submit a ranked list of all hospitals
where they would accept an internship, and each hospital submits a ranked list of doctors they would
accept as interns. The NRMP then computes an assignment of interns to hospitals that satisﬁes the
following stability requirement. For simplicity, let’s assume that there are n doctors and n hospitals; each
hospital offers exactly one internship; each doctor ranks all hospitals and vice versa; and ﬁnally, there
are no ties in the doctors’ and hospitals’ rankings.16 We say that a matching of doctors to hospitals is
unstable if there are two doctors α and β and two hospitals A and B , such that
• α is assigned to A, and β is assigned to B ;
• α prefers B to A, and B prefers α to β .
In other words, α and B would both be happier with each other than with their current assignment. The
goal of the Resident Match is a stable matching, in which no doctor or hospital has an incentive to cheat
the system. At ﬁrst glance, it is not clear that a stable matching exists!
16 In reality, most hospitals offer multiple internships, Each doctor ranks only a subset of the hospitals and vice versa, and
there are typically more internships than interested doctors. And then it starts getting complicated. 8 Algorithms Lecture 0: Introduction In 1952, the NRMP adopted the “Boston Pool” algorithm to assign interns, so named because it
had been previously used by a regional clearinghouse in the Boston area. The algorithm is often
inappropriately attributed to David Gale and Lloyd Shapley, who formally analyzed the algorithm and
ﬁrst proved that it computes a stable matching in 1962.17 The algorithm proceeds in rounds until every
position has been ﬁlled. In each round:
1. Some unassigned hospital offers its position to the best doctor (according to the hospital’s preference list) who has not already rejected it.
2. Each doctor is always assigned to the best hospital (according to the doctor’s preference list) that
has made her an offer so far. Thus, whenever a doctor receives an offer that she likes more than
her current assignment, her assignment changes.
For example, suppose there are three doctors α, β , γ,
following preference lists:
αβγδAB
AABDδβ
BDABγδ
CCCCβα
DBDAαγ δ, and three hospitals A, B , C , D with the
C
δ
α
β
γ D
γ
β
α
δ The algorithm might proceed as follows:
1. A makes an offer to δ.
2. B makes an offer to β .
3. C makes an offer to δ, who rejects her earlier offer from A.
4. D makes an offer to γ. (From this point on, because there is only one unmatched hospital, the
algorithm has no more choices.)
5. A makes an offer to γ, who rejects her earlier offer from D.
6. D makes an offer to β , who rejects her earlier offer from B .
7. B makes an offer to δ, who rejects her earlier offer from C .
8. C makes an offer to α.
At this point we have the assignment (α, C ), (β , D), (γ, A), (δ, B ). You can verify by brute force that this
matching is stable, despite the fact that no doctor was hired by her favorite hospital, and no hospital
hired its favorite doctor.
Analyzing the algorithm is straightforward. Since each hospital makes an offer to each doctor at most
once, the algorithm requires at most n2 rounds. In an actual implementation, each doctor and hospital
can be identiﬁed by a unique integer between 1 and n, and the preference lists can be represented as two
arrays DocPref[1 .. n][1 .. n] and HosPref[1 .. n][1 .. n], where DocPref[α][ r ] represents the r th hospital
17
Gale and Shapely used the metaphor of college admissions. The “Gale-Shapely algorithm” is a prime instance of Stigler’s
Law of Eponymy: No scientiﬁc discovery is named after its original discoverer. In his 1980 paper that gives the law its name,
the statistician Stephen Stigler claimed that this law was ﬁrst proposed by sociologist Robert K. Merton, although similar
sentiments were previously attributed to Vladimir Arnol’d in the 1970’s (“Discoveries are rarely attributed to the correct
person.”), Carl Boyer in 1968 (“Clio, the muse of history, often is ﬁckle in attaching names to theorems!”), Alfred North
Whitehead in 1917 (“Everything of importance has been said before by someone who did not discover it.”), and even Stephen’s
father George Stigler in 1966 (“If we should ever encounter a case where a theory is named for the correct man, it will be
noted.”)! The law was dubbed the Zeroth Theorem of the History of Science by historian Ernst Peter Fischer in 2006 (“[E]ine
Entdeckung (Regel, Gesetzmässigkeit, Einsicht), die nach einer Person benannt ist, nicht von dieser Person herrührt.”) . 9 Algorithms Lecture 0: Introduction in doctor α’s preference list, and HosPref[A][ r ] represents the r th doctor in hospital A’s preference list.
With the input in this form, the Boston Pool algorithm can be implemented to run in O(n2 ) time; we
leave the details as an exercise for the reader. A somewhat harder exercise is to prove that there are
inputs (and choices of who makes offers when) that force Ω(n2 ) rounds before the algorithm terminates.
But why is it correct? Gale and Shapely prove that the Boston Pool algorithm always computes a
stable matching as follows. The algorithm must terminate, because each hospital makes an offer to each
doctor at most once. When the algorithm terminates, every internship has been ﬁlled. Now suppose in
the ﬁnal matching, doctor α is assigned to hospital A but prefers B . Since every doctor accepts the best
offer she receives, α received no offer she liked more than A. In particular, B never made an offer to α.
On the other hand, B made offers to every doctor they like more than β . Thus, B prefers β to α, and so
there is no instability.
Surprisingly, the correctness of the algorithm does not depend on which hospital makes its offer
in which round. In fact, there is a stronger sense in which the order of offers doesn’t matter—no
matter which unassigned hospital makes an offer in each round, the algorithm always computes the same
matching! Let’s say that α is a feasible doctor for A if there is a stable matching that assigns doctor α to
hospital A, and let best(A) be the highest-ranked feasible doctor on A’s preference list.
Lemma 1. The Boston Pool algorithm assigns best(A) to A, for every hospital A.
Proof: In the ﬁnal matching, no hospital A is assigned a doctor they prefer to best(A), because that
would create an instability (by deﬁnition of ‘feasible’ and ‘best’). A hospital A can only be assigned a
doctor they like less than best(A) if best(A) rejects their offer.
So consider the ﬁrst round where some hospital A is rejected by its best choice α = best(A). In that
round, α got an offer from another hospital B that α ranks higher than A. Now, B must rank α at least as
highly as its best choice best(B ), because this is the ﬁrst round where a hospital is rejected by its best
choice. Thus, B must rank α higher than any doctor that is feasible for B .
Now consider the stable matching where α is assigned to A. On the one hand, α prefers B to A. On
the other hand, because this is a stable matching, B is assigned a doctor β that is feasible for B , which
implies that B prefers α to β . But this means the matching is unstable, and we have a contradiction.
In other words, from the hospitals’ point of view, the Boston Pool algorithm computes the best
possible stable matching. It turns out that this is also the worst possible matching from the doctors’
viewpoint! Let worst(α) be the lowest-ranked feasible hospital on doctor α’s preference list.
Lemma 2. The Boston Pool algorithm assigns α to worst(α), for every doctor α.
Proof: Suppose the Boston Pool algorithm assigns doctor α to hospital A; the previous lemma implies
that α = best(A). To prove the lemma, we need to show that A = worst(α).
Consider an arbitrary stable matching where A is not matched with α but with another doctor β .
Then A must prefer α to β . Since this matching is stable, α must therefore prefer her current assignment
to A. This argument works for any stable assignment, so α prefers every other feasible match to A; in
other words, A = worst(α).
In 1998, the National Residency Matching Program reversed its matching algorithm, so that potential
residents offer to work for hospitals, and each hospital accepts its best offer. Thus, the new algorithm
computes the best possible stable matching for the doctors, and the worst possible stable matching for
the hospitals. As far as I know, the precise effect of this change on the patients is an open problem. 10 Algorithms 0.6 Lecture 0: Introduction Why are we here, anyway? This class is ultimately about learning two skills that are crucial for all computer scientists.
1. Reasoning: How to think about abstract computation.
2. Communication: How to talk about abstract computation.
The ﬁrst goal of this course is to help you develop algorithmic intuition. How do various algorithms
really work? When you see a problem for the ﬁrst time, how should you attack it? How do you tell
which techniques will work at all, and which ones will work best? How do you judge whether one
algorithm is better than another? How do you tell whether you have the best possible solution? These
are not easy questions. Anyone who says differently is selling something.
Our second main goal is to help you develop algorithmic language. It’s not enough just to understand
how to solve a problem; you also have to be able to explain your solution to somebody else. I don’t mean
just how to turn your algorithms into working code—despite what many students (and inexperienced
programmers) think, ‘somebody else’ is not just a computer. Nobody programs alone. Code is read far
more often than it is written, or even compiled. Perhaps more importantly in the short term, explaining
something to somebody else is one of the best ways to clarify your own understanding. As Albert Einstein
(or was it Richard Feynman?) apocryphally put it, “You do not really understand something unless you
can explain it to your grandmother."
Along the way, you’ll pick up a bunch of algorithmic facts—mergesort runs in Θ(n log n) time; the
amortized time to search in a splay tree is O(log n); greedy algorithms usually don’t produce optimal
solutions; the traveling salesman problem is NP-hard—but these aren’t the point of the course. You
can always look up mere facts in a textbook or on the web, provided you have enough intuition and
experience to know what to look for. That’s why we let you bring cheat sheets to the exams; we don’t
want you wasting your study time trying to memorize all the facts you’ve seen.
You’ll also practice a lot of algorithm design and analysis skills—ﬁnding useful (counter)examples,
developing induction proofs, solving recurrences, using big-Oh notation, using probability, giving
problems crisp mathematical descriptions, and so on. These skills are incredibly useful, and it’s impossible
to develop good intuition and good communication skills without them, but they aren’t the main point
of the course either. At this point in your educational career, you should be able to pick up most of those
skills on your own, once you know what you’re trying to do.
Unfortunately, there is no systematic procedure—no algorithm—to determine which algorithmic
techniques are most effective at solving a given problem, or ﬁnding good ways to explain, analyze,
optimize, or implement a given algorithm. Like many other activities (music, writing, juggling, acting,
martial arts, sports, cooking, programming, teaching, etc.), the only way to master these skills is to make
them your own, through practice, practice, and more practice. You can only develop good problemsolving skills by solving problems. You can only develop good communication skills by communicating.
Good intuition is the product of experience, not its replacement. We can’t teach you how to do well in
this class. All we can do (and what we will do) is lay out some fundamental tools, show you how to use
them, create opportunities for you to practice with them, and give you honest feedback, based on our
own hard-won experience and intuition. The rest is up to you.
Good algorithms are extremely useful, elegant, surprising, deep, even beautiful. But most importantly,
algorithms are fun!! I hope this course will inspire at least some you to come play! 11 Algorithms Lecture 0: Introduction Boethius the algorist versus Pythagoras the abacist.
from Margarita Philosophica by Gregor Reisch (1503) 12 Algorithms Lecture 0: Introduction Exercises
0. Describe and analyze an algorithm that determines, given a legal arrangement of standard pieces
on a standard chess board, which player will win at chess from the given starting position if both
players play perfectly. [Hint: There is a one-line solution!]
1. The traditional Devonian/Cornish drinking song “The Barley Mow” has the following pseudolyrics18 , where container[i ] is the name of a container that holds 2i ounces of beer. One version
of the song uses the following containers: nipperkin, gill pot, half-pint, pint, quart, pottle, gallon,
half-anker, anker, ﬁrkin, half-barrel, barrel, hogshead, pipe, well, river, and ocean. (Every container
in this list is twice as big as its predecessor, except that a ﬁrkin is actually 2.25 ankers, and the last
three units are just silly.)
BARLEYMOW(n):
“Here’s a health to the barley-mow, my brave boys,”
“Here’s a health to the barley-mow!”
“We’ll drink it out of the jolly brown bowl,”
“Here’s a health to the barley-mow!”
“Here’s a health to the barley-mow, my brave boys,”
“Here’s a health to the barley-mow!” for i ← 1 to n “We’ll drink it out of the container[i ], boys,”
“Here’s a health to the barley-mow!” for j ← i downto 1
“The container[ j ],”
“And the jolly brown bowl!”
“Here’s a health to the barley-mow!”
“Here’s a health to the barley-mow, my brave boys,”
“Here’s a health to the barley-mow!” (a) Suppose each container name container[i ] is a single word, and you can sing four words
a second. How long would it take you to sing BARLEYMOW(n)? (Give a tight asymptotic
bound.)
(b) If you want to sing this song for n > 20, you’ll have to make up your own container names.
To avoid repetition, these names will get progressively longer as n increases19 . Suppose
container[n] has Θ(log n) syllables, and you can sing six syllables per second. Now how long
would it take you to sing BARLEYMOW(n)? (Give a tight asymptotic bound.)
(c) Suppose each time you mention the name of a container, you actually drink the corresponding
amount of beer: one ounce for the jolly brown bowl, and 2i ounces for each container[i ].
Assuming for purposes of this problem that you are at least 21 years old, exactly how many
ounces of beer would you drink if you sang BARLEYMOW(n)? (Give an exact answer, not just
an asymptotic bound.) 2. Describe and analyze the Boston Pool stable matching algorithm in more detail, so that the
worst-case running time is O(n2 ), as claimed earlier in the notes.
18
19 Pseudolyrics are to lyrics as pseudocode is to code.
“We’ll drink it out of the hemisemidemiyottapint, boys!” 13 Algorithms Lecture 0: Introduction 3. Consider a generalization of the stable matching problem, where some doctors do not rank all
hospitals and some hospitals do not rank all doctors, and a doctor can be assigned to a hospital
only if each appears in the other’s preference list. In this case, there are three additional unstable
situations:
• A hospital prefers an unmatched doctor to its assigned match.
• A doctor prefers an unmatched hospital to its assigned match.
• An unmatched doctor and an unmatched hospital each appear in the other’s preference list.
Describe and analyze an efﬁcient algorithm that computes a stable matching in this setting.
Note that a stable matching may leave some doctors and hospitals unmatched, even though
their preference lists are non-empty. For example, if every doctor lists Harvard as their only
acceptable hospital, and every hospital lists Doogie as their only acceptable intern, then only
Doogie and Harvard will be matched.
4. Recall that the input to the Huntington-Hill apportionment algorithm APPORTIONCONGRESS is an
array P [1 .. n], where P [i ] is the population of the i th state, and an integer R, the total number
of representatives to be allotted. The output is an array r [1 .. n], where r [i ] is the number of
representatives allotted to the i th state by the algorithm.
n Let P = i =1 P [i ] denote the total population of the country, and let ri∗ = R · P [i ]/ P denote
the ideal number of representatives for the i th state.
(a) Prove that r [i ] ≥ ri∗ for all i .
(b) Describe and analyze an algorithm that computes exactly the same congressional apportionment as APPORTIONCONGRESS in O(n log n) time. (Recall that the running time of APPORTIONCONGRESS depends on R, which could be arbitrarily larger than n.)
(c) If a state’s population is small relative to the other states, its ideal number ri∗ of representatives could be close to zero; thus, tiny states are over-represented by the HuntingtonHill apportionment process. Surprisingly, this can also be true of very large states. Let
α = (1 + 2)/2 ≈ 1.20710678119. Prove that for any > 0, there is an input to APPORTION∗
CONGRESS with maxi P [i ] = P [1], such that r [1] > (α − ) r1 .
(d) Can you improve the constant α in the previous question? c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 14 Algorithms Lecture 1: Recursion
Our life is frittered away by detail. Simplify, simplify.
— Henry David Thoreau
The control of a large force is the same principle as the control of a few men:
it is merely a question of dividing up their numbers.
— Sun Zi, The Art of War (c. 400 C.E.), translated by Lionel Giles (1910)
Nothing is particularly hard if you divide it into small jobs.
— Henry Ford 1 Recursion 1.1 Simplify and delegate Reduction is the single most common technique used in designing algorithms. Reducing one problem X
to another problem (or set of problems) Y means to write an algorithm for X , using an algorithm or Y
as a subroutine or black box.
For example, the congressional apportionment algorithm described in Lecture 0 reduces the problem
of apportioning Congress to the problem of maintaining a priority queue under the operations INSERT
and EXTRACTMAX. Those data structure operations are black boxes; the apportionment algorithm does
not depend on any speciﬁc implementation. Conversely, when we design a particular priority queue
data structure, we typically neither know nor care how our data structure will be used. Whether or not
the Census Bureau plans to use our code to apportion Congress is completely irrelevant to our design
choices. As a general rule, when we design algorithms, we may not know—and we should not care—how
the basic building blocks we use are implemented, or how your algorithm might be used as a basic
building block to solve a bigger problem.
A particularly powerful kind of reduction is recursion, which can be deﬁned loosely as follows:
• If the given instance of the problem is small or simple enough, just solve it.
• Otherwise, reduce the problem to one or more simpler instances of the same problem.
If the self-reference is confusing, it’s helpful to imagine that someone else is going to solve the simpler
problems, just as you would assume for other types of reductions. Your only task is to simplify the
original problem, or to solve it directly when simpliﬁcation is either unnecessary or impossible. The
Recursion Fairy will magically take care of the simpler subproblems.1
There is one mild technical condition that must be satisﬁed in order for any recursive method to work
correctly, namely, that there is no inﬁnite sequence of reductions to ‘simpler’ and ‘simpler’ subproblems.
Eventually, the recursive reductions must stop with an elementary base case that can be solved by some
other method; otherwise, the recursive algorithm will never terminate. This ﬁniteness condition is
almost always satisﬁed trivially, but we should always be wary of ‘obvious’ recursive algorithms that
actually recurse forever.2
1 I used to refer to ‘elves’ instead of the Recursion Fairy, referring to the traditional fairy tale about an old shoemaker who
leaves his work unﬁnished when he goes to bed, only to discover upon waking that elves have ﬁnished everything overnight.
Someone more entheogenically experienced than I might recognize them as Terence McKenna’s ‘self-adjusting machine elves’.
2
All too often, ‘obvious’ is a synonym for ‘false’. 1 Algorithms 1.2 Lecture 1: Recursion Tower of Hanoi The Tower of Hanoi puzzle was ﬁrst published by the mathematician François Éduoard Anatole Lucas in
1883, under the pseudonym ‘N. Claus (de Siam)’ (an anagram of ‘Lucas d’Amiens’). The following year,
Henri de Parville described the puzzle with the following remarkable story:3
In the great temple at Benares beneath the dome which marks the centre of the world, rests a brass plate
in which are ﬁxed three diamond needles, each a cubit high and as thick as the body of a bee. On one of
these needles, at the creation, God placed sixty-four discs of pure gold, the largest disc resting on the brass
plate, and the others getting smaller and smaller up to the top one. This is the Tower of Bramah. Day and
night unceasingly the priests transfer the discs from one diamond needle to another according to the ﬁxed and
immutable laws of Bramah, which require that the priest on duty must not move more than one disc at a time
and that he must place this disc on a needle so that there is no smaller disc below it. When the sixty-four discs
shall have been thus transferred from the needle on which at the creation God placed them to one of the other
needles, tower, temple, and Brahmins alike will crumble into dust, and with a thunderclap the world will vanish. Of course, being good computer scientists, we read this story and immediately substitute n for the
hardwired constant sixty-four. How can we move a tower of n disks from one needle to another, using a
third needles as an occasional placeholder, never placing any disk on top of a smaller disk? The Tower of Hanoi puzzle The trick to solving this puzzle is to think recursively. Instead of trying to solve the entire puzzle all
at once, let’s concentrate on moving just the largest disk. We can’t move it at the beginning, because
all the other disks are covering it; we have to move those n − 1 disks to the third needle before we can
move the nth disk. And then after we move the nth disk, we have to move those n − 1 disks back on top
of it. So now all we have to ﬁgure out is how to. . .
STOP!! That’s it! We’re done! We’ve successfully reduced the n-disk Tower of Hanoi problem
to two instances of the (n − 1)-disk Tower of Hanoi problem, which we can gleefully hand off to the
Recursion Fairy (or, to carry the original story further, to the junior monks at the temple). recursion recursion The Tower of Hanoi algorithm; ignore everything but the bottom disk
3 This English translation is from W. W. Rouse Ball and H. S. M. Coxeter’s book Mathematical Recreations and Essays. 2 Algorithms Lecture 1: Recursion Our algorithm does make one subtle but important assumption: There is a largest disk. In other
words, our recursive algorithm works for any n ≥ 1, but it breaks down when n = 0. We must handle
that base case directly. Fortunately, the monks at Benares, being good Buddhists, are quite adept at
moving zero disks from one needle to another. The base case for the Tower of Hanoi algorithm. There is no spoon. While it’s tempting to think about how all those smaller disks get moved—in other words, what
happens when the recursion is unfolded—it’s not necessary. In fact, for more complicated problems,
unfolding the recursive calls is merely distracting. Our only task is to reduce the problem to one or more
simpler instances, or to solve the problem directly if such a reduction is impossible. Our algorithm is
trivially correct when n = 0. For any n ≥ 1, the Recursion Fairy correctly moves (or more formally, the
inductive hypothesis implies that our algorithm correctly moves) the top n − 1 disks, so our algorithm is
clearly correct.
Here’s the recursive Hanoi algorithm in more typical pseudocode.
HANOI(n, src, dst, tmp):
if n > 0
HANOI(n, src, tmp, dst)
move disk n from src to dst
HANOI(n, tmp, dst, src) Let T (n) denote the number of moves required to transfer n disks—the running time of our algorithm.
Our vacuous base case implies that T (0) = 0, and the more general recursive algorithm implies that
T (n) = 2 T (n − 1) + 1 for any n ≥ 1. The annihilator method lets us quickly derive a closed form solution
T (n ) = 2n − 1. In particular, moving a tower of 64 disks requires 264 − 1 = 18,446,744,073,709,551,615
individual moves. Thus, even at the impressive rate of one move per second, the monks at Benares will
be at work for approximately 585 billion years before tower, temple, and Brahmins alike will crumble
into dust, and with a thunderclap the world will vanish. 1.3 MergeSort Mergesort is one of the earliest algorithms proposed for sorting. According to Donald Knuth, it was
suggested by John von Neumann as early as 1945.
1. Divide the array A[1 .. n] into two subarrays A[1 .. m] and A[m + 1 .. n], where m = n/2 .
2. Recursively mergesort the subarrays A[1 .. m] and A[m + 1 .. n].
3. Merge the newly-sorted subarrays A[1 .. m] and A[m + 1 .. n] into a single sorted list.
Input:
Divide:
Recurse:
Merge: S
S
I
A O
O
N
E R
R
O
G T
T
S
I I
I
R
L N
N
T
M G
G
A
N E
E
E
O A Mergesort example. 3 X
X
G
P A
A
L
S M
M
M
R P
P
P
T L
L
X
X Algorithms Lecture 1: Recursion The ﬁrst step is completely trivial—we only need to compute the median index m—and we can
delegate the second step to the Recursion Fairy. All the real work is done in the ﬁnal step; the two
sorted subarrays A[1 .. m] and A[m + 1 .. n] can be merged using a simple linear-time algorithm. Here’s
a complete speciﬁcation of the Mergesort algorithm; for simplicity, we separate out the merge step as a
subroutine.
MERGE(A[1 .. n], m):
i ← 1; j ← m + 1
for k ← 1 to n
if j > n
B [k] ← A[i ];
else if i > m
B [k] ← A[ j ];
else if A[i ] < A[ j ]
B [k] ← A[i ];
else
B [k] ← A[ j ]; MERGESORT(A[1 .. n]):
if (n > 1)
m ← n/ 2
MERGESORT(A[1 .. m])
MERGESORT(A[m + 1 .. n])
MERGE(A[1 .. n], m) i ← i+1
j ← j+1
i ← i+1
j ← j+1 for k ← 1 to n
A[k] ← B [k] To prove that the algorithm is correct, we use our old friend induction. We can prove that MERGE
is correct using induction on the total size of the two subarrays A[i .. m] and A[ j .. n] left to be merged
into B [k .. n]. The base case, where at least one subarray is empty, is straightforward; the algorithm just
copies it into B . Otherwise, the smallest remaining element is either A[i ] or A[ j ], since both subarrays
are sorted, so B [k] is assigned correctly. The remaining subarrays—either A[i + 1 .. m] and A[ j .. n], or
A[i .. m] and A[ j + 1 .. n]—are merged correctly into B [k + 1 .. n] by the inductive hypothesis.4 This
completes the proof.
Now we can prove MERGESORT correct by another round of straightforward induction. The base
cases n ≤ 1 are trivial. Otherwise, by the inductive hypothesis, the two smaller subarrays A[1 .. m] and
A[m + 1 .. n] are sorted correctly, and by our earlier argument, merged into the correct sorted output.
What’s the running time? Since we have a recursive algorithm, we’re going to get a recurrence of
some sort. MERGE clearly takes linear time, since it’s a simple for-loop with constant work per iteration.
We get the following recurrence for MERGESORT:
T (1) = O(1), T ( n) = T n/ 2 +T n/ 2 + O(n). Aside: Domain Transformations. Except for the ﬂoor and ceiling, this recurrence falls ﬁrmly into the
“all levels equal” case of the recursion tree method, or its corollary, the Master Theorem. If we simply
ignore the ﬂoor and ceiling, the method suggests the solution T (n) = O(n log n). We can easily check
that this answer is correct using induction, but there is a simple method for solving recurrences like this
directly, called domain transformation.
First we overestimate the time bound, once by pretending that the two subproblem sizes are equal,
and again to eliminate the ceiling:
T ( n) ≤ 2 T n/ 2 + O(n) ≤ 2 T (n/2 + 1) + O(n). Now we deﬁne a new function S (n) = T (n + α), where α is a constant chosen so that S (n) satisﬁes the
familiar recurrence S (n) ≤ 2S (n/2) + O(n). To ﬁgure out the appropriate value for α, we compare two
4 “The inductive hypothesis” is just a technical nickname for our friend the Recursion Fairy. 4 Algorithms Lecture 1: Recursion versions of the recurrence for T (n + α):
S (n) ≤ 2S (n/2) + O(n) =⇒ T (n + α) ≤ 2 T (n/2 + α) + O(n) T (n) ≤ 2 T (n/2 + 1) + O(n) =⇒ T (n + α) ≤ 2 T ((n + α)/2 + 1) + O(n + α) For these two recurrences to be equal, we need n/2 + α = (n + α)/2 + 1, which implies that α = 2. The
recursion tree method tells us that S (n) = O(n log n), so
T (n) = S (n − 2) = O((n − 2) log(n − 2)) = O(n log n).
We can use domain transformations to remove ﬂoors, ceilings, and lower order terms from any recurrence.
But now that we realize this, we don’t need to bother grinding through the details ever again! 1.4 Quicksort Quicksort was discovered by Tony Hoare in 1962. In this algorithm, the hard work is splitting the array
into subsets so that merging the ﬁnal result is trivial.
1. Choose a pivot element from the array.
2. Split the array into three subarrays containing the items less than the pivot, the pivot itself, and
the items bigger than the pivot.
3. Recursively quicksort the ﬁrst and last subarray.
Input:
Choose a pivot:
Partition:
Recurse: S O R T I N G E X A M P L S O R T I N G E X A M P L M
A A
E E
G G
I I
L L
M N
N R
O X
P O
S S
R P
T T
X A Quicksort example. Here’s a more formal speciﬁcation of the Quicksort algorithm. The separate PARTITION subroutine
takes the original position of the pivot element as input and returns the post-partition pivot position as
output. QUICKSORT(A[1 .. n]):
if (n > 1)
Choose a pivot element A[ p]
k ← PARTITION(A, p)
QUICKSORT(A[1 .. k − 1])
QUICKSORT(A[k + 1 .. n]) PARTITION(A[1 .. n], p):
if ( p = n)
swap A[ p] ↔ A[n]
i ← 0; j ← n
while (i < j )
repeat i ← i + 1 until (i = j or A[i ] ≥ A[n])
repeat j ← j − 1 until (i = j or A[ j ] ≤ A[n])
if (i < j )
swap A[i ] ↔ A[ j ]
if (i = n)
swap A[i ] ↔ A[n]
return i Just as we did for mergesort, we need two induction proofs to show that QUICKSORT is correct—weak
induction to prove that PARTITION correctly partitions the array, and then straightforward strong induction
to prove that QUICKSORT correctly sorts assuming PARTITION is correct. I’ll leave the gory details as an
exercise for the reader.
5 Algorithms Lecture 1: Recursion The analysis is also similar to mergesort. PARTITION runs in O(n) time: j − i = n at the beginning,
j − i = 0 at the end, and we do a constant amount of work each time we increment i or decrement j .
For QUICKSORT, we get a recurrence that depends on k, the rank of the chosen pivot:
T (n) = T (k − 1) + T (n − k) + O(n)
If we could choose the pivot to be the median element of the array A, we would have k = n/2 , the two
subproblems would be as close to the same size as possible, the recurrence would become
T ( n) = 2 T n/ 2 − 1 + T n/ 2 + O(n) ≤ 2 T (n/2) + O(n), and we’d have T (n) = O(n log n) by the recursion tree method.
In fact, it is possible to locate the median element in an unsorted array in linear time. However, the
algorithm is fairly complicated, and the hidden constant in the O() notation is quite large. So in practice,
programmers settle for something simple, like choosing the ﬁrst or last element of the array. In this case,
k can be anything from 1 to n, so we have
T (n) = max T (k − 1) + T (n − k) + O(n)
1≤k≤n In the worst case, the two subproblems are completely unbalanced—either k = 1 or k = n—and the
recurrence becomes T (n) ≤ T (n − 1) + O(n). The solution is T (n) = O(n2 ). Another common heuristic
is ‘median of three’—choose three elements (usually at the beginning, middle, and end of the array),
and take the middle one as the pivot. Although this is better in practice than just choosing one element,
we can still have k = 2 or k = n − 1 in the worst case. With the median-of-three heuristic, the recurrence
becomes T (n) ≤ T (1) + T (n − 2) + O(n), whose solution is still T (n) = O(n2 ).
Intuitively, the pivot element will ‘usually’ fall somewhere in the middle of the array, say between
n/10 and 9n/10. This suggests that the average-case running time is O(n log n). Although this intuition
is correct, we are still far from a proof that quicksort is usually efﬁcient. We will formalize this intuition
about average cases in a later lecture. 1.5 The Pattern Both mergesort and and quicksort follow the same general three-step pattern of all divide and conquer
algorithms:
1. Divide the problem into several smaller independent subproblems.
2. Delegate each subproblem to the Recursion Fairy to get a sub-solution.
3. Combine the sub-solutions together into the ﬁnal solution.
If the size of any subproblem falls below some constant threshold, the recursion bottoms out. Hopefully,
at that point, the problem is trivial, but if not, we switch to a different algorithm instead.
Proving a divide-and-conquer algorithm correct usually involves strong induction. Analyzing the
running time requires setting up and solving a recurrence, which often (but unfortunately not always!)
can be solved using recursion trees (or, if you insist, the Master Theorem), perhaps after a simple domain
transformation. 6 Algorithms 1.6 Lecture 1: Recursion Median Selection So, how do we ﬁnd the median element of an array in linear time? The following algorithm was
discovered by Manuel Blum, Bob Floyd, Vaughan Pratt, Ron Rivest, and Bob Tarjan in the early 1970s.
Their algorithm actually solves the more general problem of selecting the kth largest element in an array,
using the following recursive divide-and-conquer strategy. The subroutine PARTITION is the same as the
one used in QUICKSORT.
SELECT(A[1 .. n], k):
if n ≤ 25
use brute force
else
m ← n/5
for i ← 1 to m
B [i ] ← SELECT(A[5i − 4 .. 5i ], 3)
mom ← SELECT(B [1 .. m], m/2 ) 〈〈Brute force!〉〉
〈〈Recursion!〉〉 r ← PARTITION(A[1 .. n], mom)
if k < r
return SELECT(A[1 .. r − 1], k)
else if k > r
return SELECT(A[ r + 1 .. n], k − r )
else
return mom 〈〈Recursion!〉〉
〈〈Recursion!〉〉 If the input array is too large to handle by brute force, we divide it into n/5 blocks, each containing
exactly 5 elements, except possibly the last. (If the last block isn’t full, just throw in a few ∞s.) We ﬁnd
the median of each block by brute force and collect those medians into a new array. Then we recursively
compute the median of the new array (the median of medians — hence ‘mom’) and use it to partition
the input array. Finally, either we get lucky and the median-of-medians is the kth largest element of A,
or we recursively search one of the two subarrays.
The key insight is that these two subarrays cannot be too large or too small. The median-of-medians
is larger than n/5 /2 − 1 ≈ n/10 medians, and each of those medians is larger than two other
elements in its block. In other words, the median-of-medians is larger than at least 3n/10 elements in
the input array. Symmetrically, mom is smaller than at least 3n/10 input elements. Thus, in the worst
case, the ﬁnal recursive call searches an array of size 7n/10.
We can visualize the algorithm’s behavior by drawing the input array as a 5 × n/5 grid, which
each column represents ﬁve consecutive elements. For purposes of illustration, imagine that we sort
every column from top down, and then we sort the columns by their middle element. (Let me emphasize
that the algorithm doesn’t actually do this!) In this arrangement, the median-of-medians is the element
closest to the center of the grid. Visualizing the median of medians The left half of the ﬁrst three rows of the grid contains 3n/10 elements, each of which is smaller
than the median-of-medians. If the element we’re looking for is larger than the median-of-medians,
7 Algorithms Lecture 1: Recursion our algorithm will throw away everything smaller than the median-of-median, including those 3n/10
elements, before recursing. A symmetric argument applies when our target element is smaller than the
median-of-medians. Discarding approximately 3/10 of the array We conclude that the worst-case running time of the algorithm obeys the following recurrence:
T (n) ≤ O(n) + T (n/5) + T (7n/10).
The recursion tree method implies the solution T (n) = O(n).
A ﬁner analysis reveals that the hidden constants are quite large, even if we count only comparisons;
this is not a practical algorithm for small inputs. (In particular, mergesort uses fewer comparisons in
the worst case when n < 4,000,000.) Selecting the median of 5 elements requires 6 comparisons, so
we need 6n/5 comparisons to set up the recursive subproblem. We need another n − 1 comparisons to
partition the array after the recursive call returns. So the actual recurrence is
T (n) ≤ 11n/5 + T (n/5) + T (7n/10).
The recursion tree method implies the upper bound
T ( n) ≤ 1.7 5 i 9 11n
i ≥0 10 = 11n
5 · 10 = 22n. Multiplication Adding two n-digit numbers takes O(n) time by the standard iterative ‘ripple-carry’ algorithm, using a
lookup table for each one-digit addition. Similarly, multiplying an n-digit number by a one-digit number
takes O(n) time, using essentially the same algorithm.
What about multiplying two n-digit numbers? At least in the United States, every grade school
student (supposedly) learns to multiply by breaking the problem into n one-digit multiplications and n
additions:
31415962
× 27182818
251327696
31415962
251327696
62831924
251327696
31415962
219911734
62831924
853974377340916 8 Algorithms Lecture 1: Recursion We could easily formalize this algorithm as a pair of nested for-loops. The algorithm runs in O(n2 )
time—altogether, there are O(n2 ) digits in the partial products, and for each digit, we spend constant
time.
We can get a more efﬁcient algorithm by exploiting the following identity:
(10m a + b)(10m c + d ) = 102m ac + 10m ( bc + ad ) + bd
Here is a divide-and-conquer algorithm that computes the product of two n-digit numbers x and y ,
based on this formula. Each of the four sub-products e, f , g , h is computed recursively. The last line does
not involve any multiplications, however; to multiply by a power of ten, we just shift the digits and ﬁll
in the right number of zeros.
MULTIPLY( x , y, n):
if n = 1
return x · y
else
m ← n/ 2
a ← x /10m ; b ← x mod 10m
d ← y /10m ; c ← y mod 10m
e ← MULTIPLY(a, c , m)
f ← MULTIPLY( b, d , m)
g ← MULTIPLY( b, c , m)
h ← MULTIPLY(a, d , m)
return 102m e + 10m ( g + h) + f You can easily prove by induction that this algorithm is correct. The running time for this algorithm is
given by the recurrence
T (n) = 4 T ( n/2 ) + O(n),
T (1) = 1,
which solves to T (n) = O(n2 ) by the recursion tree method (after a simple domain transformation).
Hmm. . . I guess this didn’t help after all.
But there’s a trick, ﬁrst published by Anatoli˘ Karatsuba in 1962.5 We can compute the middle
ı
coefﬁcient bc + ad using only one recursive multiplication, by exploiting yet another bit of algebra:
ac + bd − (a − b)(c − d ) = bc + ad
This trick lets use replace the last three lines in the previous algorithm as follows:
FASTMULTIPLY( x , y, n):
if n = 1
return x · y
else
m ← n/ 2
a ← x /10m ; b ← x mod 10m
d ← y /10m ; c ← y mod 10m
e ← FASTMULTIPLY(a, c , m)
f ← FASTMULTIPLY( b, d , m)
g ← FASTMULTIPLY(a − b, c − d , m)
return 102m e + 10m (e + f − g ) + f
5 However, the same basic trick was used non-recursively by Gauss in the 1800s to multiply complex numbers using only
three real multiplications. 9 Algorithms Lecture 1: Recursion The running time of Karatsuba’s FASTMULTIPLY algorithm is given by the recurrence
T (n) ≤ 3 T ( n/2 ) + O(n), T (1) = 1. After a domain transformation, we can plug this into a recursion tree to get the solution T (n) = O(nlg 3 ) =
O(n1.585 ), a signiﬁcant improvement over our earlier quadratic-time algorithm.6
Of course, in practice, all this is done in binary instead of decimal.
We can take this idea even further, splitting the numbers into more pieces and combining them in
more complicated ways, to get even faster multiplication algorithms. Ultimately, this idea leads to the
development of the Fast Fourier transform, a more complicated divide-and-conquer algorithm that can
be used to multiply two n-digit numbers in O(n log n) time.7 We’ll talk about Fast Fourier transforms in
detail in another lecture. 1.8 Exponentiation Given a number a and a positive integer n, suppose we want to compute a n . The standard naïve method
is a simple for-loop that does n − 1 multiplications by a:
SLOWPOWER(a, n):
x←a
for i ← 2 to n
x ← x ·a
return x This iterative algorithm requires n multiplications.
Notice that the input a could be an integer, or a rational, or a ﬂoating point number. In fact,
it doesn’t need to be a number at all, as long as it’s something that we know how to multiply. For
example, the same algorithm can be used to compute powers modulo some ﬁnite number (an operation
commonly used in cryptography algorithms) or to compute powers of matrices (an operation used to
evaluate recurrences and to compute shortest paths in graphs). All we really require is that a belong
to a multiplicative group.8 Since we don’t know what kind of things we’re multiplying, we can’t know
how long a multiplication takes, so we’re forced analyze the running time in terms of the number of
multiplications.
There is a much faster divide-and-conquer method, using the following simple recursive formula:
an = a n/2 ·a n/2 . What makes this approach more efﬁcient is that once we compute the ﬁrst factor a
the second factor a n/2 using at most one more multiplication. n/2 , we can compute Karatsuba actually proposed an algorithm based on the formula (a + c )( b + d ) − ac − bd = bc + ad . This algorithm also
runs in O(nlg 3 ) time, but the actual recurrence is a bit messier: a − b and c − d are still m-digit numbers, but a + b and c + d
might have m + 1 digits. The simpliﬁcation presented here is due to Donald Knuth.
7
This fast algorithm for multiplying integers using FFTs was discovered by Arnold Schönhange and Volker Strassen in 1971.
The O(n log n) running time requires the standard assumption that O(log n)-bit integer arithmetic can be performed in constant
time; the number of bit operations is O(n log n log log n).
8
A multiplicative group (G , ⊗) is a set G and a function ⊗ : G × G → G , satisfying three axioms:
1. There is a unit element 1 ∈ G such that 1 ⊗ g = g ⊗ 1 for any element g ∈ G .
2. Any element g ∈ G has a inverse element g −1 ∈ G such that g ⊗ g −1 = g −1 ⊗ g = 1
3. The function is associative: for any elements f , g , h ∈ G , we have f ⊗ ( g ⊗ h) = ( f ⊗ g ) ⊗ h.
6 10 Algorithms Lecture 1: Recursion
FASTPOWER(a, n):
if n = 1
return a
else
x ← FASTPOWER(a, n/2 )
if n is even
return x · x
else
return x · x · a The total number of multiplications satisﬁes the recurrence T (n) ≤ T ( n/2 ) + 2, with the base case
T (1) = 0. After a domain transformation, recursion trees give us the solution T (n) = O(log n).
Incidentally, this algorithm is asymptotically optimal—any algorithm for computing a n must perform
at least Ω(log n) multiplications. In fact, when n is a power of two, this algorithm is exactly optimal.
However, there are slightly faster methods for other values of n. For example, our divide-and-conquer
algorithm computes a15 in six multiplications (a15 = a7 · a7 · a; a7 = a3 · a3 · a; a3 = a · a · a), but only
ﬁve multiplications are necessary (a → a2 → a3 → a5 → a10 → a15 ). Nobody knows of an efﬁcient
algorithm that always uses the minimum possible number of multiplications. Exercises
1. (a) Professor George O’Jungle has a 27-node binary tree, in which every node is labeled with a
unique letter of the Roman alphabet or the character &. Preorder and postorder traversals of
the tree visit the nodes in the following order:
• Preorder: I Q J H L E M V O T S B R G Y Z K C A & F P N U D W X
• Postorder: H E M L J V Q S G Y R Z B T C P U D N F W & X A K O I
Draw George’s binary tree.
(b) Describe and analyze a recursive algorithm for reconstructing a binary tree, given its preorder
and postorder node sequences.
(c) Describe and analyze a recursive algorithm for reconstructing a binary tree, given its preorder
and inorder node sequences.
2. Consider a 2n × 2n chessboard with one (arbitrarily chosen) square removed.
(a) Prove that any such chessboard can be tiled without gaps or overlaps by L-shaped pieces,
each composed of 3 squares.
(b) Describe and analyze an algorithm to compute such a tiling, given the integer n and two
n-bit integers representing the row and column of the missing square. The output is a list of
the positions and orientations of (4n − 1)/3 tiles. Your algorithm should run in O(4n ) time.
3. Prove that the recursive Tower of Hanoi algorithm is exactly equivalent to each of the following
non-recursive algorithms; in other words, prove that all three algorithms move the same disks,
to and from the same needles, in the same order. The needles are labeled 0, 1, and 2, and our
problem is to move a stack of n disks from needle 0 to needle 2 (as shown on page 2).
(a) Follow these four rules:
• Never move the same disk twice in a row.
11 Algorithms Lecture 1: Recursion
• If n is even, always move the smallest disk forward (· · · → 0 → 1 → 2 → 0 → · · · ).
• If n is odd, always move the smallest disk backward (· · · → 0 → 2 → 1 → 0 → · · · ).
• When there is no move that satisﬁes the other rules, the puzzle is solved. (b) Let ρ (n) denote the smallest integer k such that n/2k is not an integer. For example,
ρ (42) = 2, because 42/21 is an integer but 42/22 is not. (Equivalently, ρ (n) is one more
than the position of the least signiﬁcant 1 bit in the binary representation of n.) The function
ρ (n) is sometimes called the ‘ruler’ function, because its behavior resembles the marks on a
ruler:
1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, . . . .
Here’s the non-recursive algorithm in one line:
In step i , move disk ρ (i ) forward if n − i is even, backward if n − i is odd.
When this rule requires us to move disk n + 1, the algorithm ends.
4. Consider the following restricted variants of the Tower of Hanoi puzzle. In each problem, the
needles are numbered 0, 1, and 2, as in problem 3, and your task is to move a stack of n disks
from needle 1 to needle 2.
(a) Suppose you are forbidden to move any disk directly between needle 1 and needle 2; every
move must involve needle 0. Describe an algorithm to solve this version of the puzzle in as
few moves as possible. Exactly how many moves does your algorithm make?
(b) Suppose you are only allowed to move disks from needle 0 to needle 2, from needle 2 to
needle 1, or from needle 1 to needle 0. Equivalently, Suppose the needles are arranged
in a circle and numbered in clockwise order, and you are only allowed to move disks
counterclockwise. Describe an algorithm to solve this version of the puzzle in as few moves
as possible. Exactly how many moves does your algorithm make? 0 1 5 2 6 3 7 4 8 A top view of the ﬁrst eight moves in a counterclockwise Towers of Hanoi solution (c) Finally, suppose your only restriction is that you may never move a disk directly from needle
1 to needle 2. Describe an algorithm to solve this version of the puzzle in as few moves as
possible. How many moves does your algorithm make? [Hint: This is considerably harder to
analyze than the other two variants.] 12 Algorithms Lecture 1: Recursion 5. Most graphics hardware includes support for a low-level operation called blit, or block transfer,
which quickly copies a rectangular chunk of a pixel map (a two-dimensional array of pixel values)
from one location to another. This is a two-dimensional version of the standard C library function
memcpy().
Suppose we want to rotate an n × n pixel map 90◦ clockwise. One way to do this, at least when
n is a power of two, is to split the pixel map into four n/2 × n/2 blocks, move each block to its
proper position using a sequence of ﬁve blits, and then recursively rotate each block. Alternately,
we could ﬁrst recursively rotate the blocks and then blit them into place. CA
DB AB
CD AB
CD BD
AC
Two algorithms for rotating a pixel map.
Black arrows indicate blitting the blocks into place; white arrows indicate recursively rotating the blocks. The ﬁrst rotation algorithm (blit then recurse) in action. (a) Prove that both versions of the algorithm are correct when n is a power of two.
(b) Exactly how many blits does the algorithm perform when n is a power of two?
(c) Describe how to modify the algorithm so that it works for arbitrary n, not just powers of two.
How many blits does your modiﬁed algorithm perform?
(d) What is your algorithm’s running time if a k × k blit takes O(k2 ) time?
(e) What if a k × k blit takes only O(k) time?
6. An inversion in an array A[1 .. n] is a pair of indices (i , j ) such that i < j and A[i ] > A[ j ]. The
n
number of inversions in an n-element array is between 0 (if the array is sorted) and 2 (if the
array is sorted backward).
Describe and analyze an algorithm to count the number of inversions in an n-element array in
O(n log n) time. [Hint: Modify mergesort.]
13 Algorithms Lecture 1: Recursion 7. (a) Prove that the following algorithm actually sorts its input.
STOOGESORT(A[0 .. n − 1]) :
if n = 2 and A[0] > A[1]
swap A[0] ↔ A[1]
else if n > 2
m = 2 n/ 3
STOOGESORT(A[0 .. m − 1])
STOOGESORT(A[n − m .. n − 1])
STOOGESORT(A[0 .. m − 1]) (b) Would STOOGESORT still sort correctly if we replaced m = 2n/3 with m = 2n/3 ? Justify
your answer.
(c) State a recurrence (including the base case(s)) for the number of comparisons executed by
STOOGESORT.
(d) Solve the recurrence, and prove that your solution is correct. [Hint: Ignore the ceiling.]
(e) Prove that the number of swaps executed by STOOGESORT is at most n
2 . 8. Suppose you are given a stack of n pancakes of different sizes. You want to sort the pancakes so
that smaller pancakes are on top of larger pancakes. The only operation you can perform is a
ﬂip—insert a spatula under the top k pancakes, for some integer k between 1 and n, and ﬂip them
all over. Flipping the top three pancakes. (a) Describe an algorithm to sort an arbitrary stack of n pancakes using as few ﬂips as possible.
Exactly how many ﬂips does your algorithm perform in the worst case?
(b) Now suppose one side of each pancake is burned. Describe an algorithm to sort an arbitrary
stack of n pancakes, so that the burned side of every pancake is facing down, using as few
ﬂips as possible. Exactly how many ﬂips does your algorithm perform in the worst case?
9. You are a contestant on the hit game show “Beat Your Neighbors!” You are presented with an
m × n grid of boxes, each containing a unique number. It costs $100 to open a box. Your goal is to
ﬁnd a box whose number is larger than its neighbors in the grid (above, below, left, and right). If
you spend less money than any of your opponents, you win a week-long trip for two to Las Vegas
and a year’s supply of Rice-A-RoniTM , to which you are hopelessly addicted.
(a) Suppose m = 1. Describe an algorithm that ﬁnds a number that is bigger than any of its
neighbors. How many boxes does your algorithm open in the worst case?
(b) Suppose m = n. Describe an algorithm that ﬁnds a number that is bigger than any of its
neighbors. How many boxes does your algorithm open in the worst case?
(c) Prove that your solution to part (b) is optimal up to a constant factor. 14 Algorithms Lecture 1: Recursion 10. (a) Suppose we are given two sorted arrays A[1 .. n] and B [1 .. n] and an integer k. Describe
an algorithm to ﬁnd the kth smallest element in the union of A and B in Θ(log n) time. For
example, if k = 1, your algorithm should return the smallest element of A ∪ B ; if k = n, your
algorithm should return the median of A ∪ B .) You can assume that the arrays contain no
duplicate elements. [Hint: First solve the special case k = n.]
(b) Now suppose we are given three sorted arrays A[1 .. n], B [1 .. n], and C [1 .. n], and an
integer k. Describe an algorithm to ﬁnd the kth smallest element in A in O(log n) time.
(c) Finally, suppose we are given a two dimensional array A[1 .. m][1 .. n] in which every row
A[i ][ ] is sorted, and an integer k. Describe an algorithm to ﬁnd the kth smallest element in
A as quickly as possible. How does the running time of your algorithm depend on m? [Hint:
Use the linear-time SELECT algorithm as a subroutine.]
11. (a) Describe and analyze an algorithm to sort an array A[1 .. n] by calling a subroutine SQRTSORT(k),
which sorts the subarray A k + 1 .. k + n in place, given an arbitrary integer k between 0
and n − n as input. (To simplify the problem, assume that n is an integer.) Your algorithm
is only allowed to inspect or modify the input array by calling SQRTSORT; in particular, your
algorithm must not directly compare, move, or copy array elements. How many times does
your algorithm call SQRTSORT in the worst case?
(b) Prove that your algorithm from part (a) is optimal up to constant factors. In other words, if
f (n) is the number of times your algorithm calls SQRTSORT, prove that no algorithm can sort
using o( f (n)) calls to SQRTSORT.
(c) Now suppose SQRTSORT is implemented recursively, by calling your sorting algorithm from
part (a). For example, at the second level of recursion, the algorithm is sorting arrays roughly
of size n1/4 . What is the worst-case running time of the resulting sorting algorithm? (To
k
simplify the analysis, assume that the array size n has the form 22 , so that repeated square
roots are always integers.)
12. Suppose we have n points scattered inside a two-dimensional box. A kd-tree recursively subdivides
the points as follows. First we split the box into two smaller boxes with a vertical line, then we
split each of those boxes with horizontal lines, and so on, always alternating between horizontal
and vertical splits. Each time we split a box, the splitting line partitions the rest of the interior
points as evenly as possible by passing through a median point inside the box (not on its boundary).
If a box doesn’t contain any points, we don’t split it any more; these ﬁnal empty boxes are called
cells. A kd-tree for 15 points. The dashed line crosses the four shaded cells. 15 Algorithms Lecture 1: Recursion (a) How many cells are there, as a function of n? Prove your answer is correct.
(b) In the worst case, exactly how many cells can a horizontal line cross, as a function of n?
Prove your answer is correct. Assume that n = 2k − 1 for some integer k.
(c) Suppose we have n points stored in a kd-tree. Describe and analyze an algorithm that counts
the number of points above a horizontal line (such as the dashed line in the ﬁgure) as quickly
as possible. [Hint: Use part (b).]
(d) Describe an analyze an efﬁcient algorithm that counts, given a kd-tree storing n points, the
number of points that lie inside a rectangle R with horizontal and vertical sides. [Hint: Use
part (c).]
13. You are at a political convention with n delegates, each one a member of exactly one political
party. It is impossible to tell which political party any delegate belongs to; in particular, you will
be summarily ejected from the convention if you ask. However, you can determine whether any
two delegates belong to the same party or not by introducing them to each other—members of the
same party always greet each other with smiles and friendly handshakes; members of different
parties always greet each other with angry stares and insults.
(a) Suppose a majority (more than half) of the delegates are from the same political party.
Describe an efﬁcient algorithm that identiﬁes a member (any member) of the majority party.
(b) Now suppose exactly k political parties are represented at the convention and one party
has a plurality: more delegates belong to that party than to any other. Present a practical
procedure to pick a person from the plurality political party as parsimoniously as possible.
(Please.)
14. The median of a set of size n is its n/2 th largest element, that is, the element that is as close
as possible to the middle of the set in sorted order. In this lecture, we saw a fairly complicated
algorithm to compute the median in O(n) time.
During your lifelong quest for a simpler linear-time median-ﬁnding algorithm, you meet and
befriend the Near-Middle Fairy. Given any set X , the Near-Middle Fairy can ﬁnd an element m ∈ X
that is near the middle of X in O(1) time. Speciﬁcally, at least a third of the elements of X are
smaller than m, and at least a third of the elements of X are larger than m.
Describe and analyze a simple recursive algorithm to ﬁnd the median of a set in O(n) time if
you are allowed to ask the Near-Middle Fairy for help.
15. For this problem, a subtree of a binary tree means any connected subgraph. A binary tree is
complete if every internal node has two children, and every leaf has exactly the same depth.
Describe and analyze a recursive algorithm to compute the largest complete subtree of a given
binary tree. Your algorithm should return the root and the depth of this subtree. The largest complete subtree of this binary tree has depth 2. 16 Algorithms Lecture 1: Recursion 16. Consider the following classical recursive algorithm for computing the factorial n! of a non-negative
integer n:
FACTORIAL(n):
if n = 0
return 0
else
return n · FACTORIAL(n − 1) (a) How many multiplications does this algorithm perform?
(b) How many bits are required to write n! in binary? Express your answer in the form Θ( f (n)),
for some familiar function f (n). [Hint: Use Stirling’s approximation: n! ≈ 2πn · (n/e)n .]
(c) Your answer to (b) should convince you that the number of multiplications is not a good
estimate of the actual running time of FACTORIAL. The grade-school multiplication algorithm
takes O(k · l ) time to multiply a k-digit number and an l -digit number. What is the running
time of FACTORIAL if we use this multiplication algorithm as a subroutine?
(d) The following algorithm also computes n!, but groups the multiplication differently:
FACTORIAL2(n, m):
〈〈Compute n!/(n − m)!〉〉
if m = 0
return 1
else if m = 1
return n
else
return FACTORIAL2(n, m/2 ) · FACTORIAL2(n − m/2 , m/2 ) What is the running time of FACTORIAL2(n, n) if we use grade-school multiplication? [Hint:
Ignore the ﬂoors and ceilings.]
(e) Describe and analyze a variant of Karastuba’s algorithm that can multiply any k-digit number
and any l -digit number, where k ≥ l , in O(k · l lg 3−1 ) = O(k · l 0.585 ) time.
(f) What are the running times of FACTORIAL(n) and FACTORIAL2(n, n) if we use the modiﬁed
Karatsuba multiplication from part (e)? c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 17 Algorithms Non-Lecture A: Fast Fourier Transforms
Calvin: Here’s another math problem I can’t ﬁgure out. What’s 9+4?
Hobbes: Ooh, that’s a tricky one. You have to use calculus and imaginary
numbers for this.
Calvin: IMAGINARY NUMBERS?!
Hobbes: You know, eleventeen, thirty-twelve, and all those. It’s a little
confusing at ﬁrst.
Calvin: How did YOU learn all this? You’ve never even gone to school!
Hobbes: Instinct. Tigers are born with it.
— “Calvin and Hobbes” (January 6, 1998)
It needs evaluation
So let the games begin
A heinous crime, a show of force
A murder would be nice, of course
— “Bad Horse’s Letter”, Dr. Horrible’s Sing-Along Blog (2008) A Fast Fourier Transforms A.1 Polynomials In this lecture we’ll talk about algorithms for manipulating polynomials: functions of one variable built
from additions subtractions, and multiplications (but no divisions). The most common representation
for a polynomial p( x ) is as a sum of weighted powers of a variable x :
n aj x j. p( x ) =
j =0 The numbers a j are called coefﬁcients. The degree of the polynomial is the largest power of x ; in the
example above, the degree is n. Any polynomial of degree n can be speciﬁed by a sequence of n + 1
coefﬁcients. Some of these coefﬁcients may be zero, but not the nth coefﬁcient, because otherwise the
degree would be less than n.
Here are three of the most common operations that are performed with polynomials:
• Evaluate: Give a polynomial p and a number x , compute the number p( x ).
• Add: Give two polynomials p and q, compute a polynomial r = p + q, so that r ( x ) = p( x ) + q( x )
for all x . If p and q both have degree n, then their sum p + q also has degree n.
• Multiply: Give two polynomials p and q, compute a polynomial r = p · q, so that r ( x ) = p( x ) · q( x )
for all x . If p and q both have degree n, then their product p · q has degree 2n.
Suppose we represent a polynomial of degree n as an array of n + 1 coefﬁcients P [0 .. n], where
P [ j ] is the coefﬁcient of the x j term. We learned simple algorithms for all three of these operations in
high-school algebra:
EVALUATE( P [0 .. n], x ):
X ← 1 〈〈X = x j 〉〉
y ←0
for j ← 0 to n
y ← y + P [ j] · X
X ←X·x
return y ADD( P [0 .. n], Q[0 .. n]):
for j ← 0 to n
R[ j ] ← P [ j ] + Q [ j ]
return R[0 .. n] 1 MULTIPLY( P [0 .. n], Q[0 .. m]):
for j ← 0 to n + m
R[ j ] ← 0
for j ← 0 to n
for k ← 0 to m
R[ j + k] ← R[ j + k] + P [ j ] · Q[k]
return R[0 .. n + m] Algorithms Non-Lecture A: Fast Fourier Transforms EVALUATE uses O(n) arithmetic operations.1 This is the best we can hope for, but we can cut the
number of multiplications in half using Horner’s rule:
p( x ) = a0 + x (a1 + x (a2 + . . . + x an )).
HORNER( P [0 .. n], x ):
y ← P [ n]
for i ← n − 1 downto 0
y ← x · y + P [i ]
return y The addition algorithm also runs in O(n) time, and this is clearly the best we can do.
The multiplication algorithm, however, runs in O(n2 ) time. In the previous lecture, we saw a
divide and conquer algorithm (due to Karatsuba) for multiplying two n-bit integers in only O(nlg 3 )
steps; precisely the same algorithm can be applied here. Even cleverer divide-and-conquer strategies
lead to multiplication algorithms whose running times are arbitrarily close to linear—O(n1+ ) for your
favorite value e > 0—but with great cleverness comes great confusion. These algorithms are difﬁcult to
understand, even more difﬁcult to implement correctly, and not worth the trouble in practice thanks to
large constant factors. A.2 Alternate Representations Part of what makes multiplication so much harder than the other two operations is our input representation. Coefﬁcients vectors are the most common representation for polynomials, but there are at least
two other useful representations.
A.2.1 Roots The Fundamental Theorem of Algebra states that every polynomial p of degree n has exactly n roots
r1 , r2 , . . . rn such that p( r j ) = 0 for all j . Some of these roots may be irrational; some of these roots may
by complex; and some of these roots may be repeated. Despite these complications, this theorem implies
a unique representation of any polynomial of the form
n p( x ) = s (x − rj)
j =1 where the r j ’s are the roots and s is a scale factor. Once again, to represent a polynomial of degree n, we
need a list of n + 1 numbers: one scale factor and n roots.
Given a polynomial in this root representation, we can clearly evaluate it in O(n) time. Given two
polynomials in root representation, we can easily multiply them in O(n) time by multiplying their scale
factors and just concatenating the two root sequences.
Unfortunately, if we want to add two polynomials in root representation, we’re pretty much out of
luck. There’s essentially no correlation between the roots of p, the roots of q, and the roots of p + q.
We could convert the polynomials to the more familiar coefﬁcient representation ﬁrst—this takes O(n2 )
I’m going to assume in this lecture that each arithmetic operation takes O(1) time. This may not be true in practice; in fact,
one of the most powerful applications of FFTs is fast integer multiplication. One of the fastest integer multiplication algorithms,
due to Schönhage and Strassen, multiplies two n-bit binary numbers using O(n log n log log n log log log n log log log log n · · · )
bit operations. The algorithm uses an n-element Fast Fourier Transform, which requires several O(log n)-nit integer multiplications. These smaller multiplications are carried out recursively (of course!), which leads to the cascade of logs in the running
time. Needless to say, this is a can of worms.
1 2 Algorithms Non-Lecture A: Fast Fourier Transforms time using the high-school algorithms—but there’s no easy way to convert the answer back. In fact, for
most polynomials of degree 5 or more in coefﬁcient form, it’s impossible to compute roots exactly.2
A.2.2 Samples Our third representation for polynomials comes from a different consequence of the Fundamental
Theorem of Algebra. Given a list of n + 1 pairs {( x 0 , y0 ), ( x 1 , y1 ), . . . , ( x n , yn ) }, there is exactly one
polynomial p of degree n such that p( x j ) = y j for all j . This is just a generalization of the fact that
any two points determine a unique line, since a line is (the graph of) a polynomial of degree 1. We say
that the polynomial p interpolates the points ( x j , y j ). As long as we agree on the sample locations x j in
advance, we once again need exactly n + 1 numbers to represent a polynomial of degree n.
Adding or multiplying two polynomials in this sample representation is easy, as long as they use
the same sample locations x j . To add the polynomials, just add their sample values. To multiply two
polynomials, just multiply their sample values; however, if we’re multiplying two polynomials of degree
n, we need to start with 2n + 1 sample values for each polynomial, since that’s how many we need to
uniquely represent the product polynomial. Both algorithms run in O(n) time.
Unfortunately, evaluating a polynomial in this representation is no longer trivial. The following
formula, due to Lagrange, allows us to compute the value of any polynomial of degree n at any point,
given a set of n + 1 samples. n−1
n−1
(x − xk)
yj
k= j =
p( x ) =
yj
( x − x k ) k= j ( x j − x k )
k= j ( x j − x k ) k= j
j =0
j =0
Hopefully it’s clear that formula actually describes a polynomial, since each term in the rightmost sum
is written as a scaled product of monomials. It’s also not hard to check that p( x j ) = y j for all j . As I
mentioned earlier, the fact that this is the only polynomial that interpolates the points {( x j , y j )} is an
easy consequence of the Fundamental Theorem of Algebra. We can easily transform Lagrange’s formula
into an O(n2 )-time algorithm.
A.2.3 Summary We ﬁnd ourselves in the following frustrating situation. We have three representations for polynomials
and three basic operations. Each representation allows us to almost trivially perform a different pair of
operations in linear time, but the third takes at least quadratic time, if it can be done at all!
evaluate multiply coefﬁcients O ( n) O ( n) O (n 2 ) roots + scale O ( n) ∞ O ( n) samples A.3 add O (n 2 ) O ( n) O ( n) Converting Between Representations What we need are fast algorithms to convert quickly from one representation to another. That way,
when we need to perform an operation that’s hard for our default representation, we can switch to a
different representation that makes the operation easy, perform that operation, and then switch back.
2 This is where numerical analysis comes from. 3 Algorithms Non-Lecture A: Fast Fourier Transforms This strategy immediately rules out the root representation, since (as I mentioned earlier) ﬁnding roots
of polynomials is impossible in general, at least if we’re interested in exact results.
So how do we convert from coefﬁcients to samples and back? Clearly, once we choose our sample
positions x j , we can compute each sample value y j = p( x j ) in O(n) time from the coefﬁcients using
Horner’s rule. So we can convert a polynomial of degree n from coefﬁcients to samples in O(n2 ) time.
The Lagrange formula gives us an explicit conversion algorithm from the sample representation back to
the more familiar coefﬁcient representation. If we use the naïve algorithms for adding and multiplying
polynomials (in coefﬁcient form), this conversion takes O(n3 ) time.
We can improve the cubic running time by observing that both conversion problems boil down to
computing the product of a matrix and a vector. The explanation will be slightly simpler if we assume
the polynomial has degree n − 1, so that n is the number of coefﬁcients or samples. Fix a sequence
j
x 0 , x 1 , . . . , x n−1 of sample positions, and let V be the n × n matrix where vi j = x i (indexing rows and
columns from 0 to n − 1): n
2
1
x0
x0
· · · x 0 −1 n
2
1
x1
x1
· · · x 1 −1 n
2
x2
x2
· · · x 2 −1 .
V = 1 .
.
.
.
..
.
.
.
.
.
.
.
.
.
n−1
2
1 x n−1 x n−1 · · · x n−1
The matrix V is called a Vandermonde matrix. The vector of coefﬁcients a = (a0 , a1 , . . . , an−1 ) and the
vector of sample values y = ( y0 , y1 , . . . , yn−1 ) are related by the matrix equation
V a = y,
or in more detail, 1 1 1 .
.
.
1 x0 2
x0 ··· n
x 0 −1 x1 2
x1 ··· x2
.
.
. 2
x2
.
.
. ···
..
. n
x 1 −1 a1 y1 n
x 2 −1 a2 = y2 . . . . . . . . . . n−1
yn−1
x n−1
an−1 x n−1 2
x n−1 · · · a0 y0 Given this formulation, we can clearly transform any coefﬁcient vector a into the corresponding
sample vector y in O(n2 ) time.
Conversely, if we know the sample values y , we can recover the coefﬁcients by solving a system of n
linear equations in n unknowns; this takes O(n3 ) time if we use Gaussian elimination. But we can speed
this up by implicitly hard-coding the sample positions into the algorithm, To convert from samples to
coefﬁcients, we can simply multiply the sample vector by the inverse of V , again in O(n2 ) time.
a = V −1 y
Computing V −1 would take O(n3 ) time if we had to do it from scratch using Gaussian elimination, but
because we ﬁxed the set of sample positions in advance, the matrix V −1 can be written directly into the
algorithm.3
So we can convert from coefﬁcients to sample value and back in O(n2 ) time, which is pointless,
because we can add, multiply, or evaluate directly in either representation in O(n2 ) time. But wait!
Actually, it is possible to invert an n × n matrix in o(n3 ) time, using fast matrix multiplication algorithms that closely
resemble Karatsuba’s sub-quadratic divide-and-conquer algorithm for integer/polynomial multiplication.
3 4 Algorithms Non-Lecture A: Fast Fourier Transforms There’s a degree of freedom we haven’t exploited—We get to choose the sample positions! Our
conversion algorithm may be slow only because we’re trying to be too general. Perhaps, if we choose
a set of sample points with just the right kind of recursive structure, we can do the conversion more
quickly. In fact, there is a set of sample points that’s perfect for the job. A.4 The Discrete Fourier Transform Given a polynomial of degree n − 1, we’d like to ﬁnd n sample points that are somehow as symmetric as
possible. The most natural choice for those n points are the nth roots of unity; these are the roots of the
polynomial x n − 1 = 0. These n roots are spaced exactly evenly around the unit circle in the complex
plane.4 Every nth root of unity is a power of the primitive root
ωn = e2πi /n = cos 2π
n + i sin 2π
n . A typical nth root of unity has the form
j
ωn = e(2πi /n) j = cos 2π
n j + i sin 2π
n j. These complex numbers have several useful properties for any integers n and k:
• There are only n different nth roots of unity: ωk = ωk mod n .
n
n
k
• If n is even, then ωn+n/2 = −ωk ; in particular, ωn/2 = −ω0 = −1.
n
n
n • 1/ωk = ω−k = ωk = (ωn )k , where the bar represents complex conjugation: a + bi = a − bi
n
n
n
• ωn = ωk . Thus, every nth root of unity is also a (kn)th root of unity.
kn
If we sample a polynomial of degree n − 1 at the nth roots of unity, the resulting list of sample values
is called the discrete Fourier transform of the polynomial (or more formally, of the coefﬁcient vector).
Thus, given an array P [0 .. n − 1] of coefﬁcients, the discrete Fourier transform computes a new vector
P ∗ [0 .. n − 1] where
n−1
j
P ∗ [ j ] = p(ωn ) = jk
P [ k ] · ωn
k =0 We can obviously compute P ∗ in O(n2 ) time, but the structure of the nth roots of unity lets us do better.
But before we describe that faster algorithm, let’s think about how we might invert this transformation.
Recall that transforming coefﬁcients into sample values is a linear transformation; the sample vector
is the product of a Vandermonde matrix V and the coefﬁcient vector. For the discrete Fourier transform,
each entry in V is an nth root of unity; speciﬁcally,
jk
v jk = ωn
4 In this lecture, i always represents the square root of −1. Computer scientists are used to thinking of i as an integer index
into a sequence, an array, or a for-loop, but we obviously can’t do that here. The physicist’s habit of using j = −1 just delays
◦
the problem (How do physicists write quaternions?), and typographical tricks like I or i or Mathematica’s ı are just stupid.
ı 5 Algorithms Non-Lecture A: Fast Fourier Transforms for all integers j and k. Thus, 1
1
1
1
1 ω
2
ωn
ω3
n n 1 ω2
ω4
ω6
n
n
n
V =
1 ω3
ω6
ω9
n
n
n .
.
.
.
.
.
.
.
.
.
.
. n−1
2(n−1)
3(n−1)
1 ωn
ωn
ωn ···
···
···
···
..
.
··· 1 ωn−1 n ω2(n−1) n ω3(n−1) n .
. . 2
ω(n−1)
n To invert the discrete Fourier transform, converting sample values back to coefﬁcients, we just have
to multiply P ∗ by the inverse matrix V −1 . But the following amazing fact implies that this is almost the
same as multiplying by V itself:
Claim: V −1 = V /n
Proof: Let W = V /n. We just have to show that M = V W is the identity matrix. We can compute a
single entry in M as follows:
n−1 n−1 v jl · w l k = m jk =
l =0
j −k If j = k, then ωn jl
ωn lk · ωn / n = l =0 = ω0 = 1, so
n
m jk = 1 n−1
n 1= l =0 1 n −1
n n
n jl
ωn −lk l =0 1 n −1 j − k l
=
(ω )
n l =0 n = 1, and if j = k, we have a geometric series
n −1
j
(ωn−k )l = m jk =
l =0 j −k (ωn )n − 1
j −k ωn −1 = (ωn ) j −k − 1
n
j −k ωn −1 = 1 j −k − 1
j −k ωn −1 = 0. That’s it!
jk − jk In other words, if W = V −1 then w jk = v jk /n = ωn /n = ωn /n. What this means for us computer
scientists is that any algorithm for computing the discrete Fourier transform can be easily modiﬁed to
compute the inverse transform as well. A.5 Divide and Conquer! The symmetry in the roots of unity also allow us to compute the discrete Fourier transform efﬁciently
using a divide and conquer strategy. The basic structure of the algorithm is almost the same as
MergeSort, and the O(n log n) running time will ultimately follow from the same recurrence. The Fast
Fourier Transform algorithm, popularized by Cooley and Tukey in 19655 , assumes that n is a power of
two; if necessary, we can just pad the coefﬁcient vector with zeros.
5 Actually, the FFT algorithm was previously published by Runge and König in 1924, and again by Yates in 1932, and again
by Stumpf in 1937, and again by Danielson and Lanczos in 1942. So of course it’s often called the Coley-Tukey algorithm. But
the algorithm was ﬁrst used by Gauss in the 1800s for calculating the paths of asteroids from a ﬁnite number of equally-spaced
observations. By hand. Fourier himself always did it the hard way.
Cooley and Tukey apparently developed their algorithm to help detect Soviet nuclear tests without actually visiting Soviet
nuclear facilities, by interpolating off-shore seismic readings. Without their rediscovery of the FFT algorithm, the nuclear test
ban treaty would never have been ratiﬁed, and we’d all be speaking Russian, or more likely, whatever language radioactive
glass speaks. 6 Algorithms Non-Lecture A: Fast Fourier Transforms Let p( x ) be a polynomial of degree n − 1, represented by an array P [0 .. n − 1] of coefﬁcients. The
FFT algorithm begins by splitting p into two smaller polynomials u and v , each with degree n/2 − 1.
The coefﬁcients of u are precisely the the even-degree coefﬁcients of p; the coefﬁcients of v are the
odd-degree coefﬁcients of p. For example, if p( x ) = 3 x 3 − 4 x 2 + 7 x + 5, then u( x ) = −4 x + 5 and
v ( x ) = 3 x + 7. These three polynomials are related by the equation
p( x ) = u( x 2 ) + x · v ( x 2 ).
In particular, if x is an nth root of unity, we have
p(ωk ) = u(ω2k ) + ωk · v (ω2k ).
n
n
n
n
Now we can exploit those roots of unity again. Since n is a power of two, n must be even, so we
k mod n/2
have ω2k = ωk /2 = ωn/2
. In other words, the values of p at the nth roots of unity depend on the
n
n
values of u and v at (n/2)th roots of unity.
k mod n/2 p(ωk ) = u(ωn/2
n k mod n/2 ) + ωk · v (ωn/2
n ). But those are just coefﬁcients in the DFTs of u and v ! We conclude that the DFT coefﬁcients of P are
deﬁned by the following recurrence:
P ∗ [k] = U ∗ [k mod n/2] + ωk · V ∗ [k mod n/2]
n
Once the Recursion Fairy give us U ∗ and V ∗ , we can compute P ∗ in linear time. The base case for the
recurrence is n = 1: if p( x ) has degree 0, then P ∗ [0] = P [0].
Here’s the complete FFT algorithm, along with its inverse.
FFT( P [0 .. n − 1]):
if n = 1
return P INVERSEFFT( P ∗ [0 .. n − 1]):
if n = 1
return P for j ← 0 to n/2 − 1
U [ j ] ← P [2 j ]
V [ j ] ← P [2 j + 1] for j ← 0 to n/2 − 1
U ∗ [ j ] ← P ∗ [2 j ]
V ∗ [ j ] ← P ∗ [ 2 j + 1] U ∗ ← FFT(U [0 .. n/2 − 1])
V ∗ ← FFT(V [0 .. n/2 − 1]) U ← INVERSEFFT(U [0 .. n/2 − 1])
V ← INVERSEFFT(V [0 .. n/2 − 1]) π
π
ωn ← cos( 2n ) + i sin( 2n )
ω←1 π
π
ωn ← cos( 2n ) − i sin( 2n )
ω←1 for j ← 0 to n/2 − 1
P ∗[ j]
← U ∗[ j] + ω · V ∗[ j]
∗
P [ j + n/ 2] ← U ∗ [ j ] − ω · V ∗ [ j ]
ω ← ω · ωn for j ← 0 to n/2 − 1
P [ j]
← 2(U [ j ] + ω · V [ j ])
P [ j + n/2] ← 2(U [ j ] − ω · V [ j ])
ω ← ω · ωn return P ∗ [0 .. n − 1] return P [0 .. n − 1] The overall running time of this algorithm satisﬁes the recurrence T (n) = Θ(n) + 2 T (n/2), which as
we all know solves to T (n) = Θ(n log n). A.6 Fast Multiplication Given two polynomials p and q, each represented by an array of coefﬁcients, we can multiply them in
Θ(n log n) arithmetic operations as follows. First, pad the coefﬁcient vectors and with zeros until the
7 Algorithms Non-Lecture A: Fast Fourier Transforms size is a power of two greater than or equal to the sum of the degrees. Then compute the DFTs of each
coefﬁcient vector, multiply the sample values one by one, and compute the inverse DFT of the resulting
sample vector.
FFTMULTIPLY( P [0 .. n − 1], Q[0 .. m − 1]):
← lg(n + m)
for j ← n to 2 − 1
P [ j] ← 0
for j ← m to 2 − 1
Q[ j ] ← 0
P ∗ ← F F T (P )
Q∗ ← F F T (Q)
for j ← 0 to 2 − 1
R∗ [ j ] ← P ∗ [ j ] · Q∗ [ j ]
return INVERSEFFT(R∗ ) A.7 Inside the FFT FFTs are often implemented in hardware as circuits. To see the recursive structure of the circuit, let’s
connect the top-level inputs and outputs to the inputs and outputs of the recursive calls. On the left we
split the input P into two recursive inputs U and V . On the right, we combine the outputs U ∗ and V ∗ to
obtain the ﬁnal output P ∗ .
000 FFT(n/2) V* 110 001 001 011 101 101 011 111 V 110 110 P* 010 101 P 100 100 100 U* 010 010
011 FFT(n/2) 000 001 U 000 111 111 bit reversal permutation butterfly network The recursive structure of the FFT algorithm. If we expand this recursive structure completely, we see that the circuit splits naturally into two
parts. The left half computes the bit-reversal permutation of the input. To ﬁnd the position of P [k] in
this permutation, write k in binary, and then read the bits backward. For example, in an 8-element
bit-reversal permutation, P [3] = P [0112 ] ends up in position 6 = 1102 . The right half of the FFT circuit
is a butterﬂy network. Butterﬂy networks are often used to route between processors in massively-parallel
computers, since they allow any processor to communicate with any other in only O(log n) steps. 8 Algorithms Non-Lecture A: Fast Fourier Transforms Caveat Lector! This presentation is appropriate for graduate students or undergrads with strong
math backgrounds, but it leaves most undergrads confused. You may ﬁnd it less confusing to approach
the material in the opposite order, as follows:
First, any polynomial can be split into even-degree and odd-degree parts: p( x ) = peven ( x 2 ) + x · podd ( x 2 ).
We can evaluate p( x ) by recursively evaluating peven ( x 2 ) and podd ( x 2 ) and doing O(1) arithmetic
operations.
Now suppose our task is to evaluate the degree-n polynomial p( x ) at n different points x , as
quickly as possible. To exploit the even/odd recursive structure, we must choose the n evaluation
points carefully. Call a set X of n values delicious if one of the following conditions holds: • X has one element.
• The set X 2 = { x 2 | x ∈ X } has only n/2 elements and is delicious.
Clearly the size of any delicious set is a power of 2. If someone magically handed us a delicious
set X , we could compute { p( x ) | x ∈ X } in O(n log n) time using the even/odd recursive structure. Bit
reversal permutation, blah blah blah, butterﬂy network, yadda yadda yadda.
If n is a power of two, then the set of integers {0, 1, . . . , n − 1} is delicious, provided we perform
all arithmetic modulo n . But that only tells us p( x ) mod n, and we want the actual value of p( x ).
Of course, we can use larger moduli: {0, c , 2c , . . . , (n − 1)c } is delicious mod cn. We can avoid modular
arithmetic entirely by using complex roots of unity—the set {e2πi (k/n) | k = 0, 1, . . . , n − 1} is delicious!
The sequence of values p(e2πi (k/n) ) is called the discrete Fourier transform of p.
Finally, to invert this transformation from coefﬁcients to values, we repeat exactly the same
procedure, using the same delicious set but in the opposite order. Blardy blardy, linear algebra, hi
dee hi dee hi dee ho. Exercises
1. For any two sets X and Y of integers, the Minkowski sum X + Y is the set of all pairwise sums
{ x + y | x ∈ X , y ∈ Y }.
(a) Describe an analyze and algorithm to compute the number of elements in X + Y in O(n2 log n)
time. [Hint: The answer is not always n2 .]
(b) Describe and analyze an algorithm to compute the number of elements in X + Y in O( M log M )
time, where M is the largest absolute value of any element of X ∪ Y . [Hint: What’s this
lecture about?] 2. (a) Describe an algorithm that determines whether a given set of n integers contains two elements
whose sum is zero, in O(n log n) time.
(b) Describe an algorithm that determines whether a given set of n integers contains three
elements whose sum is zero, in O(n2 ) time.
(c) Now suppose the input set X contains only integers between − 10000n and 10000n. Describe
an algorithm that determines whether X contains three elements whose sum is zero, in
O(n log n) time. [Hint: Hint.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 9 Algorithms Lecture 2: Backtracking
’Tis a lesson you should heed,
Try, try again;
If at ﬁrst you don’t succeed,
Try, try again;
Then your courage should appear,
For, if you will persevere,
You will conquer, never fear;
Try, try again.
— Thomas H. Palmer, The Teacher’s Manual: Being an Exposition
of an Efﬁcient and Economical System of Education
Suited to the Wants of a Free People (1840)
When you come to a fork in the road, take it.
— Yogi Berra 2 Backtracking In this lecture, I want to describe another recursive algorithm strategy called backtracking. A backtracking
algorithm tries to build a solution to a computational problem incrementally. Whenever the algorithm
needs to decide between two alternatives to the next component of the solution, it simply tries both
options recursively. 2.1 n -Queens
The prototypical backtracking problem is the classical n-Queens Problem, ﬁrst proposed by German
chess enthusiast Max Bezzel in 1848 for the standard 8 × 8 board, and both solved and generalized to
larger boards by Franz Nauck in 1850. The problem is to place n queens on an n × n chessboard, so that
no two queens can attack each other. For readers not familiar with the rules of chess, this means that no
two queens are in the same row, column, or diagonal. ♕
♛
♕
♛
♕
♕
♕
♛
♕
♛
♕
♕
One solution to the 8 queens problem, represented by the array [4,7,3,8,2,5,1,6] Obviously, in any solution to the n-Queens problem, there is exactly one queen in each column. So
we will represent our possible solutions using an array Q[1 .. n], where Q[i ] indicates the square in
column i that contains a queen, or 0 if no queen has yet been placed in column i . To ﬁnd a solution,
we will put queens on the board row by row, starting at the top. A partial solution is an array Q[1 .. n]
whose ﬁrst r − 1 entries are positive and whose last n − r + 1 entries are all zeros, for some integer r .
The following recursive algorithm recursively enumerates all complete n-queens solutions that are
consistent with a given partial solution. The input parameter r is the ﬁrst empty row. Thus, to compute
all n-queens solutions with no restrictions, we would call RECURSIVENQUEENS(Q[1 .. n], 1). 1 Algorithms Lecture 2: Backtracking
RECURSIVENQUEENS(Q[1 .. n], r ):
if r = n + 1
print Q
else
for j ← 1 to n
legal ← TRUE
for i ← 1 to r − 1
if (Q[i ] = j ) or (Q[i ] = r − j + i ) or (Q[i ] = r + i − j )
legal ← FALSE
if legal
Q[ r ] ← j
RECURSIVENQUEENS(Q[1 .. n], r + 1) Like most recursive algorithms, the execution of a backtracking algorithm can be illustrated using a
recursion tree. The root of the recursion tree corresponds to the original invocation of the algorithm;
edges in the tree correspond to recursive calls. A path from the root down to any node shows the history
of a partial solution to the n-Queens problem, as queens are added to successive rows. The leaves
correspond to partial solutions that cannot be extended, either because there is already a queen on every
row, or because every position in the next empty row is in the same row, column, or diagonal as an
existing queen. ♕ ♕ ♕
♛ ♕ ♕ ♕
♛ ♕
♛ ♕ ♕ ♕ ♕
♛ ♕
♛ ♕
♛ ♕
♛
♕ ♕ ♕
♕
♛ ♕
♛ ♕
♛ ♕ ♕ ♕
♛ ♕
♛
♕
♛ ♕ ♕
♛ ♕
♛ ♕ ♕
♕ ♕
♛ ♕ ♕
♛
♕
♛ ♕ The complete recursion tree for our algorithm for the 4-queens problem. 2.2 Subset Sum Let’s consider a more complicated problem, called SUBSETSUM: Given a set X of positive integers and
target integer T , is there a subset of elements in X that add up to T ? Notice that there can be more
than one such subset. For example, if X = {8, 6, 7, 5, 3, 10, 9} and T = 15, the answer is TRUE, thanks to 2 Algorithms Lecture 2: Backtracking the subsets {8, 7} or {7, 5, 3} or {6, 9} or {5, 10}. On the other hand, if X = {11, 6, 5, 1, 7, 13, 12} and
T = 15, the answer is FALSE.
There are two trivial cases. If the target value T is zero, then we can immediately return TRUE,
because the elements of the empty set add up to zero.1 On the other hand, if T < 0, or if T = 0 but the
set X is empty, then we can immediately return FALSE.
For the general case, consider an arbitrary element x ∈ X . (We’ve already handled the case where X
is empty.) There is a subset of X that sums to T if and only if one of the following statements is true:
• There is a subset of X that includes x and whose sum is T .
• There is a subset of X that excludes x and whose sum is T .
In the ﬁrst case, there must be a subset of X \ { x } that sums to T − x ; in the second case, there must
be a subset of X \ { x } that sums to T . So we can solve SUBSETSUM(X , T ) by reducing it to two simpler
instances: SUBSETSUM(X \ { x }, T − x ) and SUBSETSUM(X \ { x }, T ). Here’s how the resulting recusive
algorithm might look if X is stored in an array.
SUBSETSUM(X [1 .. n], T ):
if T = 0
return TRUE
else if T < 0 or n = 0
return FALSE
else
return SUBSETSUM(X [2 .. n], T ) ∨ SUBSETSUM(X [2 .. n], T − X [1]) Proving this algorithm correct is a straightforward exercise in induction. If T = 0, then the elements
of the empty subset sum to T , so TRUE is the correct output. Otherwise, if T is negative or the set X is
empty, then no subset of X sums to T , so FALSE is the correct output. Otherwise, if there is a subset that
sums to T , then either it contains X [1] or it doesn’t, and the Recursion Fairy correctly checks for each of
those possibilities. Done.
The running time T (n) clearly satisﬁes the recurrence T (n) ≤ 2 T (n − 1) + O(1), so the running time
is T (n) = O(2n ) by the annihilator method.
Here is a similar recursive algorithm that actually constructs a subset of X that sums to T , if one
exists. This algorithm also runs in O(2n ) time.
CONSTRUCTSUBSET(X [1 .. n], T ):
if T = 0
return ∅
if T < 0 or n = 0
return NONE
Y ← CONSTRUCTSUBSET(X [2 .. n], T )
if Y = NONE
return Y
Y ← CONSTRUCTSUBSET(X [2 .. n], T − X [1])
if Y = NONE
return Y ∪ {X [1]}
return NONE These two algorithms are examples of a general algorithmic technique called backtracking. You can
imagine the algorithm searching through a binary tree of recursive possibilities like a maze, trying to
1 There’s no base case like the vacuous base case! 3 Algorithms Lecture 2: Backtracking ﬁnd a hidden treasure ( T = 0), and backtracking whenever it reaches a dead end ( T < 0 or n = 0).
For some problems, there are tricks that allow the recursive algorithm to recognize some branches as
dead ends without exploring them directly, thereby speeding up the algorithm; two such problems are
described later in these notes. Alas, SUBSETSUM is not one of the those problems; in the worst case, our
algorithm explicitly considers every subset of X .2 2.3 Longest Increasing Subsequence Now suppose we are given a sequence of integers, and we want to ﬁnd the longest subsequence whose
elements are in increasing order. More concretely, the input is an array A[1 .. n] of integers, and we want
to ﬁnd the longest sequence of indices 1 ≤ i1 < i2 < · · · ik ≤ n such that A[i j ] < A[i j +1 ] for all j .
To derive a recursive algorithm for this problem, we start with a recursive deﬁnition of the kinds of
objects we’re playing with: sequences and subsequences.
A sequence of integers is either empty
or an integer followed by a sequence of integers.
This deﬁnition suggests the following strategy for devising a recursive algorithm. If the input
sequence is empty, there’s nothing to do. Otherwise, we should ﬁgure out what to do with the ﬁrst
element of the input sequence, and let the Recursion Fairy take care of everything else. We can formalize
this strategy somewhat by giving a recursive deﬁnition of subsequence (using array notation to represent
sequences):
The only subsequence of the empty sequence is the empty sequence.
A subsequence of A[1 .. n] is either a subsequence of A[2 .. n]
or A[1] followed by a subsequence of A[2 .. n].
We’re not just looking for just any subsequence, but a longest subsequence with the property that
elements are in increasing order. So let’s try to add those two conditions to our deﬁnition. (I’ll omit the
familiar vacuous base case.)
The LIS of A[1 .. n] is
either the LIS of A[2 .. n]
or A[1] followed by the LIS of A[2 .. n] with elements larger than A[1],
whichever is longer.
This deﬁnition is correct, but it’s not quite recursive—we’re deﬁning ‘longest increasing subsequence’
in terms of the different ‘longest increasing subsequence with elements larger than x ’, which we haven’t
properly deﬁned yet. Fortunately, this second object has a very similar recursive deﬁnition. (Again, I’m
omitting the vacuous base case.)
If A[1] ≤ x , the LIS of A[1 .. n] with elements larger than x is
the LIS of A[2 .. n] with elements larger than x .
Otherwise, the LIS of A[1 .. n] with elements larger than x is
either the LIS of A[2 .. n] with elements larger than x
or A[1] followed by the LIS of A[2 .. n] with elements larger than A[1],
whichever is longer.
2 Indeed, because SUBSETSUM is NP-hard, there is almost certainly no way to solve it in subexponential time. 4 Algorithms Lecture 2: Backtracking The longest increasing subsequence without restrictions can now be redeﬁned as the longest increasing subsequence with elements larger than −∞. Rewriting this recursive deﬁnition into pseudocode
gives us the following recursive algorithm. LIS(A[1 .. n]):
return LISBIGGER(−∞, A[1 .. n]) LISBIGGER(prev, A[1 .. n]):
if n = 0
return 0
else
max ← LISBIGGER(prev, A[2 .. n])
if A[1] > pr ev
L ← 1 + LISBIGGER(A[1], A[2 .. n])
if L > max
max ← L
return max The running time of this algorithm satisﬁes the recurrence T (n) ≤ 2 T (n − 1) + O(1), so as usual the
annihilator method implies that T (n) = O(2n ). We really shouldn’t be surprised by this running time; in
the worst case, the algorithm examines each of the 2n subsequences of the input array.
The following alternative strategy avoids deﬁning a new object with the ‘larger than x ’ constraint.
We still only have to decide whether to include or exclude the ﬁrst element A[1]. We consider the
case where A[1] is excluded exactly the same way, but to consider the case where A[1] is included, we
remove any elements of A[2 .. n] that are larger than A[1] before we recurse. This new strategy gives us
the following algorithm: FILTER(A[1 .. n], x ):
j←1
for i ← 1 to n
if A[i ] > x
B [ j ] ← A[i ]; j ← j + 1
return B [1 .. j ] LIS(A[1 .. n]):
if n = 0
return 0
else
max ← LIS(prev, A[2 .. n])
L ← 1 + LIS(A[1], FILTER(A[2 .. n], A[1]))
if L > max
max ← L
return max The FILTER subroutine clearly runs in O(n) time, so the running time of LIS satisﬁes the recurrence
T (n) ≤ 2 T (n − 1) + O(n), which solves to T (n) ≤ O(2n ) by the annihilator method. This upper bound
pessimistically assumes that FILTER never actually removes any elements; indeed, if the input sequence
is sorted in increasing order, this assumption is correct. 2.4 Optimal Binary Search Trees Our next example combines recursive backtracking with the divide-and-conquer strategy.
Hopefully you remember that the cost of a successful search in a binary search tree is proportional to
the number of ancestors of the target node.3 As a result, the worst-case search time is proportional to
the depth of the tree. To minimize the worst-case search time, the height of the tree should be as small
as possible; ideally, the tree is perfectly balanced.
In many applications of binary search trees, it is more important to minimize the total cost of several
searches than to minimize the worst-case cost of a single search. If x is a more ‘popular’ search target
3 An ancestor of a node v is either the node itself or an ancestor of the parent of v . A proper ancestor of v is either the parent
of v or a proper ancestor of the parent of v . 5 Algorithms Lecture 2: Backtracking than y , we can save time by building a tree where the depth of x is smaller than the depth of y , even if
that means increasing the overall depth of the tree. A perfectly balanced tree is not the best choice if
some items are signiﬁcantly more popular than others. In fact, a totally unbalanced tree of depth Ω(n)
might actually be the best choice!
This situation suggests the following problem. Suppose we are given a sorted array of n keys A[1 .. n]
and an array of corresponding access frequencies f [1 .. n]. Over the lifetime of the search tree, we will
search for the key A[i ] exactly f [i ] times. Our task is to build the binary search tree that minimizes the
total search time.
Before we think about how to solve this problem, we should ﬁrst come up with a good recursive
deﬁnition of the function we are trying to optimize! Suppose we are also given a binary search tree T
with n nodes. Let vi denote the node that stores A[i ], and let r be the index of the root node. Ignoring
constant factors, the cost of searching for A[i ] is the number of nodes on the path from the root vr to vi .
Thus, the total cost of performing all the binary searches is given by the following expression:
n Cost( T, f [1 .. n]) = f [i ] · #nodes between vr and vi
i =1 Every search path includes the root node vr . If i < r , then all other nodes on the search path to vi are in
the left subtree; similarly, if i > r , all other nodes on the search path to vi are in the right subtree. Thus,
we can partition the cost function into three parts as follows:
r −1 Cost( T, f [1 .. n]) = f [i ] · #nodes between left( vr ) and vi
i =1
n + f [i ]
i =1
n + f [i ] · #nodes between right( vr ) and vi
i = r +1 Now the ﬁrst and third summations look exactly like our original expression (*) for Cost( T, f [1 .. n]).
Simple substitution gives us our recursive deﬁnition for Cost:
n Cost( T, f [1 .. n]) = Cost(left( T ), f [1 .. r − 1]) + f [i ] + Cost(right( T ), f [ r + 1 .. n])
i =1 The base case for this recurrence is, as usual, n = 0; the cost of performing no searches in the empty tree
is zero.
Now our task is to compute the tree Topt that minimizes this cost function. Suppose we somehow
magically knew that the root of Topt is vr . Then the recursive deﬁnition of Cost( T, f ) immediately implies
that the left subtree left( Topt ) must be the optimal search tree for the keys A[1 .. r − 1] and access
frequencies f [1 .. r − 1]. Similarly, the right subtree right( Topt ) must be the optimal search tree for the
keys A[ r + 1 .. n] and access frequencies f [ r + 1 .. n]. Once we choose the correct key to store at
the root, the Recursion Fairy will automatically construct the rest of the optimal tree for us. More
formally, let OptCost( f [1 .. n]) denote the total cost of the optimal search tree for the given frequency
counts. We immediately have the following recursive deﬁnition.
n OptCost( f [1 .. n]) = min 1≤ r ≤n OptCost( f [1 .. r − 1]) + f [i ] + OptCost( f [ r + 1 .. n])
i =1 6 Algorithms Lecture 2: Backtracking Again, the base case is OptCost( f [1 .. 0]) = 0; the best way to organize no keys, which we will plan to
search zero times, is by storing them in the empty tree!
This recursive deﬁnition can be translated mechanically into a recursive algorithm, whose running
time T (n) satisﬁes the recurrence
n T (n) = Θ(n) + T ( k − 1) + T ( n − k ) .
k =1
n The Θ(n) term comes from computing the total number of searches i =1 f [i ].
Yeah, that’s one ugly recurrence, but it’s actually easier to solve than it looks. To transform it into a
more familiar form, we regroup and collect identical terms, subtract the recurrence for T (n − 1) to get
rid of the summation, and then regroup again.
n−1 T (n) = Θ(n) + 2 T (k)
k =0
n−2 T (n − 1) = Θ(n − 1) + 2 T (k)
k =0 T (n) − T (n − 1) = Θ(1) + 2 T (n − 1)
T (n) = 3 T (n − 1) + Θ(1)
The solution T (n ) = Θ(3n ) now follows from the annihilator method.
Let me emphasize that this recursive algorithm does not examine all possible binary search trees.
The number of binary search trees with n nodes satisﬁes the recurrence
n−1 N ( n) = N ( r − 1) · N ( n − r ) ,
r =1 which has the closed-from solution N (n) = Θ(4n / n). Our algorithm saves considerable time by
searching independently for the optimal left and right subtrees. A full enumeration of binary search trees
would consider all possible pairings of left and right subtrees; hence the product in the recurrence for
N (n). Exercises
1. (a) Let A[1 .. m] and B [1 .. n] be two arbitrary arrays. A common subsequence of A and B is both a
subsequence of A and a subsequence of B . Give a simple recursive deﬁnition for the function
lcs(A, B ), which gives the length of the longest common subsequence of A and B .
(b) Let A[1 .. m] and B [1 .. n] be two arbitrary arrays. A common supersequence of A and B is
another sequence that contains both A and B as subsequences. Give a simple recursive deﬁnition for the function scs(A, B ), which gives the length of the shortest common supersequence
of A and B .
(c) Call a sequence X [1 .. n] oscillating if X [i ] < X [i + 1] for all even i , and X [i ] > X [i + 1] for
all odd i . Give a simple recursive deﬁnition for the function los(A), which gives the length of
the longest oscillating subsequence of an arbitrary array A of integers.
(d) Give a simple recursive deﬁnition for the function sos(A), which gives the length of the
shortest oscillating supersequence of an arbitrary array A of integers.
7 Algorithms Lecture 2: Backtracking (e) Call a sequence X [1 .. n] accelerating if 2 · X [i ] < X [i − 1] + X [i + 1] for all i . Give a simple
recursive deﬁnition for the function lxs(A), which gives the length of the longest accelerating
subsequence of an arbitrary array A of integers. For more backtracking exercises, see the next two lecture notes! c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 8 Algorithms Non-Lecture B: Efﬁcient Exponential-Time Algorithms
Wouldn’t the sentence “I want to put a hyphen between the words Fish and
And and And and Chips in my Fish-And-Chips sign.” have been clearer if
quotation marks had been placed before Fish, and between Fish and and,
and and and And, and And and and, and and and And, and And and and,
and and and Chips, as well as after Chips?1
— unknown, from the original FreeBSD ‘fortune’ ﬁle B Efﬁcient Exponential-Time Algorithms In another lecture note, we discuss the class of NP-hard problems. For every problem in this class, the
fastest algorithm anyone knows has an exponential running time. Moreover, there is very strong evidence
(but alas, no proof) that it is impossible to solve any NP-hard problem in less than exponential time—it’s
not that we’re all stupid; the problems really are that hard! Unfortunately, an enormous number of
problems that arise in practice are NP-hard; for some of these problems, even approximating the right
answer is NP-hard.
Suppose we absolutely have to ﬁnd the exact solution to some NP-hard problem. A polynomial-time
algorithm is almost certainly out of the question; the best running time we can hope for is exponential.
But which exponential? An algorithm that runs in O(1.5n ) time, while still unusable for large problems,
is still signiﬁcantly better than an algorithm that runs in O(2n ) time!
For most NP-hard problems, the only approach that is guaranteed to ﬁnd an optimal solution is
recursive backtracking. The most straightforward version of this approach is to recursively generate all
possible solutions and check each one: all satisfying assignments, or all vertex colorings, or all subsets,
or all permutations, or whatever. However, most NP-hard problems have some additional structure that
allows us to prune away most of the branches of the recursion tree, thereby drastically reducing the
running time. B.1 3SAT Let’s consider the mother of all NP-hard problems: 3SAT. Given a boolean formula in conjunctive normal
form, with at most three literals in each clause, our task is to determine whether any assignment of
values of the variables makes the formula true. Yes, this problem is NP-hard, which means that a
polynomial algorithm is almost certainly impossible. Too bad; we have to solve the problem anyway.
The trivial solution is to try every possible assignment. We’ll evaluate the running time of our
3SAT algorithms in terms of the number of variables in the formula, so let’s call that n. Provided any
clause appears in our input formula at most once—a condition that we can easily enforce in polynomial
time—the overall input size is O(n3 ). There are 2n possible assignments, and we can evaluate each
assignment in O(n3 ) time, so the overall running time is O(2n n3 ).
Since polynomial factors like n3 are essentially noise when the overall running time is exponential,
from now on I’ll use poly(n) to represent some arbitrary polynomial in n; in other words, poly(n) = nO(1) .
For example, the trivial algorithm for 3SAT runs in time O(2n poly(n)).
We can make this algorithm smarter by exploiting the special recursive structure of 3CNF formulas:
A 3CNF formula is either nothing
or a clause with three literals ∧ a 3CNF formula
1 If you ever decide to read this sentence out loud, be sure to pause brieﬂy between ‘Fish and and’ and ‘and and and And’,
‘and and and And’ and ‘and And and and’, ‘and And and and’ and ‘and and and And’, ‘and and and And’ and ‘and And and and’,
and ‘and And and and’ and ‘and and and Chips’! 1 Algorithms Non-Lecture B: Efﬁcient Exponential-Time Algorithms Suppose we want to decide whether some 3CNF formula Φ with n variables is satisﬁable. Of course this
is trivial if Φ is the empty formula, so suppose
Φ = ( x ∨ y ∨ z) ∧ Φ
for some literals x , y, z and some 3CNF formula Φ . By distributing the ∧ across the ∨s, we can rewrite
Φ as follows:
Φ = ( x ∧ Φ ) ∨ ( y ∧ Φ ) ∨ (z ∧ Φ )
For any boolean formula Ψ and any literal x , let Ψ| x (pronounced “sigh given eks") denote the simpler
boolean formula obtained by assuming x is true. It’s not hard to prove by induction (hint, hint) that
x ∧ Ψ = x ∧ Ψ| x , which implies that
Φ = ( x ∧ Φ | x ) ∨ ( y ∧ Φ | y ) ∨ (z ∧ Φ |z ).
Thus, in any satisfying assignment for Φ, either x is true and Φ | x is satisﬁable, or y is true and Φ | y is
satisﬁable, or z is true and Φ |z is satisﬁable. Each of the smaller formulas has at most n − 1 variables. If
we recursively check all three possibilities, we get the running time recurrence
T (n) ≤ 3 T (n − 1) + poly(n),
whose solution is O(3n poly(n)). So we’ve actually done worse!
But these three recursive cases are not mutually exclusive! If Φ | x is not satisﬁable, then x must be
false in any satisfying assignment for Φ. So instead of recursively checking Φ | y in the second step, we
¯
¯
can check the even simpler formula Φ | x y . Similarly, if Φ | x y is not satisﬁable, then we know that y
¯y
must be false in any satisfying assignment, so we can recursively check Φ | x ¯ z in the third step.
3SAT(Φ):
if Φ = ∅
return TRUE
( x ∨ y ∨ z) ∧ Φ ← Φ
if 3SAT(Φ| x )
return TRUE
¯
if 3SAT(Φ| x y )
return TRUE
¯y
return 3SAT(Φ| x ¯ z ) The running time off this algorithm obeys the recurrence
T (n) = T (n − 1) + T (n − 2) + T (n − 3) + poly(n),
where poly(n) denotes the polynomial time required to simplify boolean formulas, handle control ﬂow,
move stuff into and out of the recursion stack, and so on. The annihilator method gives us the solution
T (n) = O(λn poly(n)) = O(1.83928675522n )
where λ ≈ 1.83928675521 . . . is the largest root of the characteristic polynomial r 3 − r 2 − r − 1. (Notice
that we cleverly eliminated the polynomial noise by increasing the base of the exponent ever so slightly.)
We can improve this algorithm further by eliminating pure literals from the formula before recursing.
¯
A literal x is pure in if it appears in the formula Φ but its negation x does not. It’s not hard to prove (hint, 2 Algorithms Non-Lecture B: Efﬁcient Exponential-Time Algorithms hint) that if Φ has a satisfying assignment, then it has a satisfying assignment where every pure literal is
¯
true. If Φ = ( x ∨ y ∨ z ) ∧ Φ has no pure literals, then some in Φ contains the literal x , so we can write
¯
Φ = ( x ∨ y ∨ z) ∧ ( x ∨ u ∨ v) ∧ Φ
¯
for some literals u and v (each of which might be y , ¯ , z , or z ). It follows that the ﬁrst recursive formula
y
Φ| x has contains the clause (u ∨ v ). We can recursively eliminate the variables u and v just as we tested
the variables y and x in the second and third cases of our previous algorithm:
¯
Φ| x = (u ∨ v ) ∧ Φ | x = (u ∧ Φ | xu) ∨ ( v ∧ Φ | x u v ).
Here is our new faster algorithm:
3SAT(Φ):
if Φ = ∅
return TRUE
if Φ has a pure literal x
return 3SAT(Φ| x )
¯
( x ∨ y ∨ z) ∧ ( x ∨ u ∨ v) ∧ Φ ← Φ
if 3SAT(Φ| xu)
return TRUE
¯
if 3SAT(Φ| x u v )
return TRUE
¯
if 3SAT(Φ| x y )
return TRUE
¯y
return 3SAT(Φ| x ¯ z ) The running time T (n) of this new algorithm satisﬁes the recurrence
T (n) = 2 T (n − 2) + 2 T (n − 3) + poly(n),
and the annihilator method implies that
T (n) = O(µn poly(n)) = O(1.76929235425n )
where µ ≈ 1.76929235424 . . . is the largest root of the characteristic polynomial r 3 − 2 r − 2.
Naturally, this approach can be extended much further. As of 2004, the fastest (deterministic)
algorithm for 3SAT runs in O(1.473n ) time2 , but there is absolutely no reason to believe that this is the
best possible. B.2 Maximum Independent Set Now suppose we are given an undirected graph G and are asked to ﬁnd the size of the largest independent
set, that is, the largest subset of the vertices of G with no edges between them. Once again, we have
an obvious recursive algorithm: Try every subset of nodes, and return the largest subset with no edges.
Expressed recursively, the algorithm might look like this.
2 Tobias Brueggemann and Walter Kern. An improved deterministic local search algorithm for 3-SAT. Theoretical Computer
Science 329(1–3):303–313, 2004. 3 Algorithms Non-Lecture B: Efﬁcient Exponential-Time Algorithms
MAXIMUMINDSETSIZE(G ):
if G = ∅
return 0
else
v ← any node in G
withv ← 1 + MAXIMUMINDSETSIZE(G \ N ( v ))
withoutv ← MAXIMUMINDSETSIZE(G \ { v })
return max{withv, withoutv}. Here, N ( v ) denotes the neighborhood of v : The set containing v and all of its neighbors. Our
algorithm is exploiting the fact that if an independent set contains v , then by deﬁnition it contains none
of v ’s neighbors. In the worst case, v has no neighbors, so G \ { v } = G \ N ( v ). Thus, the running time of
this algorithm satisﬁes the recurrence T (n) = 2 T (n − 1) + poly(n) = O(2n poly(n)). Surprise, surprise.
This algorithm is mirroring a crude recursive upper bound for the number of maximal independent
sets in a graph; an independent set is maximal if every vertex in G is either already in the set or a
neighbor of a vertex in the set. If the graph is non-empty, then every maximal independent set either
includes or excludes each vertex. Thus, the number of maximal independent sets satisﬁes the recurrence
M (n) ≤ 2 M (n − 1), with base case M (1) = 1. The annihilator method gives us M (n) ≤ 2n − 1. The only
subset that we aren’t counting with this upper bound is the empty set!
We can improve this upper bound by more carefully examining the worst case of the recurrence. If
v has no neighbors, then N ( v ) = { v }, and both recursive calls consider a graph with n − 1 nodes. But
in this case, v is in every maximal independent set, so one of the recursive calls is redundant. On the
other hand, if v has at least one neighbor, then G \ N ( v ) has at most n − 2 nodes. So now we have the
following recurrence.
M (n) ≤ max M ( n − 1)
M ( n − 1) + M ( n − 2) = O(1.61803398875n ) The upper bound is derived by solving each case separately using the annihilator method and taking the
worst of the two cases. The ﬁrst case gives us M (n) = O(1); the second case yields our old friends the
Fibonacci numbers.
We can improve this bound even more by examining the new worst case: v has exactly one
neighbor w . In this case, either v or w appears in any maximal independent set. Thus, instead of
recursively searching in G \ { v }, we should recursively search in G \ N (w ), which has at most n − 1 nodes.
On the other hand, if G has no nodes with degree 1, then G \ N ( v ) has at most n − 3 nodes. M ( n − 1) 2 M ( n − 2)
M (n) ≤ max
= O(1.46557123188n ) M ( n − 1) + M ( n − 3)
The base of the exponent is the largest root of the characteristic polynomial r 3 − r 2 − 1. The second case
n
implies a bound of O( 2 ) = O(1.41421356237n ), which is smaller.
We can apply this improvement technique one more time. If G has a node v with degree 3 or more,
then G \ N ( v ) has at most n − 4 nodes. Otherwise (since we have already considered nodes of degree 0
and 1), every node in the graph has degree 2. Let u, v, w be a path of three nodes in G (possibly with u
adjacent to w ). In any maximal independent set, either v is present and u, w are absent, or u is present
and its two neighbors are absent, or w is present and its two neighbors are absent. In all three cases, we
recursively count maximal independent sets in a graph with n − 3 nodes. 4 Algorithms Non-Lecture B: Efﬁcient Exponential-Time Algorithms M ( n − 1) 2 M ( n − 2) M (n) ≤ max
= O(3n/3 ) = O(1.44224957031n ) M (n − 1) + M (n − 4) 3 M ( n − 3) The third case implies a bound of O(1.3802775691n ), where the base is the largest root of r 4 − r 3 − 1.
Unfortunately, we cannot apply the same improvement trick again. A graph consisting of n/3
triangles (cycles of length three) has exactly 3n/3 maximal independent sets, so our upper bound is tight
in the worst case.
Now from this recurrence, we can derive an efﬁcient algorithm to compute the largest independent
set in G in O(3n/3 poly(n)) = O(1.44224957032n ) time.
MAXIMUMINDSETSIZE(G ):
if G = ∅
return 0
else if G has a node v with degree 0
return 1 + MAXIMUMINDSETSIZE(G \ { v })
else if G has a node v with degree 1
w ← v ’s neighbor
withv ← 1 + MAXIMUMINDSETSIZE(G \ N ( v ))
withw ← 1 + MAXIMUMINDSETSIZE(G \ N (w ))
return max{withv, withw} 〈〈n − 1〉〉 〈〈n − 2〉〉
〈〈≤ n − 2〉〉 else if G has a node v with degree greater than 2
withv ← 1 + MAXIMUMINDSETSIZE(G \ N ( v )) 〈〈≤ n − 4〉〉
withoutv ← MAXIMUMINDSETSIZE(G \ { v })
〈〈≤ n − 1〉〉
return max{withv, withoutv}
else 〈〈every node in G has degree 2〉〉
v ← any node; u, w ← v ’s neighbors
withu ← 1 + MAXIMUMINDSETSIZE(G \ N (u))
withv ← 1 + MAXIMUMINDSETSIZE(G \ N ( v ))
withw ← 1 + MAXIMUMINDSETSIZE(G \ N (w ))
return max{withu, withv, withw} 〈〈≤ n − 3〉〉
〈〈≤ n − 3〉〉
〈〈≤ n − 3〉〉 Exercises
1. Describe an algorithm to solve 3SAT in time O(φ n poly(n)), where φ = (1 + 5)/2 ≈ 1.618034.
[Hint: Prove that in each recursive call, either you have just eliminated a pure literal, or the
formula has a clause with at most two literals. What recurrence leads to this running time?] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 5 Algorithms Lecture 3: Dynamic Programming
Those who cannot remember the past are doomed to repeat it.
— George Santayana, The Life of Reason, Book I:
Introduction and Reason in Common Sense (1905)
The 1950s were not good years for mathematical research. We had a very interesting gentleman
in Washington named Wilson. He was secretary of Defense, and he actually had a pathological
fear and hatred of the word ‘research’. I’m not using the term lightly; I’m using it precisely. His
face would suffuse, he would turn red, and he would get violent if people used the term ‘research’
in his presence. You can imagine how he felt, then, about the term ‘mathematical’. The RAND
Corporation was employed by the Air Force, and the Air Force had Wilson as its boss, essentially.
Hence, I felt I had to do something to shield Wilson and the Air Force from the fact that I was really
doing mathematics inside the RAND Corporation. What title, what name, could I choose?
— Richard Bellman, on the origin of his term ‘dynamic programming’ (1984)
If we all listened to the professor, we may be all looking for professor jobs.
— Pittsburgh Steelers’ head coach Bill Cowher, responding to
David Romer’s dynamic-programming analysis of football strategy (2003) 3 Dynamic Programming 3.1 Fibonacci Numbers The Fibonacci numbers Fn , named after Leonardo Fibonacci Pisano1 , the mathematician who popularized
‘algorism’ in Europe in the 13th century, are deﬁned as follows: F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for
all n ≥ 2. The recursive deﬁnition of Fibonacci numbers immediately gives us a recursive algorithm for
computing them:
RECFIBO(n):
if (n < 2)
return n
else
return RECFIBO(n − 1) + RECFIBO(n − 2) How long does this algorithm take? Except for the recursive calls, the entire algorithm requires only
a constant number of steps: one comparison and possibly one addition. If T (n) represents the number
of recursive calls to RECFIBO, we have the recurrence
T (0) = 1, T (1) = 1, T (n) = T (n − 1) + T (n − 2) + 1. This looks an awful lot like the recurrence for Fibonacci numbers! The annihilator method gives us an
asymptotic bound of Θ(φ n ), where φ = ( 5 + 1)/2 ≈ 1.61803398875, the so-called golden ratio, is
the largest root of the polynomial r 2 − r − 1. But it’s fairly easy to prove (hint, hint) the exact solution
T (n ) = 2 Fn +1 − 1. In other words, computing Fn using this algorithm takes more than twice as many
steps as just counting to Fn !
Another way to see this is that the RECFIBO is building a big binary tree of additions, with nothing
but zeros and ones at the leaves. Since the eventual output is Fn , our algorithm must call RECRIBO(1)
(which returns 1) exactly Fn times. A quick inductive argument implies that RECFIBO(0) is called exactly
Fn−1 times. Thus, the recursion tree has Fn + Fn−1 = Fn+1 leaves, and therefore, because it’s a full binary
tree, it must have 2 Fn+1 − 1 nodes.
1 literally, “Leonardo, son of Bonacci, of Pisa” 1 Algorithms 3.2 Lecture 3: Dynamic Programming Memo(r)ization and Dynamic Programming The obvious reason for the recursive algorithm’s lack of speed is that it computes the same Fibonacci numbers over and over and over. A single call to RECURSIVEFIBO(n) results in one recursive
call to RECURSIVEFIBO(n − 1), two recursive calls to RECURSIVEFIBO(n − 2), three recursive calls to
RECURSIVEFIBO(n − 3), ﬁve recursive calls to RECURSIVEFIBO(n − 4), and in general, Fk−1 recursive calls
to RECURSIVEFIBO(n − k), for any 0 ≤ k < n. For each call, we’re recomputing some Fibonacci number
from scratch.
We can speed up the algorithm considerably just by writing down the results of our recursive calls
and looking them up again if we need them later. This process is called memoization.2
MEMFIBO(n):
if (n < 2)
return n
else
if F [n] is undeﬁned
F [n] ← MEMFIBO(n − 1) + MEMFIBO(n − 2)
return F [n] If we actually trace through the recursive calls made by MEMFIBO, we ﬁnd that the array F [ ] is
ﬁlled from the bottom up: ﬁrst F [2], then F [3], and so on, up to F [n]. Once we see this pattern, we
can replace the recursion with a simple for-loop that ﬁlls the array in order, instead of relying on the
complicated recursion to do it for us. This gives us our ﬁrst explicit dynamic programming algorithm.
ITERFIBO(n):
F [0] ← 0
F [1] ← 1
for i ← 2 to n
F [ i ] ← F [ i − 1] + F [ i − 2]
return F [n] ITERFIBO clearly takes only O(n) time and O(n) space to compute Fn , an exponential speedup over
our original recursive algorithm.
We can reduce the space to O(1) by noticing that we never need more than the last two elements of
the array:
ITERFIBO2(n):
prev ← 1
curr ← 0
for i ← 1 to n
next ← curr + prev
prev ← curr
curr ← next
return curr (This algorithm uses the non-standard but perfectly consistent base case F−1 = 1 so that ITERFIBO2(0)
returns the correct value 0.)
Even this algorithm can be improved further. There’s a faster algorithm deﬁned in terms of matrix
multiplication, using the following wonderful fact:
01
11
2 x
y = y
x+y “My name is Elmer J. Fudd, millionaire. I own a mansion and a yacht.” 2 Algorithms Lecture 3: Dynamic Programming In other words, multiplying a two-dimensional vector by the matrix 0 1 does exactly the same thing
11
as one iteration of the inner loop of ITERFIBO2. This might lead us to believe that multiplying by the
matrix n times is the same as iterating the loop n times:
01
11 n 1
=
0 Fn−1
.
Fn A quick inductive argument proves this fact. So if we want the nth Fibonacci number, we just have to
compute the nth power of the matrix 0 1 .
11
If we use repeated squaring, computing the nth power of something requires only O(log n) multiplications. In this case, that means O(log n) 2 × 2 matrix multiplications, each of which reduces to a
constant number of integer multiplications and additions. Thus, we can compute Fn in only O(log n)
integer arithmetic operations.
This is an exponential speedup over the standard iterative algorithm, which was already an exponential speedup over our original recursive algorithm. Right? 3.3 Uh. . . wait a minute. Well, not exactly. Fibonacci numbers grow exponentially fast. The nth Fibonacci number is approximately
n log10 φ ≈ n/5 decimal digits long, or n log2 φ ≈ 2n/3 bits. So we can’t possibly compute Fn in
logarithmic time — we need Ω(n) time just to write down the answer!
I’ve been cheating by assuming we can do arbitrary-precision arithmetic in constant time. If we use
fast Fourier transforms, multiplying two n-digit numbers takes O(n log n) time. Thus, the matrix-based
algorithm’s actual running time is given by the recurrence
T (n) = T ( n/2 ) + O(n log n),
which solves to T (n) = O(n log n) by the Master Theorem.
Is this slower than our “linear-time” iterative algorithm? No! Addition isn’t free, either. Adding two
n-digit numbers takes O(n) time, so the running time of the iterative algorithm is O(n2 ). (Do you see
why?) Our matrix algorithm really is faster than our iterative algorithm, but not exponentially faster.
In the original recursive algorithm, the extra cost of arbitrary-precision arithmetic is overwhelmed
by the huge number of recursive calls. The correct recurrence is
T (n) = T (n − 1) + T (n − 2) + O(n),
which still has the solution O(φ n ), by the annihilator method. 3.4 The Pattern: Smart Recursion In a nutshell, dynamic programming is recursion without repetition. Developing a dynamic programming
algorithm almost always requires two distinct stages.
1. Formulate the problem recursively. Write down a recursive formula or algorithm for the whole
problem in terms of the answers to smaller subproblems. This is the hard part.
2. Build solutions to your recurrence from the bottom up. Write an algorithm that starts with
the base cases of your recurrence and works its way up to the ﬁnal solution, by considering
intermediate subproblems in the correct order. This stage can be broken down into several smaller,
relatively mechanical steps:
3 Algorithms Lecture 3: Dynamic Programming (a) Identify the subproblems. What are all the different ways can your recursive algorithm call
itself, starting with some initial input? For example, the argument to RECFIBO is always an
integer between 0 and n.
(b) Choose a data structure to memoize intermediate results. For most problems, each
recursive subproblem can be identiﬁed by a few integers, so you can use a multidimensional
array. For some problems, however, a more complicated data structure is required.
(c) Analyze running time and space. The number of possible distinct subproblems determines
the space complexity of your memoized algorithm. To compute the time complexity, add up
the running times of all possible subproblems, ignoring the recursive calls. For example, if
we already know Fi −1 and Fi −2 , we can compute Fi in O(1) time, so computing the ﬁrst n
Fibonacci numbers takes O(n) time.
(d) Identify dependencies between subproblems. Except for the base cases, every recursive
subproblem depends on other subproblems—which ones? Draw a picture of your data
structure, pick a generic element, and draw arrows from each of the other elements it
depends on. Then formalize your picture.
(e) Find a good evaluation order. Order the subproblems so that each subproblem comes after
the subproblems it depends on. Typically, this means you should consider the base cases
ﬁrst, then the subproblems that depends only on base cases, and so on. More formally, the
dependencies you identiﬁed in the previous step deﬁne a partial order over the subproblems;
in this step, you need to ﬁnd a linear extension of that partial order. Be careful!
(f) Write down the algorithm. You know what order to consider the subproblems, and you
know how to solve each subproblem. So do that! If your data structure is an array, this
usually means writing a few nested for-loops around your original recurrence.
Of course, you have to prove that each of these steps is correct. If your recurrence is wrong, or if you try
to build up answers in the wrong order, your algorithm won’t work!
Dynamic programming algorithms store the solutions of intermediate subproblems, often but not
always in some kind of array or table. Many algorithms students make the mistake of focusing on the
table (because tables are easy and familiar) instead of the much more important (and difﬁcult) task of
ﬁnding a correct recurrence.
Dynamic programming is not about ﬁlling in tables; it’s about smart recursion.
As long as we memoize the correct recurrence, an explicit table isn’t really necessary, but if the recursion
is incorrect, nothing works. 3.5 Edit Distance The edit distance between two words—sometimes also called the Levenshtein distance—is the minimum
number of letter insertions, letter deletions, and letter substitutions required to transform one word into
another. For example, the edit distance between FOOD and MONEY is at most four:
FOOD → MOOD → MON∧D → MONED → MONEY A better way to display this editing process is to place the words one above the other, with a gap in
the ﬁrst word for every insertion, and a gap in the second word for every deletion. Columns with two
different characters correspond to substitutions. Thus, the number of editing steps is just the number of
columns that don’t contain the same character twice.
4 Algorithms Lecture 3: Dynamic Programming
F
M O
O O
N E D
Y It’s fairly obvious that you can’t get from FOOD to MONEY in three steps, so their edit distance is exactly
four. Unfortunately, this is not so easy in general. Here’s a longer example, showing that the distance
between ALGORITHM and ALTRUISTIC is at most six. Is this optimal?
A
A L
L G O
T R
R U I
I S T
T H
I M
C To develop a dynamic programming algorithm to compute the edit distance between two strings,
we ﬁrst need to develop a recursive deﬁnition. Our gap representation for edit sequences has a crucial
“optimal substructure” property. Suppose we have the gap representation for the shortest edit sequence
for two strings. If we remove the last column, the remaining columns must represent the shortest
edit sequence for the remaining substrings. We can easily prove this by contradiction. If the substrings
had a shorter edit sequence, we could just glue the last column back on and get a shorter edit sequence
for the original strings. Once we ﬁgure out what should go in the last column, the Recursion Fairy will
magically give us the rest of the optimal gap representation.
So let’s recursively deﬁne the edit distance between two strings A[1 .. m] and B [1 .. n], which we
denote by Edit(A[1 .. m], B [1 .. n]). If neither string is empty, there are three possibilities for the last
column in the shortest edit sequence:
• Insertion: The last entry in the bottom row is empty. In this case, the edit distance is equal
to Edit(A[1 .. m − 1], B [1 .. n]) + 1. The +1 is the cost of the ﬁnal insertion, and the recursive
expression gives the minimum cost for the other columns.
• Deletion: The last entry in the top row is empty. In this case, the edit distance is equal to
Edit(A[1 .. m], B [1 .. n − 1]) + 1. The +1 is the cost of the ﬁnal deletion, and the recursive
expression gives the minimum cost for the other columns.
• Substitution: Both rows have characters in the last column. If the characters are the same, the
substitution is free, so the edit distance is equal to Edit(A[1 .. m − 1], B [1 .. n − 1]). If the characters
are different, then the edit distance is equal to Edit(A[1 .. m − 1], B [1 .. n − 1]) + 1.
The edit distance between A and B is the smallest of these three possibilities:3 Edit(A[1 .. m], B [1 .. n − 1]) + 1 Edit(A[1 .. m], B [1 .. n]) = min Edit(A[1 .. m − 1], B [1 .. n]) + 1
Edit(A[1 .. m − 1], B [1 .. n − 1]) + A[m] = B [n] This recurrence has two easy base cases. The only way to convert the empty string into a string of n
characters is by performing n insertions. Similarly, the only way to convert a string of m characters into
the empty string is with m deletions, Thus, if denotes the empty string, we have
Edit(A[1 .. m], ) = m, Edit( , B [1 .. n]) = n. Both of these expressions imply the trivial base case Edit( , ) = 0.
3 Once again, I’m using Iverson’s bracket notation P to denote the indicator variable for the logical proposition P , which
has value 1 if P is true and 0 if P is false. 5 Algorithms Lecture 3: Dynamic Programming Now notice that the arguments to our recursive subproblems are always preﬁxes of the original
strings A and B . Thus, we can simplify our notation considerably by using the lengths of the preﬁxes,
instead of the preﬁxes themselves, as the arguments to our recursive function. So let’s write Edit(i , j ) as
shorthand for Edit(A[1 .. i ], B [1 .. j ]). This function satisﬁes the following recurrence:
if j = 0 Edit(i , j ) = i j if i = 0 Edit(i − 1, j ) + 1, min Edit(i , j − 1) + 1, Edit(i − 1, j − 1) + A[i ] = B [ j ] otherwise The edit distance between the original strings A and B is just Edit(m, n).
This recurrence translates directly into a recursive algorithm. Just out of curiosity, we can analyze
the running time of this algorithm by solving the following recurrence:4
T ( m , n) = O(1) if n = 0 or m = 0, T (m, n − 1) + T (m − 1, n) + T (n − 1, m − 1) + O(1) otherwise. I don’t know of a general closed-form solution for this mess, but we can derive an upper bound by
deﬁning a new function
O(1) T (N ) = max T (n, m) = if N = 0, 2 T (N − 1) + T (N − 2) + O(1) otherwise. n + m= N The annihilator method implies that T (N ) = O((1 + 2)N ). Thus, the running time of our recursive
edit-distance algorithm is at most T (n + m) = O((1 + 2)n+m ).
We can dramatically reduce the running time of this algorithm down by memoization. Because each
recursive subproblem can be identiﬁed by two indices i and j , we can store the intermediate values in a
two-dimensional array Edit[0 .. m , 0 .. n ]. Note that the index ranges start at zero to accommodate the
base cases. Since there are Θ(mn) entries in the table, our memoized algorithm uses Θ(mn ) space. Since
each entry in the table can be computed in Θ(1) time once we know its predecessors, our memoized
algorithm runs in Θ(mn ) time.
It’s not immediately clear what order the recursive algorithm ﬁlls the rest of the table; all we can tell
from the recurrence is that each entry Edit[i , j ] is ﬁlled in after the neighboring entries directly above,
directly to the left, and both above and to the left. But just this partial information is enough to give us
an explicit evaluation order. If we ﬁll in our table in the standard row-major order—row by row from
top down, each row from left to right—then whenever we reach an entry in the table, the entries it
depends on are already available.
j i Dependencies in the memoization table for edit distance, and a legal evaluation order
4 You can skip this step. Since we’re going to memoize this algorithm, the running time of the non-memoized algorithm is
not particularly important. In particular, we won’t ask you do do this on homeworks or exams. 6 Algorithms Lecture 3: Dynamic Programming Putting everything together, we obtain the following dynamic programming algorithm:
EDITDISTANCE(A[1 .. m], B [1 .. n]):
for j ← 1 to n
Edit[0, j ] ← j
for i ← 1 to m
Edit[i , 0] ← i
for j ← 1 to n
if A[i ] = B [ j ]
Edit[i , j ] ← min Edit[i − 1, j ] + 1, Edit[i , j − 1] + 1, Edit[i − 1, j − 1]
else
Edit[i , j ] ← min Edit[i − 1, j ] + 1, Edit[i , j − 1] + 1, Edit[i − 1, j − 1] + 1
return Edit[m, n] Here’s the resulting table for ALGORITHM → ALTRUISTIC. Bold numbers indicate places where
characters in the two strings are equal. The arrows represent the predecessor(s) that actually deﬁne
each entry. Each direction of arrow corresponds to a different edit operation: horizontal=deletion,
vertical=insertion, and diagonal=substitution. Bold diagonal arrows indicate “free” substitutions of
a letter for itself. Any path of arrows from the top left corner to the bottom right corner of this table
represents an optimal edit sequence between the two strings. (There can be many such paths.) Moreover,
since we can compute these arrows in a post-processing phase from the values stored in the table, we
can reconstruct the actual optimal editing sequence in O(n + m) additional time. A
L
T
R
U
I
S
T
I
C ALGORITHM
0 → 1 →2→ 3 → 4 → 5 →6→ 7 → 8 → 9
↓
1 0 →1→ 2 → 3 → 4 →5→ 6 → 7 → 8
↓↓
2 1 0→ 1 → 2 → 3 → 4 → 5 → 6 → 7
↓↓↓
3 2 1 1 → 2 → 3 → 4 →4→ 5 → 6
↓↓↓↓
4 3 2 2 2 2 →3→ 4 → 5 → 6
↓↓↓↓↓↓
5 4 3 3 3 3 3→ 4 → 5 → 6
↓↓↓↓↓↓
6 5 4 4 4 4 3→ 4 → 5 → 6
↓↓↓↓↓↓↓
7655554456
↓↓↓↓↓↓↓
8 7 6 6 6 6 5 4→ 5 → 6
↓↓↓↓↓↓↓↓
9 8 7 7 7 7 6 5 5→ 6
↓↓↓↓↓↓↓↓↓
10 9 8 8 8 8 7 6 6 6 The edit distance between ALGORITHM and ALTRUISTIC is indeed six. There are three paths through
this table from the top left to the bottom right, so there are three optimal edit sequences:
A
A L
L G
T A
A L
L G A
A L
L G
T O
R R
U I
I T
T S H
I M
C O
T R
R U I
I S T
T H
I M
C O R
R U I
I S T
T H
I M
C 7 Algorithms 3.6 Lecture 3: Dynamic Programming Warning: Greed is Stupid If we’re very very very very lucky, we can bypass all the recurrences and tables and so forth, and solve the
problem using a greedy algorithm. The general greedy strategy is look for the best ﬁrst step, take it, and
then continue. While this approach seems very natural, it almost never works; optimization problems
that can be solved correctly by a greedy algorithm are very rare. Nevertheless, for many problems that
should be solved by dynamic programming, many students’ ﬁrst intuition is to apply a greedy strategy.
For example, a greedy algorithm for the edit distance problem might look for the longest common
substring of the two strings, match up those substrings (since those substitutions don’t cost anything),
and then recursively look for the edit distances between the left halves and right halves of the strings.
If there is no common substring—that is, if the two strings have no characters in common—the edit
distance is clearly the length of the larger string. If this sounds like a stupid hack to you, pat yourself on
the back. It isn’t even close to the correct solution.
Everyone should tattoo the following sentence on the back of their hands, right under all the rules
about logarithms and big-Oh notation: Greedy algorithms almost never work! Use dynamic programming instead!
What, never? No, never! What, never? Well. . . hardly ever.5
A different lecture note describes the effort required to prove that greedy algorithms are correct, in
the rare instances when they are. You will not receive any credit for any greedy algorithm for any
problem in this class without a formal proof of correctness. We’ll push through the formal proofs
for several greedy algorithms later in semester. 3.7 Dynamic Programming on Trees So far, all of our dynamic programming example use a multidimensional array to store the results of
recursive subproblems. However, as the next example shows, this is not always the most appropriate
date structure to use.
A independent set in a graph is a subset of the vertices that have no edges between them. Finding
the largest independent set in an arbitrary graph is extremely hard; in fact, this is one of the canonical
NP-hard problems described in another lecture note. But from some special cases of graphs, we can ﬁnd
the largest independent set efﬁciently. In particular, when the input graph is a tree (a connected and
acyclic graph) with n vertices, we can compute the largest independent set in O(n) time.
In the recursion notes, we saw a recursive algorithm for computing the size of the largest independent
set in an arbitrary graph:
MAXIMUMINDSETSIZE(G ):
if G = ∅
return 0
v ← any node in G
withv ← 1 + MAXIMUMINDSETSIZE(G \ N ( v ))
withoutv ← MAXIMUMINDSETSIZE(G \ { v })
return max{withv, withoutv}.
5 Greedy hardly ever ever works! Then give three cheers, and one cheer more, for the hardy Captain of the Pinafore! Then
give three cheers, and one cheer more, for the Captain of the Pinafore! 8 Algorithms Lecture 3: Dynamic Programming Here, N ( v ) denotes the neighborhood of v : the set containing v and all of its neighbors. As we observed
in the other lecture notes, this algorithm has a worst-case running time of O(2n poly(n)), where n is the
number of vertices in the input graph.
Now suppose we require that the input graph is a tree; we will call this tree T instead of G from now
on. We need to make a slight change to the algorithm to make it truly recursive. The subgraphs T \ { v }
and T \ N ( v ) are forests, which may have more than one component. But the largest independent set
in a disconnected graph is just the union of the largest independent sets in its components, so we can
separately consider each tree in these forests. Fortunately, this has the added beneﬁt of making the
recursive algorithm more efﬁcient, especially if we can choose the node v such that the trees are all
signiﬁcantly smaller than T . Here is the modiﬁed algorithm:
MAXIMUMINDSETSIZE( T ):
if T = ∅
return 0
v ← any node in T
withv ← 1
for each tree T in T \ N ( v )
withv ← withv + MAXIMUMINDSETSIZE( T )
withoutv ← 0
for each tree T in T \ { v }
withoutv ← withoutv + MAXIMUMINDSETSIZE( T )
return max{withv, withoutv}. Now let’s try to memoize this algorithm. Each recursive subproblem considers a subtree (that is, a
connected subgraph) of the original tree T . Unfortunately, a single tree T can have exponentially many
subtrees, so we seem to be doomed from the start!
Fortunately, there’s a degree of freedom that we have not yet exploited: We get to choose the vertex v .
We need a recipe—an algorithm!—for choosing v in each subproblem that limits the number of different
subproblems the algorithm considers. To make this work, we impose some additional structure on the
original input tree. Speciﬁcally, we declare one of the vertices of T to be the root, and we orient all the
edges of T away from that root. Then we let v be the root of the input tree; this choice guarantees that
each recursive subproblem considers a rooted subtree of T . Each vertex in T is the root of exactly one
subtree, so now the number of distinct subproblems is exactly n. We can further simplify the algorithm
by only passing a single node instead of the entire subtree:
MAXIMUMINDSETSIZE( v ):
withv ← 1
for each grandchild x of v
withv ← withv + MAXIMUMINDSETSIZE( x )
withoutv ← 0
for each child w of v
withoutv ← withoutv + MAXIMUMINDSETSIZE(w )
return max{withv, withoutv}. What data structure should we use to store intermediate results? The most natural choice is the tree
itself! Speciﬁcally, for each node v , we store the result of MAXIMUMINDSETSIZE( v ) in a new ﬁeld v. MIS.
(We could use an array, but then we’d have to add a new ﬁeld to each node anyway, pointing to the
corresponding array entry. Why bother?)
What’s the running time of the algorithm? The non-recursive time associated with each node v is
proportional to the number of children and grandchildren of v ; this number can be very different from
9 Algorithms Lecture 3: Dynamic Programming one vertex to the next. But we can turn the analysis around: Each vertex contributes a constant amount
of time to its parent and its grandparent! Since each vertex has at most one parent and at most one
grandparent, the total running time is O(n).
What’s a good order to consider the subproblems? The subproblem associated with any node v
depends on the subproblems associated with the children and grandchildren of v . So we can visit the
nodes in any order, provided that all children are visited before their parent. In particular, we can use a
straightforward post-order traversal.
Here is the resulting dynamic programming algorithm. Yes, it’s still recursive. I’ve swapped the
evaluation of the with- v and without- v cases; we need to visit the kids ﬁrst anyway, so why not consider
the subproblem that depends directly on the kids ﬁrst?
MAXIMUMINDSETSIZE( v ):
withoutv ← 0
for each child w of v
withoutv ← withoutv + MAXIMUMINDSETSIZE(w )
withv ← 1
for each grandchild x of v
withv ← withv + x . MIS
v. MIS ← max{withv, withoutv}
return v. MIS Another option is to store two values for each rooted subtree: the size of the largest independent set
that includes the root, and the size of the largest independent set that excludes the root. This gives us an
even simpler algorithm, with the same O(n) running time.
MAXIMUMINDSETSIZE( v ):
v. MISno ← 0
v. MISyes ← 1
for each child w of v
v. MISno ← v. MISno + MAXIMUMINDSETSIZE(w )
v. MISyes ← v. MISyes + w. MISno
return max{ v. MISyes, v. MISno} 3.8 Optimal Binary Search Trees In an earlier lecture, we developed a recursive algorithm for the optimal binary search tree problem.
We are given a sorted array A[1 .. n] of search keys and an array f [1 .. n] of frequency counts, where
f [i ] is the number of searches to A[i ]. Our task is to construct a binary search tree for that set such that
the total cost of all the searches is as small as possible. We developed the following recurrence for this
problem:
n OptCost( f [1 .. n]) = min 1≤ r ≤n OptCost( f [1 .. r − 1]) + f [i ] + OptCost( f [ r + 1 .. n])
i =1 To put this recurrence in more standard form, ﬁx the frequency array f , and let OptCost(i , j ) denote the
total search time in the optimal search tree for the subarray A[i .. j ]. To simplify notation a bit, let F (i , j )
denote the total frequency count for all the keys in the interval A[i .. j ]:
j F (i , j ) = f [k]
k=i 10 Algorithms Lecture 3: Dynamic Programming We can now write
OptCost(i , j ) = if j < i 0
F (i , j ) + min OptCost(i , r − 1) + OptCost( r + 1, j )
i≤r ≤ j otherwise The base case might look a little weird, but all it means is that the total cost for searching an empty set
of keys is zero.
The algorithm will be somewhat simpler and more efﬁcient if we precompute all possible values of
F (i , j ) and store them in an array. Computing each value F (i , j ) using a separate for-loop would O(n3 )
time. A better approach is to turn the recurrence
F (i , j ) = f [i ] if i = j F ( i , j − 1) + f [ j ] otherwise into the following O(n2 )-time dynamic programming algorithm:
INITF( f [1 .. n]):
for i ← 1 to n
F [ i , i − 1] ← 0
for j ← i to n
F [i , j ] ← F [i , j − 1] + f [i ] This will be used as an initialization subroutine in our ﬁnal algorithm.
So now let’s compute the optimal search tree cost OptCost(1, n) from the bottom up. We can store
all intermediate results in a table OptCost[1 .. n, 0 .. n]. Only the entries OptCost[i , j ] with j ≥ i − 1 will
actually be used. The base case of the recurrence tells us that any entry of the form OptCost[i , i − 1] can
immediately be set to 0. For any other entry OptCost[i , j ], we can use the following algorithm fragment,
which comes directly from the recurrence:
COMPUTEOPTCOST(i , j ):
OptCost[i , j ] ← ∞
for r ← i to j
t mp ← OptCost[i , r − 1] + OptCost[ r + 1, j ]
if OptCost[i , j ] > t mp
OptCost[i , j ] ← t mp
OptCost[i , j ] ← OptCost[i , j ] + F [i , j ] The only question left is what order to ﬁll in the table.
Each entry OptCost[i , j ] depends on all entries OptCost[i , r − 1] and OptCost[ r + 1, j ] with i ≤ k ≤ j .
In other words, every entry in the table depends on all the entries directly to the left or directly below.
In order to ﬁll the table efﬁciently, we must choose an order that computes all those entries before
OptCost[i , j ]. There are at least three different orders that satisfy this constraint. The one that occurs to
most people ﬁrst is to scan through the table one diagonal at a time, starting with the trivial base cases
OptCost[i , i − 1]. The complete algorithm looks like this:
OPTIMALSEARCHTREE( f [1 .. n]):
INITF( f [1 .. n])
for i ← 1 to n
OptCost[i , i − 1] ← 0
for d ← 0 to n − 1
for i ← 1 to n − d
COMPUTEOPTCOST(i , i + d )
return OptCost[1, n] 11 Algorithms Lecture 3: Dynamic Programming We could also traverse the array row by row from the bottom up, traversing each row from left to
right, or column by column from left to right, traversing each columns from the bottom up. These two
orders give us the following algorithms:
OPTIMALSEARCHTREE2( f [1 .. n]):
INITF( f [1 .. n])
for i ← n downto 1
OptCost[i , i − 1] ← 0
for j ← i to n
COMPUTEOPTCOST(i , j )
return OptCost[1, n] OPTIMALSEARCHTREE3( f [1 .. n]):
INITF( f [1 .. n])
for j ← 0 to n
OptCost[ j + 1, j ] ← 0
for i ← j downto 1
COMPUTEOPTCOST(i , j )
return OptCost[1, n] Three different evaluation orders for the table OptCost[i , j ]. No matter which of these orders we actually use, the resulting algorithm runs in Θ(n 3 ) time and
uses Θ(n 2 ) space.
We could have predicted these space and time bounds directly from the original recurrence.
OptCost(i , j ) = if j = i − i 0
F (i , j ) + min OptCost(i , r − 1) + OptCost( r + 1, j )
i≤r ≤ j otherwise First, the function has two arguments, each of which can take on any value between 1 and n, so we
probably need a table of size O(n2 ). Next, there are three variables in the recurrence (i , j , and r ), each
of which can take any value between 1 and n, so it should take us O(n3 ) time to ﬁll the table. 3.9 More Examples We’ve already seen two other examples of recursive algorithms that we can signiﬁcantly speed up via
dynamic programming.
3.9.1 Subset Sum Recall that the Subset Sum problem asks, given a set X of positive integers (represented as an array
X [1 .. n] and an integer T , whether any subset of X sums to T . In that lecture, we developed a recursive
algorithm which can be reformulated as follows. Fix the original input array X [1 .. n] and the original
target sum T , and deﬁne the boolean function
S (i , t ) = some subset of X [i .. n] sums to t .
Our goal is to compute S (1, T ), using the recurrence TRUE
S (i , t ) = if t = 0, FALSE
if t < 0 or i > n, S (i + 1, t ) ∨ S (i + 1, t − X [i ]) otherwise.
12 Algorithms Lecture 3: Dynamic Programming Observe that there are only nT possible values for the input parameters that lead to the interesting case
of this recurrence, so storing the results of all such subproblems requires O (mn ) space. If S (i + 1, t )
and S (i + 1, t − X [i ]) are already known, we can compute S (i , t ) in constant time, so memoizing this
recurrence gives us and algorithm that runs in O (n T ) time.6 To turn this into an explicit dynamic
programming algorithm, we only need to consider the subproblems S (i , t ) in the proper order:
SUBSETSUM(X [1 .. n], T ):
S [n + 1, 0] ← TRUE
for t ← 1 to T
S [n + 1, t ] ← FALSE
for i ← n downto 1
S [i , 0] = TRUE
for t ← 1 to X [i ] − 1
S [i , t ] ← S [i + 1, t ]
〈〈Avoid the case t < 0〉〉
for t ← X [i ] to T
S [i , t ] ← S [i + 1, t ] ∨ S [i + 1, t − X [i ]]
return S [1, T ] This direct algorithm clearly always uses O (n T ) time and space. In particular, if T is signiﬁcantly larger
than 2n , this algorithm is actually slower than our naïve recursive algorithm. Dynamic programming
isn’t always an improvement!
3.9.2 Longest Increasing Subsequence We also developed a recurrence for the longest increasing subsequence problem. Fix the original input
array A[1 .. n] with a sentinel value A[0] = −∞. Let L (i , j ) denote the length of the longest increasing
subsequence of A[ j .. n] with all elements larger than A[i ]. Our goal is to compute L (0, 1) − 1. (The −1
removes the sentinel −∞.) For any i < j , our recurrence can be stated as follows: if j > n
0
L (i , j ) = L (i , j + 1) max{ L (i , j + 1), 1 + L ( j , j + 1)} if A[i ] ≥ A[ j ]
otherwise The recurrence suggests that our algorithm should use O(n2 ) time and space, since the input parameters
i and j each can take n different values. To get an explicit dynamic programming algorithm, we only
need to ensure that both L (i , j + 1) and L ( j , j + 1) are considered before L (i , j ), for all i and j .
LIS(A[1 .. n]):
A[0] ← −∞
for i ← 0 to n
L [ i , n + 1] ← 0 〈〈Add a sentinel〉〉
〈〈Base cases〉〉 for j ← n downto 1
for i ← 0 to j − 1
if A[i ] ≥ A[ j ]
L [ i , j ] ← L [ i , j + 1]
else
L [i , j ] ← max{ L [i , j + 1], 1 + L [ j , j + 1]}
return L [0, 1] − 1 〈〈Don’t count the sentinel〉〉 Even though SubsetSum is NP-complete, this bound does not imply that P=NP, because T is not necessary bounded by a
polynomial function of the input size.
6 13 Algorithms Lecture 3: Dynamic Programming As predicted, this algorithm clearly uses O (n 2 ) time and space. We can reduce the space to O(n) by
only maintaining the two most recent columns of the table, L [·, j ] and L [·, j + 1].
This is not the only recursive strategy we could use for computing longest increasing subsequences.
Here is another recurrence that gives us the O(n) space bound for free. Let L (i ) denote the length of
the longest increasing subsequence of A[i .. n] that starts with A[i ]. Our goal is to compute L (0) − 1. To
deﬁne L (i ) recursively, we only need to specify the second element in subsequence; the Recursion Fairy
will do the rest.
L (i ) = 1 + max L ( j ) | j > i and A[ j ] > A[i ]
Here, I’m assuming that max ∅ = 0, so that the base case is L (n) = 1 falls out of the recurrence
automatically. Memoizing this recurrence requires O(n) space, and the resulting algorithm runs in O(n2 )
time. To transform this into a dynamic programming algorithm, we only need to guarantee that L ( j ) is
computed before L (i ) whenever i < j .
LIS2(A[1 .. n]):
A[0] = −∞ 〈〈Add a sentinel〉〉 for i ← n downto 0
L [i ] ← 1
for j ← i + 1 to n
if A[ j ] > A[i ] and 1 + L [ j ] > L [i ]
L [i ] ← 1 + L [ j ]
return L [0] − 1 〈〈Don’t count the sentinel〉〉 Exercises
1. Suppose you are given an array A[1 .. n] of integers. Describe and analyze an algorithm that
ﬁnds the largest sum of of elements in a contiguous subarray A[i .. j ]. For example, if the array
contains the numbers (−6, 12, −7, 0, 14, −7, 5), then the largest sum of any contiguous subarray is
19 = 12 − 7 + 0 + 14.
−6 12 −7 0 14 −7 5
19 2. This series of exercises asks you to develop efﬁcient algorithms to ﬁnd optimal subsequences of various kinds. A subsequence is anything obtained from a sequence by extracting a subset of elements,
but keeping them in the same order; the elements of the subsequence need not be contiguous in
the original sequence. For example, the strings C, DAMN, YAIOAI, and DYNAMICPROGRAMMING are all
subsequences of the sequence DYNAMICPROGRAMMING.
(a) Let A[1 .. m] and B [1 .. n] be two arbitrary arrays. A common subsequence of A and B is
another sequence that is a subsequence of both A and B . Describe an efﬁcient algorithm to
compute the length of the longest common subsequence of A and B .
(b) Let A[1 .. m] and B [1 .. n] be two arbitrary arrays. A common supersequence of A and B is
another sequence that contains both A and B as subsequences. Describe an efﬁcient algorithm
to compute the length of the shortest common supersequence of A and B .
(c) Call a sequence X [1 .. n] of numbers oscillating if X [i ] < X [i + 1] for all even i , and
X [i ] > X [i + 1] for all odd i . Describe an efﬁcient algorithm to compute the length of the
longest oscillating subsequence of an arbitrary array A of integers. 14 Algorithms Lecture 3: Dynamic Programming (d) Describe an efﬁcient algorithm to compute the length of the shortest oscillating supersequence
of an arbitrary array A of integers.
(e) Call a sequence X [1 .. n] of numbers accelerating if 2 · X [i ] < X [i − 1] + X [i + 1] for all i .
Describe an efﬁcient algorithm to compute the length of the longest accelerating subsequence
of an arbitrary array A of integers.
(f) Recall that a sequence X [1 .. n] of numbers is increasing if X [i ] < X [i + 1] for all i . Describe an efﬁcient algorithm to compute the length of the longest common increasing subsequence of two given arrays of integers. For example, 〈1, 4, 5, 6, 7, 9〉 is the longest common increasing subsequence of the sequences 〈3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3〉 and
〈1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5〉.
3. You and your eight-year-old nephew Elmo decide to play a simple card game. At the beginning
of the game, the cards are dealt face up in a long row. Each card is worth a different number
of points. After all the cards are dealt, you and Elmo take turns removing either the leftmost or
rightmost card from the row, until all the cards are gone. At each turn, you can decide which of
the two cards to take. The winner of the game is the player that has collected the most points
when the game ends.
Having never taken an algorithms class, Elmo follows the obvious greedy strategy—when it’s
his turn, Elmo always takes the card with the higher point value. Your task is to ﬁnd a strategy
that will beat Elmo whenever possible. (It might seem mean to beat up on a little kid like this, but
Elmo absolutely hates it when grown-ups let him win.)
(a) Prove that you should not also use the greedy strategy. That is, show that there is a game
that you can win, but only if you do not follow the same greedy strategy as Elmo.
(b) Describe and analyze an algorithm to determine, given the initial sequence of cards, the
maximum number of points that you can collect playing against Elmo.
4. A palindrome is any string that is exactly the same as its reversal, like I, or DEED, or RACECAR, or
AMANAPLANACATACANALPANAMA.
(a) Describe and analyze an algorithm to ﬁnd the length of the longest subsequence of a given
string that is also a palindrome. For example, the longest palindrome subsequence of
MAHDYNAMICPROGRAMZLETMESHOWYOUTHEM is MHYMRORMYHM, so given that string as input, your
algorithm should output the number 11.
(b) Any string can be decomposed into a sequence of palindromes. For example, the string
BUBBASEESABANANA (‘Bubba sees a banana.’) can be broken into palindromes in the following
ways (and many others):
BUB + BASEESAB
B + U + BB + A + SEES
B + U + BB + A + SEES
B+U+B+B+A+S+E+E+S +
+
+
+ ANANA
ABA + NAN + A
A + B + ANANA
A+B+A+N+A+N+A Describe and analyze an efﬁcient algorithm to ﬁnd the smallest number of palindromes that
make up a given input string. For example, given the input string BUBBASEESABANANA, your
algorithm would return the integer 3. 15 Algorithms Lecture 3: Dynamic Programming 5. Suppose you have a subroutine QUALITY that can compute the ‘quality’ of any given string A[1 .. k]
in time O(k). For example, the quality of a string might be 1 if the string is a Québecois curse
word, and 0 otherwise.
Given an arbitrary input string T [1 .. n], we would like to break it into contiguous substrings,
such that the total quality of all the substrings is as large as possible. For example, the string
SAINTCIBIOREDESACRAMENTDECRISSE can be decomposed into the substrings SAINT + CIBIORE
+ DE + SACRAMENT + DE + CRISSE, of which three (or possibly four) are sacres.
Describe an algorithm that breaks a string into substrings of maximum total quality, using the
QUALITY subroutine.
6. Suppose you are given an m × n bitmap, represented by an array M [1 .. m, 1 .. n] whose entries
are all 0 or 1. A solid block is a subarray of the form M [i .. i , j .. j ] in which every bit is equal to 1.
Describe and analyze an efﬁcient algorithm to ﬁnd a solid block in M with maximum area.
7. Consider two horizontal lines 1 and 2 in the plane. Suppose we are given (the x -coordinates
of) n distinct points a1 , a2 , . . . , an on 1 and n distinct points b1 , b2 , . . . , bn on 2 . (The points ai
and b j are not necessarily indexed in order from left to right, or in the same order.) Design an
algorithm to compute the largest set S of non-intersecting line segments satisfying to the following
restrictions:
(a) Each segment in S connects some point ai to the corresponding point bi .
(b) No two segments in S intersect.
8. You are a bus driver with a soda fountain machine in the back and a bus full of very hyper students,
who are drinking more soda as they ride along the highway. Your goal is to drop the students
off as quickly as possible. More speciﬁcally, each minute that a student is on your bus, he drinks
another ounce of soda. Your goal is to drop the students off quickly, so that in total they drink as
little soda as possible.
You know how many students will get off of the bus at each exit. Your bus begins partway
along the highway (probably not at either end), and moves at a constant speed of 37.4 miles per
hour. You must drive the bus along the highway; however, you may drive forward to one exit then
backward to an exit in the other direction, switching as often as you like. (You can stop the bus,
drop off students, and turn around instantaneously.)
Describe an efﬁcient algorithm to drop the students off so that they drink as little soda as
possible. The input to the algorithm should be: the bus route (a list of the exits, together with the
travel time between successive exits), the number of students you will drop off at each exit, and
the current location of your bus (which you may assume is an exit).
9. In a previous life, you worked as a cashier in the lost Antarctican colony of Nadira, spending the
better part of your day giving change to your customers. Because paper is a very rare and valuable
resource in Antarctica, cashiers were required by law to use the fewest bills possible whenever
they gave change. Thanks to the numerological predilections of one of its founders, the currency 16 Algorithms Lecture 3: Dynamic Programming of Nadira, called Dream Dollars, was available in the following denominations: $1, $4, $7, $13,
$28, $52, $91, $365.7
(a) The greedy change algorithm repeatedly takes the largest bill that does not exceed the target
amount. For example, to make $122 using the greedy algorithm, we ﬁrst take a $91 bill,
then a $28 bill, and ﬁnally three $1 bills. Give an example where this greedy algorithm uses
more Dream Dollar bills than the minimum possible.
(b) Describe and analyze a recursive algorithm that computes, given an integer k, the minimum
number of bills needed to make k Dream Dollars. (Don’t worry about making your algorithm
fast; just make sure it’s correct.)
(c) Describe a dynamic programming algorithm that computes, given an integer k, the minimum
number of bills needed to make k Dream Dollars. (This one needs to be fast.)
10. What excitement! The Champaign Spinners and the Urbana Dreamweavers have advanced to
meet each other in the World Series of Basket-weaving! The World Champions will be decided
by a best-of- 2n − 1 series of head-to-head weaving matches, and the ﬁrst to win n matches will
take home the coveted Golden Basket (for example, a best-of-7 series requiring four match wins,
but we will keep the generalized case). We know that for any given match there is a constant
probability p that Champaign will win, and a subsequent probability q = 1 − p that Urbana will
win.
Let P (i , j ) be the probability that Champaign will win the series given that they still need i
more victories, whereas Urbana needs j more victories for the championship. P (0, j ) = 1 for any j ,
because Champaign needs no more victories to win. Similarly, P (i , 0) = 0 for any i , as Champaign
cannot possibly win if Urbana already has. P (0, 0) is meaningless. Champaign wins any particular
match with probability p and loses with probability q, so
P (i , j ) = p · P (i − 1, j ) + q · P (i , j − 1)
for any i ≥ 1 and j ≥ 1.
Describe and analyze an efﬁcient algorithm that computes the probability that Champaign will
win the series (that is, calculate P (n, n)), given the parameters n, p, and q as input.
11. Vankin’s Mile is a solitaire game played on an n × n square grid. The player starts by placing a
token on any square of the grid. Then on each turn, the player moves the token either one square
to the right or one square down. The game ends when player moves the token off the edge of the
board. Each square of the grid has a numerical value, which could be positive, negative, or zero.
The player starts with a score of zero; whenever the token lands on a square, the player adds its
value to his score. The object of the game is to score as many points as possible.
For example, given the grid below, the player can score 8 − 6 + 7 − 3 + 4 = 10 points by placing
the initial token on the 8 in the second row, and then moving down, down, right, down, down.
(This is not the best possible score for these values.)
7
For more details on the history and culture of Nadira, including images of the various denominations of Dream Dollars, see
http://www.dream-dollars.com. 17 Algorithms Lecture 3: Dynamic Programming -1 7 -4 -9 5 -2 -7 4 7 -8 1 10 -5 8 -6 0 -6 -6 7 ⇓ ⇓
7⇒ -3
⇓ -6 4 ⇓ -3
-9 Describe and analyze an efﬁcient algorithm to compute the maximum possible score for a
game of Vankin’s Mile, given the n × n array of values as input.
12. A shufﬂe of two strings X and Y is formed by interspersing the characters into a new string, keeping
the characters of X and Y in the same order. For example, ‘bananaananas’ is a shufﬂe of ‘banana’
and ‘ananas’ in several different ways.
banana ananas ban ana
ana
nas b an an a na
a na s The strings ‘prodgyrnamammiincg’ and ‘dyprongarmammicing’ are both shufﬂes of ‘dynamic’ and
‘programming’:
d y nam
ic
pro g r
ammi n g dy nam
ic
pro g r amm ing Given three strings A[1 .. m], B [1 .. n], and C [1 .. m + n], describe and analyze an algorithm to
determine whether C is a shufﬂe of A and B .
13. Suppose we want to display a paragraph of text on a computer screen. The text consists of n
words, where the i th word is pi pixels wide. We want to break the paragraph into several lines,
each exactly P pixels long. Depending on which words we put on each line, we will need to insert
different amounts of white space between the words. The paragraph should be fully justiﬁed,
meaning that the ﬁrst word on each line starts at its leftmost pixel, and except for the last line,
the last character on each line ends at its rightmost pixel. There must be at least one pixel of
white-space between any two words on the same line.
Deﬁne the slop of a paragraph layout as the sum over all lines, except the last, of the cube of
the number of extra white-space pixels in each line (not counting the one pixel required between
every adjacent pair of words). Speciﬁcally, if a line contains words i through j , then the amount of
j
extra white space on that line is P − j + i − k=i pk . Describe a dynamic programming algorithm
to print the paragraph with minimum slop.
14. Let P be a set of points in the plane in convex position. Intuitively, if a rubber band were wrapped
around the points, then every point would touch the rubber band. More formally, for any point p
in P , there is a line that separates p from the other points in P . Moreover, suppose the points are
indexed P [1], P [2], . . . , P [n] in counterclockwise order around the ‘rubber band’, starting with
the leftmost point P [1].
This problem asks you to solve a special case of the traveling salesman problem, where the
salesman must visit every point in P , and the cost of moving from one point p ∈ P to another point
q ∈ P is the Euclidean distance | pq|.
18 Algorithms Lecture 3: Dynamic Programming (a) Describe a simple algorithm to compute the shortest cyclic tour of P .
(b) A simple tour is one that never crosses itself. Prove that the shortest tour of P must be simple.
(c) Describe and analyze an efﬁcient algorithm to compute the shortest tour of P that starts at
the leftmost point P [1] and ends at the rightmost point P [ r ].
15. Describe and analyze an algorithm to solve the traveling salesman problem in O(2n poly(n)) time.
Given an undirected n-vertex graph G with weighted edges, your algorithm should return the
weight of the lightest cycle in G that visits every vertex exactly once, or ∞ if G has no such cycles.
[Hint: The obvious recursive algorithm takes O(n!) time.]
16. Recall that a subtree of a (rooted, ordered) binary tree T consists of a node and all its descendants.
Design and analyze an efﬁcient algorithm to compute the largest common subtree of two given
binary trees T1 and T2 ; this is the largest subtree of T1 that is isomorphic to a subtree in T2 .
The contents of the nodes are irrelevant; we are only interested in matching the underlying
combinatorial structure. Two binary trees, with their largest common subtree emphasized 17. Suppose we need to distribute a message to all the nodes in a rooted tree. Initially, only the root
node knows the message. In a single round, any node that knows the message can forward it
to at most one of its children. Design an algorithm to compute the minimum number of rounds
required for the message to be delivered to all nodes. A message being distributed through a tree in ﬁve rounds. 18. A company is planning a party for its employees. The employees in the company are organized
into a strict hierarchy, that is, a tree with the company president at the root. The organizers of
the party have assigned a real number to each employee measuring how ‘fun’ the employee is. In
order to keep things social, there is one restriction on the guest list: an employee cannot attend
the party if their immediate supervisor is present. On the other hand, the president of the company
must attend the party, even though she has a negative fun rating; it’s her company, after all. Give
an algorithm that makes a guest list for the party that maximizes the sum of the ‘fun’ ratings of the
guests.
19 Algorithms Lecture 3: Dynamic Programming 19. Scientists have branched out from the bizarre planet of Yggdrasil to study the vodes which have
settled on Ygdrasil’s moon, Xryltcon. All vodes on Xryltcon are descended from the ﬁrst vode to
arrive there, named George. Each vode has a color, either cyan, magenta, or yellow, but breeding
patterns are not the same as on Yggdrasil; every vode, regardless of color, has either two children
(with arbitrary colors) or no children.
George and all his descendants are alive and well, and they are quite excited to meet the
scientists who wish to study them. Unsurprisingly, these vodes have had some strange mutations
in their isolation on Xryltcon. Each vode has a weirdness rating; weirder vodes are more interesting
to the visiting scientists. (Some vodes even have negative weirdness ratings; they make other
vodes more boring just by standing next to them.)
Also, Xryltconian society is strictly governed by a number of sacred cultural traditions.
• No cyan vode may be in the same room as its non-cyan children (if it has any).
• No magenta vode may be in the same room as its parent (if it has one).
• Each yellow vode must be attended at all times by its grandchildren (if it has any).
• George must be present at any gathering of more than ﬁfty vodes.
The scientists have exactly one chance to study a group of vodes in a single room. You are
given the family tree of all the vodes on Xryltcon, along with the weirdness value of each vode.
Design and analyze an efﬁcient algorithm to decide which vodes the scientists should invite to
maximize the sum of the weirdness values of the vodes in the room. Be careful to respect all of
the vodes’ cultural taboos.
20. Oh, no! You have been appointed as the gift czar for Giggle, Inc.’s annual mandatory holiday
party! The president of the company, who is certiﬁably insane, has declared that every Giggle
employee must receive one of three gifts: (1) an all-expenses-paid six-week vacation anywhere in
the world, (2) an all-the-pancakes-you-can-eat breakfast for two at Jumping Jack Flash’s Flapjack
Stack Shack, or (3) a burning paper bag full of dog poop. Corporate regulations prohibit any
employee from receiving the same gift as his/her direct supervisor. Any employee who receives a
better gift than his/her direct supervisor will almost certainly be ﬁred in a ﬁt of jealousy. How do
you decide what gifts everyone gets if you want to minimize the number of people that get ﬁred?
1
2 3 3
2 1 3
2 1 1 3 3 2
1 2
3 3 3 1
2 2 1 A tree labeling with cost 9. Bold nodes have smaller labels than their parents.
This is not the optimal labeling for this tree. More formally, suppose you are given a rooted tree T , representing the company hierarchy.
You want to label each node in T with an integer 1, 2, or 3, such that every node has a different
20 Algorithms Lecture 3: Dynamic Programming label from its parent.. The cost of an labeling is the number of nodes that have smaller labels than
their parents. Describe and analyze an algorithm to compute the minimum cost of any labeling of
the given tree T . (Your algorithm does not have to compute the actual best labeling—just its cost.)
21. Every year, as part of its annual meeting, the Antarctican Snail Lovers of Upper Glacierville hold a
Round Table Mating Race. Several high-quality breeding snails are placed at the edge of a round
table. The snails are numbered in order around the table from 1 to n. During the race, each snail
wanders around the table, leaving a trail of slime behind it. The snails have been specially trained
never to fall off the edge of the table or to cross a slime trail, even their own. If two snails meet,
they are declared a breeding pair, removed from the table, and whisked away to a romantic hole
in the ground to make little baby snails. Note that some snails may never ﬁnd a mate, even if the
race goes on forever.
For every pair of snails, the Antarctican SLUG race organizers have posted a monetary reward,
to be paid to the owners if that pair of snails meets during the Mating Race. Speciﬁcally, there
is a two-dimensional array M [1 .. n, 1 .. n] posted on the wall behind the Round Table, where
M [i , j ] = M [ j , i ] is the reward to be paid if snails i and j meet.
Describe and analyze an algorithm to compute the maximum total reward that the organizers
could be forced to pay, given the array M as input.
1
8 2 8
7 1
6 7 3
52 4 6 3 4
5 The end of a typical Antarctican SLUG race. Snails 6 and 8 never ﬁnd mates.
The organizers must pay M [3, 4] + M [2, 5] + M [1, 7]. 22. Ribonucleic acid (RNA) molecules are long chains of millions of nucleotides or bases of four
different types: adenine (A), cytosine (C), guanine (G), and uracil (U). The sequence of an RNA
molecule is a string b[1 .. n], where each character b[i ] ∈ {A, C , G , U } corresponds to a base. In
addition to the chemical bonds between adjacent bases in the sequence, hydrogen bonds can form
between certain pairs of bases. The set of bonded base pairs is called the secondary structure of the
RNA molecule.
We say that two base pairs (i , j ) and (i , j ) with i < j and i < j overlap if i < i < j < j or
i < i < j < j . In practice, most base pairs are non-overlapping. Overlapping base pairs create
so-called pseudoknots in the secondary structure, which are essential for some RNA functions, but
are more difﬁcult to predict.
Suppose we want to predict the best possible secondary structure for a given RNA sequence.
We will adopt a drastically simpliﬁed model of secondary structure:
21 Algorithms
•
•
•
• Lecture 3: Dynamic Programming Each base can be paired with at most one other base.
Only A-U pairs and C-G pairs can bond.
Pairs of the form (i , i + 1) and (i , i + 2) cannot bond.
Overlapping base pairs cannot bond. The last restriction allows us to visualize RNA secondary structure as a sort of fat tree.
(a) Describe and analyze an algorithm that computes the maximum possible number of bonded
base pairs in a secondary structure for a given RNA sequence.
(b) A gap in a secondary structure is a maximal substring of unpaired bases. Large gaps lead to
chemical instabilities, so secondary structures with smaller gaps are more likely. To account
for this preference, let’s deﬁne the score of a secondary structure to be the sum of the squares
of the gap lengths.8 Describe and analyze an algorithm that computes the minimum possible
score of a secondary structure for a given RNA sequence.
CU
G
A
U U U
G
C A
U G
C U
A GU A C CA C UACUCAU
G C
UUACA CAUGAGUA AAUGU U
A A
A
G U
C U
U C
U A
A A
A
U CU C GG A
U
U Example RNA secondary structure with 21 base pairs, indicated by heavy red lines.
Gaps are indicated by dotted curves. This structure has score 22 + 22 + 82 + 12 + 72 + 42 + 72 = 187 23. Let
= {A1 , A2 , . . . , An } be a ﬁnite set of strings over some ﬁxed alphabet Σ. An edit center for
is a string C ∈ Σ∗ such that the maximum edit distance from C to any string in
is as small as
possible. The edit radius of
is the maximum edit distance from an edit center to a string in .
A set of strings may have several edit centers, but its edit radius is unique.
EditRadius( ) = min max Edit(A, C )
∗
C ∈Σ A∈ EditCenter( ) = arg min max Edit(A, C )
C ∈ Σ∗ A∈ (a) Describe and analyze an efﬁcient algorithm to compute the edit radius of three given strings.
(b) Describe and analyze an efﬁcient algorithm to approximate the edit radius of an arbitrary
set of strings within a factor of 2. (Computing the edit radius exactly is NP-hard unless the
number of strings is ﬁxed.)
8 This score function has absolutely no connection to reality; I just made it up. Real RNA structure prediction requires much
more complicated scoring functions. 22 Algorithms Lecture 3: Dynamic Programming 24. Let D[1 .. n] be an array of digits, each an integer between 0 and 9. An digital subsequence of
D is an sequence of positive integers composed in the usual way from disjoint substrings of D.
For example, 3, 4, 5, 6, 8, 9, 32, 38, 46, 64, 83, 279 is an increasing digital subsequence of the ﬁrst
several digits of π:
3 , 1, 4 , 1, 5 , 9, 2, 6 , 5, 3, 5, 8 , 9 , 7, 9, 3, 2 , 3, 8 , 4, 6 , 2, 6, 4 , 3, 3, 8, 3 , 2, 7, 9
The length of a digital subsequence is the number of integers it contains, not the number of digits;
the preceding example has length 12.
Describe and analyze an efﬁcient algorithm to compute the longest increasing digital subsequence of D. [Hint: Be careful about your computational assumptions. How long does it take to
compare two k-digit numbers?] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 23 Algorithms Non-Lecture C: Advanced Dynamic Programming Tricks
Ninety percent of science ﬁction is crud.
But then, ninety percent of everything is crud,
and it’s the ten percent that isn’t crud that is important.
— [Theodore] Sturgeon’s Law (1953) C Advanced Dynamic Programming Tricks Dynamic programming is a powerful technique for efﬁciently solving recursive problems, but it’s hardly
the end of the story. In many cases, once we have a basic dynamic programming algorithm in place,
we can make further improvements to bring down the running time or the space usage. We saw one
example in the Fibonacci number algorithm. Buried inside the naïve iterative Fibonacci algorithm is a
recursive problem—computing a power of a matrix—that can be solved more efﬁciently by dynamic
programming techniques—in this case, repeated squaring. C.1 Saving Space: Divide and Conquer Just as we did for the Fibonacci recurrence, we can reduce the space complexity of our edit distance
algorithm from O(mn) to O(m + n) by only storing the current and previous rows of the memoization
table. This ‘sliding window’ technique provides an easy space improvement for most (but not all)
dynamic programming algorithm.
Unfortunately, this technique seems to be useful only if we are interested in the cost of the optimal
edit sequence, not if we want the optimal edit sequence itself. By throwing away most of the table, we
apparently lose the ability to walk backward through the table to recover the optimal sequence.
However, if we throw away most of the rows in the table, it seems we no longer have enough
information to reconstruct the actual editing sequence. Now what?
Fortunately for memory-misers, in 1975 Dan Hirshberg discovered a simple divide-and-conquer
strategy that allows us to compute the optimal edit sequence in O(mn) time, using just O(m + n) space.
The trick is to record not just the edit distance for each pair of preﬁxes, but also a single position in the
middle of the editing sequence for that preﬁx. Speciﬁcally, the optimal editing sequence that transforms
A[1 .. m] into B [1 .. n] can be split into two smaller editing sequences, one transforming A[1 .. m/2] into
B [1 .. h] for some integer h, the other transforming A[m/2 + 1 .. m] into B [h + 1 .. n]. To compute this
breakpoint h, we deﬁne a second function Half(i , j ) as follows: ∞
if i < m/2 j
if i = m/2 Half(i , j ) = Half(i − 1, j )
if i > m/2 and Edit(i , j ) = Edit(i − 1, j ) + 1 Half(i , j − 1) if i > m/2 and Edit(i , j ) = Edit(i , j − 1) + 1 Half(i − 1, j − 1) otherwise
A simple inductive argument implies that Half(m, n) is the correct value of h. We can easily modify our
earlier algorithm so that it computes Half(m, n) at the same time as the edit distance Edit(m, n), all in
O(mn) time, using only O(m) space.
Now, to compute the optimal editing sequence that transforms A into B , we recursively compute
the optimal subsequences. The recursion bottoms out when one string has only constant length, in
which case we can determine the optimal editing sequence by our old dynamic programming algorithm. 1 Algorithms Non-Lecture C: Advanced Dynamic Programming Tricks Overall the running time of our recursive algorithm satisﬁes the following recurrence: if m ≤ 1
O(n)
T (m, n) = O(m)
if n ≤ 1 O(mn) + T (m/2, h) + T (m/2, n − h) otherwise It’s easy to prove inductively that T (m, n) = O(mn), no matter what the value of h is. Speciﬁcally, the
entire algorithm’s running time is at most twice the time for the initial dynamic programming phase.
T (m, n) ≤ αmn + T (m/2, h) + T (m/2, n − h)
≤ αmn + 2αmh/2 + 2αm(n − h)/2 [inductive hypothesis] = 2αmn
A similar inductive argument implies that the algorithm uses only O(n + m) space.
Hirschberg’s divide-and-conquer trick can be applied to almost any dynamic programming problem
to obtain an algorithm to construct an optimal structure (in this case, the cheapest edit sequence) within
the same space and time bounds as computing the cost of that optimal structure (in this case, edit
distance). For this reason, we will almost always ask you for algorithms to compute the cost of some
optimal structure, not the optimal structure itself. C.2 Saving Time: Sparseness In many applications of dynamic programming, we are faced with instances where almost every recursive
subproblem will be resolved exactly the same way. We call such instances sparse. For example, we might
want to compute the edit distance between two strings that have few characters in common, which
means there are few “free” substitutions anywhere in the table. Most of the table has exactly the same
structure. If we can reconstruct the entire table from just a few key entries, then why compute the entire
table?
To better illustrate how to exploit sparseness, let’s consider a simpliﬁcation of the edit distance
problem, in which substitutions are not allowed (or equivalently, where a substitution counts as
two operations instead of one). Now our goal is to maximize the number of “free” substitutions, or
equivalently, to ﬁnd the longest common subsequence of the two input strings.
Let LCS(i , j ) denote the length of the longest common subsequence of two ﬁxed strings A[1 .. m] and
B [1 .. n]. This function can be deﬁned recursively as follows: if i = 0 or j = 0
0
LCS(i , j ) = LCS(i − 1, j − 1) max LCS(i , j − 1), LCS(i − 1, j ) if A[i ] = B [ j ]
otherwise This recursive deﬁnition directly translates into an O(mn)-time dynamic programming algorithm.
Call an index pair (i , j ) a match point if A[i ] = B [ j ]. In some sense, match points are the only
‘interesting’ locations in the memoization table; given a list of the match points, we could easily
reconstruct the entire table. 2 Algorithms Non-Lecture C: Advanced Dynamic Programming Tricks «
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2
33
33
34
34
34 0000
1111
2222
3333
3333
3333
4444
4445
5555 0
1
2
3
3
3
4
5
5 012223455555
012223455555
012223455556 The LCS memoization table for ALGORITHMS and ALTRUISTIC; the brackets « and » are sentinel characters. More importantly, we can compute the LCS function directly from the list of match points using the
following recurrence: if i = j = 0
0
LCS(i , j ) = max LCS(i , j ) | A[i ] = B [ j ] and i < i and j < j + 1 max LCS(i , j ) | A[i ] = B [ j ] and i ≤ i and j ≤ j if A[i ] = B [ j ]
otherwise (Notice that the inequalities are strict in the second case, but not in the third.) To simplify boundary
issues, we add unique sentinel characters A[0] = B [0] and A[m + 1] = B [n + 1] to both strings. This
ensures that the sets on the right side of the recurrence equation are non-empty, and that we only have
to consider match points to compute LCS(m, n) = LCS(m + 1, n + 1) − 1.
If there are K match points, we can actually compute them all in O(m log m + n log n + K ) time. Sort
the characters in each input string, but remembering the original index of each character, and then
essentially merge the two sorted arrays, as follows:
FINDMATCHES(A[1 .. m], B [1 .. n]):
for i ← 1 to m: I [i ] ← i
for j ← 1 to n: J [ j ] ← j
sort A and permute I to match
sort B and permute J to match
i ← 1; j ← 1
while i < m and j < n
if A[i ] < B [ j ]
i ← i+1
else if A[i ] > B [ j ]
j ← j+1
else
〈〈Found a match!〉〉
ii ← i
while A[ii ] = A[i ]
jj ← j
while B [ j j ] = B [ j ]
report ( I [ii ], J [ j j ])
jj ← jj +1
ii ← i + 1
i ← ii ; j ← j j 3 Algorithms Non-Lecture C: Advanced Dynamic Programming Tricks To efﬁciently evaluate our modiﬁed recurrence, we once again turn to dynamic programming. We
consider the match points in lexicographic order—the order they would be encountered in a standard
row-major traversal of the m × n table—so that when we need to evaluate LCS(i , j ), all match points
(i , j ) with i < i and j < j have already been evaluated.
SPARSELCS(A[1 .. m], B [1 .. n]):
Match[1 .. K ] ← FINDMATCHES(A, B )
Match[K + 1] ← (m + 1, n + 1)
〈〈Add end sentinel〉〉
Sort M lexicographically
for k ← 1 to K
(i , j ) ← Match[k]
LCS[k] ← 1
〈〈From start sentinel〉〉
for ← 1 to k − 1
(i , j ) ← Match[ ]
if i < i and j < j
LCS[k] ← min{LCS[k], 1 + LCS[ ]}
return LCS[K + 1] − 1 The overall running time of this algorithm is O(m log m + n log n + K 2 ). So as long as K = o( mn), this
algorithm is actually faster than naïve dynamic programming. C.3 Saving Time: Monotonicity Recall the optimal binary search tree problem from the previous lecture. Given an array F [1 .. n] of
access frequencies for n items, the problem it to compute the binary search tree that minimizes the cost
of all accesses. A straightforward dynamic programming algorithm solves this problem in O(n3 ) time.
As for longest common subsequence problem, the algorithm can be improved by exploiting some
structure in the memoization table. In this case, however, the relevant structure isn’t in the table of costs,
but rather in the table used to reconstruct the actual optimal tree. Let OptRoot[i , j ] denote the index of
the root of the optimal search tree for the frequencies F [i .. j ]; this is always an integer between i and j .
Donald Knuth proved the following nice monotonicity property for optimal subtrees: If we move either
end of the subarray, the optimal root moves in the same direction or not at all. More formally:
OptRoot[i , j − 1] ≤ OptRoot[i , j ] ≤ OptRoot[i + 1, j ] for all i and j .
This (nontrivial!) observation leads to the following more efﬁcient algorithm:
FASTEROPTIMALSEARCHTREE( f [1 .. n]):
INITF( f [1 .. n])
for i ← 1 downto n
OptCost[i , i − 1] ← 0
OptRoot[i , i − 1] ← i
for d ← 0 to n
for i ← 1 to n
COMPUTECOSTANDROOT(i , i + d )
return OptCost[1, n] COMPUTECOSTANDROOT(i , j ):
OptCost[i , j ] ← ∞
for r ← OptRoot[i , j − 1] to OptRoot[i + 1, j ]
tmp ← OptCost[i , r − 1] + OptCost[ r + 1, j ]
if OptCost[i , j ] > tmp
OptCost[i , j ] ← tmp
OptRoot[i , j ] ← r
OptCost[i , j ] ← OptCost[i , j ] + F [i , j ] It’s not hard to see that the loop index r increases monotonically from 1 to n during each iteration
of the outermost for loop of FASTEROPTIMALSEARCHTREE. Consequently, the total cost of all calls to
COMPUTECOSTANDROOT is only O(n2 ).
If we formulate the problem slightly differently, this algorithm can be improved even further. Suppose
we require the optimum external binary tree, where the keys A[1 .. n] are all stored at the leaves, and
4 Algorithms Non-Lecture C: Advanced Dynamic Programming Tricks intermediate pivot values are stored at the internal nodes. An algorithm due to Te Ching Hu and Alan
Tucker1 computes the optimal binary search tree in this setting in only O(n log n) time! Exercises
1. Describe an algorithm to compute the edit distance between two strings A[1 .. m] and B [1 ... n] in
O(m log m + n log n + K 2 ) time, where K is the number of match points. [Hint: Use the FINDMATCHES
algorithm on page 3 as a subroutine.] 2. (a) Describe an algorithm to compute the longest increasing subsequence of a string X [1 .. n] in
O(n log n) time.
(b) Using your solution to part (a) as a subroutine, describe an algorithm to compute the longest
common subsequence of two strings A[1 .. m] and B [1 ... n] in O(m log m + n log n + K log K )
time, where K is the number of match points.
3. Describe an algorithm to compute the edit distance between two strings A[1 .. m] and B [1 ... n] in
O(m log m + n log n + K log K ) time, where K is the number of match points. [Hint: Combine your
answers for problems 1 and 2(b).]
4. Let T be an arbitrary rooted tree, where each vertex is labeled with a positive integer. A subset S
of the nodes of T is heap-ordered if it satisﬁes two properties:
• S contains a node that is an ancestor of every other node in S .
• For any node v in S , the label of v is larger than the labels of any ancestor of v in S .
3
1 4 9 2 3
2 5
3 1 5
5 8 9 8 4 2 7 7 9 6
3 6
9 A heap-ordered subset of nodes in a tree. (a) Describe an algorithm to ﬁnd the largest heap-ordered subset S of nodes in T that has the
heap property in O(n2 ) time.
1
T. C. Hu and A. C. Tucker, Optimal computer search trees and variable length alphabetic codes, SIAM J. Applied Math.
21:514–532, 1971. For a slightly simpler algorithm with the same running time, see A. M. Garsia and M. L. Wachs, A new
algorithms for minimal binary search trees, SIAM J. Comput. 6:622–642, 1977. The original correctness proofs for both
algorithms are rather intricate; for simpler proofs, see Marek Karpinski, Lawrence L. Larmore, and Wojciech Rytter, Correctness
of constructing optimal alphabetic trees revisited, Theoretical Computer Science, 180:309-324, 1997. 5 Algorithms Non-Lecture C: Advanced Dynamic Programming Tricks (b) Modify your algorithm from part (a) so that it runs in O(n log n) time when T is either a
linked list. [Hint: This special case is equivalent to a problem you’ve seen before.]
(c) Describe an algorithm to ﬁnd the largest subset S of nodes in T that has the heap property,
in O(n log n) time. [Hint: Find an algorithm to merge two sorted lists of lengths k and in
k+
O(log k ) time.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 6 Algorithms Lecture 4: Greedy Algorithms
The point is, ladies and gentleman, greed is good. Greed works, greed is right.
Greed clariﬁes, cuts through, and captures the essence of the evolutionary spirit.
Greed in all its forms, greed for life, money, love, knowledge has marked the
upward surge in mankind. And greed—mark my words—will save not only Teldar
Paper but the other malfunctioning corporation called the USA.
— Michael Douglas as Gordon Gekko, Wall Street (1987)
There is always an easy solution to every human problem—
neat, plausible, and wrong.
— H. L. Mencken, “The Divine Afﬂatus”,
New York Evening Mail (November 16, 1917) 4
4.1 Greedy Algorithms
Storing Files on Tape Suppose we have a set of n ﬁles that we want to store on a tape. In the future, users will want to
read those ﬁles from the tape. Reading a ﬁle from tape isn’t like reading from disk; ﬁrst we have to
fast-forward past all the other ﬁles, and that takes a signiﬁcant amount of time. Let L [1 .. n] be an array
listing the lengths of each ﬁle; speciﬁcally, ﬁle i has length L [i ]. If the ﬁles are stored in order from 1
to n, then the cost of accessing the kth ﬁle is
k cost(k) = L [ i ].
i =1 The cost reﬂects the fact that before we read ﬁle k we must ﬁrst scan past all the earlier ﬁles on the tape.
If we assume for the moment that each ﬁle is equally likely to be accessed, then the expected cost of
searching for a random ﬁle is
n E[cost] = cost(k) k =1 n n k = L [i ] k = 1 i =1 n . If we change the order of the ﬁles on the tape, we change the cost of accessing the ﬁles; some ﬁles
become more expensive to read, but others become cheaper. Different ﬁle orders are likely to result in
different expected costs. Speciﬁcally, let π(i ) denote the index of the ﬁle stored at position i on the tape.
Then the expected cost of the permutation π is
n k E[cost(π)] =
k = 1 i =1 L [π(i )]
n . Which order should we use if we want the expected cost to be as small as possible? The answer is
intuitively clear; we should store the ﬁles in order from shortest to longest. So let’s prove this.
Lemma 1. E[cost(π)] is minimized when L [π(i )] ≤ L [π(i + 1)] for all i .
Proof: Suppose L [π(i )] > L [π(i + 1)] for some i . To simplify notation, let a = π(i ) and b = π(i + 1).
If we swap ﬁles a and b, then the cost of accessing a increases by L [ b], and the cost of accessing b
decreases by L [a]. Overall, the swap changes the expected cost by ( L [ b] − L [a])/n. But this change is
an improvement, because L [ b] < L [a]. Thus, if the ﬁles are out of order, we can improve the expected
cost by swapping some mis-ordered adjacent pair.
1 Algorithms Lecture 4: Greedy Algorithms This example gives us our ﬁrst greedy algorithm. To minimize the total expected cost of accessing
the ﬁles, we put the ﬁle that is cheapest to access ﬁrst, and then recursively write everything else; no
backtracking, no dynamic programming, just make the best local choice and blindly plow ahead. If
we use an efﬁcient sorting algorithm, the running time is clearly O(n log n), plus the time required to
actually write the ﬁles. To prove the greedy algorithm is actually correct, we simply prove that the
output of any other algorithm can be improved by some sort of swap.
Let’s generalize this idea further. Suppose we are also given an array f [1 .. n] of access frequencies
for each ﬁle; ﬁle i will be accessed exactly f [i ] times over the lifetime of the tape. Now the total cost of
accessing all the ﬁles on the tape is
n Σcost(π) = k f [π(k)] · n L [π(i )]
i =1 k =1 k = f [π(k)] · L [π(i )] .
k =1 i = 1 Now what order should store the ﬁles if we want to minimize the total cost?
We’ve already proved that if all the frequencies are equal, then we should sort the ﬁles by increasing
size. If the frequencies are all different but the ﬁle lengths L [i ] are all equal, then intuitively, we should
sort the ﬁles by decreasing access frequency, with the most-accessed ﬁle ﬁrst. In fact, this is not hard to
prove by modifying the proof of Lemma 1. But what if the sizes and the frequencies are both different?
In this case, we should sort the ﬁles by the ratio L / f .
Lemma 2. Σcost(π) is minimized when L [π(i )]
F [π(i )] ≤ L [π(i + 1)]
F [π(i + 1)] for all i . Proof: Suppose L [π(i )]/ F [π(i )] > L [π(i + 1)]/ F [π(i + i )] for some i . To simplify notation, let a = π(i )
and b = π(i + 1). If we swap ﬁles a and b, then the cost of accessing a increases by L [ b], and the cost
of accessing b decreases by L [a]. Overall, the swap changes the total cost by L [ b] F [a] − L [a] F [a]. But
this change is an improvement, since
L [a]
F [a] > L [ b]
F [ b] =⇒ L [ b] F [a] − L [a] F [a] < 0. Thus, if the ﬁles are out of order, we can improve the total cost by swapping some mis-ordered adjacent
pair. 4.2 Scheduling Classes The next example is slightly less trivial. Suppose you decide to drop out of computer science at the last
minute and change your major to Applied Chaos. The Applied Chaos department has all of its classes on
the same day every week, referred to as “Soberday" by the students (but interestingly, not by the faculty).
Every class has a different start time and a different ending time: AC 101 (‘Toilet Paper Landscape
Architecture’) starts at 10:27pm and ends at 11:51pm; AC 666 (‘Immanentizing the Eschaton’) starts
at 4:18pm and ends at 7:06pm, and so on. In the interests of graduating as quickly as possible, you
want to register for as many classes as you can. (Applied Chaos classes don’t require any actual work.)
The University’s registration computer won’t let you register for overlapping classes, and no one in the
department knows how to override this ‘feature’. Which classes should you take?
More formally, suppose you are given two arrays S [1 .. n] and F [1 .. n] listing the start and ﬁnish
times of each class. Your task is to choose the largest possible subset X ∈ {1, 2, . . . , n} so that for any
pair i , j ∈ X , either S [i ] > F [ j ] or S [ j ] > F [i ]. We can illustrate the problem by drawing each class as a
rectangle whose left and right x -coordinates show the start and ﬁnish times. The goal is to ﬁnd a largest
subset of rectangles that do not overlap vertically.
2 Algorithms Lecture 4: Greedy Algorithms A maximal conﬂict-free schedule for a set of classes. This problem has a fairly simple recursive solution, based on the observation that either you take
class 1 or you don’t. Let B4 be the set of classes that end before class 1 starts, and let L8 be the set of
classes that start later than class 1 ends:
B4 = {i | 2 ≤ i ≤ n and F [i ] < S [1]} L8 = {i | 2 ≤ i ≤ n and S [i ] > F [1]} If class 1 is in the optimal schedule, then so are the optimal schedules for B4 and L8 , which we can ﬁnd
recursively. If not, we can ﬁnd the optimal schedule for {2, 3, . . . , n} recursively. So we should try both
choices and take whichever one gives the better schedule. Evaluating this recursive algorithm from the
bottom up gives us a dynamic programming algorithm that runs in O(n2 ) time. I won’t bother to go
through the details, because we can do better.1
Intuitively, we’d like the ﬁrst class to ﬁnish as early as possible, because that leaves us with the most
remaining classes. If this greedy strategy works, it suggests the following very simple algorithm. Scan
through the classes in order of ﬁnish time; whenever you encounter a class that doesn’t conﬂict with
your latest class so far, take it! The same classes sorted by ﬁnish times and the greedy schedule. We can write the greedy algorithm somewhat more formally as follows. (Hopefully the ﬁrst line is
understandable.)
GREEDYSCHEDULE(S [1 .. n], F [1 .. n]):
sort F and permute S to match
count ← 1
X [count ] ← 1
for i ← 2 to n
if S [i ] > F [X [count ]]
count ← count + 1
X [count ] ← i
return X [1 .. count ]
1 But you should still work out the details yourself. The dynamic programming algorithm can be used to ﬁnd the “best”
schedule for any deﬁnition of “best”, but the greedy algorithm I’m about to describe only works that “best” means “biggest”.
Also, you need the practice. 3 Algorithms Lecture 4: Greedy Algorithms This algorithm clearly runs in O(n log n) time.
To prove that this algorithm actually gives us a maximal conﬂict-free schedule, we use an exchange
argument, similar to the one we used for tape sorting. We are not claiming that the greedy schedule is
the only maximal schedule; there could be others. (See the ﬁgures on the previous page.) All we can
claim is that at least one of the maximal schedules is the one that the greedy algorithm produces.
Lemma 3. At least one maximal conﬂict-free schedule includes the class that ﬁnishes ﬁrst.
Proof: Let f be the class that ﬁnishes ﬁrst. Suppose we have a maximal conﬂict-free schedule X that
does not include f . Let g be the ﬁrst class in X to ﬁnish. Since f ﬁnishes before g does, f cannot conﬂict
with any class in the set S \ { g }. Thus, the schedule X = X ∪ { f } \ { g } is also conﬂict-free. Since X has
the same size as X , it is also maximal.
To ﬁnish the proof, we call on our old friend, induction.
Theorem 4. The greedy schedule is an optimal schedule.
Proof: Let f be the class that ﬁnishes ﬁrst, and let L be the subset of classes the start after f ﬁnishes.
The previous lemma implies that some optimal schedule contains f , so the best schedule that contains
f is an optimal schedule. The best schedule that includes f must contain an optimal schedule for the
classes that do not conﬂict with f , that is, an optimal schedule for L . The greedy algorithm chooses f
and then, by the inductive hypothesis, computes an optimal schedule of classes from L .
The proof might be easier to understand if we unroll the induction slightly.
Proof: Let 〈 g1 , g2 , . . . , g k 〉 be the sequence of classes chosen by the greedy algorithm. Suppose we have
a maximal conﬂict-free schedule of the form
〈 g1 , g2 , . . . , g j −1 , c j , c j +1 , . . . , cm 〉,
where the classes ci are different from the classes chosen by the greedy algorithm. By construction,
the j th greedy choice g j does not conﬂict with any earlier class g1 , g2 , . . . , g j −1 , and since our schedule
is conﬂict-free, neither does c j . Moreover, g j has the earliest ﬁnish time among all classes that don’t
conﬂict with the earlier classes; in particular, g j ﬁnishes before c j . This implies that g j does not conﬂict
with any of the later classes c j +1 , . . . , cm . Thus, the schedule
〈 g1 , g2 , . . . , g j −1 , g j , c j +1 , . . . , cm 〉,
is conﬂict-free. (This is just a generalization of Lemma 3, which considers the case j = 1.) By induction,
it now follows that there is an optimal schedule 〈 g1 , g2 , . . . , g k , ck+1 , . . . , cm 〉 that includes every class
chosen by the greedy algorithm. But this is impossible unless k = m; if there were a class ck+1 that does
not conﬂict with g k , the greedy algorithm would choose more than k classes. 4.3 General Structure The basic structure of this correctness proof is exactly the same as for the tape-sorting problem: an
inductive exchange argument.
• Assume that there is an optimal solution that is different from the greedy solution.
• Find the ‘ﬁrst’ difference between the two solutions.
4 Algorithms Lecture 4: Greedy Algorithms • Argue that we can exchange the optimal choice for the greedy choice without degrading the
solution.
This argument implies by induction that there is an optimal solution that contains the entire greedy
solution. Sometimes, as in the scheduling problem, an additional step is required to show no optimal
solution strictly improves the greedy solution. 4.4 Huffman codes A binary code assigns a string of 0s and 1s to each character in the alphabet. A binary code is preﬁx-free
if no code is a preﬁx of any other. 7-bit ASCII and Unicode’s UTF-8 are both preﬁx-free binary codes.
Morse code is a binary code, but it is not preﬁx-free; for example, the code for S (· · ·) includes the code
for E (·) as a preﬁx. Any preﬁx-free binary code can be visualized as a binary tree with the encoded
characters stored at the leaves. The code word for any symbol is given by the path from the root to the
corresponding leaf; 0 for left, 1 for right. The length of a codeword for a symbol is the depth of the
corresponding leaf. (Note that the code tree is not a binary search tree. We don’t care at all about the
sorted order of symbols at the leaves. (In fact. the symbols may not have a well-deﬁned order!)
Suppose we want to encode messages in an n-character alphabet so that the encoded message is as
short as possible. Speciﬁcally, given an array frequency counts f [1 .. n], we want to compute a preﬁx-free
binary code that minimizes the total encoded length of the message:2
n f [i ] · depth(i ).
i =1 In 1952, David Huffman developed the following greedy algorithm to produce such an optimal code:
HUFFMAN: Merge the two least frequent letters and recurse.
For example, suppose we want to encode the following helpfully self-descriptive sentence, discovered by
Lee Sallows:3
This sentence contains three a’s, three c’s, two d’s, twenty-six e’s, ﬁve f’s, three g’s,
eight h’s, thirteen i’s, two l’s, sixteen n’s, nine o’s, six r’s, twenty-seven s’s, twenty-two t’s,
two u’s, ﬁve v’s, eight w’s, four x’s, ﬁve y’s, and only one z. To keep things simple, let’s forget about the forty-four spaces, nineteen apostrophes, nineteen commas,
three hyphens, and one period, and just encode the letters. Here’s the frequency table:
A C D 3 3 E 2 F G H 26 5 3 I 8 L N O R S T U V W X Y Z 13 2 16 9 6 27 22 2 5 8 451 Huffman’s algorithm picks out the two least frequent letters, breaking ties arbitrarily—in this case, say, Z
and D—and merges them together into a single new character D with frequency 3. This new character
Z
becomes an internal node in the code tree we are constructing, with Z and D as its children; it doesn’t
matter which child is which. The algorithm then recursively constructs a Huffman code for the new
frequency table
A C F G H 3 3 26 5 E 3 8 N O R U V W X Y D
Z 13 2 16 I L 9 6 27 22 2 5 8 45 3 2 S T This looks almost exactly like the cost of a binary search tree, but the optimization problem is very different: code trees are
not search trees!
3
A. K. Dewdney. Computer recreations. Scientiﬁc American, October 1984. Douglas Hofstadter published a few earlier
examples of Lee Sallows’ self-descriptive sentences in his Scientiﬁc American column in January 1982. 5 Algorithms Lecture 4: Greedy Algorithms After 19 merges, all 20 characters have been merged together. The record of merges gives us our code
tree. The algorithm makes a number of arbitrary choices; as a result, there are actually several different
Huffman codes. One such code is shown below.
170
59 111 S
27 32 60
N
16 16
H
8 51 39 W
8 21
T
22 17
O
9 8
X
4 10
F
5 11
V
5 Y
5 I
13 12
R
6 6
A
3 4
L
2 E
26 25 6
C
3 G
3 U
2 3
D
2 Z
1 A Huffman code for Lee Sallows’ self-descriptive sentence; the numbers are frequencies for merged characters For example, the code for A is 110000, and the code for S is 00. The encoded message starts like this:
1001 0100 1101 00 00 111 011 1001 111 011 110001 111 110001 10001 011 1001 110000 1101
T H I SS E N T E N C E C O N T A I ··· Here is the list of costs for encoding each character, along with that character’s contribution to the total
length of the encoded message:
char.
freq.
depth
total A C D E F G H I L N O R S T U V W X Y Z 3
3
2 26 5
3
8 13 2 16 9
6 27 22 2
5
8
4
51
6
6
7
3
5
6
4
4
7
3
4
4
2
4
7
5
4
6
57
18 18 14 78 25 18 32 52 14 48 36 24 54 88 14 25 32 24 25 7 Altogether, the encoded message is 646 bits long. Different Huffman codes would assign different codes,
possibly with different lengths, to various characters, but the overall length of the encoded message is
the same for any Huffman code: 646 bits.
Given the simple structure of Huffman’s algorithm, it’s rather surprising that it produces an optimal
preﬁx-free binary code. Encoding Lee Sallows’ sentence using any preﬁx-free code requires at least 646
bits! Fortunately, the recursive structure makes this claim easy to prove using an exchange argument,
similar to our earlier optimality proofs. We start by proving that the algorithm’s very ﬁrst choice is
correct.
Lemma 5. Let x and y be the two least frequent characters (breaking ties between equally frequent
characters arbitrarily). There is an optimal code tree in which x and y are siblings.
Proof: I’ll actually prove a stronger statement: There is an optimal code in which x and y are siblings
and have the largest depth of any leaf.
Let T be an optimal code tree, and suppose this tree has depth d . Since T is a full binary tree, it has
at least two leaves at depth d that are siblings. (Verify this by induction!) Suppose those two leaves are
not x and y , but some other characters a and b.
6 Algorithms Lecture 4: Greedy Algorithms Let T be the code tree obtained by swapping x and a. The depth of x increases by some amount ∆,
and the depth of a decreases by the same amount. Thus,
cost( T ) = cost( T ) − ( f [a] − f [ x ])∆.
By assumption, x is one of the two least frequent characters, but a is not, which implies that f [a] ≥ f [ x ].
Thus, swapping x and a does not increase the total cost of the code. Since T was an optimal code
tree, swapping x and a does not decrease the cost, either. Thus, T is also an optimal code tree (and
incidentally, f [a] actually equals f [ x ]).
Similarly, swapping y and b must give yet another optimal code tree. In this ﬁnal optimal code tree,
x and y as maximum-depth siblings, as required.
Now optimality is guaranteed by our dear friend the Recursion Fairy! Essentially we’re relying on the
following recursive deﬁnition for a full binary tree: either a single node, or a full binary tree where some
leaf has been replaced by an internal node with two leaf children.
Theorem 6. Huffman codes are optimal preﬁx-free binary codes.
Proof: If the message has only one or two different characters, the theorem is trivial.
Otherwise, let f [1 .. n] be the original input frequencies, where without loss of generality, f [1] and
f [2] are the two smallest. To keep things simple, let f [n + 1] = f [1] + f [2]. By the previous lemma,
we know that some optimal code for f [1 .. n] has characters 1 and 2 as siblings.
Let T be the Huffman code tree for f [3 .. n + 1]; the inductive hypothesis implies that T is an
optimal code tree for the smaller set of frequencies. To obtain the ﬁnal code tree T , we replace the leaf
labeled n + 1 with an internal node with two children, labelled 1 and 2. I claim that T is optimal for the
original frequency array f [1 .. n].
To prove this claim, we can express the cost of T in terms of the cost of T as follows. (In these
equations, depth(i ) denotes the depth of the leaf labelled i in either T or T ; if the leaf appears in both
T and T , it has the same depth in both trees.)
n cost( T ) = f [i ] · depth(i )
i =1
n+1 = f [i ] · depth(i ) + f [1] · depth(1) + f [2] · depth(2) − f [n + 1] · depth(n + 1)
i =3 = cost( T ) + f [1] · depth(1) + f [2] · depth(2) − f [n + 1] · depth(n + 1)
= cost( T ) + ( f [1] + f [2]) · depth( T ) − f [n + 1] · (depth( T ) − 1)
= cost( T ) + f [1] + f [2]
This equation implies that minimizing the cost of T is equivalent to minimizing the cost of T ; in
particular, attaching leaves labeled 1 and 2 to the leaf in T labeled n + 1 gives an optimal code tree for
the original frequencies.
To actually implement Huffman codes efﬁciently, we keep the characters in a min-heap, where the
priority of each character is its frequency. We can construct the code tree by keeping three arrays of
indices, listing the left and right children and the parent of each node. The root of the tree is the node
with index 2n − 1. 7 Algorithms Lecture 4: Greedy Algorithms
BUILDHUFFMAN( f [1 .. n]):
for i ← 1 to n
L [i ] ← 0; R[i ] ← 0
INSERT(i , f [i ])
for i ← n to 2n − 1
x ← EXTRACTMIN( )
y ← EXTRACTMIN( )
f [i ] ← f [ x ] + f [ y ]
L [i ] ← x ; R[i ] ← y
P [ x ] ← i; P [ y ] ← i
INSERT(i , f [i ])
P [ 2 n − 1] ← 0 The algorithm performs O(n) min-heap operations. If we use a balanced binary tree as the heap, each
operation requires O(log n) time, so the total running time of BUILDHUFFMAN is O(n log n).
Finally, here are simple algorithms to encode and decode messages:
HUFFMANENCODE(A[1 .. k]):
m←1
for i ← 1 to k
HUFFMANENCODEONE(A[i ]) HUFFMANDECODE(B [1 .. m]):
k←1
v ← 2n − 1
for i ← 1 to m
if B [i ] = 0
v ← L[v]
else
v ← R[ v ] HUFFMANENCODEONE( x ):
if x < 2n − 1
HUFFMANENCODEONE( P [ x ])
if x = L [ P [ x ]]
B [ m] ← 0
else
B [ m] ← 1
m← m+1 if L [ v ] = 0
A[k] ← v
k ← k+1
v ← 2n − 1 Exercises
1. Let X be a set of n intervals on the real line. A subset of intervals Y ⊆ X is called a tiling path if the
intervals in Y cover the intervals in X , that is, any real value that is contained in some interval in
X is also contained in some interval in Y . The size of a tiling cover is just the number of intervals.
Describe and analyze an algorithm to compute the smallest tiling path of X as quickly as possible.
Assume that your input consists of two arrays X L [1 .. n] and X R [1 .. n], representing the left and
right endpoints of the intervals in X . If you use a greedy algorithm, you must prove that it is
correct. A set of intervals. The seven shaded intervals form a tiling path. 2. Let X be a set of n intervals on the real line. We say that a set P of points stabs X if every interval
in X contains at least one point in P . Describe and analyze an efﬁcient algorithm to compute the
smallest set of points that stabs X . Assume that your input consists of two arrays X L [1 .. n] and
8 Algorithms Lecture 4: Greedy Algorithms X R [1 .. n], representing the left and right endpoints of the intervals in X . As usual, If you use a
greedy algorithm, you must prove that it is correct. A set of intervals stabbed by four points (shown here as vertical segments) 3. Let X be a set of n intervals on the real line. A proper coloring of X assigns a color to each interval,
so that any two overlapping intervals are assigned different colors. Describe and analyze an
efﬁcient algorithm to compute the minimum number of colors needed to properly color X . Assume
that your input consists of two arrays L [1 .. n] and R[1 .. n], where L [i ] and R[i ] are the left and
right endpoints of the i th interval. As usual, if you use a greedy algorithm, you must prove that it
is correct.
1
2 2 3 1
5 5 4 4
3 5
4 3 3
1
2 A proper coloring of a set of intervals using ﬁve colors. 4. Suppose you are standing in a ﬁeld surrounded by several large balloons. You want to use your
brand new Acme Brand Zap-O-MaticTM to pop all the balloons, without moving from your current
location. The Zap-O-MaticTM shoots a high-powered laser beam, which pops all the balloons it hits.
Since each shot requires enough energy to power a small country for a year, you want to ﬁre as
few shots as possible. Nine balloons popped by 4 shots of the Zap-O-MaticTM The minimum zap problem can be stated more formally as follows. Given a set C of n circles in the
plane, each speciﬁed by its radius and the ( x , y ) coordinates of its center, compute the minimum
number of rays from the origin that intersect every circle in C . Your goal is to ﬁnd an efﬁcient
algorithm for this problem.
9 Algorithms Lecture 4: Greedy Algorithms (a) Suppose it is possible to shoot a ray that does not intersect any balloons. Describe and
analyze a greedy algorithm that solves the minimum zap problem in this special case. [Hint:
See Exercise 2.]
(b) Describe and analyze a greedy algorithm whose output is within 1 of optimal. That is, if m is
the minimum number of rays required to hit every balloon, then your greedy algorithm must
output either m or m + 1. (Of course, you must prove this fact.)
(c) Describe an algorithm that solves the minimum zap problem in O(n2 ) time.
(d) Describe an algorithm that solves the minimum zap problem in O(n log n) time.
Assume you have a subroutine INTERSECTS( r, c ) that determines whether a ray r intersects a circle!c
in O(1) time. It’s not that hard to write this subroutine, but it’s not the interesting part of the
problem. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 10 Algorithms Non-Lecture D: Matroids
The problem is that we attempt to solve the simplest questions cleverly,
thereby rendering them unusually complex.
One should seek the simple solution.
— Anton Pavlovich Chekhov (c. 1890)
I love deadlines. I like the whooshing sound they make as they ﬂy by.
— Douglas Adams D Matroids D.1 Deﬁnitions Many problems that can be correctly solved by greedy algorithms can be described in terms of an abstract
combinatorial object called a matroid. Matroids were ﬁrst described in 1935 by the mathematician
Hassler Whitney as a combinatorial generalization of linear independence of vectors—‘matroid’ means
‘something sort of like a matrix’.
A matroid M is a ﬁnite collection of ﬁnite sets that satisﬁes three axioms:
• Non-emptiness: The empty set is in M. (Thus, M is not itself empty.)
• Heredity: If a set X is an element of M, then any subset of X is also in M.
• Exchange: If X and Y are two sets in M and |X | > |Y |, then there is an element x ∈ X \ Y such
that Y ∪ { x } is in M.
The sets in M are typically called independent sets; for example, we would say that any subset of an
independent set is independent. The union of all sets in M is called the ground set. An independent set is
called a basis if it is not a proper subset of another independent set. The exchange property implies that
every basis of a matroid has the same cardinality. The rank of a subset X of the ground set is the size
of the largest independent subset of X . A subset of the ground set that is not in M is called dependent
(surprise, surprise). Finally, a dependent set is called a circuit if every proper subset is independent.
Most of this terminology is justiﬁed by Whitney’s original example:
Linear matroid: Let A be any n × m matrix. A subset I ⊆ {1, 2, . . . , n} is independent if and
only if the corresponding subset of columns of A is linearly independent.
The heredity property follows directly from the deﬁnition of linear independence; the exchange property
is implied by an easy dimensionality argument. A basis in any linear matroid is also a basis (in the
linear-algebra sense) of the vector space spanned by the columns of A. Similarly, the rank of a set of
indices is precisely the rank (in the linear-algebra sense) of the corresponding set of column vectors.
Here are several other examples of matroids; some of these we will see again later. I will leave the
proofs that these are actually matroids as exercises for the reader.
• Uniform matroid Uk ,n : A subset X ⊆ {1, 2, . . . , n} is independent if and only if |X | ≤ k. Any
subset of {1, 2, . . . , n} of size k is a basis; any subset of size k + 1 is a circuit.
• Graphic/cycle matroid M(G ): Let G = (V, E ) be an arbitrary undirected graph. A subset of E is
independent if it deﬁnes an acyclic subgraph of G . A basis in the graphic matroid is a spanning
tree of G ; a circuit in this matroid is a cycle in G . 1 Algorithms Non-Lecture D: Matroids • Cographic/cocycle matroid M∗ (G ): Let G = (V, E ) be an arbitrary undirected graph. A subset
I ⊆ E is independent if the complementary subgraph (V, E \ I ) of G is connected. A basis in this
matroid is the complement of a spanning tree; a circuit in this matroid is a cocycle—a minimal set
of edges that disconnects the graph.
• Matching matroid: Let G = (V, E ) be an arbitrary undirected graph. A subset I ⊆ V is independent
if there is a matching in G that covers I .
• Disjoint path matroid: Let G = (V, E ) be an arbitrary directed graph, and let s be a ﬁxed vertex
of G . A subset I ⊆ V is independent if and only if there are edge-disjoint paths from s to each
vertex in I .
Now suppose each element of the ground set of a matroid M is given an arbitrary non-negative
weight. The matroid optimization problem is to compute a basis with maximum total weight. For
example, if M is the cycle matroid for a graph G , the matroid optimization problem asks us to ﬁnd the
maximum spanning tree of G . Similarly, if M is the cocycle matroid for G , the matroid optimization
problem seeks (the complement of) the minimum spanning tree.
The following natural greedy strategy computes a basis for any weighted matroid:
GREEDYBASIS(M, w ):
X [1 .. n] ← M (the ground set)
sort X in decreasing order of weight w
G←∅
for i ← 1 to n
if G ∪ {X [i ]} ∈ M
add X [i ] to G
return G Suppose we can test in F (n) whether a given subset of the ground set is independent. Then this algorithm
runs in O(n log n + n · F (n)) time.
Theorem 1. For any matroid M and any weight function w , GREEDYBASIS(M, w ) returns a maximumweight basis of M.
Proof: Let G = { g1 , g2 , . . . , g k } be the independent set returned by GREEDYBASIS(M, w ). If any other
element could be added to G to get a larger independent set, the greedy algorithm would have added it.
Thus, G is a basis.
For purposes of deriving a contradiction, suppose there is an independent set H = {h1 , h2 , . . . , h }
such that
k w( gi ) <
i =1 w (hi ).
j =1 Without loss of generality, we assume that H is a basis. The exchange property now implies that k = .
Now suppose the elements of G and H are indexed in order of decreasing weight. Let i be the
smallest index such that w ( g i ) < w (hi ), and consider the independent sets
G i −1 = { g 1 , g 2 , . . . , g i −1 } and H i = {h1 , h2 , . . . , hi −1 , hi }. By the exchange property, there is some element h j ∈ H i such that Gi −1 ∪ {h j } is an independent set. We
have w (h j ) ≥ w (hi ) > w ( g i ). Thus, the greedy algorithm considers and rejects the heavier element h j
before it considers the lighter element g i . But this is impossible—the greedy algorithm accepts elements
in decreasing order of weight.
2 Algorithms Non-Lecture D: Matroids We now immediately have a correct greedy optimization algorithm for any matroid. Returning to
our examples:
• Linear matroid: Given a matrix A, compute a subset of vectors of maximum total weight that span
the column space of A.
• Uniform matroid: Given a set of weighted objects, compute its k largest elements.
• Cycle matroid: Given a graph with weighted edges, compute its maximum spanning tree. In this
setting, the greedy algorithm is better known as Kruskal’s algorithm.
• Cocycle matroid: Given a graph with weighted edges, compute its minimum spanning tree.
• Matching matroid: Given a graph, determine whether it has a perfect matching.
• Disjoint path matroid: Given a directed graph with a special vertex s, ﬁnd the largest set of
edge-disjoint paths from s to other vertices.
The exchange condition for matroids turns out to be crucial for the success of this algorithm. A subset
system is a ﬁnite collection S of ﬁnite sets that satisﬁes the heredity condition—If X ∈ S and Y ⊆ X , then
Y ∈ S—but not necessarily the exchange condition.
Theorem 2. For any subset system S that is not a matroid, there is a weight function w such that
GREEDYBASIS(S, w ) does not return a maximum-weight set in S.
Proof: Let X and Y be two sets in S that violate the exchange property—|X | > |Y |, but for any element
x ∈ X \ Y , the set Y ∪ { x } is not in S. Let m = |Y |. We deﬁne a weight function as follows:
• Every element of Y has weight m + 2.
• Every element of X \ Y has weight m + 1.
• Every other element of the ground set has weight zero.
With these weights, the greedy algorithm will consider and accept every element of Y , then consider
and reject every element of X , and ﬁnally consider all the other elements. The algorithm returns a set
with total weight m(m + 2) = m2 + 2m. But the total weight of X is at least (m + 1)2 = m2 + 2m + 1.
Thus, the output of the greedy algorithm is not the maximum-weight set in S.
Recall the Applied Chaos scheduling problem considered in the previous lecture note. There is a
natural subset system associated with this problem: A set of classes is independent if and only if not
two classes overlap. (This is just the graph-theory notion of ‘independent set’!) This subset system is
not a matroid, because there can be maximal independent sets of different sizes, which violates the
exchange property. If we consider a weighted version of the class scheduling problem, say where each
class is worth a different number of hours, Theorem 2 implies that the greedy algorithm will not always
ﬁnd the optimal schedule. (In fact, there’s an easy counterexample with only two classes!) However,
Theorem 2 does not contradict the correctness of the greedy algorithm for the original unweighted
problem, however; that problem uses a particularly lucky choice of weights (all equal). 3 Algorithms D.2 Non-Lecture D: Matroids Scheduling with Deadlines Suppose you have n tasks to complete in n days; each task requires your attention for a full day. Each
task comes with a deadline, the last day by which the job should be completed, and a penalty that you
must pay if you do not complete each task by its assigned deadline. What order should you perform
your tasks in to minimize the total penalty you must pay?
More formally, you are given an array D[1 .. n] of deadlines an array P [1 .. n] of penalties. Each
deadline D[i ] is an integer between 1 and n, and each penalty P [i ] is a non-negative real number.
A schedule is a permutation of the integers {1, 2, . . . , n}. The scheduling problem asks you to ﬁnd a
schedule π that minimizes the following cost:
n cost(π) := P [i ] · [π(i ) > D[i ]].
i =1 This doesn’t look anything like a matroid optimization problem. For one thing, matroid optimization
problems ask us to ﬁnd an optimal set; this problem asks us to ﬁnd an optimal permutation. Surprisingly,
however, this scheduling problem is actually a matroid optimization in disguise! For any schedule π, call
tasks i such that π(i ) > D[i ] late, and all other tasks on time. The following trivial observation is the key
to revealing the underlying matroid structure.
The cost of a schedule is determined by the subset of tasks that are on time.
Call a subset X of the tasks realistic if there is a schedule π in which every task in X is on time. We
can precisely characterize the realistic subsets as follows. Let X ( t ) denote the subset of tasks in X whose
deadline is on or before t :
X ( t ) := { i ∈ X | D [ i ] ≤ t } .
In particular, X (0) = ∅ and X (n) = X .
Lemma 3. Let X {1, 2, . . . , n} be an arbitrary subset of the n tasks. X is realistic if and only if
|X ( t )| ≤ t for every integer t .
Proof: Let π be a schedule in which every task in X is on time. Let i t be the t th task in X to be
completed. On the one hand, we have π(i t ) ≥ t , since otherwise, we could not have completed t − 1
other jobs in X before i t . On the other hand, π(i t ) ≤ D[i ], because i t is on time. We conclude that
D[i t ] ≥ t , which immediately implies that |X ( t )| ≤ t .
Now suppose |X ( t )| ≤ t for every integer t . If we perform the tasks in X in increasing order of
deadline, then we complete all tasks in X with deadlines t or less by day t . In particular, for any i ∈ X ,
we perform task i on or before its deadline D[i ]. Thus, X is realistic.
We can deﬁne a canonical schedule for any set X as follows: execute the tasks in X in increasing
deadline order, and then execute the remaining tasks in any order. The previous proof implies that a
set X is realistic if and only if every task in X is on time in the canonical schedule for X . Thus, our
scheduling problem can be rephrased as follows:
Find a realistic subset X such that i ∈X P [i ] is maximized. So we’re looking for optimal subsets after all.
Lemma 4. The collection of realistic sets of jobs forms a matroid.
4 Algorithms Non-Lecture D: Matroids Proof: The empty set is vacuously realistic, and any subset of a realistic set is clearly realistic. Thus, to
prove the lemma, it sufﬁces to show that the exchange property holds. Let X and Y be realistic sets of
jobs with |X | > |Y |.
Let t ∗ be the largest integer such that |X ( t ∗ )| ≤ |Y ( t ∗ )|. This integer must exist, because |X (0)| =
0 ≤ 0 = |Y (0)| and |X (n)| = |X | > |Y | = |Y (n)|. By deﬁnition of t ∗ , there are more tasks with deadline
t ∗ + 1 in X than in Y . Thus, we can choose a task j in X \ Y with deadline t ∗ + 1; let Z = Y ∪ { j }.
Let t be an arbitrary integer. If t ≤ t ∗ , then | Z ( t )| = |Y ( t )| ≤ t , because Y is realistic. On the other
hand, if t > t ∗ , then | Z ( t )| = |Y ( t )| + 1 ≤ |X ( t )| < t by deﬁnition of t ∗ and because X is realistic. The
previous lemma now implies that Z is realistic. This completes the proof of the exchange property.
This lemma implies that our scheduling problem is a matroid optimization problem, so the greedy
algorithm ﬁnds the optimal schedule.
GREEDYSCHEDULE( D[1 .. n], P [1 .. n]):
Sort P in increasing order, and permute D to match
j←0
for i ← 1 to n
X [ j + 1] ← i
if X [1 .. j + 1] is realistic
j ← j+1
return the canonical schedule for X [1 .. j ] To turn this outline into a real algorithm, we need a procedure to test whether a given subset of jobs
is realistic. Lemma 9 immediately suggests the following strategy to answer this question in O(n) time.
REALISTIC?(X [1 .. m], D[1 .. n]):
〈〈X is sorted by increasing deadline: i ≤ j =⇒ D[X [i ]] ≤ D[X [ j ]]〉〉
N ←0
j←0
for t ← 1 to n
if D[X [ j ]] = t
N ← N + 1; j ← j + 1
〈〈Now N = |X ( t )|〉〉
if N > t
return FALSE
return TRUE If we use this subroutine, GREEDYSCHEDULE runs in O(n2 ) time. By using some appropriate data structures,
the running time can be reduced to O(n log n); details are left as an exercise for the reader. Exercises
1. Prove that for any graph G , the ‘graphic matroid’ M(G ) is in fact a matroid. (This problem is really
asking you to prove that Kruskal’s algorithm is correct!)
2. Prove that for any graph G , the ‘cographic matroid’ M∗ (G ) is in fact a matroid.
3. Prove that for any graph G , the ‘matching matroid’ of G is in fact a matroid. [Hint: What is the
symmetric difference of two matchings?]
5 Algorithms Non-Lecture D: Matroids 4. Prove that for any directed graph G and any vertex s of G , the resulting ‘disjoint path matroid’ of G
is in fact a matroid. [Hint: This question is much easier if you’re already familiar with maximum
ﬂows.]
5. Let G be an undirected graph. A set of cycles {c1 , c2 , . . . , ck } in G is called redundant if every edge
in G appears in an even number of ci ’s. A set of cycles is independent if it contains no redundant
subset. A maximal independent set of cycles is called a cycle basis for G .
(a) Let C be any cycle basis for G . Prove that for any cycle γ in G , there is a subset A ⊆ C such
that A ∩ {γ} is redundant. In other words, γ is the ‘exclusive or’ of the cycles in A.
(b) Prove that the set of independent cycle sets form a matroid.
(c) Now suppose each edge of G has a weight. Deﬁne the weight of a cycle to be the total weight
of its edges, and the weight of a set of cycles to be the total weight of all cycles in the set.
(Thus, each edge is counted once for every cycle in which it appears.) Describe and analyze
an efﬁcient algorithm to compute the minimum-weight cycle basis in G .
6. Describe a modiﬁcation of GREEDYSCHEDULE that runs in O(n log n) time. [Hint: Store X in an
appropriate data structure that supports the operations “Is X ∪ {i } realistic?” and “Add i to X ” in
O(log n) time each.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 6 Algorithms Lecture 5: Randomized Algorithms
The ﬁrst nuts and bolts appeared in the middle 1400’s. The bolts were just screws with straight sides
and a blunt end. The nuts were hand-made, and very crude. When a match was found between a
nut and a bolt, they were kept together until they were ﬁnally assembled.
In the Industrial Revolution, it soon became obvious that threaded fasteners made it easier to
assemble products, and they also meant more reliable products. But the next big step came in 1801,
with Eli Whitney, the inventor of the cotton gin. The lathe had been recently improved. Batches of
bolts could now be cut on different lathes, and they would all ﬁt the same nut.
Whitney set up a demonstration for President Adams, and Vice-President Jefferson. He had
piles of musket parts on a table. There were 10 similar parts in each pile. He went from pile to
pile, picking up a part at random. Using these completely random parts, he quickly put together a
working musket.
— Karl S. Kruszelnicki (‘Dr. Karl’), Karl Trek, December 1997
Dr [John von] Neumann in his Theory of Games and Economic Behavior introduces the cut-up method
of random action into game and military strategy: Assume that the worst has happened and act
accordingly. If your strategy is at some point determined. . . by random factor your opponent will gain
no advantage from knowing your strategy since he cannot predict the move. The cut-up method
could be used to advantage in processing scientiﬁc data. How many discoveries have been made by
accident? We cannot produce accidents to order.
— William S. Burroughs, "The Cut-Up Method of Brion Gysin"
in The Third Mind by William S. Burroughs and Brion Gysin (1978) 5 Randomized Algorithms 5.1 Nuts and Bolts Suppose we are given n nuts and n bolts of different sizes. Each nut matches exactly one bolt and vice
versa. The nuts and bolts are all almost exactly the same size, so we can’t tell if one bolt is bigger than
the other, or if one nut is bigger than the other. If we try to match a nut witch a bolt, however, the nut
will be either too big, too small, or just right for the bolt.
Our task is to match each nut to its corresponding bolt. But before we do this, let’s try to solve some
simpler problems, just to get a feel for what we can and can’t do.
Suppose we want to ﬁnd the nut that matches a particular bolt. The obvious algorithm — test every
nut until we ﬁnd a match — requires exactly n − 1 tests in the worst case. We might have to check every
bolt except one; if we get down the the last bolt without ﬁnding a match, we know that the last nut is
the one we’re looking for.1
Intuitively, in the ‘average’ case, this algorithm will look at approximately n/2 nuts. But what exactly
does ‘average case’ mean? 5.2 Deterministic vs. Randomized Algorithms Normally, when we talk about the running time of an algorithm, we mean the worst-case running time.
This is the maximum, over all problems of a certain size, of the running time of that algorithm on that
input:
Tworst-case (n) = max T (X ).
|X |=n On extremely rare occasions, we will also be interested in the best-case running time:
Tbest-case (n) = min T (X ).
|X |=n 1 “Whenever you lose something, it’s always in the last place you look. So why not just look there ﬁrst?” 1 Algorithms Lecture 5: Randomized Algorithms The average-case running time is best deﬁned by the expected value, over all inputs X of a certain size, of
the algorithm’s running time for X :2
Taverage-case (n) = E [ T (X )] =
|X |=n T ( x ) · Pr[X ].
|X |=n The problem with this deﬁnition is that we rarely, if ever, know what the probability of getting any
particular input X is. We could compute average-case running times by assuming a particular probability
distribution—for example, every possible input is equally likely—but this assumption doesn’t describe
reality very well. Most real-life data is decidedly non-random (or at least random in some unpredictable
way).
Instead of considering this rather questionable notion of average case running time, we will make a
distinction between two kinds of algorithms: deterministic and randomized. A deterministic algorithm
is one that always behaves the same way given the same input; the input completely determines the
sequence of computations performed by the algorithm. Randomized algorithms, on the other hand,
base their behavior not only on the input but also on several random choices. The same randomized
algorithm, given the same input multiple times, may perform different computations in each invocation.
This means, among other things, that the running time of a randomized algorithm on a given input
is no longer ﬁxed, but is itself a random variable. When we analyze randomized algorithms, we are
typically interested in the worst-case expected running time. That is, we look at the average running time
for each input, and then choose the maximum over all inputs of a certain size:
Tworst-case expected (n) = max E[ T (X )].
|X |=n It’s important to note here that we are making no assumptions about the probability distribution of
possible inputs. All the randomness is inside the algorithm, where we can control it! 5.3 Back to Nuts and Bolts Let’s go back to the problem of ﬁnding the nut that matches a given bolt. Suppose we use the same
algorithm as before, but at each step we choose a nut uniformly at random from the untested nuts.
‘Uniformly’ is a technical term meaning that each nut has exactly the same probability of being chosen.3
So if there are k nuts left to test, each one will be chosen with probability 1/k. Now what’s the expected
number of comparisons we have to perform? Intuitively, it should be about n/2, but let’s formalize our
intuition.
Let T (n) denote the number of comparisons our algorithm uses to ﬁnd a match for a single bolt
out of n nuts.4 We still have some simple base cases T (1) = 0 and T (2) = 1, but when n > 2, T (n) is
a random variable. T (n) is always between 1 and n − 1; it’s actual value depends on our algorithm’s
random choices. We are interested in the expected value or expectation of T (n), which is deﬁned as
follows:
n−1 E[ T (n)] = k · Pr[ T (n) = k]
k =1 The notation E[ ] for expectation has nothing to do with the shift operator E used in the annihilator method for solving
recurrences!
3
This is what most people think ‘random’ means, but they’re wrong.
4
Note that for this algorithm, the input is completely speciﬁed by the number n. Since we’re choosing the nuts to test at
random, even the order in which the nuts and bolts are presented doesn’t matter. That’s why I’m using the simpler notation
T (n) instead of T (X ).
2 2 Algorithms
Lecture 5: Randomized Algorithms If the target nut is the kth nut tested, our algorithm performs min{k, n − 1} comparisons. In particular,
if the target nut is the last nut chosen, we don’t actually test it. Because we choose the next nut to
test uniformly at random, the target nut is equally likely—with probability exactly 1/n—to be the ﬁrst,
second, third, or kth bolt tested, for any k. Thus:
Pr[ T (n) = k] = 1/ n if k < n − 1, 2/ n if k = n − 1. Plugging this into the deﬁnition of expectation gives us our answer.
n−2 E[ T (n)] =
k =1
n−1 =
k =1 =
= k
n
k
n + 2(n − 1) + n
n−1 n( n − 1)
2n
n+1
2 − n
+1− 1
n 1
n We can get exactly the same answer by thinking of this algorithm recursively. We always have to
perform at least one test. With probability 1/n, we successfully ﬁnd the matching nut and halt. With the
remaining probability 1 − 1/n, we recursively solve the same problem but with one fewer nut. We get
the following recurrence for the expected number of tests:
T (1) = 0, E[ T (n)] = 1 + n−1
n E[ T (n − 1)] To get the solution, we deﬁne a new function t (n) = n E[ T (n)] and rewrite:
t (1) = 0, t (n) = n + t (n − 1) This recurrence translates into a simple summation, which we can easily solve.
n t ( n) = k= n(n + 1)
2 k =2 =⇒ E[ T (n)] = 5.4 t ( n)
n = n+1
2 − −1 1
n Finding All Matches Not let’s go back to the problem introduced at the beginning of the lecture: ﬁnding the matching nut
for every bolt. The simplest algorithm simply compares every nut with every bolt, for a total of n2
comparisons. The next thing we might try is repeatedly ﬁnding an arbitrary matched pair, using our very
ﬁrst nuts and bolts algorithm. This requires
n (i − 1) =
i =1 n2 − n
2 comparisons in the worst case. So we save roughly a factor of two over the really stupid algorithm. Not
very exciting.
3 Algorithms Lecture 5: Randomized Algorithms Here’s another possibility. Choose a pivot bolt, and test it against every nut. Then test the matching
pivot nut against every other bolt. After these 2n − 1 tests, we have one matched pair, and the remaining
nuts and bolts are partitioned into two subsets: those smaller than the pivot pair and those larger than
the pivot pair. Finally, recursively match up the two subsets. The worst-case number of tests made by
this algorithm is given by the recurrence
T (n) = 2n − 1 + max { T (k − 1) + T (n − k)}
1≤k≤n = 2 n − 1 + T ( n − 1)
Along with the trivial base case T (0) = 0, this recurrence solves to
n (2n − 1) = n2 . T ( n) =
i =1 In the worst case, this algorithm tests every nut-bolt pair! We could have been a little more clever—for
example, if the pivot bolt is the smallest bolt, we only need n − 1 tests to partition everything, not
2n − 1—but cleverness doesn’t actually help that much. We still end up with about n2 /2 tests in the
worst case.
However, since this recursive algorithm looks almost exactly like quicksort, and everybody ‘knows’
that the ‘average-case’ running time of quicksort is Θ(n log n), it seems reasonable to guess that the
average number of nut-bolt comparisons is also Θ(n log n). As we shall see shortly, if the pivot bolt is
always chosen uniformly at random, this intuition is exactly right. 5.5 Reductions to and from Sorting The second algorithm for mathing up the nuts and bolts looks exactly like quicksort. The algorithm not
only matches up the nuts and bolts, but also sorts them by size.
In fact, the problems of sorting and matching nuts and bolts are equivalent, in the following sense.
If the bolts were sorted, we could match the nuts and bolts in O(n log n) time by performing a binary
search with each nut. Thus, if we had an algorithm to sort the bolts in O(n log n) time, we would
immediately have an algorithm to match the nuts and bolts, starting from scratch, in O(n log n) time.
This process of assuming a solution to one problem and using it to solve another is called reduction—we
can reduce the matching problem to the sorting problem in O(n log n) time.
There is a reduction in the other direction, too. If the nuts and bolts were matched, we could sort
them in O(n log n) time using, for example, merge sort. Thus, if we have an O(n log n) time algorithm
for either sorting or matching nuts and bolts, we automatically have an O(n log n) time algorithm for the
other problem.
Unfortunately, since we aren’t allowed to directly compare two bolts or two nuts, we can’t use
heapsort or mergesort to sort the nuts and bolts in O(n log n) worst case time. In fact, the problem
of sorting nuts and bolts deterministically in O(n log n) time was only ‘solved’ in 19955 , but both the
algorithms and their analysis are incredibly technical and the constant hidden in the O(·) notation is
quite large.
Reductions will come up again later in the course when we start talking about lower bounds and
NP-completeness.
5
János Komlós, Yuan Ma, and Endre Szemerédi, Sorting nuts and bolts in O(n log n) time, SIAM J. Discrete Math 11(3):347–
372, 1998. See also Phillip G. Bradford, Matching nuts and bolts optimally, Technical Report MPI-I-95-1-025, Max-Planck-Institut
für Informatik, September 1995. Bradford’s algorithm is slightly simpler. 4 Algorithms 5.6 Lecture 5: Randomized Algorithms Recursive Analysis Intuitively, we can argue that our quicksort-like algorithm will usually choose a bolt of approximately
median size, and so the average numbers of tests should be O(n log n). We can now ﬁnally formalize
this intuition. To simplify the notation slightly, I’ll write T (n) in place of E[ T (n)] everywhere.
Our randomized matching/sorting algorithm chooses its pivot bolt uniformly at random from the
set of unmatched bolts. Since the pivot bolt is equally likely to be the smallest, second smallest, or kth
smallest for any k, the expected number of tests performed by our algorithm is given by the following
recurrence:
T ( n) = 2 n − 1 + E k T ( k − 1) + T ( n − k )
= 2n − 1 + 1 n n k =1 T ( k − 1) + T ( n − k ) The base case is T (0) = 0. (We can save a few tests by setting T (1) = 0 instead of 1, but the analysis
will be easier if we’re a little stupid.)
Yuck. At this point, we could simply guess the solution, based on the incessant rumors that quicksort
runs in O(n log n) time in the average case, and prove our guess correct by induction. A similar inductive
proof appears in [CLR, pp. 166–167], but it was removed from the new edition [CLRS]. That’s okay;
nobody ever really understood that proof anyway. (See Section 5.8 below for details.)
However, if we’re only interested in asymptotic bounds, we can afford to be a little conservative.
What we’d really like is for the pivot bolt to be the median bolt, so that half the bolts are bigger and half
the bolts are smaller. This isn’t very likely, but there is a good chance that the pivot bolt is close to the
median bolt. Let’s say that a pivot bolt is good if it’s in the middle half of the ﬁnal sorted set of bolts, that
is, bigger than at least n/4 bolts and smaller than at least n/4 bolts. If the pivot bolt is good, then the
worst split we can have is into one set of 3n/4 pairs and one set of n/4 pairs. If the pivot bolt is bad,
then our algorithm is still better than starting over from scratch. Finally, a randomly chosen pivot bolt is
good with probability 1/2.
These simple observations give us the following simple recursive upper bound for the expected
running time of our algorithm:
T (n) ≤ 2n − 1 + 1
2 T 3n
4 +T n
4 + 1
2 · T ( n) A little algebra simpliﬁes this even further:
T ( n) ≤ 4 n − 2 + T 3n
4 +T n
4 We can solve this recurrence using the recursion tree method, giving us the unsurprising upper bound
T (n) = O(n log n). A similar argument gives us the matching lower bound T (n) = Ω(n log n).
Unfortunately, while this argument is convincing, it is not a formal proof, because it relies on the
unproven assumption that T (n) is a convex function, which means that T (n + 1) + T (n − 1) ≥ 2 T (n) for
all n. T (n) is actually convex, but we never proved it. Convexity follows form the closed-form solution
of the recurrence, but using that fact would be circular logic. Sadly, formally proving convexity seems to
be almost as hard as solving the recurrence. If we want a proof of the expected cost of our algorithm,
we need another way to proceed. 5 Algorithms 5.7 Lecture 5: Randomized Algorithms Iterative Analysis By making a simple change to our algorithm, which has no effect on the number of tests, we can analyze
it much more directly and exactly, without solving a recurrence or relying on hand-wavy intuition.
The recursive subproblems solved by quicksort can be laid out in a binary tree, where each node
corresponds to a subset of the nuts and bolts. In the usual recursive formulation, the algorithm partitions
the nuts and bolts at the root, then the left child of the root, then the leftmost grandchild, and so forth,
recursively sorting everything on the left before starting on the right subproblem.
But we don’t have to solve the subproblems in this order. In fact, we can visit the nodes in the
recursion tree in any order we like, as long as the root is visited ﬁrst, and any other node is visited after
its parent. Thus, we can recast quicksort in the following iterative form. Choose a pivot bolt, ﬁnd its
match, and partition the remaining nuts and bolts into two subsets. Then pick a second pivot bolt and
partition whichever of the two subsets contains it. At this point, we have two matched pairs and three
subsets of nuts and bolts. Continue choosing new pivot bolts and partitioning subsets, each time ﬁnding
one match and increasing the number of subsets by one, until every bolt has been chosen as the pivot.
At the end, every bolt has been matched, and the nuts and bolts are sorted.
Suppose we always choose the next pivot bolt uniformly at random from the bolts that haven’t been
pivots yet. Then no matter which subset contains this bolt, the pivot bolt is equally likely to be any bolt
in that subset. That implies (by induction) that our randomized iterative algorithm performs exactly the
same set of tests as our randomized recursive algorithm, but possibly in a different order.
Now let Bi denote the i th smallest bolt, and N j denote the j th smallest nut. For each i and j , deﬁne
an indicator variable X i j that equals 1 if our algorithm compares Bi with N j and zero otherwise. Then
the total number of nut/bolt comparisons is exactly
n n T ( n) = Xi j.
i =1 j =1 We are interested in the expected value of this double summation: n E[ T (n)] = E n i =1 j =1 n Xi j = n E[X i j ].
i =1 j =1 This equation uses a crucial property of random variables called linearity of expectation: for any
random variables X and Y , the sum of their expectations is equal to the expectation of their sum:
E [ X + Y ] = E [ X ] + E [ Y ].
To analyze our algorithm, we only need to compute the expected value of each X i j . By deﬁnition of
expectation,
E[X i j ] = 0 · Pr[X i j = 0] + 1 · Pr[X i j = 1] = Pr[X i j = 1],
so we just need to calculate Pr[X i j = 1] for all i and j .
First let’s assume that i < j . The only comparisons our algorithm performs are between some pivot
bolt (or its partner) and a nut (or bolt) in the same subset. The only thing that can prevent us from
comparing Bi and N j is if some intermediate bolt Bk , with i < k < j , is chosen as a pivot before Bi or B j .
In other words:
Our algorithm compares Bi and N j if and only if the ﬁrst pivot chosen
from the set {Bi , Bi +1 , . . . , B j } is either Bi or B j . 6 Algorithms Lecture 5: Randomized Algorithms Since the set {Bi , Bi +1 , . . . , B j } contains j − i + 1 bolts, each of which is equally likely to be chosen ﬁrst,
we immediately have
2
E[ X i j ] =
for all i < j .
j−i+1
Symmetric arguments give us E[X i j ] = i −2+1 for all i > j . Since our algorithm is a little stupid, every
j
bolt is compared with its partner, so X ii = 1 for all i . (In fact, if a pivot bolt is the only bolt in its subset,
we don’t need to compare it against its partner, but this improvement complicates the analysis.)
Putting everything together, we get the following summation.
n n E[ T (n)] = E[X i j ]
i =1 j =1
n = n n E[X ii ] + 2
i =1 E[ X i j ]
i =1 j = i +1 n n = n+4
i =1 j = i +1 1
j−i+1 This is quite a bit simpler than the recurrence we got before. With just a few more lines of algebra, we
can turn it into an exact, closed-form expression for the expected number of comparisons.
n n−i +1 E[ T (n)] = n + 4
i =1 n n−k+1 = n+4
k =2 [substitute k = j − i + 1] k j =2 = n+4
k =2
n 1 1 [reorder summations] k i =1 n−k+1
k
n = n+4 1 ( n − 1) k k =2 n − 1
k =2 = n + 4((n + 1)(H n − 1) − (n − 1))
= 4nH n − 7n + 4H n
Sure enough, it’s Θ(n log n). 5.8 Masochistic Analysis If we’re feeling particularly masochistic, we can actually solve the recurrence directly, all the way to an
exact closed-form solution. I’m including this only to show you it can be done; this won’t be on the test.
First we simplify the recurrence slightly by combining symmetric terms.
T (n) = 2n − 1 +
= 2n − 1 + 1 n T ( k − 1) + T ( n − k ) n k =1
2 n −1
n k =0 T (k) 7 Algorithms Lecture 5: Randomized Algorithms We then convert this ‘full history’ recurrence into a ‘limited history’ recurrence by shifting and subtracting
away common terms. (I call this “Magic step #1”.) To make this step slightly easier, we ﬁrst multiply
both sides of the recurrence by n to get rid of the fractions.
n −1
2 n T ( n) = 2 n − n + 2 T (k)
k =0
n−2 (n − 1) T (n − 1) = 2(n − 1)2 − (n − 1) + 2 T (k)
k =0 2n2 −5n+3 n T ( n) − ( n − 1) T ( n − 1) = 4 n − 3 + 2 T ( n − 1)
3 n+1
T ( n) = 4 − +
T ( n − 1)
n
n
To solve this limited-history recurrence, we deﬁne a new function t (n) = T (n)/(n + 1). (I call this “Magic
step #2”.) This gives us an even simpler recurrence for t (n) in terms of t (n − 1):
t ( n) =
=
= T (n)
n+1
1
n+1
4 4− 3
n
3 + ( n + 1) T ( n − 1)
n + t ( n − 1)
n( n + 1)
3
=
− + t ( n − 1)
n+1 n
n+1
7 − 1
I used the technique of partial fractions (remember calculus?) to replace n(n1 1) with 1 − n+1 in the last
+
n
step. The base case for this recurrence is t (0) = 0. Once again, we have a recurrence that translates
directly into a summation, which we can solve with just a few lines of algebra.
n t (n) =
i =1
n =7
i =1 7
i+1
1
i+1 − 3
i
n −3
i =1 1
i = 7(H n+1 − 1) − 3H n
7
= 4H n − 7 +
n+1
The last step uses the recursive deﬁnition of the harmonic numbers: H n+1 = H n + 1
.
n+1 Finally, substituting T (n) = (n + 1) t (n) and simplifying gives us the exact solution to the original recurrence.
T (n) = 4(n + 1)H n − 7(n + 1) + 7 = 4nH n − 7n + 4H n
Surprise, surprise, we get exactly the same solution! 8 Algorithms Lecture 5: Randomized Algorithms Exercises
Unless a problem speciﬁcally states otherwise, you can assume a function RANDOM(k ) that
returns, given any positive integer k , an integer chosen independently and uniformly at
random from the set {1, 2, . . . , k }, in O (1) time. For example, to perform a fair coin ﬂip,
one could call RANDOM(2). 1. Consider the following randomized algorithm for choosing the largest bolt. Draw a bolt uniformly
at random from the set of n bolts, and draw a nut uniformly at random from the set of n nuts.
If the bolt is smaller than the nut, discard the bolt, draw a new bolt uniformly at random from
the unchosen bolts, and repeat. Otherwise, discard the nut, draw a new nut uniformly at random
from the unchosen nuts, and repeat. Stop either when every nut has been discarded, or every bolt
except the one in your hand has been discarded.
What is the exact expected number of nut-bolt tests performed by this algorithm? Prove your
answer is correct. [Hint: What is the expected number of unchosen nuts and bolts when the
algorithm terminates?]
2. Consider the following algorithm for ﬁnding the smallest element in an unsorted array:
RANDOMMIN(A[1 .. n]):
min ← ∞
for i ← 1 to n in random order
if A[i ] < min
min ← A[i ] ( )
return min (a) In the worst case, how many times does RANDOMMIN execute line ( )?
(b) What is the probability that line ( ) is executed during the nth iteration of the for loop?
(c) What is the exact expected number of executions of line ( )?
3. Let S be a set of n points in the plane. A point p in S is called Pareto-optimal if no other point in S
is both above and to the right of p.
(a) Describe and analyze a deterministic algorithm that computes the Pareto-optimal points in S
in O(n log n) time.
(b) Suppose each point in S is chosen independently and uniformly at random from the unit
square [0, 1] × [0, 1]. What is the exact expected number of Pareto-optimal points in S ?
4. Suppose we want to write an efﬁcient function RANDOMPERMUTATION(n) that returns a permutation
of the integers 〈1, . . . , n〉 chosen uniformly at random.
(a) Prove that the following algorithm is not correct. [Hint: Consider the case n = 3.]
RANDOMPERMUTATION(n):
for i ← 1 to n
π[i ] ← i
for i ← 1 to n
swap π[i ] ↔ π[RANDOM(n)] 9 Algorithms Lecture 5: Randomized Algorithms (b) Consider the following implementation of RANDOMPERMUTATION.
RANDOMPERMUTATION(n):
for i ← 1 to n
π[i ] ← NULL
for i ← 1 to n
j ← RANDOM(n)
while (π[ j ] != NULL)
j ← RANDOM(n)
π[ j ] ← i
return π Prove that this algorithm is correct. Analyze its expected runtime.
(c) Consider the following partial implementation of RANDOMPERMUTATION.
RANDOMPERMUTATION(n):
for i ← 1 to n
A[i ] ← RANDOM(n)
π ← SOMEFUNCTION(A)
return π Prove that if the subroutine SOMEFUNCTION is deterministic, then this algorithm cannot be
correct. [Hint: There is a one-line proof.]
(d) Consider a correct implementation of RANDOMPERMUTATION(n) with the following property:
whenever it calls RANDOM(k), the argument k is at most m. Prove that this algorithm always
n log n
calls RANDOM at least Ω( log m ) times.
(e) Describe and analyze an implementation of RANDOMPERMUTATION that runs in expected
worst-case time O(n).
5. Consider the following randomized algorithm for generating biased random bits. The subroutine
FAIRCOIN returns either 0 or 1 with equal probability; the random bits returned by FAIRCOIN are
mutually independent.
ONEINTHREE:
if FAIRCOIN = 0
return 0
else
return 1 − ONEINTHREE (a) Prove that ONEINTHREE returns 1 with probability 1/3.
(b) What is the exact expected number of times that this algorithm calls FAIRCOIN?
(c) Now suppose you are given a subroutine ONEINTHREE that generates a random bit that is
equal to 1 with probability 1/3. Describe a FAIRCOIN algorithm that returns either 0 or 1 with
equal probability, using ONEINTHREE as a subroutine. For this question, your only source
of randomness is ONEINTHREE; in particular, you may not use the RANDOM function.
(d) What is the exact expected number of times that your FAIRCOIN algorithm calls ONEINTHREE? 10 Algorithms Lecture 5: Randomized Algorithms 6. Suppose n lights labeled 0, . . . , n − 1 are placed clockwise around a circle. Initially, every light is
off. Consider the following random process.
LIGHT THECIRCLE(n):
k←0
turn on light 0
while at least one light is off
with probability 1/2
k ← (k + 1) mod n
else
k ← (k − 1) mod n
if light k is off, turn it on (a) Let p(i , n) be the probability that light i is the last to be turned on by LIGHT THECIRCLE(n, 0).
For example, p(0, 2) = 0 and p(1, 2) = 1. Find an exact closed-form expression for p(i , n) in
terms of n and i . Prove your answer is correct.
(b) Give the tightest upper bound you can on the expected running time of this algorithm.
7. Consider a random walk on a path with vertices numbered 1, 2, . . . , n from left to right. At each
step, we ﬂip a coin to decide which direction to walk, moving one step left or one step right with
equal probability. The random walk ends when we fall off one end of the path, either by moving
left from vertex 1 or by moving right from vertex n.
(a) Prove that the probability that the walk ends by falling off the right end of the path is exactly
1/(n + 1).
(b) Prove that if we start at vertex k, the probability that we fall off the right end of the path is
exactly k/(n + 1).
(c) Prove that if we start at vertex 1, the expected number of steps before the random walk ends
is exactly n.
(d) Suppose we start at vertex n/2 instead. State and prove a tight Θ-bound on the expected
length of the random walk in this case.
8. A data stream is an extremely long sequence of items that you can only read only once, in order. A
good example of a data stream is the sequence of packets that pass through a router. Data stream
algorithms must process each item in the stream quickly, using very little memory; there is simply
too much data to store, and it arrives too quickly for any complex computations. Every data stream
algorithm looks roughly like this:
DOSOMETHINGINTERESTING(stream S ):
repeat
x ← next item in S
〈〈do something fast with x 〉〉
until S ends
return 〈〈something〉〉 Describe and analyze an algorithm that chooses one element uniformly at random from a data
stream, without knowing the length of the stream in advance. Your algorithm should spend O(1)
time per stream element and use O(1) space (not counting the stream itself).
11 Algorithms Lecture 5: Randomized Algorithms 9. The following randomized algorithm, sometimes called ‘one-armed quicksort’, selects the r th smallest element in an unsorted array A[1 .. n]. For example, to ﬁnd the smallest element, you would
call RANDOMSELECT(A, 1); to ﬁnd the median element, you would call RANDOMSELECT(A, n/2 ).
Recall from lecture that PARTITION splits the array into three parts by comparing the pivot element
A[ p] to every other element of the array, using n − 1 comparisons altogether, and returns the new
index of the pivot element. The subroutine RANDOM(n) returns an integer chosen uniformly at
random between 1 and n, in O(1) time.
RANDOMSELECT(A[1 .. n], r ) :
k ← PARTITION(A[1 .. n], RANDOM(n))
if r < k
return RANDOMSELECT(A[1 .. k − 1], r )
else if r > k
return RANDOMSELECT(A[k + 1 .. n], r − k)
else
return A[k] (a) State a recurrence for the expected running time of RANDOMSELECT, as a function of n and r .
(b) What is the exact probability that RANDOMSELECT compares the i th smallest and j th smallest
elements in the input array? The correct answer is a simple function of i , j , and r . [Hint:
Check your answer by trying a few small examples.]
(c) Show that for any n and r , the expected running time of RANDOMSELECT is Θ(n). You can use
either the recurrence from part (a) or the probabilities from part (b). For extra credit, ﬁnd
the exact expected number of comparisons, as a function of n and r .
(d) What is the expected number of times that RANDOMSELECT calls itself recursively?
10. Let M [1 .. n][1 .. n] be an n × n matrix in which every row and every column is sorted. Such an
array is called totally monotone. No two elements of M are equal.
(a) Describe and analyze an algorithm to solve the following problem in O(n) time: Given indices
i , j , i , j as input, compute the number of elements of M smaller than M [i ][ j ] and larger
than M [i ][ j ].
(b) Describe and analyze an algorithm to solve the following problem in O(n) time: Given indices
i , j , i , j as input, return an element of M chosen uniformly at random from the elements
smaller than M [i ][ j ] and larger than M [i ][ j ]. Assume the requested range is always
non-empty.
(c) Describe and analyze a randomized algorithm to compute the median element of M in
O(n log n) expected time.
11. Clock Solitaire is played with a standard deck of 52 cards, containing 13 ranks of cards in four
different suits. To set up the game, deal the cards face down into 13 piles of four cards each,
one in each of the ‘hour’ positions of a clock and one in the center. Each pile corresponds to a
particular rank—A through Q in clockwise order for the hour positions, and K for the center. To
start the game, turn over a card in the center pile. Then repeatedly turn over a card in the pile
corresponding to the value of the previous card. The game ends when you try to turn over a card
from a pile whose four cards are already face up. (This is always the center pile—why?) You win
if and only if every card is face up when the game ends.
12 Algorithms Lecture 5: Randomized Algorithms What is the exact probability that you win a game of Clock Solitaire, assuming that the cards
are permuted uniformly at random before they are dealt into their piles?
12. Suppose we have a circular linked list of numbers, implemented as a pair of arrays, one storing
the actual numbers and the other storing successor pointers. Speciﬁcally, let X [1 .. n] be an array
of n distinct real numbers, and let N [1 .. n] be an array of indices with the following property: If
X [i ] is the largest element of X , then X [N [i ]] is the smallest element of X ; otherwise, X [N [i ]] is
the smallest element of X that is larger than X [i ]. For example:
i1
2
3
4
5
6
7
8
9
X [i ] 83 54 16 31 45 99 78 62 27
N [i ] 6
8
9
5
2
3
1
7
4
Describe and analyze a randomized algorithm that determines whether a given number x appears
in the array X in O( n) expected time. Your algorithm may not modify the arrays X and N.
13. Death knocks on your door one cold blustery morning and challenges you to a game. Death knows
that you are an algorithms student, so instead of the traditional game of chess, Death presents you
with a complete binary tree with 4n leaves, each colored either black or white. There is a token at
the root of the tree. To play the game, you and Death will take turns moving the token from its
current node to one of its children. The game will end after 2n moves, when the token lands on a
leaf. If the ﬁnal leaf is black, you die; if it’s white, you will live forever. You move ﬁrst, so Death
gets the last turn.
∨
∧ ∧
∨ ∨
∧ ∧ ∧ ∨
∧ ∧ ∨
∧ ∧ ∧ You can decide whether it’s worth playing or not as follows. Imagine that the nodes at even
levels (where it’s your turn) are OR gates, the nodes at odd levels (where it’s Death’s turn) are
AND gates. Each gate gets its input from its children and passes its output to its parent. White and
black stand for TRUE and FALSE. If the output at the top of the tree is TRUE, then you can win and
live forever! If the output at the top of the tree is FALSE, you should challenge Death to a game of
Twister instead.
(a) Describe and analyze a deterministic algorithm to determine whether or not you can win.
[Hint: This is easy!]
(b) Unfortunately, Death won’t give you enough time to look at every node in the tree. Describe
a randomized algorithm that determines whether you can win in O(3n ) expected time. [Hint:
Consider the case n = 1.]
(c) Describe and analyze a randomized algorithm that determines whether you can win in O(c n )
expected time, for some constant c < 3. [Hint: You may not need to change your algorithm
from part (b) at all!]
13 Algorithms Lecture 5: Randomized Algorithms 14. A majority tree is a complete binary tree with depth n, where every leaf is labeled either 0 or 1.
The value of a leaf is its label; the value of any internal node is the majority of the values of
its three children. Consider the problem of computing the value of the root of a majority tree,
given the sequence of 3n leaf labels as input. For example, if n = 2 and the leaves are labeled
1, 0, 0, 0, 1, 0, 1, 1, 1, the root has value 0.
0 0 0 1 100 010 111 A majority tree with depth n = 2. (a) Prove that any deterministic algorithm that computes the value of the root of a majority tree
must examine every leaf. [Hint: Consider the special case n = 1. Recurse.]
(b) Describe and analyze a randomized algorithm that computes the value of the root in worstcase expected time O(c n ) for some constant c < 3. [Hint: Consider the special case n = 1.
Recurse.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 14 Algorithms Lecture 6: Treaps and Skip Lists
I thought the following four [rules] would be enough, provided that I made a ﬁrm and constant
resolution not to fail even once in the observance of them. The ﬁrst was never to accept anything
as true if I had not evident knowledge of its being so. . . . The second, to divide each problem I
examined into as many parts as was feasible, and as was requisite for its better solution. The
third, to direct my thoughts in an orderly way. . . establishing an order in thought even when the
objects had no natural priority one to another. And the last, to make throughout such complete
enumerations and such general surveys that I might be sure of leaving nothing out.
— René Descartes, Discours de la Méthode (1637) What is luck?
Luck is probability taken personally.
It is the excitement of bad math.
— Penn Jillette (2001), quoting Chip Denman (1998) 6 Randomized Binary Search Trees In this lecture, we consider two randomized alternatives to balanced binary search tree structures such
as AVL trees, red-black trees, B-trees, or splay trees, which are arguably simpler than any of these
deterministic structures. 6.1 Treaps A treap is a binary tree in which every node has both a search key and a priority, where the inorder
sequence of search keys is sorted and each node’s priority is smaller than the priorities of its children.1
In other words, a treap is simultaneously a binary search tree for the search keys and a (min-)heap for
the priorities. In our examples, we will use letters for the search keys and numbers for the priorities.
M
1
H
2
G
7
A
9 T
3
I
4 R
5
L
8 O
6 A treap. The top half of each node shows its search key and the bottom half shows its priority. I’ll assume from now on that all the keys and priorities are distinct. Under this assumption, we can
easily prove by induction that the structure of a treap is completely determined by the search keys and
priorities of its nodes. Since it’s a heap, the node v with highest priority must be the root. Since it’s also
a binary search tree, any node u with key(u) < key( v ) must be in the left subtree, and any node w with
key(w ) > key( v ) must be in the right subtree. Finally, since the subtrees are treaps, by induction, their
structures are completely determined. The base case is the trivial empty treap.
Another way to describe the structure is that a treap is exactly the binary tree that results by inserting
the nodes one at a time into an initially empty tree, in order of increasing priority, using the usual
insertion algorithm. This is also easy to prove by induction.
1 Sometimes I hate English. Normally, ‘higher priority’ means ‘more important’, but ‘ﬁrst priority’ is also more important
than ‘second priority’. Maybe ‘posteriority’ would be better; one student suggested ‘unimportance’. 1 Algorithms Lecture 6: Treaps and Skip Lists A third description interprets the keys and priorities as the coordinates of a set of points in the plane.
The root corresponds to a T whose joint lies on the topmost point. The T splits the plane into three parts.
The top part is (by deﬁnition) empty; the left and right parts are split recursively. This interpretation has
some interesting applications in computational geometry, which (unfortunately) we probably won’t have
time to talk about.
1
2
3
4
5
6
7
8
9
A G H I L M O R T A geometric interpretation of the same treap. Treaps were ﬁrst discovered by Jean Vuillemin in 1980, but he called them Cartesian trees.2 The word
‘treap’ was ﬁrst used by Edward McCreight around 1980 to describe a slightly different data structure,
but he later switched to the more prosaic name priority search trees.3 Treaps were rediscovered and used
to build randomized search trees by Cecilia Aragon and Raimund Seidel in 1989.4 A different kind of
randomized binary search tree, which uses random rebalancing instead of random priorities, was later
discovered and analyzed by Conrado Martínez and Salvador Roura in 1996.5 6.2 Treap Operations The search algorithm is the usual one for binary search trees. The time for a successful search is
proportional to the depth of the node. The time for an unsuccessful search is proportional to the depth
of either its successor or its predecessor.
To insert a new node z , we start by using the standard binary search tree insertion algorithm to
insert it at the bottom of the tree. At the point, the search keys still form a search tree, but the priorities
may no longer form a heap. To ﬁx the heap property, as long as z has smaller priority than its parent,
perform a rotation at z , a local operation that decreases the depth of z by one and increases its parent’s
depth by one, while maintaining the search tree property. Rotations can be performed in constant time,
since they only involve simple pointer manipulation.
y right x x
y left
A right rotation at x and a left rotation at y are inverses.
2 J. Vuillemin, A unifying look at data structures. Commun. ACM 23:229–239, 1980.
E. M. McCreight. Priority search trees. SIAM J. Comput. 14(2):257–276, 1985.
4
R. Seidel and C. R. Aragon. Randomized search trees. Algorithmica 16:464–497, 1996.
5
C. Martínez and S. Roura. Randomized binary search trees. J. ACM 45(2):288-323, 1998. The results in this paper are
virtually identical (including the constant factors!) to the corresponding results for treaps, although the analysis techniques are
quite different.
3 2 Algorithms Lecture 6: Treaps and Skip Lists The overall time to insert z is proportional to the depth of z before the rotations—we have to walk
down the treap to insert z , and then walk back up the treap doing rotations. Another way to say this is
that the time to insert z is roughly twice the time to perform an unsuccessful search for key(z ).
M
1 M
1 H
2
G
7
A
9 T
3
I
4 H
2
G
7 R
5
L
8 O
6 M
1 S
−1 A
9 T
3 L
8 S
−1 H
2 S
−1 I
4 S
−1 G
7 R
5 A
9 I
4 R
5
L
8 M
1
T
3 H
2 O
6 O
6 T
3 G
7
A
9 R
5
I
4 O
6
L
8 Left to right: After inserting (S , 10), rotate it up to ﬁx the heap property.
Right to left: Before deleting (S , 10), rotate it down to make it a leaf. Deleting a node is exactly like inserting a node, but in reverse order. Suppose we want to delete node
z . As long as z is not a leaf, perform a rotation at the child of z with smaller priority. This moves z down
a level and its smaller-priority child up a level. The choice of which child to rotate preserves the heap
property everywhere except at z . When z becomes a leaf, chop it off.
We sometimes also want to split a treap T into two treaps T< and T> along some pivot key π, so
that all the nodes in T< have keys less than π and all the nodes in T> have keys bigger then π. A simple
way to do this is to insert a new node z with key(z ) = π and priority(z ) = −∞. After the insertion, the
new node is the root of the treap. If we delete the root, the left and right sub-treaps are exactly the trees
we want. The time to split at π is roughly twice the time to (unsuccessfully) search for π.
Similarly, we may want to merge two treaps T< and T> , where every node in T< has a smaller search
key than any node in T> , into one super-treap. Merging is just splitting in reverse—create a dummy root
whose left sub-treap is T< and whose right sub-treap is T> , rotate the dummy node down to a leaf, and
then cut it off.
The cost of each of these operations is proportional to the depth of some node v in the treap.
• Search: A successful search for key k takes O(depth( v )) time, where v is the node with key( v ) = k.
For an unsuccessful search, let v − be the inorder predecessor of k (the node whose key is just barely
smaller than k), and let v + be the inorder successor of k (the node whose key is just barely larger
than k). Since the last node examined by the binary search is either v − or v + , the time for an
unsuccessful search is either O(depth( v + )) or O(depth( v − )).
• Insert/Delete: Inserting a new node with key k takes either O(depth( v + )) time or O(depth( v − ))
time, where v + and v − are the predecessor and successor of the new node. Deletion is just
insertion in reverse.
• Split/Merge: Splitting a treap at pivot value k takes either O(depth( v + )) time or O(depth( v − ))
time, since it costs the same as inserting a new dummy root with search key k and priority −∞.
Merging is just splitting in reverse.
Since the depth of a node in a treap is Θ(n) in the worst case, each of these operations has a
worst-case running time of Θ(n). 6.3 Random Priorities A randomized treap is a treap in which the priorities are independently and uniformly distributed continuous
random variables. That means that whenever we insert a new search key into the treap, we generate a
3 Algorithms Lecture 6: Treaps and Skip Lists random real number between (say) 0 and 1 and use that number as the priority of the new node. The
only reason we’re using real numbers is so that the probability of two nodes having the same priority is
zero, since equal priorities make the analysis slightly messier. In practice, we could just choose random
integers from a large range, like 0 to 231 − 1, or random ﬂoating point numbers. Also, since the priorities
are independent, each node is equally likely to have the smallest priority.
The cost of all the operations we discussed—search, insert, delete, split, join—is proportional to the
depth of some node in the tree. Here we’ll see that the expected depth of any node is O(log n), which
implies that the expected running time for any of those operations is also O(log n).
Let x k denote the node with the kth smallest search key. To analyze the expected depth, we deﬁne
an indicator variable
Aik = x i is a proper ancestor of x k .
(The superscript doesn’t mean power in this case; it just a reminder of which node is supposed to be
further up in the tree.) Since the depth of v is just the number of proper ancestors of v , we have the
following identity:
n depth( x k ) =
i =1 Aik . Now we can express the expected depth of a node in terms of these indicator variables as follows.
n E[depth( x k )] =
i =1 Pr[Aik = 1] (Just as in our analysis of matching nuts and bolts, we’re using linearity of expectation and the fact that
E[X ] = Pr[X = 1] for any indicator variable X .) So to compute the expected depth of a node, we just
have to compute the probability that some node is a proper ancestor of some other node.
Fortunately, we can do this easily once we prove a simple structural lemma. Let X (i , k) denote either
the subset of treap nodes { x i , x i +1 , . . . , x k } or the subset { x k , x k+1 , . . . , x i }, depending on whether i < k
or i > k. X (i , k) and X (k, i ) always denote prceisly the same subset, and this subset contains |k − i | + 1
nodes. X (1, n) = X (n, 1) contains all n nodes in the treap.
Lemma 1. For all i = k, x i is a proper ancestor of x k if and only if x i has the smallest priority among all
nodes in X (i , k).
Proof: If x i is the root, then it is an ancestor of x k , and by deﬁnition, it has the smallest priority of any
node in the treap, so it must have the smallest priority in X (i , k).
On the other hand, if x k is the root, then x i is not an ancestor of x k , and indeed x i does not have
the smallest priority in X (i , k) — x k does.
On the gripping hand6 , suppose some other node x j is the root. If x i and x k are in different subtrees,
then either i < j < k or i > j > k, so x j ∈ X (i , k). In this case, x i is not an ancestor of x k , and indeed x i
does not have the smallest priority in X (i , k) — x j does.
Finally, if x i and x k are in the same subtree, the lemma follows inductively (or, if you prefer,
recursively), since the subtree is a smaller treap. The empty treap is the trivial base case.
Since each node in X (i , k) is equally likely to have smallest priority, we immediately have the
probability we wanted: 1
if i < k k − i +1
[i = k]
i
Pr[Ak = 1] =
=0
if i = k
|k − i | + 1 1
if i > k
i − k +1
6 See Larry Niven and Jerry Pournelle, The Gripping Hand, Pocket Books, 1994. 4 Algorithms Lecture 6: Treaps and Skip Lists To compute the expected depth of a node x k , we just plug this probability into our formula and grind
through the algebra.
k −1 n E[depth( x k )] =
i =1 Pr[Aik = 1] = i =1
k =
j =2 n 1
k−i+1
1
j + n−k+1 +
i =2 i = k +1 1
i−k+1 1
j = H k − 1 + H n−k+1 − 1
< ln k + ln(n − k + 1) − 2
< 2 ln n − 2.
In conclusion, every search, insertion, deletion, split, and merge operation in an n-node randomized
binary search tree takes O(log n) expected time.
Since a treap is exactly the binary tree that results when you insert the keys in order of increasing
priority, a randomized treap is the result of inserting the keys in random order. So our analysis also
automatically gives us the expected depth of any node in a binary tree built by random insertions
(without using priorities). 6.4 Randomized Quicksort (Again!) We’ve already seen two completely different ways of describing randomized quicksort. The ﬁrst is the
familiar recursive one: choose a random pivot, partition, and recurse. The second is a less familiar
iterative version: repeatedly choose a new random pivot, partition whatever subset contains it, and
continue. But there’s a third way to describe randomized quicksort, this time in terms of binary search
trees.
RANDOMIZEDQUICKSORT:
T ← an empty binary search tree
insert the keys into T in random order
output the inorder sequence of keys in T Our treap analysis tells us is that this algorithm will run in O(n log n) expected time, since each key is
inserted in O(log n) expected time.
Why is this quicksort? Just like last time, all we’ve done is rearrange the order of the comparisons.
Intuitively, the binary tree is just the recursion tree created by the normal version of quicksort. In the
recursive formulation, we compare the initial pivot against everything else and then recurse. In the
binary tree formulation, the ﬁrst “pivot” becomes the root of the tree without any comparisons, but
then later as each other key is inserted into the tree, it is compared against the root. Either way, the
ﬁrst pivot chosen is compared with everything else. The partition splits the remaining items into a left
subarray and a right subarray; in the binary tree version, these are exactly the items that go into the left
subtree and the right subtree. Since both algorithms deﬁne the same two subproblems, by induction,
both algorithms perform the same comparisons.
We even saw the probability |k−1|+1 before, when we were talking about sorting nuts and bolts
i
with a variant of randomized quicksort. In the more familiar setting of sorting an array of numbers,
the probability that randomized quicksort compares the i th largest and kth largest elements is exactly
2
. The binary tree version compares x i and x k if and only if x i is an ancestor of x k or vice versa, so
|k−i |+1
the probabilities are exactly the same. 5 Algorithms 6.5 Lecture 6: Treaps and Skip Lists Skip Lists Skip lists, which were ﬁrst discovered by Bill Pugh in the late 1980’s,7 have many of the usual desirable
properties of balanced binary search trees, but their structure is very different.
At a high level, a skip list is just a sorted linked list with some random shortcuts. To do a search in a
normal singly-linked list of length n, we obviously need to look at n items in the worst case. To speed
up this process, we can make a second-level list that contains roughly half the items from the original
list. Speciﬁcally, for each item in the original list, we duplicate it with probability 1/2. We then string
together all the duplicates into a second sorted linked list, and add a pointer from each duplicate back
to its original. Just to be safe, we also add sentinel nodes at the beginning and end of both lists.
−∞ 0 1 −∞ 0 1 3
2 3 6
4 5 7 6 7 9
8 +∞ 9 +∞ A linked list with some randomly-chosen shortcuts. Now we can ﬁnd a value x in this augmented structure using a two-stage algorithm. First, we scan
for x in the shortcut list, starting at the −∞ sentinel node. If we ﬁnd x , we’re done. Otherwise, we
reach some value bigger than x and we know that x is not in the shortcut list. Let w be the largest item
less than x in the shortcut list. In the second phase, we scan for x in the original list, starting from w .
Again, if we reach a value bigger than x , we know that x is not in the data structure.
−∞ 0 1 −∞ 0 1 3
2 3 6
4 5 7 6 7 9
8 +∞ 9 +∞ Searching for 5 in a list with shortcuts. Since each node appears in the shortcut list with probability 1/2, the expected number of nodes
examined in the ﬁrst phase is at most n/2. Only one of the nodes examined in the second phase has
a duplicate. The probability that any node is followed by k nodes without duplicates is 2−k , so the
expected number of nodes examined in the second phase is at most 1 + k≥0 2−k = 2. Thus, by adding
these random shortcuts, we’ve reduced the cost of a search from n to n/2 + 2, roughly a factor of two in
savings. 6.6 Recursive Random Shortcuts Now there’s an obvious improvement—add shortcuts to the shortcuts, and repeat recursively. That’s
exactly how skip lists are constructed. For each node in the original list, we ﬂip a coin over and over
until we get tails. For each heads, we make a duplicate of the node. The duplicates are stacked up in
levels, and the nodes on each level are strung together into sorted linked lists. Each node v stores a
search key (key( v )), a pointer to its next lower copy (down( v )), and a pointer to the next node in its
level (right( v )).
7 William Pugh. Skip lists: A probabilistic alternative to balanced trees. Commun. ACM 33(6):668–676, 1990. 6 Algorithms Lecture 6: Treaps and Skip Lists
−∞ +∞ 7 +∞ 7 +∞ 6 7 +∞ 6 7 6 7 −∞ −∞ 1 −∞ 1 −∞ 0 1 −∞ 0 1 3
2 3 4 5 9
8 +∞ 9 +∞ A skip list is a linked list with recursive random shortcuts. The search algorithm for skip lists is very simple. Starting at the leftmost node L in the highest level,
we scan through each level as far as we can without passing the target value x , and then proceed down
to the next level. The search ends when we either reach a node with search key x or fail to ﬁnd x on the
lowest level.
SKIPLISTFIND( x , L ):
v←L
while ( v = NULL and key( v ) = x )
if key(right( v )) > x
v ← down( v )
else
v ← right( v )
return v
−∞ +∞ 7 +∞ 7 +∞ 6 7 +∞ 6 7 6 7 −∞ −∞ 1 −∞ 1 −∞ 0 1 −∞ 0 1 3
2 3 4 5 9
8 +∞ 9 +∞ Searching for 5 in a skip list. Intuitively, since each level of the skip lists has about half the number of nodes as the previous level,
the total number of levels should be about O(log n). Similarly, each time we add another level of random
shortcuts to the skip list, we cut the search time roughly in half, except for a constant overhead, so after
O(log n) levels, we should have a search time of O(log n). Let’s formalize each of these two intuitive
observations. 6.7 Number of Levels The actual values of the search keys don’t affect the skip list analysis, so let’s assume the keys are the
integers 1 through n. Let L ( x ) be the number of levels of the skip list that contain some search key x , not
counting the bottom level. Each new copy of x is created with probability 1/2 from the previous level,
essentially by ﬂipping a coin. We can compute the expected value of L ( x ) recursively—with probability 7 Algorithms Lecture 6: Treaps and Skip Lists 1/2, we ﬂip tails and L ( x ) = 0; and with probability 1/2, we ﬂip heads, increase L ( x ) by one, and
recurse:
1
1
E [ L ( x )] = · 0 + 1 + E [ L ( x )]
2
2
Solving this equation gives us E [L( x )] = 1.
In order to analyze the expected worst-case cost of a search, however, we need a bound on the
number of levels L = max x L ( x ). Unfortunately, we can’t compute the average of a maximum the way we
would compute the average of a sum. Instead, we will derive a stronger result, showing that the depth is
O(log n) with high probability. ‘High probability’ is a technical term that means the probability is at
least 1 − 1/nc for some constant c ≥ 1; the hidden constant in the O(log n) bound could depend on c .
In order for a search key x to appear on level , it must have ﬂipped heads in a row when it was
inserted, so Pr[L( x ) ≥ ] = 2− . The skip list has at least levels if and only if L ( x ) ≥ for at least one
of the n search keys.
Pr[ L ≥ ] = Pr ( L (1) ≥ ) ∨ ( L (2) ≥ ) ∨ · · · ∨ ( L (n) ≥ )
Using the union bound — Pr[A ∨ B ] ≤ P r [A] + Pr[B ] for any random events A and B — we can simplify
this as follows:
n
n
Pr[ L ≥ ] ≤
Pr[ L ( x ) ≥ ] = n · Pr[ L ( x ) ≥ ] =
.
2
x =1
When ≤ lg n, this bound is trivial. However, for any constant c > 1, we have a strong upper bound
Pr[ L ≥ c lg n] ≤ 1
n c −1 . We conclude that with high probability, a skip list has O (log n ) levels.
This high-probability bound indirectly implies a bound on the expected number of levels. Some
simple algebra gives us the following alternate deﬁnition for expectation:
E[ L ] = · Pr[ L = ] =
≥0 Clearly, if < , then Pr[ L ( x ) ≥ ] > Pr[ L ( x ) ≥
number of levels as follows:
E[ L ( x )] = Pr[ L ≥ ]
≥1
. So we can derive an upper bound on the expected Pr[ L ≥ ]
≥1
lg n = Pr[ L ≥ ] + Pr[ L ≥ ] =1 ≥lg n+1 lg n n 1+ ≤
=1 ≥lg n+1 2 1 = lg n +
i ≥1 [i = − lg n] 2i = lg n + 2
So in expectation, a skip list has at most two more levels than an ideal version where each level contains
exactly half the nodes of the next level below.
8 Algorithms 6.8 Lecture 6: Treaps and Skip Lists Logarithmic Search Time It’s a little easier to analyze the cost of a search if we imagine running the algorithm backwards.
D N I F T S I L P I K S takes the output from S KIP L IST F IND as input and traces back through the data structure to
the upper left corner. Skip lists don’t really have up and left pointers, but we’ll pretend that they do so
we don’t have to write ‘) v (nwod ← v ’ or ‘) v (thgir ← v ’.8
D N I F T S I L P I K S( v ): while ( v = L )
if up( v ) exists
v ← up( v )
else
v ← left( v ) Now for every node v in the skip list, up( v ) exists with probability 1/2. So for purposes of analysis,
D N I F T S I L P I K S is equivalent to the following algorithm:
FLIPWALK( v ):
while ( v = L )
if COINFLIP = HEADS
v ← up( v )
else
v ← left( v ) Obviously, the expected number of heads is exactly the same as the expected number of TAILs. Thus,
the expected running time of this algorithm is twice the expected number of upward jumps. Since we
already know that the number of upward jumps is O(log n) with high probability, we can conclude that
the worst-case search time is O(log n) with high probability (and therefore in expectation). Exercises
1. Prove that the expected number of proper descendants of any node in a treap is exactly equal to
the expected depth of that node.
2. What is the exact expected number of leaves in an n-node treap? [Hint: What is the probability
that in an n-node treap, the node with kth smallest search key is a leaf?]
3. Consider an n-node treap T . As in the lecture notes, we identify nodes in T by the ranks of their
search keys. Thus, ‘node 5’ means the node with the 5th smallest search key. Let i , j , k be integers
such that 1 ≤ i ≤ j ≤ k ≤ n.
(a) What is the exact probability that node j is a common ancestor of node i and node k?
(b) What is the exact expected length of the unique path from node i to node k in T ?
4. Recall that a priority search tree is a binary tree in which every node has both a search key and
a priority, arranged so that the tree is simultaneously a binary search tree for the keys and a
min-heap for the priorities. A heater is a priority search tree in which the priorities are given by
8 .terces seirevocsid sih peek ot detnaw eh esuaceb ton tub ,gnitirw-rorrim gnisu seton sih lla etorw icniV ad odranoeL
!dnah thgir sih ni sitirhtra dab yllaer dah tsuj eH 9 Algorithms Lecture 6: Treaps and Skip Lists the user, and the search keys are distributed uniformly and independently at random in the real
interval [0, 1]. Intuitively, a heater is a sort of dual treap.
The following problems consider an n-node heater T whose node priorities are the integers
from 1 to n. We identify nodes in T by their priorities; thus, ‘node 5’ means the node in T with
priority 5. The min-heap property implies that node 1 is the root of T . Finally, let i and j be
integers with 1 ≤ i < j ≤ n.
(a) Prove that in a random permutation of the (i + 1)-element set {1, 2, . . . , i , j }, elements i and j
are adjacent with probability 2/(i + 1).
(b) Prove that node i is an ancestor of node j with probability 2/(i + 1). [Hint: Use part (a)!]
(c) What is the probability that node i is a descendant of node j ? [Hint: Don’t use part (a)!]
(d) What is the exact expected depth of node j ?
(e) Describe and analyze an algorithm to insert a new item into a heater. Express the expected
running time of the algorithm in terms of the rank of the newly inserted item.
(f) Describe an algorithm to delete the minimum-priority item (the root) from an n-node heater.
What is the expected running time of your algorithm?
5. A meldable priority queue stores a set of keys from some totally-ordered universe (such as the
integers) and supports the following operations:
• MAKEQUEUE: Return a new priority queue containing the empty set.
• FINDMIN(Q): Return the smallest element of Q (if any).
• DELETEMIN(Q): Remove the smallest element in Q (if any).
• INSERT(Q, x ): Insert element x into Q, if it is not already there.
• DECREASEKEY(Q, x , y ): Replace an element x ∈ Q with a smaller key y . (If y > x , the
operation fails.) The input is a pointer directly to the node in Q containing x .
• DELETE(Q, x ): Delete the element x ∈ Q. The input is a pointer directly to the node in Q
containing x .
• MELD(Q 1 , Q 2 ): Return a new priority queue containing all the elements of Q 1 and Q 2 ; this
operation destroys Q 1 and Q 2 .
A simple way to implement such a data structure is to use a heap-ordered binary tree, where
each node stores a key, along with pointers to its parent and two children. MELD can be implemented using the following randomized algorithm:
MELD(Q 1 , Q 2 ):
if Q 1 is empty return Q 2
if Q 2 is empty return Q 1
if key(Q 1 ) > key(Q 2 )
swap Q 1 ↔ Q 2
with probability 1/2
left(Q 1 ) ← MELD(left(Q 1 ), Q 2 )
else
right(Q 1 ) ← MELD(right(Q 1 ), Q 2 )
return Q 1 10 Algorithms Lecture 6: Treaps and Skip Lists (a) Prove that for any heap-ordered binary trees Q 1 and Q 2 (not just those constructed by the
operations listed above), the expected running time of MELD(Q 1 , Q 2 ) is O(log n), where
n = |Q 1 | + |Q 2 |. [Hint: How long is a random root-to-leaf path in an n-node binary tree if
each left/right choice is made with equal probability?]
(b) Prove that MELD(Q 1 , Q 2 ) runs in O(log n) time with high probability.
(c) Show that each of the other meldable priority queue operations can be implemented with at
most one call to MELD and O(1) additional time. (This implies that every operation takes
O(log n) time with high probability.)
6. In the usual theoretical presentation of treaps, the priorities are random real numbers chosen
uniformly from the interval [0, 1], but in practice, computers only have access to random bits.
This problem asks you to analyze a modiﬁcation of treaps that takes this limitation into account.
Suppose the priority of a node v is abstractly represented as an inﬁnite sequence π v [1 .. ∞] of
random bits, which is interpreted as the rational number
∞ π v [ i ] · 2− i . priority( v ) =
i =1 However, only a ﬁnite number v of these bits are actually known at any given time. When a
node v is ﬁrst created, none of the priority bits are known: v = 0. We generate (or ‘reveal’)
new random bits only when they are necessary to compare priorities. The following algorithm
compares the priorities of any two nodes in O(1) expected time:
LARGERPRIORITY( v, w ):
for i ← 1 to ∞
if i > v
v ← i ; π v [ i ] ← R ANDOM B IT
if i > w
w ← i ; πw [i ] ← RANDOMBIT if π v [i ] > πw [i ]
return v
else if π v [i ] < πw [i ]
return w Suppose we insert n items one at a time into an initially empty treap. Let L = v v denote
the total number of random bits generated by calls to LARGERPRIORITY during these insertions.
(a) Prove that E [ L ] = Θ(n).
(b) Prove that E [ v ] = Θ(1) for any node v . [Hint: This is equivalent to part (a). Why?]
(c) Prove that E [ root ] = Θ(log n). [Hint: Why doesn’t this contradict part (b)?] 7. Any skip list L can be transformed into a binary search tree T (L) as follows. The root of T (L)
is the leftmost node on the highest non-empty level of L; the left and subtrees are constructed
recursively from the nodes to the left and to the right of the root. Let’s call the resulting tree T (L)
a skip list tree. 11 Algorithms Lecture 6: Treaps and Skip Lists (a) Show that any search in T (L) is no more expensive than the corresponding search in L.
(Searching in T (L) could be considerably cheaper—why?)
(b) Describe an algorithm to insert a new search key into a skip list tree in O(log n) expected
time. Inserting key x into T (L) should produce exactly the same tree as inserting x into L
and then transforming L into a tree. [Hint: You will need to maintain some additional
information in the tree nodes.]
(c) Describe an algorithm to delete a search key from a skip list tree in O(log n) expected time.
Again, deleting key x from T (L) should produce exactly the same tree as deleting x from L
and then transforming L into a tree. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 12 Algorithms Non-Lecture E: Tail Inequalities
If you hold a cat by the tail
you learn things you cannot learn any other way.
— Mark Twain E Tail Inequalities The simple recursive structure of skip lists made it relatively easy to derive an upper bound on the
expected worst-case search time, by way of a stronger high-probability upper bound on the worst-case
search time. We can prove similar results for treaps, but because of the more complex recursive structure,
we need slightly more sophisticated probabilistic tools. These tools are usually called tail inequalities;
intuitively, they bound the probability that a random variable with a bell-shaped distribution takes a
value in the tails of the distribution, far away from the mean. E.1 Markov’s Inequality Perhaps the simplest tail inequality was named after the Russian mathematician Andrey Markov; however,
in strict accordance with Stigler’s Law of Eponymy, it ﬁrst appeared in the works of Markov’s probability
teacher, Pafnuty Chebyshev.1
Markov’s Inequality. Let X be a non-negative integer random variable. For any t > 0, we have
Pr[X ≥ t ] ≤ E[X ]/ t .
Proof: The inequality follows from the deﬁnition of expectation by simple algebraic manipulation.
∞ E[ X ] = k · Pr[X = k] [deﬁnition of E[X ]] k =0
∞ = Pr[X ≥ k] [algebra] Pr[X ≥ k] [since t < ∞] Pr[X ≥ t ] [since k < t ] k =0
t −1 ≥
k =0
t −1 ≥
k =0 = t · Pr[X ≥ t ] [algebra] Unfortunately, the bounds that Markov’s inequality implies (at least directly) are often very weak,
even useless. (For example, Markov’s inequality implies that with high probability, every node in an
n-node treap has depth O(n2 log n). Well, duh!) To get stronger bounds, we need to exploit some
additional structure in our random variables. 1
The closely related tail bound traditionally called Chebyshev’s inequality was actually discovered by the French statistician
Irénée-Jules Bienaymé, a friend and colleague of Chebyshev’s. 1 Algorithms E.2 Non-Lecture E: Tail Inequalities Sums of Indicator Variables A set of random variables X 1 , X 2 , . . . , X n are said to be mutually independent if and only if n n (X i = x i ) = Pr Pr[X i = x i ]
i =1 i =1 for all possible values x 1 , x 2 , . . . , x n . For examples, different ﬂips of the same fair coin are mutually
independent, but the number of heads and the number of tails in a sequence of n coin ﬂips are not
independent (since they must add to n). Mutual independence of the X i ’s implies that the expectation of
the product of the X i ’s is equal to the product of the expectations: n n Xi = E
i =1 E[ X i ] .
i =1 Moreover, if X 1 , X 2 , . . . , X n are independent, then for any function f , the random variables f (X 1 ),
f (X 2 ), . . . , f (X n ) are also mutually independent.
n
Suppose X = i =1 X i is the sum of n mutually independent random indicator variables X i . For
each i , let pi = Pr[X i = 1], and let µ = E[X ] = i E[X i ] = i pi .
eδ Chernoff Bound (Upper Tail). Pr[X > (1 + δ)µ] < µ for any δ > 0. (1 + δ)1+δ Proof: The proof is fairly long, but it replies on just a few basic components: a clever substitution,
Markov’s inequality, the independence of the X i ’s, The World’s Most Useful Inequality e x > 1 + x , a tiny
bit of calculus, and lots of high-school algebra.
We start by introducing a variable t , whose role will become clear shortly.
P r [X > (1 + δ)µ] = Pr[e t X > e t (1+δ)µ ]
To cut down on the superscripts, I’ll usually write exp( x ) instead of e x in the rest of the proof. Now
apply Markov’s inequality to the right side of this equation:
P r [X > (1 + δ)µ] < E[exp( t X )]
exp( t (1 + δ)µ) . We can simplify the expectation on the right using the fact that the terms X i are independent.
E[exp( t X )] = E exp t Xi =E i We can bound the individual expectations E e t X i exp( t X i ) =
i E[exp( t X i )]
i using The World’s Most Useful Inequality: E[exp( t X i )] = pi e t + (1 − pi ) = 1 + (e t − 1) pi < exp (e t − 1) pi
This inequality gives us a simple upper bound for E[e t X ]:
exp((e t − 1) pi ) < exp E[exp( t X )] <
i (e t − 1) pi = exp((e t − 1)µ)
i 2 Algorithms Non-Lecture E: Tail Inequalities Substituting this back into our original fraction from Markov’s inequality, we obtain
P r [X > (1 + δ)µ] < E[exp( t X )]
exp( t (1 + δ)µ) < exp((e t − 1)µ)
exp( t (1 + δ)µ) = exp(e t − 1 − t (1 + δ)) µ Notice that this last inequality holds for all possible values of t . To obtain the ﬁnal tail bound, we will
choose t to make this bound as tight as possible. To minimize e t − 1 − t − t δ, we take its derivative with
respect to t and set it to zero:
d
dt (e t − 1 − t (1 + δ)) = e t − 1 − δ = 0. (And you thought calculus would never be useful!) This equation has just one solution t = ln(1 + δ).
Plugging this back into our bound gives us
P r [X > (1 + δ)µ] < exp(δ − (1 + δ) ln(1 + δ)) µ = µ eδ
(1 + δ)1+δ And we’re done!
This form of the Chernoff bound can be a bit clumsy to use. A more complicated argument gives us
the bound
Pr[X > (1 + δ)µ] < e−µδ 2 /3 for any 0 < δ < 1. A similar argument gives us an inequality bounding the probability that X is signiﬁcantly smaller
than its expected value:
Chernoff Bound (Lower Tail). Pr[X < (1 − δ)µ] < E.3 eδ
(1 − δ)1−δ µ < e−µδ 2 /2 for any δ > 0. Back to Treaps In our analysis of randomized treaps, we deﬁned the indicator variable Aik to have the value 1 if and only
if the node with the i th smallest key (‘node i ’) was a proper ancestor of the node with the kth smallest
key (‘node k’). We argued that
[i = k]
Pr[Aik = 1] =
,
|k − i | + 1
and from this we concluded that the expected depth of node k is
n E[depth(k)] =
i =1 Pr[Aik = 1] = H k + H n−k − 2 < 2 ln n. To prove a worst-case expected bound on the depth of the tree, we need to argue that the maximum
depth of any node is small. Chernoff bounds make this argument easy, once we establish that the
relevant indicator variables are mutually independent.
Lemma 1. For any index k, the k − 1 random variables Aik with i < k are mutually independent. Similarly,
for any index k, the n − k random variables Aik with i > k are mutually independent. 3 Algorithms Non-Lecture E: Tail Inequalities Proof: To simplify the notation, we explicitly consider only the case k = 1, although the argument
generalizes easily to other values of k. Fix n − 1 arbitrary indicator values x 2 , x 3 , . . . , x n . We prove the
lemma by induction on n, with the vacuous base case n = 1. The deﬁnition of conditional probability
gives us n−1 n (Ai1 Pr (Aik = x i ) ∧ An = x n 1 = x i ) = Pr i =2 i =2 n−1 (Aik = x i ) An = x n · Pr An = x n
1
1 = Pr i =2 Now recall that
= 1 if and only if node n has the smallest priority, and the other n − 2 indicator
variables Ai1 depend only on the order of the priorities of nodes 1 through n − 1. There are exactly (n − 1)!
permutations of the n priorities in which the nth priority is smallest, and each of these permutations is
equally likely. Thus, An
1 n−1 n −1 (Aik = x i ) An = x n = Pr 1 Pr i =2 (Aik = x i )
i =2 A2 , . . . , An−1
1
1 The inductive hypothesis implies that the variables n−1 n−1 (Aik = x i ) = Pr are mutually independent, so i =2 i =2 Pr Ai1 = x i . We conclude that Pr n (Ai1
i =2 n −1 n−1 = x i ) = Pr An = x n ·
1 Pr
i =2 Ai1 = xi =
i =1 Pr Ai1 = x i , or in other words, that the indicator variables are mutually independent.
Theorem 2. The depth of a randomized treap with n nodes is O(log n) with high probability.
Proof: First let’s bound the probability that the depth of node k is at most 8 ln n. There’s nothing special
about the constant 8 here; I’m being generous to make the analysis easier.
The depth is a sum of n indicator variables Aik , as i ranges from 1 to n. Our Observation allows
us to partition these variables into two mutually independent subsets. Let d< (k) = i <k Aik and
d> (k) = i <k Aik , so that depth(k) = d< (k) + d> (k). If depth(k) > 8 ln n, then either d< (k) > 4 ln n or
d> (k) > 4 ln n.
Chernoff’s inequality, with µ = E[d< (k)] = H k − 1 < ln n and δ = 3, bounds the probability that
d< (k) > 4 ln n as follows.
Pr[d< (k) > 4 ln n] < Pr[d< (k) > 4µ] < e3
44 µ < e3
44 ln n = nln(e 3 /44 ) = n3−4 ln 4 < 1
n2 . (The last step uses the fact that 4 ln 4 ≈ 5.54518 > 5.) The same analysis implies that Pr[d> (k) >
4 ln n] < 1/n2 . These inequalities imply the crude bound Pr[depth(k) > 4 ln n] < 2/n2 .
Now consider the probability that the treap has depth greater than 10 ln n. Even though the
distributions of different nodes’ depths are not independent, we can conservatively bound the probability
of failure as follows: n
n
2
Pr max depth(k) > 8 ln n = Pr (depth(k) > 8 ln n) ≤
Pr[depth(k) > 8 ln n] < .
k
n
k =1
k =1 4 Algorithms Non-Lecture E: Tail Inequalities This argument implies more generally that for any constant c , the depth of the treap is greater
than c ln n with probability at most 2/nc ln c −c . We can make the failure probability an arbitrarily small
polynomial by choosing c appropriately.
This lemma implies that any search, insertion, deletion, or merge operation on an n-node treap
requires O(log n) time with high probability. In particular, the expected worst-case time for each of these
operations is O(log n). Exercises
1. Prove that for any integer k such that 1 < k < n, the n − 1 indicator variables Aik with i = k are
not mutually independent. [Hint: Consider the case n = 3.]
2. Recall from Exercise 1 in the previous note that the expected number of descendants of any node
in a treap is O(log n). Why doesn’t the Chernoff-bound argument for depth imply that, with high
probability, every node in a treap has O(log n) descendants? The conclusion is clearly bogus—Every
treap has a node with n descendants!—but what’s the hole in the argument?
3. A heater is a sort of dual treap, in which the priorities of the nodes are given, but their search
keys are generate independently and uniformly from the unit interval [0, 1]. You can assume all
priorities and keys are distinct.
(a) Prove that for any r , the node with the r th smallest priority has expected depth O(log r ).
(b) Prove that an n-node heater has depth O(log n) with high probability.
(c) Describe algorithms to perform the operations INSERT and DELETEMIN in a heater. What are
the expected worst-case running times of your algorithms? In particular, can you express the
expected running time of INSERT in terms of the priority rank of the newly inserted item? c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 5 Algorithms Lecture 7: Hash Tables
Calvin: There! I ﬁnished our secret code!
Hobbes: Let’s see.
Calvin: I assigned each letter a totally random number, so the code will be hard to
crack. For letter “A”, you write 3,004,572,688. “B” is 28,731,569 1 2 .
/
Hobbes: That’s a good code all right.
Calvin: Now we just commit this to memory.
Calvin: Did you ﬁnish your map of our neighborhood?
Hoobes: Not yet. How many bricks does the front walk have?
— Bill Watterson, “Calvin and Hobbes” (August 23, 1990) 7
7.1 Hash Tables
Introduction A hash table is a data structure for storing a set of items, so that we can quickly determine whether an
item is or is not in the set. The basic idea is to pick a hash function h that maps every possible item x to
a small integer h( x ). Then we store x in slot h( x ) in an array. The array is the hash table.
Let’s be a little more speciﬁc. We want to store a set of n items. Each item is an element of some
ﬁnite1 set U called the universe; we use u to denote the size of the universe, which is just the number of
items in U. A hash table is an array T [1 .. m], where m is another positive integer, which we call the
table size. Typically, m is much smaller than u. A hash function is any function of the form
h : U → {0, 1, . . . , m − 1},
mapping each possible item in U to a slot in the hash table. We say that an item x hashes to the slot
T [h( x )].
Of course, if u = m, then we can always just use the trivial hash function h( x ) = x . In other words,
use the item itself as the index into the table. This is called a direct access table, or more commonly, an
array. In most applications, though, the universe of possible keys is orders of magnitude too large for
this approach to be practical. Even when it is possible to allocate enough memory, we usually need to
store only a small fraction of the universe. Rather than wasting lots of space, we should make m roughly
equal to n, the number of items in the set we want to maintain.
What we’d like is for every item in our set to hash to a different position in the array. Unfortunately,
unless m = u, this is too much to hope for, so we have to deal with collisions. We say that two items x
and y collide if the have the same hash value: h( x ) = h( y ). Since we obviously can’t store two items
in the same slot of an array, we need to describe some methods for resolving collisions. The two most
common methods are called chaining and open addressing. 7.2 Chaining In a chained hash table, each entry T [i ] is not just a single item, but rather (a pointer to) a linked list of
all the items that hash to T [i ]. Let ( x ) denote the length of the list T [h( x )]. To see if an item x is in
the hash table, we scan the entire list T [h( x )]. The worst-case time required to search for x is O(1) to
compute h( x ) plus O(1) for every element in T [h( x )], or O(1 + ( x )) overall. Inserting and deleting x
also take O(1 + ( x )) time.
1
This ﬁniteness assumption is necessary for several of the technical details to work out, but can be ignored in practice.
To hash elements from an inﬁnite universe (for example, the positive integers), pretend that the universe is actually ﬁnite
but very very large. In fact, in real practice, the universe actually is ﬁnite but very very large. For example, on most modern
computers, there are only 264 integers (unless you use a big integer package like GMP in which case the number of integers is
,
32
closer to 22 .) 1 Algorithms Lecture 7: Hash Tables
GH RO M AL I T
S A chained hash table with load factor 1. In the worst case, every item would be hashed to the same value, so we’d get just one long list of n
items. In principle, for any deterministic hashing scheme, a malicious adversary can always present a set
of items with exactly this property. In order to defeat such malicious behavior, we’d like to use a hash
function that is as random as possible. Choosing a truly random hash function is completely impractical,
but there are several heuristics for producing hash functions that behave randomly, or at least close to
randomly on real data. Thus, we will analyze the performance as though our hash function were truly
random. More formally, we make the following assumption.
Simple uniform hashing assumption: If x = y then Pr h( x ) = h( y ) = 1/m.
In the next section, I’ll describe a small set of functions with the property that a random hash function
in this set satisﬁes the simple uniform hashing assumption. Most actual implementations of has tables
use deterministic hash functions. These clearly violate the uniform hashing assumption—the collision
probability is either 0 or 1, depending on the pair of items! Nevertheless, it is common practice to adopt
the uniform hashing assumption as a convenient ﬁction for purposes of analysis.
Let’s compute the expected value of ( x ) under this assumption; this will immediately imply a bound
on the expected time to search for an item x . To be concrete, let’s suppose that x is not already stored in
the hash table. For all items x and y , we deﬁne the indicator variable
C x , y = h( x ) = h( y ) .
(In case you’ve forgotten the bracket notation, C x , y = 1 if h( x ) = h( y ) and C x , y = 0 if h( x ) = h( y ).)
Since the length of T [h( x )] is precisely equal to the number of items that collide with x , we have
(x) = Cx, y .
y ∈T We can rewrite the simple uniform hashing assumption as follows:
x = y =⇒ E[C x , y ] = Pr[C x , y = 1] = 1
m . Now we just have to grind through the deﬁnitions.
E[ ( x )] = 1 E Cx, y =
y ∈T y ∈T m = n
m We call this fraction n/m the load factor of the hash table. Since the load factor shows up everywhere,
we will give it its own symbol α.
α= 2 n
m Algorithms Lecture 7: Hash Tables Our analysis implies that the expected time for an unsuccessful search in a chained hash table is Θ(1 + α).
As long as the number if items n is only a constant factor bigger than the table size m, the search time is
a constant. A similar analysis gives the same expected time bound (with a slightly smaller constant) for
a successful search.
Obviously, linked lists are not the only data structure we could use to store the chains; any data
structure that can store a set of items will work. For example, if the universe U has a total ordering, we
can store each chain in a balanced binary search tree. This reduces the expected time for any search to
O(1 + log ( x )), and under the simple uniform hashing assumption, the expected time for any search is
O(1 + log α).
Another natural possibility is to work recursively! Speciﬁcally, for each T [i ], we maintain a hash
table Ti containing all the items with hash value i . Collisions in those secondary tables are resolved
recursively, by storing secondary overﬂow lists in tertiary hash tables, and so on. The resulting data
structure is a tree of hash tables, whose leaves correspond to items that (at some level of the tree) are
hashed without any collisions. If every hash table in this tree has size m, then the expected time for any
search is O(logm n). In particular, if we set m = n, the expected time for any search is constant. On the
other hand, there is no inherent reason to use the same hash table size everywhere; after all, hash tables
deeper in the tree are storing fewer items.
Caveat Lector!2 The preceding analysis does not imply bounds on the expected worst-case search
time is constant. The expected worst-case search time is O(1 + L ), where L = max x ( x ). Under the
uniform hashing assumption, the maximum list size L is very likely to grow faster than any constant,
unless the load factor α is signiﬁcantly smaller than 1. For example, E[ L ] = Θ(log n/ log log n) when
α = 1. We’ve stumbled on a powerful but counterintuitive fact about probability: When several individual
items are distributed independently and uniformly at random, the resulting distribution is not uniform
in the traditional sense! In a later section, I’ll describe how to achieve constant expected worst-case
search time using secondary hash tables. 7.3 Universal Hashing Now I’ll describe a method to generate random hash functions that satisfy the simple uniform hashing
assumption. We say that a set H of hash function is universal if it satisﬁes the following property: For
any items x = y , if a hash function h is chosen uniformly at random from the set H, then Pr[h( x ) =
h( y )] = 1/m. Note that this probability holds for any items x and y ; the randomness is entirely in
choosing a hash function from the set H.
To simplify the following discussion, I’ll assume that the universe U contains exactly m2 items, each
represented as a pair ( x , x ) of integers between 0 and m − 1. (Think of the items as two-digit numbers
in base m.) I will also assume that m is a prime number.
For any integers 0 ≤ a, b ≤ m − 1, deﬁne the function ha, b : U → {0, 1, . . . , m − 1} as follows:
ha, b ( x , x ) = (a x + b x ) mod m.
Then the set H = ha, b 0 ≤ a , b ≤ m − 1
of all such functions is universal. To prove this, we need to show that for any pair of distinct items
( x , x ) = ( y, y ), exactly m of the m2 functions in H cause a collision.
2 No, this is not the name of Hannibal’s brother. It’s Latin for “Reader beware!” 3 Algorithms Lecture 7: Hash Tables Choose two items ( x , x ) = ( y, y ), and assume without loss of generality3 that x = y . A function
ha, b ∈ H causes a collision between ( x , x ) and ( y, y ) if and only if
ha, b ( x , x ) = ha, b ( y , y )
(a x + b x ) mod m = (a y + b y ) mod m
ax + b x ≡ a y + b y (mod m) a( x − y ) ≡ b( y − x ) (mod m)
a≡ b( y − x )
x−y (mod m). In the last step, we are using the fact that m is prime and x − y = 0, which implies that x − y has a
unique multiplicative inverse modulo m. (For example, the multiplicative inverse of 12 modulo 17 is 10,
since 12 · 10 = 120 ≡ 1 (mod 17).) For each possible value of b, the last identity deﬁnes a unique value
of a such that ha, b causes a collision. Since there are m possible values for b, there are exactly m hash
functions ha, b that cause a collision, which is exactly what we needed to prove.
Thus, if we want to achieve the constant expected time bounds described earlier, we should choose a
random element of H as our hash function, by generating two numbers a and b uniformly at random
between 0 and m − 1. This is precisely the same as choosing a element of U uniformly at random.
One perhaps undesirable ‘feature’ of this construction is that we have a small chance of choosing the
trivial hash function h0,0 , which maps everything to 0. So in practice, if we happen to pick a = b = 0,
we should reject that choice and pick new random numbers. By taking h0,0 out of consideration, we
reduce the probability of a collision from 1/m to (m − 1)/(m2 − 1) = 1/(m + 1). In other words, the set
H \ {h0,0 } is slightly better than universal.
This construction can be easily generalized to larger universes. Suppose u = m r for some constant r ,
so that each element x ∈ U can be represented by a vector ( x 0 , x 1 , . . . , x r −1 ) of integers between 0 and
m − 1. (Think of x as an r -digit number written in base m.) Then for each vector a = (a0 , a1 , . . . , a r −1 ),
deﬁne the corresponding hash function ha as follows:
ha ( x ) = (a0 x 0 + a1 x 1 + · · · + a r −1 x r −1 ) mod m.
Then the set of all m r such functions is universal. 7.4 High Probability Bounds: Balls and Bins Although any particular search in a chained hash tables requires only constant expected time, but what
about the worst search time? Under a stronger version4 of the uniform hashing assumption, this is
equivalent to the following more abstract problem. Suppose we toss n balls independently and uniformly
at random into one of n bins. Can we say anything about the number of balls in the fullest bin?
Lemma 1. If n balls are thrown independently and uniformly into n bins, then with high probability,
the fullest bin contains O(log n/ log log n) balls.
3 ‘Without loss of generality’ is a phrase that appears (perhaps too) often in combinatorial proofs. What it means is that
we are considering one of many possible cases, but once we see the proof for one case, the proofs for all the other cases are
obvious thanks to some inherent symmetry. For this proof, we are not explicitly considering what happens when x = y and
x =y.
4
The simple uniform hashing assumption requires only pairwise independence, but the following analysis requires full
independence. 4 Algorithms Lecture 7: Hash Tables ˆ
Proof: Let X j denote the number of balls in bin j , and let X = max j X j be the maximum number of balls
in any bin. Clearly, E[X j ] = 1 for all j .
n
Now consider the probability that bin j contains exactly k balls. There are k choices for those
k balls; each chosen ball has probability 1/n of landing in bin j ; and each of the remaining balls has
probability 1 − 1/n of missing bin j . Thus,
n
n ≤ n k 1−
k 1 k!
1 < k 1 k Pr[X j = k] = 1− n n−k 1
n n−k 1
n k! This bound shrinks super-exponentially as k increases, so we can very crudely bound the probability that
bin 1 contains at least k balls as follows:
Pr[X j ≥ k] < n
k! To prove the high-probability bound, we need to choose a value for k such that n/k! ≈ 1/nc for some
constant c . Taking logs of both sides of the desired approximation and applying Stirling’s approximation,
we ﬁnd
ln k! ≈ k ln k − k ≈ (c + 1) ln n
⇐⇒ k ≈
≈ (c + 1) ln n
ln k − 1
(c + 1) ln n
ln (c +1) ln n
ln k−1 −1
(c + 1) ln n = ln ln n + ln(c + 1) − ln(ln k − 1) − 1
(c + 1) ln n
.
≈
ln ln n
We have shown (modulo some algebraic hand-waving that is easy but tedious to clean up) that
Pr X j ≥ (c + 1) ln n
ln ln n < 1
nc . This probability bound holds for every bin j . Thus, by the union bound, we conclude that
Pr max X j >
j (c + 1) ln n
ln ln n = Pr X j > (c + 1) ln n
ln ln n n Pr X j > ≤
j =1 < 1
n c −1 5 . for all j (c + 1) ln n
ln ln n Algorithms Lecture 7: Hash Tables A similar analysis shows that if we throw n balls randomly into n bins, then with high probability, at
least one bin contains Ω(log n/ log log n) balls.
However, if we make the hash table large enough, we can expect every ball to land in a different bin.
Suppose there are m bins. Let Ci j be the indicator variable that equals 1 if and only if i = j and ball i
and ball j land in the same bin, and let C = i < j Ci j be the total number of pairwise collisions. Since
n
the balls are thrown uniformly at random, the probability of a collision is exactly 1/m, so E[C ] = 2 /m.
In particular, if m = n2 , the expected number of collisions is less than 1/2.
To get a high probability bound, let X j denote the number of balls in bin j , as in the previous proof.
We can easily bound the probability that bin j is empty, by taking the two most signiﬁcant terms in a
binomial expansion:
Pr[X j = 0] = 1− 1 n n m −1 n =
i =1 i i = 1− m n
m +Θ n2 > 1− m2 n
m We can similarly bound the probability that bin j contains exactly one ball:
Pr[X j = 1] = n · 1
m 1− 1
m n−1 = n
m 1− n−1
m +Θ n2
m2 > n
m − n( n − 1)
m2 ˆ
It follows immediately that Pr[X j > 1] < n(n − 1)/m2 . The union bound now implies that Pr[X > 1] <
2+
n(n − 1)/m. If we set m = n
for any constant > 0, then the probability that no bin contains more
than one ball is at least 1 − 1/n .
Lemma 2. For any > 0, if n balls are thrown independently and uniformly into n2+ bins, then with
high probability, no bin contains more than one ball. 7.5 Perfect Hashing So far we are faced with two alternatives. If we use a small hash table, to keep the space usage down,
the resulting worst-case expected search time is Θ(log n/ log log n) with high probability, which is not
much better than a binary search tree. On the other hand, we can get constant worst-case search time,
at least in expectation, by using a table of quadratic size, but that seems unduly wasteful.
Fortunately, there is a fairly simple way to combine these two ideas to get a data structure of linear
expected size, whose expected worst-case search time is constant. At the top level, we use a hash table
of size n, but instead of linked lists, we use secondary hash tables to resolve collisions. Speciﬁcally, the
j th secondary hash table has size n2 , where n j is the number of items whose primary hash value is j .
j
The expected worst-case search time in any secondary hash table is O(1), by our earlier analysis.
Although this data structure needs signiﬁcantly more memory for each secondary structure, the
overall increase in space is insigniﬁcant, at least in expectation.
Lemma 3. Under the simple uniform hashing assumption, E[n2 ] < 2.
j
Proof: Let X i j be the indicator variable that equals 1 if item i hashes to slot j in the primary hash table. 6 Algorithms Lecture 7: Hash Tables n E[n2 ] = E j
n n n Xkj = E Xi j
i =1 n Xi jXk j
i =1 k =1 k =1 n = E[X i j X k j ]
i =1 k =1
n =
i =1 [linearity of expectation] n E[X i2j ] + 2 n E[X i j X k j ]
i =1 k = i +1 Because X i j is an indicator variable, we have X i2j = X i j , which implies that E[X i2j ] = E[X i j ] = 1/n by
the uniform hashing assumption. The uniform hashing assumption also implies that X i j and X k j are
independent whenever i = k, so E[X i j X k j ] = E[X i j ] E[X k j ] = 1/n2 . Thus,
n E[n2 ] =
j i =1 1
n n n +2
i = 1 k = i +1 1
n2 =1+2 1
=2− .
2 n2
n
n 1 This lemma implies that the expected size of our two-level hash table is O(n). By our earlier analysis,
the expected worst-case search time is O(1). 7.6 Open Addressing Another method used to resolve collisions in hash tables is called open addressing. Here, rather than
building secondary data structures, we resolve collisions by looking elsewhere in the table. Speciﬁcally,
we have a sequence of hash functions 〈h0 , h1 , h2 , . . . , hm−1 〉, such that for any item x , the probe sequence
h0 ( x ), h1 ( x ), . . . , hm−1 ( x ) is a permutation of 〈0, 1, 2, . . . , m − 1〉. In other words, different hash
functions in the sequence always map x to different locations in the hash table.
We search for x using the following algorithm, which returns the array index i if T [i ] = x , ‘absent’ if
x is not in the table but there is an empty slot, and ‘full’ if x is not in the table and there no no empty
slots.
OPENADDRESSSEARCH( x ):
for i ← 0 to m − 1
if T [hi ( x )] = x
return hi ( x )
else if T [hi ( x )] = ∅
return ‘absent’
return ‘full’ The algorithm for inserting a new item into the table is similar; only the second-to-last line is changed to
T [hi ( x )] ← x . Notice that for an open-addressed hash table, the load factor is never bigger than 1.
Just as with chaining, we’d like to pretend that the sequence of hash values is random. For purposes
of analysis, there is a stronger uniform hashing assumption that gives us constant expected search and
insertion time.
Strong uniform hashing assumption:
For any item x , the probe sequence 〈h0 ( x ), h1 ( x ), . . . , hm−1 ( x )〉 is equally
likely to be any permutation of the set {0, 1, 2, . . . , m − 1}.
7 Algorithms Lecture 7: Hash Tables Let’s compute the expected time for an unsuccessful search using this stronger assumption. Suppose
there are currently n elements in the hash table. Our strong uniform hashing assumption has two
important consequences:
• The initial hash value h0 ( x ) is equally likely to be any integer in the set {0, 1, 2, . . . , m − 1}.
• If we ignore the ﬁrst probe, the remaining probe sequence 〈h1 ( x ), h2 ( x ), . . . , hm−1 ( x )〉 is equally
likely to be any permutation of the smaller set {0, 1, 2, . . . , m − 1} \ {h0 ( x )}.
The ﬁrst sentence implies that the probability that T [h0 ( x )] is occupied is exactly n/m. The second
sentence implies that if T [h0 ( x )] is occupied, our search algorithm recursively searches the rest of the hash
table! Since the algorithm will never again probe T [h0 ( x )], for purposes of analysis, we might as well
pretend that slot in the table no longer exists. Thus, we get the following recurrence for the expected
number of probes, as a function of m and n:
E[ T (m, n)] = 1 + n
m E[ T (m − 1, n − 1)]. The trivial base case is T (m, 0) = 1; if there’s nothing in the hash table, the ﬁrst probe always hits an
empty slot. We can now easily prove by induction that
EMPHE[ T (m, n)] ≤ m/(m − n):
E[ T (m, n)] = 1 + n E[ T (m − 1, n − 1)]
m
n m−1
≤1+ ·
[induction hypothesis]
m m−n
n
m
<1+ ·
[m − 1 < m]
m m−n
m
[algebra]
=
m−n Rewriting this in terms of the load factor α = n/m, we get E[ T (m , n )] ≤ 1/(1 − α). In other words, the
expected time for an unsuccessful search is O(1), unless the hash table is almost completely full.
In practice, however, we can’t generate truly random probe sequences, so we use one of the following
heuristics:
• Linear probing: We use a single hash function h( x ), and deﬁne hi ( x ) = (h( x ) + i ) mod m. This is
nice and simple, but collisions tend to make items in the table clump together badly, so this is not
really a good idea.
• Quadratic probing: We use a single hash function h( x ), and deﬁne hi ( x ) = (h( x ) + i 2 ) mod m.
Unfortunately, for certain values of m, the sequence of hash values 〈hi ( x )〉 does not hit every
possible slot in the table; we can avoid this problem by making m a prime number. (That’s often a
good idea anyway.) Although quadratic probing does not suffer from the same clumping problems
as linear probing, it does have a weaker clustering problem: If two items have the same initial
hash value, their entire probe sequences will be the same.
• Double hashing: We use two hash functions h( x ) and h ( x ), and deﬁne hi as follows:
hi ( x ) = (h( x ) + i · h ( x )) mod m
To guarantee that this can hit every slot in the table, the stride function h ( x ) and the table size m
must be relatively prime. We can guarantee this by making m prime, but a simpler solution is to
8 Algorithms Lecture 7: Hash Tables make m a power of 2 and choose a stride function that is always odd. Double hashing avoids the
clustering problems of linear and quadratic probing. In fact, the actual performance of double
hashing is almost the same as predicted by the uniform hashing assumption, at least when m is
large and the component hash functions h and h are sufﬁciently random. This is the method of
choice!5 7.7 Deleting from an Open-Addressed Hash Table
This section assumes familiarity with amortized analysis. Deleting an item x from an open-addressed hash table is a bit more difﬁcult than in a chained hash
table. We can’t simply clear out the slot in the table, because we may need to know that T [h( x )] is
occupied in order to ﬁnd some other item!
Instead, we should delete more or less the way we did with scapegoat trees. When we delete an
item, we mark the slot that used to contain it as a wasted slot. A sufﬁciently long sequence of insertions
and deletions could eventually ﬁll the table with marks, leaving little room for any real data and causing
searches to take linear time.
However, we can still get good amortized performance by using two rebuilding rules. First, if the
number of items in the hash table exceeds m/4, double the size of the table (m ← 2m) and rehash
everything. Second, if the number of wasted slots exceeds m/2, clear all the marks and rehash everything
in the table. Rehashing everything takes m steps to create the new hash table and O(n) expected steps
to hash each of the n items. By charging a $4 tax for each insertion and a $2 tax for each deletion, we
expect to have enough money to pay for any rebuilding.
In conclusion, the expected amortized cost of any insertion or deletion is O(1), under the uniform
hashing assumption. Notice that we’re doing two very different kinds of averaging here. On the one
hand, we are averaging the possible costs of each individual search over all possible probe sequences
(‘expected’). On the other hand, we are also averaging the costs of the entire sequence of operations to
‘smooth out’ the cost of rebuilding (‘amortized’). Both randomization and amortization are necessary to
get this constant time bound. Exercises
1. Suppose we are using an open-addressed hash table of size m to store n items, where n ≤ m/2.
Assume that the strong uniform hashing assumption holds. For any i , let X i denote the number of
probes required for the i th insertion into the table, and let X = maxi X i denote the length of the
longest probe sequence.
(a) Prove that Pr[X i > k] ≤ 1/2k for all i and k.
(b) Prove that Pr[X i > 2 lg n] ≤ 1/n2 for all i .
(c) Prove that Pr[X > 2 lg n] ≤ 1/n.
(d) Prove that E[X ] = O(lg n). 5
...unless your hash tables are really huge, in which case linear probing has far better cache behavior, especially when the
load factor is small. 9 Algorithms Lecture 7: Hash Tables 2. Your boss wants you to ﬁnd a perfect hash function for mapping a known set of n items into a
table of size m. A hash function is perfect if there are no collisions; each of the n items is mapped
to a different slot in the hash table. Of course, this requires that m ≥ n. (This is a different
deﬁnition of perfect hashing than the one considered in the lecture notes.) After cursing your
algorithms instructor for not teaching you about perfect hashing, you decide to try something
simple: repeatedly pick random hash functions until you ﬁnd one that happens to be perfect. A
random hash function h satisﬁes two properties:
• Pr[h( x ) = h( y )] = 1/m for any pair of items x = y .
• Pr[h( x ) = i ] = 1/m for any item x and any integer 1 ≤ i ≤ m.
(a) Suppose you pick a random hash function h. What is the exact expected number of collisions,
as a function of n (the number of items) and m (the size of the table)? Don’t worry about
how to resolve collisions; just count them.
(b) What is the exact probability that a random hash function is perfect?
(c) What is the exact expected number of different random hash functions you have to test before
you ﬁnd a perfect hash function?
(d) What is the exact probability that none of the ﬁrst N random hash functions you try is perfect?
(e) How many random hash functions do you have to test to ﬁnd a perfect hash function with
high probability?
3. Recall that Fk denotes the kth Fibonacci number: F0 = 0, F1 = 1, and Fk = Fk−1 + Fk−2 for all
k ≥ 2. Suppose we are building a hash table of size m = Fk using the hash function
h( x ) = ( Fk−1 · x ) mod Fk
Prove that if the consecutive integers 0, 1, 2, . . . , Fk − 1 are inserted in order into an initially empty
table, each integer will be hashed into the largest contiguous empty interval in the table. In
particular, show that there are no collisions. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 10 Algorithms Non-Lecture F: Randomized Minimum Cut
Jaques: But, for the seventh cause; how did you ﬁnd the quarrel on the seventh cause?
Touchstone: Upon a lie seven times removed:–bear your body more seeming, Audrey:–as
thus, sir. I did dislike the cut of a certain courtier’s beard: he sent me word, if I said his
beard was not cut well, he was in the mind it was: this is called the Retort Courteous. If
I sent him word again ‘it was not well cut,’ he would send me word, he cut it to please
himself: this is called the Quip Modest. If again ‘it was not well cut,’ he disabled my
judgment: this is called the Reply Churlish. If again ‘it was not well cut,’ he would
answer, I spake not true: this is called the Reproof Valiant. If again ‘it was not well cut,’
he would say I lied: this is called the Counter-cheque Quarrelsome: and so to the Lie
Circumstantial and the Lie Direct.
Jaques: And how oft did you say his beard was not well cut?
Touchstone: I durst go no further than the Lie Circumstantial, nor he durst not give me the
Lie Direct; and so we measured swords and parted.
— William Shakespeare, As You Like It, Act V, Scene 4 (1600) F
F.1 Randomized Minimum Cut
Setting Up the Problem This lecture considers a problem that arises in robust network design. Suppose we have a connected
multigraph1 G representing a communications network like the UIUC telephone system, the internet, or
Al-Qaeda. In order to disrupt the network, an enemy agent plans to remove some of the edges in this
multigraph (by cutting wires, placing police at strategic drop-off points, or paying street urchins to ‘lose’
messages) to separate it into multiple components. Since his country is currently having an economic
crisis, the agent wants to remove as few edges as possible to accomplish this task.
More formally, a cut partitions the nodes of G into two nonempty subsets. The size of the cut is the
number of crossing edges, which have one endpoint in each subset. Finally, a minimum cut in G is a cut
with the smallest number of crossing edges. The same graph may have several minimum cuts. a b
d c e g f h A multigraph whose minimum cut has three edges. This problem has a long history. The classical deterministic algorithms for this problem rely on
network ﬂow techniques, which are discussed in another lecture. The fastest such algorithms (that we
will discuss) run in O(n3 ) time and are quite complex and difﬁcult to understand (unless you’re already
familiar with network ﬂows). Here I’ll describe a relatively simple randomized algorithm discovered by
David Karger when he was a Ph.D. student.2
Karger’s algorithm uses a primitive operation called collapsing an edge. Suppose u and v are vertices
that are connected by an edge in some multigraph G . To collapse the edge {u, v }, we create a new node
called uv , replace any edge of the form {u, w } or { v, w } with a new edge {uv, w }, and then delete the
original vertices u and v . Equivalently, collapsing the edge shrinks the edge down to nothing, pulling the
1
A multigraph allows multiple edges between the same pair of nodes. Everything in this lecture could be rephrased in terms
of simple graphs where every edge has a non-negative weight, but this would make the algorithms and analysis slightly more
complicated.
2
David R. Karger*. Random sampling in cut, ﬂow, and network design problems. Proc. 25th STOC, 648–657, 1994. 1 Algorithms Non-Lecture F: Randomized Minimum Cut two endpoints together. The new collapsed graph is denoted G /{u, v }. We don’t allow self-loops in our
multigraphs; if there are multiple edges between u and v , collapsing any one of them deletes them all.
a a
c d a
c b be d cd b e e
A graph G and two collapsed graphs G /{ b, e} and G /{c , d }. I won’t describe how to actually implement collapsing an edge—it will be a homework exercise later
in the course—but it can be done in O(n) time. Let’s just accept collapsing as a black box subroutine for
now.
The correctness of our algorithms will eventually boil down the following simple observation: For
any cut in G /{u, v }, there is cut in G with exactly the same number of crossing edges. In fact, in some
sense, the ‘same’ edges form the cut in both graphs. The converse is not necessarily true, however. For
example, in the picture above, the original graph G has a cut of size 1, but the collapsed graph G /{c , d }
does not.
This simple observation has two immediate but important consequences. First, collapsing an edge
cannot decrease the minimum cut size. More importantly, collapsing an edge increases the minimum cut
size if and only if that edge is part of every minimum cut. F.2 Blindly Guessing Let’s start with an algorithm that tries to guess the minimum cut by randomly collapsing edges until the
graph has only two vertices left.
GUESSMINCUT(G ):
for i ← n downto 2
pick a random edge e in G
G ← G /e
return the only cut in G
Since each collapse takes O(n) time, this algorithm runs in O(n2 ) time. Our earlier observations
imply that as long as we never collapse an edge that lies in every minimum cut, our algorithm will
actually guess correctly. But how likely is that?
Suppose G has only one minimum cut—if it actually has more than one, just pick your favorite—and
this cut has size k. Every vertex of G must lie on at least k edges; otherwise, we could separate that
vertex from the rest of the graph with an even smaller cut. Thus, the number of incident vertex-edge
pairs is at least kn. Since every edge is incident to exactly two vertices, G must have at least kn/2 edges.
That implies that if we pick an edge in G uniformly at random, the probability of picking an edge in the
minimum cut is at most 2/n. In other words, the probability that we don’t screw up on the very ﬁrst step
is at least 1 − 2/n.
Once we’ve collapsed the ﬁrst random edge, the rest of the algorithm proceeds recursively (with
independent random choices) on the remaining (n − 1)-node graph. So the overall probability P (n) that
GUESSMINCUT returns the true minimum cut is given by the following recurrence:
P ( n) ≥ n−2
n
2 · P (n − 1). Algorithms Non-Lecture F: Randomized Minimum Cut The base case for this recurrence is P (2) = 1. We can immediately expand this recurrence into a product,
most of whose factors cancel out immediately.
n
n P ( n) ≥
i =3 i−2
i = n −2 ( i − 2) i =3 i
i =1
n = n i i i =3 F.3 = 2
n(n − 1) i =3 Blindly Guessing Over and Over That’s not very good. Fortunately, there’s a simple method for increasing our chances of ﬁnding the
minimum cut: run the guessing algorithm many times and return the smallest guess. Randomized
algorithms folks like to call this idea ampliﬁcation.
KARGERMINCUT(G ):
mink ← ∞
for i ← 1 to N
X ← GUESSMINCUT(G )
if |X | < mink
mink ← |X |
minX ← X
return minX
Both the running time and the probability of success will depend on the number of iterations N , which
we haven’t speciﬁed yet.
First let’s ﬁgure out the probability that KARGERMINCUT returns the actual minimum cut. The only
way for the algorithm to return the wrong answer is if GUESSMINCUT fails N times in a row. Since each
guess is independent, our probability of success is at least
1− 1− N 2 . n( n − 1) We can simplify this using one of the most important (and easiest) inequalities known to mankind:
1 − x ≤ e− x
So our success probability is at least 1 − e−2N /n(n−1) . By making N larger, we can make this probability arbitrarily close to 1, but never equal to 1. In particular,
n
if we set N = c 2 ln n for some constant c , then KARGERMINCUT is correct with probability at least
1 − e−c ln n = 1 − 1
nc . When the failure probability is a polynomial fraction, we say that the algorithm is correct with high
probability. Thus, KARGERMINCUT computes the minimum cut of any n-node graph in O (n 4 log n ) time.
If we make the number of iterations even larger, say N = n2 (n − 1)/2, the success probability
becomes 1 − e−n . When the failure probability is exponentially small like this, we say that the algorithm
is correct with very high probability. In practice, very high probability is usually overkill; high probability
is enough. (Remember, there is a small but non-zero probability that your computer will transform itself
into a kitten before your program is ﬁnished.)
3 Algorithms F.4 Non-Lecture F: Randomized Minimum Cut Not-So-Blindly Guessing The O(n4 log n) running time is actually comparable to some of the simpler ﬂow-based algorithms, but
it’s nothing to get excited about. But we can improve our guessing algorithm, and thus decrease the
number of iterations in the outer loop, by observing that as the graph shrinks, the probability of collapsing
an edge in the minimum cut increases. At ﬁrst the probability is quite small, only 2/n, but near the end of
execution, when the graph has only three vertices, we have a 2/3 chance of screwing up!
A simple technique for working around this increasing probability of error was developed by David
Karger and Cliff Stein.3 Their idea is to group the ﬁrst several random collapses a ‘safe’ phase, so that
the cumulative probability of screwing up is small—less than 1/2, say—and a ‘dangerous’ phase, which
is much more likely to screw up.
The safe phase shrinks the graph from n nodes to n/ 2 + 1 nodes, using a sequence of n − n/ 2 − 1
random collapses. Following our earlier analysis, the probability that none of these safe collapses touches
the minimum cut is at least
n
i =n/ 2+2 i−2
i = (n/ 2)(n/ 2 + 1)
n( n − 1) = n+ 2 2(n − 1) > 1
2 . Now, to get around the danger of the dangerous phase, we use ampliﬁcation. However, instead of
running through the dangerous phase once, we run it twice and keep the best of the two answers.
Naturally, we treat the dangerous phase recursively, so we actually obtain a binary recursion tree, which
expands as we get closer to the base case, instead of a single path. More formally, the algorithm looks
like this:
CONTRACT(G , m):
for i ← n downto m
pick a random edge e in G
G ← G /e
return G BETTERGUESS(G ):
if G has more than 8 vertices
X 1 ← BETTERGUESS(CONTRACT(G , n/ 2 + 1))
X 2 ← BETTERGUESS(CONTRACT(G , n/ 2 + 1))
return min{X 1 , X 2 }
else
use brute force This might look like we’re just doing to same thing twice, but remember that CONTRACT (and thus
BETTERGUESS) is randomized. Each call to CONTRACT contracts an independent random set of edges; X 1
and X 2 are almost always different cuts.
BETTERGUESS correctly returns the minimum cut unless both recursive calls return the wrong result.
X 1 is the minimum cut of G if and only if (1) none of the min cut edges are CONTRACTed and (2)
the recursive BETTERGUESS returns the minimum cut of the CONTRACTed graph. If P (n) denotes the
probability that BETTERGUESS returns a minimum cut of an n-node graph, then X 1 is the minimum cut
with probability at least 1/2 · P (n/ 2), and X 2 is the minimum cut with the same probability. Since
these two events are independent, we have the following recurrence, with base case P (n) = 1 for all
n ≤ 6.
2
1
n
P ( n) ≥ 1 − 1 − P
+1
2
2
Using a series of transformations, Karger and Stein prove that P (n) = Ω(1/ log n). I’ve included the
proof at the end of this note.
3 ˜
David R. Karger∗ and Cliff Stein. An O(n2 ) algorithm for minimum cuts. Proc. 25th STOC, 757–765, 1993. 4 Algorithms Non-Lecture F: Randomized Minimum Cut For the running time, we get a simple recurrence that is easily solved using recursion trees or the
Master theorem (after a domain transformation to remove the +1 from the recurrence).
T ( n) = O ( n2 ) + 2 T n
2 +1 = O(n2 log n) So all this splitting and recursing has slowed down the guessing algorithm slightly, but the probability of
failure is exponentially smaller!
Let’s express the lower bound P (n) = Ω(1/ log n) explicitly as P (n) ≥ α/ ln n for some constant α.
(Karger and Klein’s proof implies α > 2). If we call BETTERGUESS N = c ln2 n times, for some new
constant c , the overall probability of success is at least
1− 1− α
ln n c ln2 n ≥ 1 − e−(c /α) ln n = 1 − 1
nc /α . By setting c sufﬁciently large, we can bound the probability of failure by an arbitrarily small polynomial
function of n. In other words, we now have an algorithm that computes the minimum cut with high
probability in only O (n 2 log3 n ) time! F.5 Solving the Karger/Stein recurrence Recall the following recurrence for the probability that BETTERGUESS successfully ﬁnds a minimum cut of
an n-node graph:
2
n
1
P ( n) ≥ 1 − 1 − P
+1
2
2
Karger and Stein solve this rather ugly recurrence through a series of functional transformations. Let
p(k) denote the probability of success at the kth level of recursion, counting upward from the base case.
This function satisﬁes the recurrence
p(k) ≥ 1 − 1 − p ( k − 1) 2 2 with base case p(0) = 1. Let ¯ (k) be the function that satisﬁes this recurrence with equality; clearly,
p
p
p(k) ≥ ¯ (k). Now consider the function z (k) = 4/ ¯ (k) − 1. Substituting ¯ (k) = 4/(z (k) + 1) into our
p
p
old recurrence implies (after a bit of algebra) that
z ( k ) = z ( k − 1) + 2 + 1
z ( k − 1) . Since clearly z (k) > 1 for all k, we have a conservative upper bound
z (k) < z (k − 1) + 2,
which implies inductively that z (k) ≤ 2k + 3, since z (0) = 3. It follows that
p(k) ≥ ¯ (k) >
p 1
2k + 4 = Ω(1/k). To compute the number of levels of recursion that BETTERGUESS executes for an n-node graph, we
solve the secondary recurrence
n
k ( n) = 1 + k
+1
2
with base cases k(n) = 0 for all n ≤ 8. After a domain transformation to remove the +1 from the right
side, the recursion tree method (or the Master theorem) implies that k(n) = Θ(log n).
We conclude that P (n ) = p (k (n )) = Ω(1/ log n ), as promised.
5 Algorithms Non-Lecture F: Randomized Minimum Cut Exercises
1. Suppose you had an algorithm to compute the minimum spanning tree of a graph in O(m) time,
where m is the number of edges in the input graph.4 Use this algorithm as a subroutine to improve
the running time of GUESSMINCUT from O(n2 ) to O(m).
2. Suppose you are given a graph G with weighted edges, and your goal is to ﬁnd a cut whose total
weight (not just number of edges) is smallest.
(a) Describe an algorithm to select a random edge of G , where the probability of choosing edge e
is proportional to the weight of e.
(b) Prove that if you use the algorithm from part (a), instead of choosing edges uniformly at
random, the probability that GUESSMINCUT returns a minimum-weight cut is still Ω(1/n2 ).
(c) What is the running time of your modiﬁed GUESSMINCUT algorithm?
3. Prove that GUESSMINCUT returns the second smallest cut in its input graph with probability Ω(1/n3 ).
(The second smallest cut could be signiﬁcantly larger than the minimum cut.) 4
In fact, there is a randomized MST algorithm (due to Philip Klein, David Karger, and Robert Tarjan) whose expected
running time is O(m). © Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 6 Algorithms Lecture 8: Amortized Analysis
The goode workes that men don whil they ben in good lif al amortised by
synne folwyng.
— Geoffrey Chaucer, “The Persones [Parson’s] Tale” (c.1400)
I will gladly pay you Tuesday for a hamburger today.
— J. Wellington Wimpy, “Thimble Theatre” (1931)
I want my two dollars!
— Johnny Gasparini [Demian Slade], “Better Off Dead” (1985) 8
8.1 Amortized Analysis
Incrementing a Binary Counter It is a straightforward exercise in induction, which often appears on Homework 0, to prove that any
non-negative integer n can be represented as the sum of distinct powers of 2. Although some students
correctly use induction on the number of bits—pulling off either the least signiﬁcant bit or the most
signiﬁcant bit in the binary representation and letting the Recursion Fairy convert the remainder—the
most commonly submitted proof uses induction on the value of the integer, as follows:
Proof: The base case n = 0 is trivial. For any n > 0, the inductive hypothesis implies that there is set of
distinct powers of 2 whose sum is n − 1. If we add 20 to this set, we obtain a multiset of powers of two
whose sum is n, which might contain two copies of 20 . Then as long as there are two copies of any 2i
in the multiset, we remove them both and insert 2i +1 in their place. The sum of the elements of the
multiset is unchanged by this replacement, because 2i +1 = 2i + 2i . Each iteration decreases the size of
the multiset by 1, so the replacement process must eventually terminate. When it does terminate, we
have a set of distinct powers of 2 whose sum is n.
This proof is describing an algorithm to increment a binary counter from n − 1 to n. Here’s a more
formal (and shorter!) description of the algorithm to add 1 to a binary counter. The input B is an
(inﬁnite) array of bits, where B [i ] = 1 if and only if 2i appears in the sum.
INCREMENT(B [0 .. ∞]):
i←0
while B [i ] = 1
B [i ] ← 0
i ← i+1
B [i ] ← 1 We’ve already argued that INCREMENT must terminate, but how quickly? Obviously, the running time
depends on the array of bits passed as input. If the ﬁrst k bits are all 1s, then INCREMENT takes Θ(k) time.
The binary representation of any positive integer n is exactly lg n + 1 bits long. Thus, if B represents
an integer between 0 and n, INCREMENT takes Θ(log n) time in the worst case. 8.2 Counting from 0 to n Now suppose we want to use INCREMENT to count from 0 to n. If we only use the worst-case running
time for each INCREMENT, we get an upper bound of O(n log n) on the total running time. Although this
bound is correct, we can do better. There are several general methods for proving that the total running
time is actually only O(n). Many of these methods are logically equivalent, but different formulations
are more natural for different problems.
1 Algorithms
8.2.1 Lecture 8: Amortized Analysis The Summation (Aggregate) Method The easiest way to get a tighter bound is to observe that we don’t need to ﬂip Θ(log n) bits every time
INCREMENT is called. The least signiﬁcant bit B [0] does ﬂip every time, but B [1] only ﬂips every other
time, B [2] ﬂips every 4th time, and in general, B [i ] ﬂips every 2i th time. If we start with an array full
of 0’s, a sequence of n INCREMENTs ﬂips each bit B [i ] exactly n/2i times. Thus, the total number of
bit-ﬂips for the entire sequence is
lg n
i =0 n
2i ∞ <
i =0 n
2i = 2 n. Thus, on average, each call to INCREMENT ﬂips only two bits, and so runs in constant time.
This ‘on average’ is quite different from the averaging we consider with randomized algorithms.
There is no probability involved; we are averaging over a sequence of operations, not the possible
running times of a single operation. This averaging idea is called amortization—the amortized time
for each INCREMENT is O(1). Amortization is a sleazy clever trick used by accountants to average large
one-time costs over long periods of time; the most common example is calculating uniform payments for
a loan, even though the borrower is paying interest on less and less capital over time. For this reason,
‘amortized cost’ is a common synonym for amortized running time.
Most textbooks call this technique for bounding amortized costs the aggregate method, or aggregate
analysis, but this is just a fancy name for adding up the costs of the individual operations and dividing
by the number of operations.
The Summation Method. Let T (n) be the worst-case running time for a sequence of n
operations. The amortized time for each operation is T (n)/n.
8.2.2 The Taxation (Accounting) Method A second method we can use to derive amortized bounds is called either the accounting method or the
taxation method. Suppose it costs us a dollar to toggle a bit, so we can measure the running time of our
algorithm in dollars. Time is money!
Instead of paying for each bit ﬂip when it happens, the Increment Revenue Service charges a twodollar increment tax whenever we want to set a bit from zero to one. One of those dollars is spent
changing the bit from zero to one; the other is stored away as credit until we need to reset the same bit
to zero. The key point here is that we always have enough credit saved up to pay for the next INCREMENT.
The amortized cost of an INCREMENT is the total tax it incurs, which is exactly 2 dollars, since each
INCREMENT changes just one bit from 0 to 1.
It is often useful to distribute the tax income to speciﬁc pieces of the data structure. For example, for
each INCREMENT, we could store one of the two dollars on the single bit that is set for 0 to 1, so that that
bit can pay to reset itself back to zero later on.
Taxation Method 1. Certain steps in the algorithm charge you taxes, so that the total cost
incurred by the algorithm is never more than the total tax you pay. The amortized cost of an
operation is the overall tax charged to you during that operation.
A different way to schedule the taxes is for every bit to charge us a tax at every operation, regardless
of whether the bit changes of not. Speciﬁcally, each bit B [i ] charges a tax of 1/2i dollars for each
INCREMENT. The total tax we are charged during each INCREMENT is i ≥0 2−i = 2 dollars. Every time a
bit B [i ] actually needs to be ﬂipped, it has collected exactly $1, which is just enough for us to pay for
the ﬂip.
2 Algorithms Lecture 8: Amortized Analysis Taxation Method 2. Certain portions of the data structure charge you taxes at each operation, so that the total cost of maintaining the data structure is never more than the total
taxes you pay. The amortized cost of an operation is the overall tax you pay during that
operation.
In both of the taxation methods, our task as algorithm analysts is to come up with an appropriate
‘tax schedule’. Different ‘schedules’ can result in different amortized time bounds. The tightest bounds
are obtained from tax schedules that just barely stay in the black.
8.2.3 The Charging Method Another common method of amortized analysis involves charging the cost of some steps to some other,
earlier steps. The method is similar to taxation, except that we focus on where each unit of tax is (or will
be) spent, rather than where is it collected. By charging the cost of some operations to earlier operations,
we are overestimating the total cost of any sequence of operations, since we pay for some charges from
future operations that may never actually occur.
For example, in our binary counter, suppose we charge the cost of clearing a bit (changing its value
from 1 to 0) to the previous operation that sets that bit (changing its value from 0 to 1). If we ﬂip k
bits during an INCREMENT, we charge k − 1 of those bit-ﬂips to earlier bit-ﬂips. Conversely, the single
operation that sets a bit receives at most one unit of charge from the next time that bit is cleared. So
instead of paying for k bit-ﬂips, we pay for at most two: one for actually setting a bit, plus at most one
charge from the future for clearing that same bit. Thus, the total amortized cost of the INCREMENT is at
most two bit-ﬂips.
The Charging Method. Charge the cost of some steps of the algorithm to earlier steps, or
to steps in some earlier operation. The amortized cost of the algorithm is its actual running
time, minus its total charges to past operations, plus its total charge from future operations.
8.2.4 The Potential Method The most powerful method (and the hardest to use) builds on a physics metaphor of ‘potential energy’.
Instead of associating costs or taxes with particular operations or pieces of the data structure, we
represent prepaid work as potential that can be spent on later operations. The potential is a function of
the entire data structure.
Let Di denote our data structure after i operations, and let Φi denote its potential. Let ci denote
the actual cost of the i th operation (which changes Di −1 into Di ). Then the amortized cost of the i th
operation, denoted ai , is deﬁned to be the actual cost plus the increase in potential:
a i = c i + Φ i − Φ i −1
So the total amortized cost of n operations is the actual total cost plus the total increase in potential:
n n n ai =
i =1 c i + Φ i − Φ i −1 =
i =1 c i + Φ n − Φ0 .
i =1 A potential function is valid if Φi − Φ0 ≥ 0 for all i . If the potential function is valid, then the total actual
cost of any sequence of operations is always less than the total amortized cost:
n n ci =
i =1 n ai − Φn ≤
i =1 ai .
i =1 3 Algorithms Lecture 8: Amortized Analysis For our binary counter example, we can deﬁne the potential Φi after the i th INCREMENT to be the
number of bits with value 1. Initially, all bits are equal to zero, so Φ0 = 0, and clearly Φi > 0 for all
i > 0, so this is a valid potential function. We can describe both the actual cost of an INCREMENT and the
change in potential in terms of the number of bits set to 1 and reset to 0.
ci = #bits changed from 0 to 1 + #bits changed from 1 to 0
Φi − Φi −1 = #bits changed from 0 to 1 − #bits changed from 1 to 0
Thus, the amortized cost of the i th INCREMENT is
ai = ci + Φi − Φi −1 = 2 × #bits changed from 0 to 1
Since INCREMENT changes only one bit from 0 to 1, the amortized cost INCREMENT is 2.
The Potential Method. Deﬁne a potential function for the data structure that is initially
equal to zero and is always nonnegative. The amortized cost of an operation is its actual
cost plus the change in potential.
For this particular example, the potential is precisely the total unspent taxes paid using the taxation
method, so not too surprisingly, we obtain precisely the same amortized cost. In general, however, there
may be no way of interpreting the change in potential as ‘taxes’. Taxation and charging are useful when
there is a convenient way to distribute costs to speciﬁc steps in the algorithm or components of the data
structure; potential arguments allow us to argue more globally when a local distribution is difﬁcult or
impossible.
Different potential functions can lead to different amortized time bounds. The trick to using the
potential method is to come up with the best possible potential function. A good potential function goes
up a little during any cheap/fast operation, and goes down a lot during any expensive/slow operation.
Unfortunately, there is no general technique for doing this other than playing around with the data
structure and trying lots of different possibilities. 8.3 Incrementing and Decrementing Now suppose we wanted a binary counter that we can both increment and decrement efﬁciently. A
standard binary counter won’t work, even in an amortized sense; if we alternate between 2k and 2k − 1,
every operation costs Θ(k) time.
A nice alternative is represent a number as a pair of bit strings ( P, N ), where for any bit position i ,
at most one of the bits P [i ] and N [i ] is equal to 1. The actual value of the counter is P − N . Here are
algorithms to increment and decrement our double binary counter.
INCREMENT( P, N ):
i←0
while P [i ] = 1
P [i ] ← 0
i ← i+1 DECREMENT( P, N ):
i←0
while N [i ] = 1
N [i ] ← 0
i ← i+1 if N [i ] = 1
N [i ] ← 0
else
P [i ] ← 1 if P [i ] = 1
P [i ] ← 0
else
N [i ] ← 1 Here’s an example of these algorithms in action. Notice that any number other than zero can be
represented in multiple (in fact, inﬁnitely many) ways.
4 Algorithms Lecture 8: Amortized Analysis P = 10001
P = 10000
P = 10000
P = 10000
P = 10011
P = 10010
P = 10001
++
−−
−−
++
++
++
N = 01100 −→ N = 01100 −→ N = 01100 −→ N = 01000 −→ N = 01001 −→ N = 01010 −→ N = 01010
P−N =7
P−N =6
P−N =7
P−N =8
P−N =7
P−N =6
P−N =5
Incrementing and decrementing a double-binary counter. Now suppose we start from (0, 0) and apply a sequence of n INCREMENTs and DECREMENTs. In the
worst case, operation takes Θ(log n) time, but what is the amortized cost? We can’t use the aggregate
method here, since we don’t know what the sequence of operations looks like.
What about the taxation method? It’s not hard to prove (by induction, of course) that after either
P [i ] or N [i ] is set to 1, there must be at least 2i operations, either INCREMENTs or DECREMENTs, before
that bit is reset to 0. So if each bit P [i ] and N [i ] pays a tax of 2−i at each operation, we will always
have enough money to pay for the next operation. Thus, the amortized cost of each operation is at most
−i
= 4.
i ≥0 2 · 2
We can get even better bounds using the potential method. Deﬁne the potential Φi to be the number
of 1-bits in both P and N after i operations. Just as before, we have
ci = #bits changed from 0 to 1 + #bits changed from 1 to 0
Φi − Φi −1 = #bits changed from 0 to 1 − #bits changed from 1 to 0
=⇒ ai = 2 × #bits changed from 0 to 1 Since each operation changes at most one bit to 1, the i th operation has amortized cost ai ≤ 2. 8.4 Gray Codes An attractive alternate solution to the increment/decrement problem was independently suggested by
several students. Gray codes (named after Frank Gray, who discovered them in the 1950s) are methods
for representing numbers as bit strings so that successive numbers differ by only one bit. For example,
here is the four-bit binary reﬂected Gray code for the integers 0 through 15:
0000, 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111, 1110, 1010, 1011, 1001, 1000
The general rule for incrementing a binary reﬂected Gray code is to invert the bit that would be set from
0 to 1 by a normal binary counter. In terms of bit-ﬂips, this is the perfect solution; each increment of
decrement by deﬁnition changes only one bit. Unfortunately, the naïve algorithm to ﬁnd the single bit
to ﬂip still requires Θ(log n) time in the worst case. Thus, so the total cost of maintaining a Gray code,
using the obvious algorithm, is the same as that of maintaining a normal binary counter.
Fortunately, this is only true of the naïve algorithm. The following algorithm, discovered by Gideon
Ehrlich1 in 1973, maintains a Gray code counter in constant worst-case time per increment! The
algorithm uses a separate ‘focus’ array F [0 .. n] in addition to a Gray-code bit array G [0 .. n − 1].
EHRLICHGRAYINCREMENT(n):
j ← F [ 0]
F [0] ← 0
if j = n
G [ n − 1] ← 1 − G [ n − 1]
else
G[ j] = 1 − G[ j]
F [ j ] ← F [ j + 1]
F [ j + 1] ← j + 1 EHRLICHGRAYINIT(n):
for i ← 0 to n − 1
G [i ] ← 0
for i ← 0 to n
F [i ] ← i 1 Gideon Ehrlich. Loopless algorithms for generating permutations, combinations, and other combinatorial conﬁgurations. J.
Assoc. Comput. Mach. 20:500–513, 1973. 5 Algorithms Lecture 8: Amortized Analysis The EHRLICHGRAYINCREMENT algorithm obviously runs in O(1) time, even in the worst case. Here’s
the algorithm in action with n = 4. The ﬁrst line is the Gray bit-vector G , and the second line shows the
focus vector F , both in reverse order:
G : 0000, 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111, 1110, 1010, 1011, 1001, 1000
F : 3210, 3211, 3220, 3212, 3310, 3311, 3230, 3213, 4210, 4211, 4220, 4212, 3410, 3411, 3240, 3214
Voodoo! I won’t explain in detail how Ehrlich’s algorithm works, except to point out the following
invariant. Let B [i ] denote the i th bit in the standard binary representation of the current number.
If B[ j ] = 0 and B[ j − 1] = 1, then F [ j ] is the smallest integer k > j such that B[k ] = 1; otherwise,
F [ j ] = j . Got that?
But wait — this algorithm only handles increments; what if we also want to decrement? Sorry, I
don’t have a clue. Extra credit, anyone? 8.5 Generalities and Warnings Although computer scientists usually apply amortized analysis to understand the efﬁciency of maintaining
and querying data structures, you should remember that amortization can be applied to any sequence of
numbers. Banks have been using amortization to calculate ﬁxed payments for interest-bearing loans for
centuries. The IRS allows taxpayers to amortize business expenses or gambling losses across several
years for purposes of computing income taxes. Some cell phone contracts let you to apply amortization
to calling time, by rolling unused minutes from one month into the next month.
It’s also important to keep in mind when you’re doing amortized analysis is that amortized time
bounds are not unique. For a data structure that supports multiple operations, different amortization
schemes can assign different costs to exactly the same algorithms. For example, consider a generic data
structure that can be modiﬁed by three algorithms: FOLD, SPINDLE, and MUTILATE. One amortization
scheme might imply that FOLD and SPINDLE each run in O(log n) amortized time, while MUTILATE
runs in O(n) amortized time. Another scheme might imply that FOLD runs in O( n) amortized time,
while SPINDLE and MUTILATE each run in O(1) amortized time. These two results are not necessarily
inconsistent! Moreover, there is no general reason to prefer one of these sets of amortized time bounds
over the other; our preference may depend on the context in which the data structure is used. Exercises
1. Suppose we are maintaining a data structure under a series of operations. Let f (n) denote the
actual running time of the nth operation. For each of the following functions f , determine the
resulting amortized cost of a single operation. (For practice, try all of the methods described in
this note.)
(a) f (n) is the largest power of 2 that divides n.
(b) f (n) = n if n is a power of 2, and f (n) = 1 otherwise.
(c) f (n) = n2 if n is a power of 2, and f (n) = 1 otherwise.
2. A multistack consists of an inﬁnite series of stacks S0 , S1 , S2 , . . . , where the i th stack Si can hold up
to 3i elements. The user always pushes and pops elements from the smallest stack S0 . However,
before any element can be pushed onto any full stack Si , we ﬁrst pop all the elements off Si and
push them onto stack Si +1 to make room. (Thus, if Si +1 is already full, we ﬁrst recursively move
6 Algorithms Lecture 8: Amortized Analysis all its members to Si +2 .) Similarly, before any element can be popped from any empty stack Si ,
we ﬁrst pop 3i elements from Si +1 and push them onto Si to make room. (Thus, if Si +1 is already
empty, we ﬁrst recursively ﬁll it by popping elements from Si +2 .) Moving a single element from
one stack to another takes O(1) time.
Here is pseudocode for the multistack operations MSPUSH and MSPOP. The internal stacks are
managed with the subroutines PUSH and POP.
MPUSH( x ) :
i←0
while Si is full
i ← i+1 MPOP( x ) :
i←0
while Si is empty
i ← i+1 while i > 0
i ← i−1
for j ← 1 to 3i
PUSH(Si +1 , POP(Si )) while i > 0
i ← i−1
for j ← 1 to 3i
PUSH(Si , POP(Si +1 )) PUSH(S0 , x ) return POP(S0 )
×9 ×3 Making room in a multistack, just before pushing on a new element. (a) In the worst case, how long does it take to push one more element onto a multistack
containing n elements?
(b) Prove that if the user never pops anything from the multistack, the amortized cost of a push
operation is O(log n), where n is the maximum number of elements in the multistack during
its lifetime.
(c) Prove that in any intermixed sequence of pushes and pops, each push or pop operation takes
O(log n) amortized time, where n is the maximum number of elements in the multistack
during its lifetime.
3. Remember the difference between stacks and queues? Good.
(a) Describe how to implement a queue using two stacks and O(1) additional memory, so that
the amortized time for any enqueue or dequeue operation is O(1). The only access you have
to the stacks is through the standard subroutines PUSH and POP.
(b) A quack is a data structure combining properties of both stacks and queues. It can be viewed
as a list of elements written left to right such that three operations are possible:
• Push: add a new item to the left end of the list;
7 Algorithms Lecture 8: Amortized Analysis
• Pop: remove the item on the left end of the list;
• Pull: remove the item on the right end of the list. Implement a quack using three stacks and O(1) additional memory, so that the amortized
time for any push, pop, or pull operation is O(1). In particular, each element pushed onto
the quack should be stored in exactly one of the three stacks. Again, you are only allowed to
access the stacks through the standard functions PUSH and POP.
4. Suppose we can insert or delete an element into a hash table in O(1) time. In order to ensure that
our hash table is always big enough, without wasting a lot of memory, we will use the following
global rebuilding rules:
• After an insertion, if the table is more than 3/4 full, we allocate a new table twice as big as
our current table, insert everything into the new table, and then free the old table.
• After a deletion, if the table is less than 1/4 full, we allocate a new table half as big as our
current table, insert everything into the new table, and then free the old table.
Show that for any sequence of insertions and deletions, the amortized time per operation is
still O(1). [Hint: Do not even look at the potential-function argument in CLRS; there is a much
easier solution!]
5. Chicago has many tall buildings, but only some of them have a clear view of Lake Michigan.
Suppose we are given an array A[1 .. n] that stores the height of n buildings on a city block,
indexed from west to east. Building i has a good view of Lake Michigan if and only if every
building to the east of i is shorter than i .
Here is an algorithm that computes which buildings have a good view of Lake Michigan. What
is the running time of this algorithm?
GOODVIEW(A[1 .. n]):
initialize a stack S
for i ← 1 to n
while (S not empty and A[i ] > A[TOP(S )])
POP(S )
PUSH(S , i )
return S 6. Design and analyze a simple data structure that maintains a list of integers and supports the
following operations.
• CREATE( ) creates and returns a new list
• PUSH( L , x ) appends x to the end of L
• POP( L ) deletes the last entry of L and returns it
• LOOKUP( L , k) returns the kth entry of L
Your solution may use these primitive data structures: arrays, balanced binary search trees, heaps,
queues, single or doubly linked lists, and stacks. If your algorithm uses anything fancier, you must
give an explicit implementation. Your data structure must support all operations in amortized
constant time. In addition, your data structure must support each LOOKUP in worst-case O(1) time.
At all times, the size of your data structure must be linear in the number of objects it stores.
8 Algorithms Lecture 8: Amortized Analysis 7. Suppose instead of powers of two, we represent integers as the sum of Fibonacci numbers. In
other words, instead of an array of bits, we keep an array of ﬁts, where the i th least signiﬁcant
ﬁt indicates whether the sum includes the i th Fibonacci number Fi . For example, the ﬁtstring
101110 F represents the number F6 + F4 + F3 + F2 = 8 + 3 + 2 + 1 = 14. Describe algorithms to
increment and decrement a single ﬁtstring in constant amortized time. [Hint: Most numbers can
be represented by more than one ﬁtstring!]
8. A doubly lazy binary counter represents any number as a weighted sum of powers of two, where
each weight is one of four values: −1, 0, 1, or 2. (For succinctness, I’ll write 1 instead of −1.) Every
integer—positive, negative, or zero—has an inﬁnite number of doubly lazy binary representations.
For example, the number 13 can be represented as 1101 (the standard binary representation), or
2101 (because 2 · 23 − 22 + 20 = 13) or 10111 (because 24 − 22 + 21 − 20 = 13) or 11200010111
(because −210 + 29 + 2 · 28 + 24 − 22 + 21 − 20 = 13).
To increment a doubly lazy binary counter, we add 1 to the least signiﬁcant bit, then carry the
rightmost 2 (if any). To decrement, we subtract 1 from the lest signiﬁcant bit, and then borrow
the rightmost 1 (if any).
LAZYINCREMENT(B [0 .. n]):
B [ 0] ← B [ 0] + 1
for i ← 1 to n − 1
if B [i ] = 2
B [i ] ← 0
B [ i + 1] ← B [ i + 1] + 1
return LAZYDECREMENT(B [0 .. n]):
B [ 0] ← B [ 0] − 1
for i ← 1 to n − 1
if B [i ] = −1
B [i ] ← 1
B [ i + 1] ← B [ i + 1] − 1
return For example, here is a doubly lazy binary count from zero up to twenty and then back down to
zero. The bits are written with the least signiﬁcant bit B [0] on the right, omitting all leading 0’s
++ ++ ++ ++ ++ ++ ++ ++ ++ ++ 0 −→ 1 −→ 10 −→ 11 −→ 20 −→ 101 −→ 110 −→ 111 −→ 120 −→ 201 −→ 210
++ ++ ++ ++ ++ ++ ++ ++ ++ ++ −− −− −− −− −− −− −− −− −− −− −→ 1011 −→ 1020 −→ 1101 −→ 1110 −→ 1111 −→ 1120 −→ 1201 −→ 1210 −→ 2011 −→ 2020
−→ 2011 −→ 2010 −→ 2001 −→ 2000 −→ 2011 −→ 2110 −→ 2101 −→ 1100 −→ 1111 −→ 1010
−− −− −− −− −− −− −− −− −− −− −→ 1001 −→ 1000 −→ 1011 −→ 1110 −→ 1101 −→ 100 −→ 111 −→ 10 −→ 1 −→ 0 Prove that for any intermixed sequence of increments and decrements of a doubly lazy binary
number, starting with 0, the amortized time for each operation is O(1). Do not assume, as in the
example above, that all the increments come before all the decrements. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 9 Algorithms Lecture 9: Scapegoat and Splay Trees
Everything was balanced before the computers went off line. Try and adjust something, and
you unbalance something else. Try and adjust that, you unbalance two more and before you
know what’s happened, the ship is out of control.
— Blake, Blake’s 7, “Breakdown” (March 6, 1978)
A good scapegoat is nearly as welcome as a solution to the problem.
— Anonymous
Let’s play.
— El Mariachi [Antonio Banderas], Desperado (1992)
CAPTAIN:
CAPTAIN:
CAPTAIN:
CAPTAIN: TAKE OFF EVERY ’ZIG’!!
YOU KNOW WHAT YOU DOING.
MOVE ’ZIG’.
FOR GREAT JUSTICE.
— Zero Wing (1992) 9 Scapegoat and Splay Trees 9.1 Deﬁnitions I’ll assume that everyone is already familiar with the standard terminology for binary search trees—node,
search key, edge, root, internal node, leaf, right child, left child, parent, descendant, sibling, ancestor,
subtree, preorder, postorder, inorder, etc.—as well as the standard algorithms for searching for a node,
inserting a node, or deleting a node. Otherwise, consult your favorite data structures textbook.
For this lecture, we will consider only full binary trees—where every internal node has exactly two
children—where only the internal nodes actually store search keys. In practice, we can represent the
leaves with null pointers.
Recall that the depth of a node is its distance from the root, and its height is the distance to the
farthest leaf in its subtree. The height (or depth) of the tree is just the height of the root. The size of a
node is the number of nodes in its subtree. The size n of the whole tree is just the total number of nodes.
A tree with height h has at most 2h leaves, so the minimum height of an n-leaf binary tree is lg n .
In the worst case, the time required for a search, insertion, or deletion to the height of the tree, so in
general we would like keep the height as close to lg n as possible. The best we can possibly do is to
have a perfectly balanced tree, in which each subtree has as close to half the leaves as possible, and both
subtrees are perfectly balanced. The height of a perfectly balanced tree is lg n , so the worst-case search
time is O(log n). However, even if we started with a perfectly balanced tree, a malicious sequence of
insertions and/or deletions could make the tree arbitrarily unbalanced, driving the search time up to
Θ(n).
To avoid this problem, we need to periodically modify the tree to maintain ‘balance’. There are
several methods for doing this, and depending on the method we use, the search tree is given a different
name. Examples include AVL trees, red-black trees, height-balanced trees, weight-balanced trees,
bounded-balance trees, path-balanced trees, B -trees, treaps, randomized binary search trees, skip lists,1
and jumplists. Some of these trees support searches, insertions, and deletions, in O(log n) worst-case
time, others in O(log n) amortized time, still others in O(log n) expected time.
In this lecture, I’ll discuss two binary search tree data structures with good amortized performance.
The ﬁrst is the scapegoat tree, discovered by Arne Andersson* in 1989 [1, 2] and independently2 by Igal
1 Yeah, yeah. Skip lists aren’t really binary search trees. Whatever you say, Mr. Picky.
The claim of independence is Andersson’s [2]. The two papers actually describe very slightly different rebalancing
algorithms. The algorithm I’m using here is closer to Andersson’s, but my analysis is closer to Galperin and Rivest’s.
2 1 Algorithms Lecture 9: Scapegoat and Splay Trees Galperin* and Ron Rivest in 1993 [10]. The second is the splay tree, discovered by Danny Sleator and
Bob Tarjan in 1981 [16, 14]. 9.2 Lazy Deletions: Global Rebuilding First let’s consider the simple case where we start with a perfectly-balanced tree, and we only want to
perform searches and deletions. To get good search and delete times, we will use a technique called
global rebuilding. When we get a delete request, we locate and mark the node to be deleted, but we don’t
actually delete it. This requires a simple modiﬁcation to our search algorithm—we still use marked nodes
to guide searches, but if we search for a marked node, the search routine says it isn’t there. This keeps
the tree more or less balanced, but now the search time is no longer a function of the amount of data
currently stored in the tree. To remedy this, we also keep track of how many nodes have been marked,
and then apply the following rule:
Global Rebuilding Rule. As soon as half the nodes in the tree have been marked, rebuild a
new perfectly balanced tree containing only the unmarked nodes.3
With this rule in place, a search takes O(log n) time in the worst case, where n is the number of
unmarked nodes. Speciﬁcally, since the tree has at most n marked nodes, or 2n nodes altogether, we
need to examine at most lg n + 1 keys. There are several methods for rebuilding the tree in O(n) time,
where n is the size of the new tree. (Homework!) So a single deletion can cost Θ(n) time in the worst
case, but only if we have to rebuild; most deletions take only O(log n) time.
We spend O(n) time rebuilding, but only after Ω(n) deletions, so the amortized cost of rebuilding the
tree is O(1) per deletion. (Here I’m using a simple version of the ‘taxation method’. For each deletion,
we charge a $1 tax; after n deletions, we’ve collected $n, which is just enough to pay for rebalancing the
tree containing the remaining n nodes.) Since we also have to ﬁnd and mark the node being ‘deleted’,
the total amortized time for a deletion is O(log n) . 9.3 Insertions: Partial Rebuilding Now suppose we only want to support searches and insertions. We can’t ‘not really insert’ new nodes
into the tree, since that would make them unavailable to the search algorithm.4 So instead, we’ll use
another method called partial rebuilding. We will insert new nodes normally, but whenever a subtree
becomes unbalanced enough, we rebuild it. The deﬁnition of ‘unbalanced enough’ depends on an
arbitrary constant α > 1.
Each node v will now also store height( v ) and size( v ). We now modify our insertion algorithm with
the following rule:
Partial Rebuilding Rule. After we insert a node, walk back up the tree updating the heights
and sizes of the nodes on the search path. If we encounter a node v where height( v ) >
α · lg(size( v )), rebuild its subtree into a perfectly balanced tree (in O(size( v )) time).
If we always follow this rule, then after an insertion, the height of the tree is at most α · lg n. Thus,
since α is a constant, the worst-case search time is O(log n). In the worst case, insertions require Θ(n)
time—we might have to rebuild the entire tree. However, the amortized time for each insertion is again
only O(log n). Not surprisingly, the proof is a little bit more complicated than for deletions.
3
Alternately: When the number of unmarked nodes is one less than an exact power of two, rebuild the tree. This rule
ensures that the tree is always exactly balanced.
4
Well, we could use the Bentley-Saxe* logarithmic method [3], but that would raise the query time to O(log2 n). 2 Algorithms Lecture 9: Scapegoat and Splay Trees Deﬁne the imbalance I ( v ) of a node v to be one less than the absolute difference between the sizes
of its two subtrees, or zero, whichever is larger:
I ( v ) = max 0, size(left( v )) − size(right( v )) − 1
A simple induction proof implies that I ( v ) = 0 for every node v in a perfectly balanced tree. So
immediately after we rebuild the subtree of v , we have I ( v ) = 0. On the other hand, each insertion into
the subtree of v increments either size(left( v )) or size(right( v )), so I ( v ) changes by at most 1.
The whole analysis boils down to the following lemma.
Lemma 1. Just before we rebuild v ’s subtree, I ( v ) = Ω(size( v )).
Before we prove this, let’s ﬁrst look at what it implies. If I ( v ) = Ω(size( v )), then Ω(size( v )) keys
have been inserted in the v ’s subtree since the last time it was rebuilt from scratch. On the other hand,
rebuilding the subtree requires O(size( v )) time. Thus, if we amortize the rebuilding cost across all the
insertions since the last rebuilding, v is charged constant time for each insertion into its subtree. Since
each new key is inserted into at most α · lg n = O(log n) subtrees, the total amortized cost of an insertion
is O (log n ).
Proof: Since we’re about to rebuild the subtree at v , we must have height( v ) > α · lg size( v ). Without
loss of generality, suppose that the node we just inserted went into v ’s left subtree. Either we just rebuilt
this subtree or we didn’t have to, so we also have height(left( v )) ≤ α · lg size(left( v )). Combining these
two inequalities with the recursive deﬁnition of height, we get
α · lg size( v ) < height( v ) ≤ height(left( v )) + 1 ≤ α · lg size(left( v )) + 1.
After some algebra, this simpliﬁes to size(left( v )) > size( v )/21/α . Combining this with the identity
size( v ) = size(left( v )) + size(right( v )) + 1 and doing some more algebra gives us the inequality
size(right( v )) < 1 − 1/21/α size( v ) − 1.
Finally, we combine these two inequalities using the recursive deﬁnition of imbalance.
I ( v ) ≥ size(left( v )) − size(right( v )) − 1 > 2/21/α − 1 size( v )
Since α is a constant bigger than 1, the factor in parentheses is a positive constant. 9.4 Scapegoat Trees Finally, to handle both insertions and deletions efﬁciently, scapegoat trees use both of the previous
techniques. We use partial rebuilding to re-balance the tree after insertions, and global rebuilding to
re-balance the tree after deletions. Each search takes O(log n) time in the worst case, and the amortized
time for any insertion or deletion is also O(log n). There are a few small technical details left (which I
won’t describe), but no new ideas are required.
Once we’ve done the analysis, we can actually simplify the data structure. It’s not hard to prove that
at most one subtree (the scapegoat) is rebuilt during any insertion. Less obviously, we can even get the
same amortized time bounds (except for a small constant factor) if we only maintain the three integers
in addition to the actual tree: the size of the entire tree, the height of the entire tree, and the number
of marked nodes. Whenever an insertion causes the tree to become unbalanced, we can compute the
sizes of all the subtrees on the search path, starting at the new leaf and stopping at the scapegoat, in
time proportional to the size of the scapegoat subtree. Since we need that much time to re-balance the
scapegoat subtree, this computation increases the running time by only a small constant factor! Thus,
unlike almost every other kind of balanced trees, scapegoat trees require only O(1) extra space.
3 Algorithms 9.5 Lecture 9: Scapegoat and Splay Trees Rotations, Double Rotations, and Splaying Another method for maintaining balance in binary search trees is by adjusting the shape of the tree
locally, using an operation called a rotation. A rotation at a node x decreases its depth by one and
increases its parent’s depth by one. Rotations can be performed in constant time, since they only involve
simple pointer manipulation.
y x right x y
left Figure 1. A right rotation at x and a left rotation at y are inverses. For technical reasons, we will need to use rotations two at a time. There are two types of double
rotations, which might be called zig-zag and roller-coaster. A zig-zag at x consists of two rotations at x ,
in opposite directions. A roller-coaster at a node x consists of a rotation at x ’s parent followed by a
rotation at x , both in the same direction. Each double rotation decreases the depth of x by two, leaves
the depth of its parent unchanged, and increases the depth of its grandparent by either one or two,
depending on the type of double rotation. Either type of double rotation can be performed in constant
time.
z z w x x w w x z Figure 2. A zig-zag at x . The symmetric case is not shown. z x
y y
x x y
z z Figure 3. A right roller-coaster at x and a left roller-coaster at z . Finally, a splay operation moves an arbitrary node in the tree up to the root through a series of double
rotations, possibly with one single rotation at the end. Splaying a node v requires time proportional to
depth( v ). (Obviously, this means the depth before splaying, since after splaying v is the root and thus has
depth zero!) 9.6 Splay Trees A splay tree is a binary search tree that is kept more or less balanced by splaying. Intuitively, after we
access any node, we move it to the root with a splay operation. In more detail:
• Search: Find the node containing the key using the usual algorithm, or its predecessor or successor
if the key is not present. Splay whichever node was found. 4 Algorithms Lecture 9: Scapegoat and Splay Trees b
a b
d c a
m k e n m f
e j
i a k h
g d
c l f b c m f l n f c a m
k
j e k n m
d c g k
l
j e l n f h l h
g b x
d j e x h
g a x j
x d n x b h
i g i i i Figure 4. Splaying a node. Irrelevant subtrees are omitted for clarity. • Insert: Insert a new node using the usual algorithm, then splay that node.
• Delete: Find the node x to be deleted, splay it, and then delete it. This splits the tree into two
subtrees, one with keys less than x , the other with keys bigger than x . Find the node w in the left
subtree with the largest key (the inorder predecessor of x in the original tree), splay it, and ﬁnally
join it to the right subtree.
x x w w w
Figure 5. Deleting a node in a splay tree. Each search, insertion, or deletion consists of a constant number of operations of the form walk down
to a node, and then splay it up to the root. Since the walk down is clearly cheaper than the splay up, all
we need to get good amortized bounds for splay trees is to derive good amortized bounds for a single
splay.
Believe it or not, the easiest way to do this uses the potential method. We deﬁne the rank of a node v
to be lg size( v ) , and the potential of a splay tree to be the sum of the ranks of its nodes:
Φ= rank( v ) =
v lg size( v )
v It’s not hard to observe that a perfectly balanced binary tree has potential Θ(n), and a linear chain of
nodes (a perfectly unbalanced tree) has potential Θ(n log n).
The amortized analysis of splay trees boils down to the following lemma. Here, rank( v ) denotes the
rank of v before a (single or double) rotation, and rank ( v ) denotes its rank afterwards. Recall that the
amortized cost is deﬁned to be the number of rotations plus the drop in potential.
The Access Lemma. The amortized cost of a single rotation at any node v is at most 1 + 3 rank ( v ) −
3 rank( v ), and the amortized cost of a double rotation at any node v is at most 3 rank ( v ) − 3 rank( v ).
Proving this lemma is a straightforward but tedious case analysis of the different types of rotations.
For the sake of completeness, I’ll give a proof (of a generalized version) in the next section.
5 Algorithms Lecture 9: Scapegoat and Splay Trees By adding up the amortized costs of all the rotations, we ﬁnd that the total amortized cost of splaying
a node v is at most 1 + 3 rank ( v ) − 3 rank( v ), where rank ( v ) is the rank of v after the entire splay.
(The intermediate ranks cancel out in a nice telescoping sum.) But after the splay, v is the root! Thus,
rank ( v ) = lg n , which implies that the amortized cost of a splay is at most 3 lg n − 1 = O(log n).
We conclude that every insertion, deletion, or search in a splay tree takes O(log n) amortized time. 9.7 Other Optimality Properties In fact, splay trees are optimal in several other senses. Some of these optimality properties follow easily
from the following generalization of the Access Lemma.
Let’s arbitrarily assign each node v a non-negative real weight w ( v ). These weights are not actually
stored in the splay tree, nor do they affect the splay algorithm in any way; they are only used to help with
the analysis. We then redeﬁne the size s( v ) of a node v to be the sum of the weights of the descendants
of v , including v itself:
s( v ) := w ( v ) + s(right( v )) + s(left( v )).
If w ( v ) = 1 for every node v , then the size of a node is just the number of nodes in its subtree, as in the
previous section. As before, we deﬁne the rank of any node v to be r ( v ) = lg s( v ), and the potential of a
splay tree to be the sum of the ranks of all its nodes:
Φ= r (v) =
v lg s( v )
v In the following lemma, r ( v ) denotes the rank of v before a (single or double) rotation, and r ( v )
denotes its rank afterwards.
The Generalized Access Lemma. For any assignment of non-negative weights to the nodes, the amortized cost of a single rotation at any node x is at most 1 + 3 r ( x ) − 3 r ( x ), and the amortized cost of a
double rotation at any node v is at most 3 r ( x ) − 3 r ( x ).
Proof: First consider a single rotation, as shown in Figure 1.
1 + Φ − Φ = 1 + r (x) + r ( y) − r(x) − r( y) [only x and y change rank] ≤ 1 + r (x) − r(x) [ r ( y ) ≤ r ( y )] ≤ 1 + 3r ( x ) − 3r ( x ) [ r ( x ) ≥ r ( x )] Now consider a zig-zag, as shown in Figure 2. Only w , x , and z change rank.
2+Φ −Φ
= 2 + r (w ) + r ( x ) + r (z ) − r (w ) − r ( x ) − r (z )
≤ 2 + r (w ) + r ( x ) + r (z ) − 2 r ( x ) [only w, x , z change rank]
[ r ( x ) ≤ r (w ) and r ( x ) = r (z )] = 2 + ( r (w ) − r ( x )) + ( r (z ) − r ( x )) + 2( r ( x ) − r ( x ))
= 2 + lg s (w ) + lg s (z ) + 2( r ( x ) − r ( x ))
s (x)
s (x)
s ( x )/2
≤ 2 + 2 lg
+ 2( r ( x ) − r ( x ))
s (x)
= 2( r ( x ) − r ( x )) [s (w ) + s (z ) ≤ s ( x ), lg is concave] ≤ 3( r ( x ) − r ( x )) [ r ( x ) ≥ r ( x )]
6 Algorithms Lecture 9: Scapegoat and Splay Trees Finally, consider a roller-coaster, as shown in Figure 3. Only x , y , and z change rank.
2+Φ −Φ
= 2 + r ( x ) + r ( y ) + r (z ) − r ( x ) − r ( y ) − r (z ) [only x , y, z change rank] ≤ 2 + r ( x ) + r (z ) − 2 r ( x ) [ r ( y ) ≤ r (z ) and r ( x ) ≥ r ( y )] = 2 + ( r ( x ) − r ( x )) + ( r (z ) − r ( x )) + 3( r ( x ) − r ( x ))
s( x ) + lg s (z ) + 3( r ( x ) − r ( x ))
s (x)
s (x)
s ( x )/2
+ 3( r ( x ) − r ( x ))
≤ 2 + 2 lg
s (x)
= 3( r ( x ) − r ( x ))
= 2 + lg This completes the proof. [s( x ) + s (z ) ≤ s ( x ), lg is concave] 5 Observe that this argument works for arbitrary non-negative vertex weights. By adding up the
amortized costs of all the rotations, we ﬁnd that the total amortized cost of splaying a node x is at most
1 + 3 r (root) − 3 r ( x ). (The intermediate ranks cancel out in a nice telescoping sum.)
This analysis has several immediate corollaries. The ﬁrst corollary is that the amortized search time
in a splay tree is within a constant factor of the search time in the best possible static binary search tree.
Thus, if some nodes are accessed more often than others, the standard splay algorithm automatically
keeps those more frequent nodes closer to the root, at least most of the time.
Static Optimality Theorem. Suppose each node x is accessed at least t ( x ) times, and let T =
The amortized cost of accessing x is O(log T − log t ( x )). x t ( x ). Proof: Set w ( x ) = t ( x ) for each node x .
For any nodes x and z , let dist( x , z ) denote the rank distance between x and y , that is, the number
of nodes y such that key( x ) ≤ key( y ) ≤ key(z ) or key( x ) ≥ key( y ) ≥ key(z ). In particular, dist( x , x ) = 1
for all x .
Static Finger Theorem. For any ﬁxed node f (‘the ﬁnger’), the amortized cost of accessing x is
O(lg dist( f , x )).
Proof: Set w ( x ) = 1/dist( x , f )2 for each node x . Then s(root) ≤
r ( x ) ≥ lg w ( x ) = −2 lg dist( f , x ). ∞
2
i =1 2/ i = π2 /3 = O(1), and Here are a few more interesting properties of splay trees, which I’ll state without proof.6 The proofs
of these properties (especially the dynamic ﬁnger theorem) are considerably more complicated than the
amortized analysis presented above.
Working Set Theorem [14]. The amortized cost of accessing node x is O(log D), where D is the
number of distinct items accessed since the last time x was accessed. (For the ﬁrst access to x , we set
D = n.)
5 This proof is essentially taken verbatim from the original Sleator and Tarjan paper. Another proof technique, which may be
more accessible, involves maintaining lg s( v ) tokens on each node v and arguing about the changes in token distribution
caused by each single or double rotation. But I haven’t yet internalized this approach enough to include it here.
6
This list and the following section are taken almost directly from Erik Demaine’s lecture notes [5]. 7 Algorithms Lecture 9: Scapegoat and Splay Trees Scanning Theorem [17]. Splaying all nodes in a splay tree in order, starting from any initial tree,
requires O(n) total rotations.
Dynamic Finger Theorem [7, 6]. Immediately after accessing node y , the amortized cost of accessing
node x is O(lg dist( x , y )). 9.8 Splay Tree Conjectures Splay trees are conjectured to have many interesting properties in addition to the optimality properties
that have been proved; I’ll describe just a few of the more important ones.
The Deque Conjecture [17] considers the cost of dynamically maintaining two ﬁngers l and r , starting
on the left and right ends of the tree. Suppose at each step, we can move one of these two ﬁngers either
one step left or one step right; in other words, we are using the splay tree as a doubly-ended queue.
Sundar* proved that the total cost of m deque operations on an n-node splay tree is O((m + n)α(m + n))
[15]. More recently, Pettie later improved this bound to O(mα∗ (n)) [13]. The Deque Conjecture states
that the total cost is actually O(m + n).
The Traversal Conjecture [14] states that accessing the nodes in a splay tree, in the order speciﬁed by
a preorder traversal of any other binary tree with the same keys, takes O(n) time. This is generalization
of the Scanning Theorem.
The Uniﬁed Conjecture [12] states that the time to access node x is O(lg min y ( D( y ) + d ( x , y ))),
where D( y ) is the number of distinct nodes accessed since the last time y was accessed. This would
immediately imply both the Dynamic Finger Theorem, which is about spatial locality, and the Working
Set Theorem, which is about temporal locality. Two other structures are known that satisfy the uniﬁed
bound [4, 12].
Finally, the most important conjecture about splay trees, and one of the most important open
problems about data structures, is that they are dynamically optimal [14]. Speciﬁcally, the cost of any
sequence of accesses to a splay tree is conjectured to be at most a constant factor more than the cost of
the best possible dynamic binary search tree that knows the entire access sequence in advance. To make
the rules concrete, we consider binary search trees that can undergo arbitrary rotations after a search;
the cost of a search is the number of key comparisons plus the number of rotations. We do not require
that the rotations be on or even near the search path. This is an extremely strong conjecture!
No dynamically optimal binary search tree is known, even in the ofﬂine setting. However, three very
similar O(log log n)-competitive binary search trees have been discovered in the last few years: Tango
trees [8], multisplay trees [18], and chain-splay trees [11]. A recently-published geometric formulation
of dynamic binary search trees [9] also offers signiﬁcant hope for future progress. References
[1] A. Andersson*. Improving partial rebuilding by using simple balance criteria. Proc. Workshop on Algorithms
and Data Structures, 393–402, 1989. Lecture Notes Comput. Sci. 382, Springer-Verlag.
[2] A. Andersson. General balanced trees. J. Algorithms 30:1–28, 1999.
[3] J. L. Bentley and J. B. Saxe*. Decomposable searching problems I: Static-to-dynamic transformation. J.
Algorithms 1(4):301–358, 1980.
˘
[4] M. Badiou* and E. D. Demaine. A simpliﬁed and dynamic uniﬁed structure. Proc. 6th Latin American Symp.
Theoretical Informatics, 466–473, 2004. Lecture Notes Comput. Sci. 2976, Springer-Verlag.
[5] J. Cohen* and E. Demaine. 6.897: Advanced data structures (Spring 2005), Lecture 3, February 8 2005.
〈http://theory.csail.mit.edu/classes/6.897/spring05/lec.html〉.
[6] R. Cole. On the dynamic ﬁnger conjecture for splay trees. Part II: The proof. SIAM J. Comput. 30(1):44–85,
2000. 8 Algorithms Lecture 9: Scapegoat and Splay Trees [7] R. Cole, B. Mishra, J. Schmidt, and A. Siegel. On the dynamic ﬁnger conjecture for splay trees. Part I: Splay
sorting log n-block sequences. SIAM J. Comput. 30(1):1–43, 2000.
˘s
[8] E. D. Demaine, D. Harmon*, J. Iacono, and M. Patra¸cu*. Dynamic optimality—almost. Proc. 45th Ann. IEEE
Symp. Foundations Comput. Sci., 484–490, 2004.
˘s
[9] E. D. Demaine, D. Harmon*, J. Iacono, D. Kane*, and M. Patra¸cu*. The geometry of binary search trees.
Proc. 20th ACM-SIAM Symp. Discrete Algorithms., 496–505, 2009.
[10] I. Galperin* and R. R. Rivest. Scapegoat trees. Proc. 4th Ann. ACM-SIAM Symp. Discrete Algorithms, 165–174,
1993.
[11] G. Georgakopolous. Chain-splay trees, or, how to achieve and prove log log N -competitiveness by splaying.
Inform. Proc. Lett. 106:37–34, 2008.
[12] J. Iacono*. Alternatives to splay trees with O(log n) worst-case access times. Proc. 12th Ann. ACM-SIAM
Symp. Discrete Algorithms, 516–522, 2001.
[13] S. Pettie. Splay trees, Davenport-Schinzel sequences, and the deque conjecture. Proc. 19th Ann. ACM-SIAM
Symp. Discrete Algorithms, 1115–1124, 2008.
[14] D. D. Sleator and R. E. Tarjan. Self-adjusting binary search trees. J. ACM 32(3):652–686, 1985.
[15] R. Sundar*. On the Deque conjecture for the splay algorithm. Combinatorica 12(1):95–124, 1992.
[16] R. E. Tarjan. Data Structures and Network Algorithms. CBMS-NSF Regional Conference Series in Applied
Mathematics 44. SIAM, 1983.
[17] R. E. Tarjan. Sequential access in splay trees takes linear time. Combinatorica 5(5):367–378, 1985.
[18] C. C. Wang*, J. Derryberry*, and D. D. Sleator. O(log log n)-competitive dynamic binary search trees. Proc.
17th Ann. ACM-SIAM Symp. Discrete Algorithms, 374–383, 2006.
*Starred authors were students at the time that the cited work was published. Exercises
1. (a) An n-node binary tree is perfectly balanced if either n ≤ 1, or its two subtrees are perfectly
balanced binary trees, each with at most n/2 nodes. Prove that I ( v ) = 0 for every node v
of any perfectly balanced tree.
(b) Prove that at most one subtree is rebalanced during a scapegoat tree insertion.
2. In a dirty binary search tree, each node is labeled either clean or dirty. The lazy deletion scheme
used for scapegoat trees requires us to purge the search tree, keeping all the clean nodes and
deleting all the dirty nodes, as soon as half the nodes become dirty. In addition, the purged tree
should be perfectly balanced.
(a) Describe and analyze an algorithm to purge an arbitrary n-node dirty binary search tree in
O(n) time. (Such an algorithm is necessary for scapegoat trees to achieve O(log n) amortized
insertion cost.)
(b) Modify your algorithm so that is uses only O(log n) space, in addition to the tree itself. Don’t
forget to include the recursion stack in your space bound.
(c) Modify your algorithm so that is uses only O(1) additional space. In particular, your algorithm
cannot call itself recursively at all.
3. Consider the following simpler alternative to splaying:
MOVETOROOT( v ):
while parent( v ) = NULL
rotate at v 9 Algorithms Lecture 9: Scapegoat and Splay Trees Prove that the amortized cost of MOVETOROOT in an n-node binary tree can be Ω(n). That is, prove
that for any integer k, there is a sequence of k MOVETOROOT operations that require Ω(kn) time to
execute.
4. Suppose we want to maintain a dynamic set of values, subject to the following operations:
• INSERT( x ): Add x to the set (if it isn’t already there).
• PRINT&DELETEBETWEEN(a, b): Print every element x in the range a ≤ x ≤ b, in increasing
order, and delete those elements from the set.
For example, if the current set is {1, 5, 3, 4, 8}, then PRINT&DELETEBETWEEN(4, 6) prints the numbers 4 and 5 and changes the set to {1, 3, 8}. For any underlying set, PRINT&DELETEBETWEEN(−∞, ∞)
prints its contents in increasing order and deletes everything.
(a) Describe and analyze a data structure that supports these operations, each with amortized
cost O(log n), where n is the maximum number of elements in the set.
(b) What is the running time of your INSERT algorithm in the worst case?
(c) What is the running time of your PRINT&DELETEBETWEEN algorithm in the worst case?
5. Let P be a set of n points in the plane. The staircase of P is the set of all points in the plane that
have at least one point in P both above and to the right. A set of points in the plane and its staircase (shaded). (a) Describe an algorithm to compute the staircase of a set of n points in O(n log n) time.
(b) Describe and analyze a data structure that stores the staircase of a set of points, and an
algorithm ABOVE?( x , y ) that returns TRUE if the point ( x , y ) is above the staircase, or FALSE
otherwise. Your data structure should use O(n) space, and your ABOVE? algorithm should
run in O(log n) time.
TRUE FALSE Two staircase queries. (c) Describe and analyze a data structure that maintains a staircase as new points are inserted.
Speciﬁcally, your data structure should support a function INSERT( x , y ) that adds the point
10 Algorithms Lecture 9: Scapegoat and Splay Trees ( x , y ) to the underlying point set and returns TRUE or FALSE to indicate whether the staircase
of the set has changed. Your data structure should use O(n) space, and your INSERT algorithm
should run in O(log n) amortized time.
TRUE! FALSE! Two staircase insertions. 6. Say that a binary search tree is augmented if every node v also stores size( v ), the number of nodes
in the subtree rooted at v .
(a) Show that a rotation in an augmented binary tree can be performed in constant time.
(b) Describe an algorithm SCAPEGOATSELECT(k) that selects the kth smallest item in an augmented
scapegoat tree in O(log n) worst-case time. (The scapegoat trees presented in these notes are
already augmented.)
(c) Describe an algorithm SPLAYSELECT(k) that selects the kth smallest item in an augmented
splay tree in O(log n) amortized time.
(d) Describe an algorithm TREAPSELECT(k) that selects the kth smallest item in an augmented
treap in O(log n) expected time.
7. Many applications of binary search trees attach a secondary data structure to each node in the tree,
to allow for more complicated searches. Let T be an arbitrary binary tree. The secondary data
structure at any node v stores exactly the same set of items as the subtree of T rooted at v . This
secondary structure has size O(size( v )) and can be built in O(size( v )) time, where size( v ) denotes
the number of descendants of v .
The primary and secondary data structures are typically deﬁned by different attributes of the
data being stored. For example, to store a set of points in the plane, we could deﬁne the primary
tree T in terms of the x -coordinates of the points, and deﬁne the secondary data structures in
terms of their y -coordinate.
Maintaining these secondary structures complicates algorithms for keeping the top-level search
tree balanced. Speciﬁcally, performing a rotation at any node v in the primary tree now requires
O(size( v )) time, because we have to rebuild one of the secondary structures (at the new child
of v ). When we insert a new item into T , we must also insert into one or more secondary data
structures.
(a) Overall, how much space does this data structure use in the worst case?
(b) How much space does this structure use if the primary search tree is perfectly balanced?
(c) Suppose the primary tree is a splay tree. Prove that the amortized cost of a splay (and
therefore of a search, insertion, or deletion) is Ω(n). [Hint: This is easy!]
(d) Now suppose the primary tree T is a scapegoat tree. How long does it take to rebuild the
subtree of T rooted at some node v , as a function of size( v )? 11 Algorithms Lecture 9: Scapegoat and Splay Trees (e) Suppose the primary tree and all the secondary trees are scapegoat trees. What is the
amortized cost of a single insertion?
(f) Finally, suppose the primary tree and every secondary tree is a treap. What is the worst-case
expected time for a single insertion?
8. Let X = 〈 x 1 , x 2 , . . . , x m 〉 be a sequence of m integers, each from the set {1, 2, . . . , n}. We can
visualize this sequence as a set of integer points in the plane, by interpreting each element x i as
the point ( x i , i ). The resulting point set, which we can also call X , has exactly one point on each
row of the n × m integer grid.
(a) Let Y be an arbitrary set of integer points in the plane. Two points ( x 1 , y1 ) and ( x 2 , y2 ) in
Y are isolated if (1) x 1 = x 2 and y1 = y2 , and (2) there is no other point ( x , y ) ∈ Y with
x 1 ≤ x ≤ x 2 and y1 ≤ y ≤ y2 . If the set Y contains no isolated pairs of points, we call Y a
commune.7
Let X be an arbitrary set of points on the n × n integer grid with exactly one point per
row. Show that there is a commune Y that contains X and consists of O(n log n) points.
(b) Consider the following model of self-adjusting binary search trees. We interpret X as a
sequence of accesses in a binary search tree. Let T0 denote the initial tree. In the i th round,
we traverse the path from the root to node x i , and then arbitrarily reconﬁgure some subtree Si
of the current search tree Ti −1 to obtain the next search tree Ti . The only restriction is that
the subtree Si must contain both x i and the root of Ti −1 . (For example, in a splay tree, Si is
the search path to x i .) The cost of the i th access is the number of nodes in the subtree Si .
Prove that the minimum cost of executing an access sequence X in this model is at least
the size of the smallest commune containing the corresponding point set X . [Hint: Lowest
common ancestor.]
(c) Suppose X is a random permutation of the integers 1, 2, . . . , n. Use the lower bound in
part (b) to prove that the expected minimum cost of executing X is Ω(n log n).
(d) Describe a polynomial-time algorithm to compute (or even approximate up to constant
factors) the smallest commune containing a given set X of integer points, with at most one
point per row. Alternately, prove that the problem is NP-hard. 7 Demaine et al. [9] refer to communes as arborally satisﬁed sets. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 12 Algorithms Lecture 10: Disjoint Sets
E pluribus unum (Out of many, one)
— Ofcial motto of the United States of America
John: Who’s your daddy? C’mon, you know who your daddy is! Who’s your daddy?
D’Argo, tell him who his daddy is!"
D’Argo: I’m your daddy.
— Farscape, “Thanks for Sharing” (June 15, 2001)
What rolls down stairs, alone or in pairs, rolls over your neighbor’s dog?
What’s great for a snack, and ﬁts on your back? It’s Log, Log, Log!
It’s Log! It’s Log! It’s big, it’s heavy, it’s wood!
It’s Log! It’s Log! It’s better than bad, it’s good!
— Ren & Stimpy, “Stimpy’s Big Day/The Big Shot" (August 11, 1991)
lyrics by John Kricfalusi
The thing’s hollow - it goes on forever - and - oh my God! - it’s full of stars!
— Capt. David Bowman’s last words(?)
2001: A Space Odyssey by Arthur C. Clarke (1968) 10 Data Structures for Disjoint Sets In this lecture, we describe some methods for maintaining a collection of disjoint sets. Each set is
represented as a pointer-based data structure, with one node per element. We will refer to the elements
as either ‘objects’ or ‘nodes’, depending on whether we want to emphasize the set abstraction or the
actual data structure. Each set has a unique ‘leader’ element, which identiﬁes the set. (Since the sets are
always disjoint, the same object cannot be the leader of more than one set.) We want to support the
following operations.
• MAKESET( x ): Create a new set { x } containing the single element x . The object x must not appear
in any other set in our collection. The leader of the new set is obviously x .
• FIND( x ): Find (the leader of) the set containing x .
• UNION(A, B ): Replace two sets A and B in our collection with their union A ∪ B . For example,
UNION(A, MAKESET( x )) adds a new element x to an existing set A. The sets A and B are speciﬁed
by arbitrary elements, so UNION( x , y ) has exactly the same behavior as UNION(FIND( x ), FIND( y )).
Disjoint set data structures have lots of applications. For instance, Kruskal’s minimum spanning
tree algorithm relies on such a data structure to maintain the components of the intermediate spanning
forest. Another application is maintaining the connected components of a graph as new vertices and
edges are added. In both these applications, we can use a disjoint-set data structure, where we maintain
a set for each connected component, containing that component’s vertices. 10.1 Reversed Trees One of the easiest ways to store sets is using trees, in which each node represents a single element of
the set. Each node points to another node, called its parent, except for the leader of each set, which
points to itself and thus is the root of the tree. MAKESET is trivial. FIND traverses parent pointers up to
the leader. UNION just redirects the parent pointer of one leader to the other. Unlike most tree data
structures, nodes do not have pointers down to their children. 1 Algorithms Lecture 10: Disjoint Sets
FIND( x ):
while x = parent( x )
x ← parent( x )
return x MAKESET( x ):
parent( x ) ← x a
b UNION( x , y ):
x ← FIND( x )
y ← FIND( y )
parent( y ) ← x p
d q p
r a c b q r d c
Merging two sets stored as trees. Arrows point to parents. The shaded node has a new parent. MAKE-SET clearly takes Θ(1) time, and UNION requires only O(1) time in addition to the two FINDs.
The running time of FIND( x ) is proportional to the depth of x in the tree. It is not hard to come up with
a sequence of operations that results in a tree that is a long chain of nodes, so that FIND takes Θ(n ) time
in the worst case.
However, there is an easy change we can make to our UNION algorithm, called union by depth, so
that the trees always have logarithmic depth. Whenever we need to merge two trees, we always make
the root of the shallower tree a child of the deeper one. This requires us to also maintain the depth of
each tree, but this is quite easy. MAKESET( x ):
parent( x ) ← x
depth( x ) ← 0 FIND( x ):
while x = parent( x )
x ← parent( x )
return x UNION( x , y )
x ← FIND( x )
y ← FIND( y )
if depth( x ) > depth( y )
parent( y ) ← x
else
parent( x ) ← y
if depth( x ) = depth( y )
depth( y ) ← depth( y ) + 1 With this new rule in place, it’s not hard to prove by induction that for any set leader x , the size of
x ’s set is at least 2depth( x ) , as follows. If depth( x ) = 0, then x is the leader of a singleton set. For any
d > 0, when depth( x ) becomes d for the ﬁrst time, x is becoming the leader of the union of two sets,
both of whose leaders had depth d − 1. By the inductive hypothesis, both component sets had at least
2d −1 elements, so the new set has at least 2d elements. Later UNION operations might add elements to
x ’s set without changing its depth, but that only helps us.
Since there are only n elements altogether, the maximum depth of any set is lg n. We conclude that
if we use union by depth, both FIND and UNION run in Θ(log n ) time in the worst case. 10.2 Shallow Threaded Trees Alternately, we could just have every object keep a pointer to the leader of its set. Thus, each set is
represented by a shallow tree, where the leader is the root and all the other elements are its children.
With this representation, MAKESET and FIND are completely trivial. Both operations clearly run in
constant time. UNION is a little more difﬁcult, but not much. Our algorithm sets all the leader pointers
in one set to point to the leader of the other set. To do this, we need a method to visit every element
in a set; we will ‘thread’ a linked list through each set, starting at the set’s leader. The two threads are
merged in the UNION algorithm in constant time.
2 Algorithms Lecture 10: Disjoint Sets b c a p a
d q p r q r b c d Merging two sets stored as threaded trees.
Bold arrows point to leaders; lighter arrows form the threads. Shaded nodes have a new leader. UNION( x , y ):
x ← FIND( x )
y ← FIND( y )
MAKESET( x ):
leader( x ) ← x
next( x ) ← x FIND( x ):
return leader( x ) y←y
leader( y ) ← x
while (next( y ) = NULL)
y ← next( y )
leader( y ) ← x
next( y ) ← next( x )
next( x ) ← y The worst-case running time of UNION is a constant times the size of the larger set. Thus, if we merge
a one-element set with another n-element set, the running time can be Θ(n). Generalizing this idea, it is
quite easy to come up with a sequence of n MAKESET and n − 1 UNION operations that requires Θ(n2 )
time to create the set {1, 2, . . . , n} from scratch.
WORSTCASESEQUENCE(n):
MAKESET(1)
for i ← 2 to n
MAKESET(i )
UNION(1, i ) We are being stupid in two different ways here. One is the order of operations in WORSTCASESEQUENCE. Obviously, it would be more efﬁcient to merge the sets in the other order, or to use some sort
of divide and conquer approach. Unfortunately, we can’t ﬁx this; we don’t get to decide how our data
structures are used! The other is that we always update the leader pointers in the larger set. To ﬁx this,
we add a comparison inside the UNION algorithm to determine which set is smaller. This requires us to
maintain the size of each set, but that’s easy. MAKEWEIGHTEDSET( x ):
leader( x ) ← x
next( x ) ← x
size( x ) ← 1 WEIGHTEDUNION( x , y )
x ← FIND( x )
y ← FIND( y )
if size( x ) > size( y )
UNION( x , y )
size( x ) ← size( x ) + size( y )
else
UNION( y , x )
size( x ) ← size( x ) + size( y ) The new WEIGHTEDUNION algorithm still takes Θ(n) time to merge two n-element sets. However, in
an amortized sense, this algorithm is much more efﬁcient. Intuitively, before we can merge two large
sets, we have to perform a large number of MAKEWEIGHTEDSET operations.
Theorem 1. A sequence of m MAKEWEIGHTEDSET operations and n WEIGHTEDUNION operations takes
O(m + n log n) time in the worst case.
3 Algorithms Lecture 10: Disjoint Sets Proof: Whenever the leader of an object x is changed by a WEIGHTEDUNION, the size of the set containing
x increases by at least a factor of two. By induction, if the leader of x has changed k times, the set
containing x has at least 2k members. After the sequence ends, the largest set contains at most n
members. (Why?) Thus, the leader of any object x has changed at most lg n times.
Since each WEIGHTEDUNION reduces the number of sets by one, there are m − n sets at the end of the
sequence, and at most n objects are not in singleton sets. Since each of the non-singleton objects had
O(log n) leader changes, the total amount of work done in updating the leader pointers is O(n log n).
The aggregate method now implies that each WEIGHTEDUNION has amortized cost O (log n ). 10.3 Path Compression Using unthreaded tress, FIND takes logarithmic time and everything else is constant; using threaded
trees, UNION takes logarithmic amortized time and everything else is constant. A third method allows us
to get both of these operations to have almost constant running time.
We start with the original unthreaded tree representation, where every object points to a parent.
The key observation is that in any FIND operation, once we determine the leader of an object x , we can
speed up future FINDs by redirecting x ’s parent pointer directly to that leader. In fact, we can change the
parent pointers of all the ancestors of x all the way up to the root; this is easiest if we use recursion for
the initial traversal up the tree. This modiﬁcation to FIND is called path compression. p
a
b q p
c r d b a q r d c
Path compression during Find(c ). Shaded nodes have a new parent. FIND( x )
if x = parent( x )
parent( x ) ← FIND(parent( x ))
return parent( x ) If we use path compression, the ‘depth’ ﬁeld we used earlier to keep the trees shallow is no longer
correct, and correcting it would take way too long. But this information still ensures that FIND runs in
Θ(log n) time in the worst case, so we’ll just give it another name: rank. The following algorithm is
usually called union by rank: MAKESET( x ):
parent( x ) ← x
rank( x ) ← 0 UNION( x , y )
x ← FIND( x )
y ← FIND( y )
if rank( x ) > rank( y )
parent( y ) ← x
else
parent( x ) ← y
if rank( x ) = rank( y )
rank( y ) ← rank( y ) + 1 4 Algorithms Lecture 10: Disjoint Sets FIND still runs in O(log n) time in the worst case; path compression increases the cost by only most a
constant factor. But we have good reason to suspect that this upper bound is no longer tight. Our new
algorithm memoizes the results of each FIND, so if we are asked to FIND the same item twice in a row,
the second call returns in constant time. Splay trees used a similar strategy to achieve their optimal
amortized cost, but our up-trees have fewer constraints on their structure than binary search trees, so
we should get even better performance.
This intuition is exactly correct, but it takes a bit of work to deﬁne precisely how much better the
performance is. As a ﬁrst approximation, we will prove below that the amortized cost of a FIND operation
is bounded by the iterated logarithm of n, denoted log∗ n, which is the number of times one must take
the logarithm of n before the value is less than 1:
lg∗ n = if n ≤ 2, 1
∗ 1 + lg (lg n) otherwise. Our proof relies on several useful properties of ranks, which follow directly from the UNION and FIND
algorithms.
• If a node x is not a set leader, then the rank of x is smaller than the rank of its parent.
• Whenever parent( x ) changes, the new parent has larger rank than the old parent.
• Whenever the leader of x ’s set changes, the new leader has larger rank than the old leader.
• The size of any set is exponential in the rank of its leader: size( x ) ≥ 2rank( x ) . (This is easy to prove
by induction, hint, hint.)
• In particular, since there are only n objects, the highest possible rank is lg n .
• For any integer r , there are at most n/2 r objects of rank r .
Only the last property requires a clever argument to prove. Fix your favorite integer r . Observe that
only set leaders can change their rank. Whenever the rank of any set leader x changes from r − 1 to r ,
mark all the objects in x ’s set. Since leader ranks can only increase over time, each object is marked
at most once. There are n objects altogether, and any object with rank r marks at least 2 r objects. It
follows that there are at most n/2 r objects with rank r , as claimed. 10.4 O (log∗ n ) Amortized Time The following analysis of path compression was discovered just a few years ago by Raimund Seidel and
Micha Sharir.1 Previous proofs2 relied on complicated charging schemes or potential-function arguments;
Seidel and Sharir’s analysis relies on a comparatively simple recursive decomposition.
Seidel and Sharir phrase their analysis in terms of two more general operations on set forests. Their
more general COMPRESS operation compresses any directed path, not just paths that lead to the root.
The new SHATTER operation makes every node on a root-to-leaf path into its own parent.
COMPRESS( x , y ):
〈〈 y must be an ancestor of x 〉〉
if x = y
COMPRESS(parent( x ), y )
parent( x ) ← parent( y )
1
2 SHATTER( x ):
if parent( x ) = x
SHATTER(parent( x ))
parent( x ) ← x Raimund Seidel and Micha Sharir. Top-down analysis of path compression. SIAM J. Computing 34(3):515–525, 2005.
Robert E. Tarjan. Efﬁciency of a good but not linear set union algorithm. J. Assoc. Comput. Mach. 22:215–225, 1975. 5 Algorithms Lecture 10: Disjoint Sets Clearly, the running time of FIND( x ) operation is dominated by the running time of COMPRESS( x , y ),
where y is the leader of the set containing x . This implies that we can prove the upper bound by
analyzing an arbitrary sequence of UNION and COMPRESS operations. Moreover, we can assume that the
arguments to each UNION operation are set leaders, so that each UNION takes only constant worst-case
time.
Finally, since each call to COMPRESS speciﬁes the top node in the path to be compressed, we can
reorder the sequence of operations, so that every UNION occurs before any COMPRESS, without changing
the number of pointer assignments. y x y
x y x x y y
y
x
x
Top row: A COMPRESS followed by a UNION. Bottom row: The same operations in the opposite order. Each UNION requires only constant time in the worst case, so we only need to analyze the amortized
cost of COMPRESS. The running time of COMPRESS is proportional to the number of parent pointer
assignments, plus O(1) overhead, so we will phrase our analysis in terms of pointer assignments. Let
T (m , n , r ) denote the worst case number of pointer assignments in any sequence of at most m COMPRESS
operations, executed on a forest of at most n nodes, with maximum rank at most r .
The following trivial upper bound will be the base case for our recursive argument.
Theorem 2. T (m, n, r ) ≤ nr
Proof: Each node can change parents at most r times, because each new parent has higher rank than
the previous parent.
Fix a forest F of n nodes with maximum rank r , and a sequence C of m COMPRESS operations on F ,
and let T ( F, C ) denote the total number of pointer assignments executed by this sequence.
Let s be an arbitrary positive rank. Partition F into two sub-forests: a ‘low’ forest F− containing all
nodes with rank at most s, and a ‘high’ forest F+ containing all nodes with rank greater than s. Since
ranks increase as we follow parent pointers, every ancestor of a high node is another high node. Let n−
and n+ denote the number of nodes in F− and F+ , respectively. Finally, let m+ denote the number of
COMPRESS operations that involve any node in F+ , and let m− = m − m+ . F F+ rank ≥ s
rank < s rank ≥ s
rank < s F–
Splitting the forest F (in this case, a single tree) into sub-forests F+ and F− at rank s. 6 Algorithms Lecture 10: Disjoint Sets Any sequence of COMPRESS operations on F can be decomposed into a sequence of COMPRESS
operations on F+ , plus a sequence of COMPRESS and SHATTER operations on F− , with the same total
cost. This requires only one small modiﬁcation to the code: We forbid any low node from having a high
parent. Speciﬁcally, if x is a low node and y is a high node, we replace any assignment parent( x ) ← y
with parent( x ) ← x . A Compress operation in F splits into a Compress operation in F+ and a Shatter operation in F− This modiﬁcation is equivalent to the following reduction:
COMPRESS( x , y, F ):
〈〈 y is an ancestor of x 〉〉
if rank( x ) > r
COMPRESS( x , y, F+ )
〈〈in C+ 〉〉
else if rank( y ) ≤ r
COMPRESS( x , y, F− )
〈〈in C− 〉〉
else
z ← highest ancestor of x in F with rank at most r
COMPRESS(parent F (z ), y, F+ )
〈〈in C+ 〉〉
SHATTER( x , z , F− )
parent(z ) ← z
(∗) The pointer assignment in the last line looks redundant, but it is actually necessary for the analysis.
Each execution of line (∗) mirrors an assignment of the form parent( x ) ← y , where x is a low node, y is a
high node, and the previous parent of x was a high node. Each of these ‘redundant’ assignments happens
immediately after a COMPRESS in the top forest, so we perform at most m+ redundant assignments.
Each node x is touched by at most one SHATTER operation, so the total number of pointer reassignments in all the SHATTER operations is at most n.
Thus, by partitioning the forest F into F+ and F− , we have also partitioned the sequence C of
COMPRESS operations into subsequences C+ and C− , with respective lengths m+ and m− , such that the
following inequality holds:
T ( F, C ) ≤ T ( F+ , C+ ) + T ( F− , C− ) + m+ + n
Since there are only n/2i nodes of any rank i , we have n+ ≤ i >s n/2i = n/2s . The number of
different ranks in F+ is r − s < r . Thus, Theorem 2 implies the upper bound
T ( F+ , C+ ) < r n/2s .
Let us ﬁx s = lg r , so that T ( F+ , C+ ) ≤ n. We can now simplify our earlier recurrence to
T ( F, C ) ≤ T ( F− , C− ) + m+ + 2n,
7 Algorithms Lecture 10: Disjoint Sets or equivalently,
T ( F, C ) − m ≤ T ( F− , C− ) − m− + 2n.
Since this argument applies to any forest F and any sequence C , we have just proved that
T (m, n, r ) ≤ T (m, n, lg r ) + 2n ,
where T (m, n, r ) = T (m, n, r ) − m. The solution to this recurrence is T (n, m, r ) ≤ 2n lg∗ r . Voilá!
Theorem 3. T (m, n, r ) ≤ m + 2n lg∗ r 10.5 Turning the Crank There is one place in the preceding analysis where we have signiﬁcant room for improvement. Recall
that we bounded the total cost of the operations on F+ using the trivial upper bound from Theorem 2.
But we just proved a better upper bound in Theorem 3! We can apply precisely the same strategy, using
Theorem 3 instead of Theorem 2, to improve the bound even more.
∗
Suppose we ﬁx s = lg∗ r , so that n+ = n/2lg r . Theorem 3 implies that
T ( F+ , C+ ) ≤ m+ + 2n lg∗ r
2lg ∗ ≤ m+ + 2n. r This implies the recurrence
T ( F, C ) ≤ T ( F− , C− ) + 2m+ + 3n,
which in turn implies that
T (m, n, r ) ≤ T (m, n, lg∗ r ) + 3n,
where T (m, n, r ) = T (m, n, r ) − 2m. The solution to this equation is T (m, n, r ) ≤ 2m + 3n lg∗∗ r ,
where lg∗∗ r is the iterated iterated logarithm of r :
lg∗∗ r = if r ≤ 2, 1
∗∗ ∗ 1 + lg (lg r ) otherwise. Naturally we can apply the same improvement strategy again, and again, as many times as we like,
each time producing a tighter upper bound. Applying the reduction c times, for any positive integer c ,
gives us
c T (m, n, r ) ≤ cm + (c + 1)n lg∗ r
where
∗c lg r = lg r if c = 0, 1
if r ≤ 2, ∗c
∗ c −1
1 + lg (lg
r ) otherwise. Each time we ‘turn the crank’, the dependence on m increases, while the dependence on n and
r decreases. For sufﬁciently large values of c , the cm term dominates the time bound, and further
iterations only make things worse. The point of diminishing returns can be estimated by the minimum
number of stars such that lg∗∗···∗ r is smaller than a constant:
c α( r ) = min c ≥ 1 lg∗ n ≤ 3 .
c (The threshold value 3 is used here because lg∗ 5 ≥ 2 for all c .) By setting c = α( r ), we obtain our ﬁnal
upper bound.
8 Algorithms Lecture 10: Disjoint Sets Theorem 4. T (m, n, r ) ≤ mα( r ) + 3n(α( r ) + 1)
We can assume without loss of generality that m ≥ n by ignoring any singleton sets, so this upper
bound can be further simpliﬁed to T (m, n, r ) = O(mα( r )) = O(mα(n)). It follows that if we use union
by rank, FIND with path compression runs in O(α(n)) amortized time .
Even this upper bound is somewhat conservative if m is larger than n. A closer estimate is given by
the function
c
α(m, n) = min c ≥ 1 log∗ (lg n) ≤ m/n .
It’s not hard to prove that if m = Θ(n), then α(m, n) = Θ(α(n)). On the other hand, if m ≥ n lg∗∗∗∗∗ n,
for any constant number of stars, then α(m, n) = O(1). So even if the number of FIND operations is only
slightly larger than the number of nodes, the amortized cost of each FIND is constant.
O(α(m, n)) is actually a tight upper bound for the amortized cost of path compression; there are no
more tricks that will improve the analysis further. More surprisingly, this is the best amortized bound we
obtain for any pointer-based data structure for maintaining disjoint sets; the amortized cost of every FIND
algorithm is at least Ω(α(m, n)). The proof of the matching lower bound is, unfortunately, far beyond
the scope of this class.3 10.6 The Ackermann Function and its Inverse The iterated logarithms that fell out of our analysis of path compression are the inverses of a hierarchy
of recursive functions deﬁned by Wilhelm Ackermann in 1928.4 c 2↑ n= 2 if n = 1 2n
if c = 0 c −1
c
2↑
(2 ↑ (n − 1)) otherwise For each ﬁxed c , the function 2 ↑c n is monotonically increasing in n, and these functions grow incredibly
faster as the index c increases. 2 ↑ n is the familiar power function 2n . 2 ↑↑ n is the tower function
22 .2
.. n 2
; this function is also sometimes called tetration. John Conway named 2 ↑↑↑ n the wower function:
2 ↑↑↑ n = 2 ↑↑ 2 ↑↑ · · · ↑↑ 2. And so on, et cetera, ad inﬁnitum.
n For any ﬁxed c , the function log∗ n is the inverse of the function 2 ↑c +1 n, the (c + 1)th row in the
Ackerman hierarchy. Thus, for any remotely reasonable values of n, say n ≤ 2256 , we have log∗ n ≤ 5,
c
log∗∗ n ≤ 4, and log∗ n ≤ 3 for any c ≥ 3.
The function α(n) is usually called the inverse Ackerman function.5 Our earlier deﬁnition is equivalent
to α(n) = min{c ≥ 1 | 2 ↑c +2 3 ≥ n}; in other words, α(n) + 2 is the inverse of the third column in the
c
Ackermann hierarchy. The function α(n) grows much more slowly than log∗ n for any ﬁxed c ; we have
c 3
Robert E. Tarjan. A class of algorithms which require non-linear time to maintain disjoint sets. J. Comput. Syst. Sci.
19:110–127, 1979.
4
Ackermann didn’t deﬁne his functions this way—I’m actually describing a slightly cleaner hierarchy deﬁned 35 years later
by R. Creighton Buck—but the exact details of the deﬁnition are surprisingly irrelevant! The mnemonic up-arrow notation for
these functions was introduced by Don Knuth in the 1970s.
5
Strictly speaking, the name ‘inverse Ackerman function’ is inaccurate. One good formal deﬁnition of the true inverse
c
˜
Ackerman function is α(n) = min c ≥ 1 lg∗ n ≤ c = min c ≥ 1 2 ↑c +2 c ≥ n . However, it’s not hard to prove that
˜
˜
α(n) ≤ α(n) ≤ α(n) + 1 for all sufﬁciently large n, so the inaccuracy is completely forgivable. As I said in the previous footnote,
the exact details of the deﬁnition are surprisingly irrelevant! 9 Algorithms Lecture 10: Disjoint Sets α(n) ≤ 3 for all even remotely imaginable values of n. Nevertheless, the function α(n) is eventually
larger than any constant, so it is not O(1).
2 ↑c n n=1 2 n=3 n=4 n=5 2n 2 4 6 8 10 2↑n 2 4 8 16 32 2 ↑↑ n 2 4 16 65536 265536 2 ↑↑↑ n
2 ↑↑↑↑ n
2 ↑↑↑↑↑ n .2
.. 2
2 2 4 65536
22 4 4 22 .2
.. 2 2 2...2 2...2 .
.. 2...2 2 65536 2 65536 22 2
..
2. 2...2 2...2 2...2
.
.. 65536 65536 22 2
2
..
2. 22 .2
.. 2.
22 2
.. 65536 65536 〈〈Yeah, right.〉〉 65536 〈〈Very funny.〉〉 〈〈Argh! My eyes!〉〉 Small (!!) values of Ackermann’s functions. 10.7 To inﬁnity. . . and beyond! Of course, one can generalize the inverse Ackermann function to functions that grow arbitrarily more
slowly, starting with the iterated inverse Ackermann function
α∗ (n) = if n ≤ 4 1
∗ 1 + α (α(n)) otherwise, then the iterated iterated inverse Ackermann function
α∗∗ (n) = if n ≤ 4 1
∗∗ 1 + α (α(n)) otherwise, and then the diagonalized inverse Ackermann function
c Head-asplode(n) = min{c ≥ 1 | α∗ n ≤ 4},
and so on forever. Fortunately(?), such functions appear extremely rarely in algorithm analysis. In fact,
the only example (as far as Jeff knows) of a naturally-occurring super-constant sub-inverse-Ackermann
function is a recent result of Seth Pettie6 , who proved that if a splay tree is used as a double-ended
queue — insertions and deletions of only smallest or largest elements — then the amortized cost of any
operation is O(α∗ (n)). Exercises
1. Consider the following solution for the union-ﬁnd problem, called union-by-weight. Each set leader
x stores the number of elements of its set in the ﬁeld weight( x ). Whenever we UNION two sets,
the leader of the smaller set becomes a new child of the leader of the larger set (breaking ties
arbitrarily). 6
Splay trees, Davenport-Schinzel sequences, and the deque conjecture. Proceedings of the 19th Annual ACM-SIAM
Symposium on Discrete Algorithms, 1115–1124, 2008. 10 Algorithms Lecture 10: Disjoint Sets MAKESET( x ):
parent( x ) ← x
weight( x ) ← 1
FIND( x ):
while x = parent( x )
x ← parent( x )
return x UNION( x , y )
x ← FIND( x )
y ← FIND( y )
if weight( x ) > weight( y )
parent( y ) ← x
weight( x ) ← weight( x ) + weight( y )
else
parent( x ) ← y
weight( x ) ← weight( x ) + weight( y ) Prove that if we use union-by-weight, the worst-case running time of FIND( x ) is O(log n),
where n is the cardinality of the set containing x .
2. Consider a union-ﬁnd data structure that uses union by depth (or equivalently union by rank)
without path compression. For all integers m and n such that m ≥ 2n, prove that there is a sequence
of n MakeSet operations, followed by m UNION and FIND operations, that require Ω(m log n) time
to execute.
3. Consider an arbitrary sequence of m MAKESET operations, followed by u UNION operations, followed
by f FIND operations, and let n = m + u + f . Prove that if we use union by rank and FIND with
path compression, all n operations are executed in O(n) time.
4. Describe and analyze a data structure to support the following operations on an array X [1 .. n] as
quickly as possible. Initially, X [i ] = 0 for all i .
• Given an index i such that X [i ] = 0, set X [i ] to 1.
• Given an index i , return X [i ].
• Given an index i , return the smallest index j ≥ i such that X [ j ] = 0, or report that no such
index exists.
For full credit, the ﬁrst two operations should run in worst-case constant time, and the amortized
cost of the third operation should be as small as possible.
5. (a) Describe and analyze an algorithm to compute the size of the largest connected component
of black pixels in an n × n bitmap B [1 .. n, 1 .. n].
For example, given the bitmap below as input, your algorithm should return the number 9,
because the largest conected black component (marked with white dots on the right) contains
nine pixels. 9 (b) Design and analyze an algorithm BLACKEN(i , j ) that colors the pixel B [i , j ] black and returns
the size of the largest black component in the bitmap. For full credit, the amortized running
time of your algorithm (starting with an all-white bitmap) must be as small as possible.
11 Algorithms Lecture 10: Disjoint Sets For example, at each step in the sequence below, we blacken the pixel marked with an X.
The largest black component is marked with white dots; the number underneath shows the
correct output of the BLACKEN algorithm. 9 14 14 16 17 (c) What is the worst-case running time of your BLACKEN algorithm?
6. Consider the following game. I choose a positive integer n and keep it secret; your goal is to
discover this integer. We play the game in rounds. In each round, you write a list of at most n
integers on the blackboard. If you write more than n numbers in a single round, you lose. (Thus,
in the ﬁrst round, you must write only the number 1; do you see why?) If n is one of the numbers
you wrote, you win the game; otherwise, I announce which of the numbers you wrote is smaller
or larger than n, and we proceed to the next round. For example:
You Me 1 It’s bigger than 1. 4, 42 It’s between 4 and 42. 8, 15, 16, 23, 30 It’s between 8 and 15. 9, 10, 11, 12, 13, 14 It’s 11; you win! Describe a strategy that allows you to win in O(α(n)) rounds! c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 12 Algorithms Non-Lecture G: String Matching
Why are our days numbered and not, say, lettered?
— Woody Allen G String Matching G.1 Brute Force The basic object that we’re going to talk about for the next two lectures is a string, which is really just
an array. The elements of the array come from a set Σ called the alphabet; the elements themselves
are called characters. Common examples are ASCII text, where each character is an seven-bit integer1 ,
strands of DNA, where the alphabet is the set of nucleotides {A, C , G , T }, or proteins, where the alphabet
is the set of 22 amino acids.
The problem we want to solve is the following. Given two strings, a text T [1 .. n] and a pattern
P [1 .. m], ﬁnd the ﬁrst substring of the text that is the same as the pattern. (It would be easy to extend
our algorithms to ﬁnd all matching substrings, but we will resist.) A substring is just a contiguous
subarray. For any shift s, let Ts denote the substring T [s .. s + m − 1]. So more formally, we want to
ﬁnd the smallest shift s such that Ts = P , or report that there is no match. For example, if the text is
the string ‘AMANAPLANACATACANALPANAMA’2 and the pattern is ‘CAN’, then the output should be 15. If the
pattern is ‘SPAM’, then the answer should be ‘none’. In most cases the pattern is much smaller than the
text; to make this concrete, I’ll assume that m < n/2.
Here’s the ‘obvious’ brute force algorithm, but with one immediate improvement. The inner while
loop compares the substring Ts with P . If the two strings are not equal, this loop stops at the ﬁrst
character mismatch.
ALMOSTBRUTEFORCE( T [1 .. n], P [1 .. m]):
for s ← 1 to n − m + 1
equal ← true
i←1
while equal and i ≤ m
if T [s + i − 1] = P [i ]
equal ← false
else
i ← i+1
if equal
return s
return ‘none’
1 Yes, seven. Most computer systems use some sort of 8-bit character set, but there’s no universally accepted standard.
Java supposedly uses the Unicode character set, which has variable-length characters and therefore doesn’t really ﬁt into our
framework. Just think, someday you’ll be able to write ‘¶ = ℵ[∞++]/ ;’ in your Java code! Joy!
2
Dan Hoey (or rather, his computer program) found the following 540-word palindrome in 1984:
A man, a plan, a caret, a ban, a myriad, a sum, a lac, a liar, a hoop, a pint, a catalpa, a gas, an oil, a bird, a yell, a vat, a caw, a pax, a wag, a tax, a nay, a ram, a cap,
a yam, a gay, a tsar, a wall, a car, a luger, a ward, a bin, a woman, a vassal, a wolf, a tuna, a nit, a pall, a fret, a watt, a bay, a daub, a tan, a cab, a datum, a gall, a
hat, a fag, a zap, a say, a jaw, a lay, a wet, a gallop, a tug, a trot, a trap, a tram, a torr, a caper, a top, a tonk, a toll, a ball, a fair, a sax, a minim, a tenor, a bass, a
passer, a capital, a rut, an amen, a ted, a cabal, a tang, a sun, an ass, a maw, a sag, a jam, a dam, a sub, a salt, an axon, a sail, an ad, a wadi, a radian, a room, a
rood, a rip, a tad, a pariah, a revel, a reel, a reed, a pool, a plug, a pin, a peek, a parabola, a dog, a pat, a cud, a nu, a fan, a pal, a rum, a nod, an eta, a lag, an eel, a
batik, a mug, a mot, a nap, a maxim, a mood, a leek, a grub, a gob, a gel, a drab, a citadel, a total, a cedar, a tap, a gag, a rat, a manor, a bar, a gal, a cola, a pap, a
yaw, a tab, a raj, a gab, a nag, a pagan, a bag, a jar, a bat, a way, a papa, a local, a gar, a baron, a mat, a rag, a gap, a tar, a decal, a tot, a led, a tic, a bard, a leg, a
bog, a burg, a keel, a doom, a mix, a map, an atom, a gum, a kit, a baleen, a gala, a ten, a don, a mural, a pan, a faun, a ducat, a pagoda, a lob, a rap, a keep, a nip,
a gulp, a loop, a deer, a leer, a lever, a hair, a pad, a tapir, a door, a moor, an aid, a raid, a wad, an alias, an ox, an atlas, a bus, a madam, a jag, a saw, a mass, an
anus, a gnat, a lab, a cadet, an em, a natural, a tip, a caress, a pass, a baronet, a minimax, a sari, a fall, a ballot, a knot, a pot, a rep, a carrot, a mart, a part, a tort,
a gut, a poll, a gateway, a law, a jay, a sap, a zag, a fat, a hall, a gamut, a dab, a can, a tabu, a day, a batt, a waterfall, a patina, a nut, a ﬂow, a lass, a van, a mow,
a nib, a draw, a regular, a call, a war, a stay, a gam, a yap, a cam, a ray, an ax, a tag, a wax, a paw, a cat, a valley, a drib, a lion, a saga, a plat, a catnip, a pooh, a
rail, a calamus, a dairyman, a bater, a canal—Panama! Peter Norvig generated even longer examples in 2002; see http://norvig.com/palindrome.html. 1 Algorithms Non-Lecture G: String Matching In the worst case, the running time of this algorithm is O((n − m)m) = O(nm), and we can actually
achieve this running time by searching for the pattern AAA...AAAB with m − 1 A’s, in a text consisting
entirely of n A’s.
In practice, though, breaking out of the inner loop at the ﬁrst mismatch makes this algorithm quite
practical. We can wave our hands at this by assuming that the text and pattern are both random. Then
on average, we perform a constant number of comparisons at each position i , so the total expected
number of comparisons is O(n). Of course, neither English nor DNA is really random, so this is only a
heuristic argument. G.2 Strings as Numbers For the rest of the lecture, let’s assume that the alphabet consists of the numbers 0 through 9, so we
can interpret any array of characters as either a string or a decimal number. In particular, let p be the
numerical value of the pattern P , and for any shift s, let t s be the numerical value of Ts :
m m 10m−i · P [i ] p= 10m−i · T [s + i − 1] ts = i =1 i =1 For example, if T = 31415926535897932384626433832795028841971 and m = 4, then t 17 =
2384.
Clearly we can rephrase our problem as follows: Find the smallest s, if any, such that p = t s . We can
compute p in O(m) arithmetic operations, without having to explicitly compute powers of ten, using
Horner’s rule:
p = P [m] + 10 P [m − 1] + 10 P [m − 2] + · · · + 10 P [2] + 10 · P [1] · · ·
We could also compute any t s in O(m) operations using Horner’s rule, but this leads to essentially the
same brute-force algorithm as before. But once we know t s , we can actually compute t s+1 in constant
time just by doing a little arithmetic — subtract off the most signiﬁcant digit T [s] · 10m−1 , shift everything
up by one digit, and add the new least signiﬁcant digit T [ r + m]:
t s+1 = 10 t s − 10m−1 · T [s] + T [s + m]
To make this fast, we need to precompute the constant 10m−1 . (And we know how to do that quickly.
Right?) So it seems that we can solve the string matching problem in O(n) worst-case time using the
following algorithm:
NUMBERSEARCH( T [1 .. n], P [1 .. m]):
σ ← 10m−1
p←0
t1 ← 0
for i ← 1 to m
p ← 10 · p + P [i ]
t 1 ← 10 · t 1 + T [i ]
for s ← 1 to n − m + 1
if p = t s
return s
t s+1 ← 10 · t s − σ · T [s] + T [s + m]
return ‘none’ Unfortunately, the most we can say is that the number of arithmetic operations is O(n). These operations
act on numbers with up to m digits. Since we want to handle arbitrarily long patterns, we can’t assume
that each operation takes only constant time!
2 Algorithms G.3 Non-Lecture G: String Matching Karp-Rabin Fingerprinting To make this algorithm efﬁcient, we will make one simple change, discovered by Richard Karp and
Michael Rabin in 1981:
Perform all arithmetic modulo some prime number q.
We choose q so that the value 10q ﬁts into a standard integer variable, so that we don’t need any fancy
long-integer data types. The values ( p mod q) and ( t s mod q) are called the ﬁngerprints of P and Ts ,
respectively. We can now compute ( p mod q) and ( t 1 mod q) in O(m) time using Horner’s rule ‘mod q’
p mod q = P [m] + · · · + 10 · P [2] + 10 · P [1] mod q mod q mod q · · · mod q and similarly, given ( t s mod q), we can compute ( t s+1 mod q) in constant time.
t s+1 mod q = 10 · t s − 10m−1 mod q · T [s] mod q mod q mod q + T [s + m] mod q Again, we have to precompute the value (10m−1 mod q) to make this fast.
If ( p mod q) = ( t s mod q), then certainly P = Ts . However, if ( p mod q) = ( t s mod q), we can’t tell
whether P = Ts or not. All we know for sure is that p and t s differ by some integer multiple of q. If
P = Ts in this case, we say there is a false match at shift s. To test for a false match, we simply do a
brute-force string comparison. (In the algorithm below, ˜ = p mod q and ˜s = t s mod q.)
p
t
KARPRABIN( T [1 .. n], P [1 .. m]:
choose a small prime q
σ ← 10m−1 mod q
˜←0
p
˜1 ← 0
t
for i ← 1 to m
˜ ← (10 · ˜ mod q) + P [i ] mod q
p
p
˜1 ← (10 · ˜1 mod q) + T [i ] mod q
t
t
for s ← 1 to n − m + 1
if ˜ = ˜s
pt
if P = Ts
〈〈brute-force O(m)-time comparison〉〉
return s
˜s+1 ← 10 · ˜s − σ · T [s] mod q mod q mod q + T [s + m] mod q
t
t
return ‘none’ The running time of this algorithm is O(n + F m), where F is the number of false matches.
Intuitively, we expect the ﬁngerprints t s to jump around between 0 and q − 1 more or less at random,
so the ‘probability’ of a false match ‘ought’ to be 1/q. This intuition implies that F = n/q ‘on average’,
which gives us an ‘expected’ running time of O(n + nm/q). If we always choose q ≥ m, this simpliﬁes to
O(n). But of course all this intuitive talk of probabilities is just frantic meaningless handwaving, since
we haven’t actually done anything random yet. G.4 Random Prime Number Facts The real power of the Karp-Rabin algorithm is that by choosing the modulus q randomly, we can actually
formalize this intuition! The ﬁrst line of KARPRABIN should really read as follows:
Let q be a random prime number less than nm2 log(nm2 ).
3 Algorithms Non-Lecture G: String Matching For any positive integer u, let π(u) denote the number of prime numbers less than u. There are
π(nm2 log nm2 ) possible values for q, each with the same probability of being chosen.
Our analysis needs two results from number theory. I won’t even try to prove the ﬁrst one, but the
second one is quite easy.
Lemma 1 (The Prime Number Theorem). π(u) = Θ(u/ log u).
Lemma 2. Any integer x has at most lg x distinct prime divisors.
Proof: If x has k distinct prime divisors, then x ≥ 2k , since every prime number is bigger than 1.
Let’s assume that there are no true matches, so p = t s for all s. (That’s the worst case for the
algorithm anyway.) Let’s deﬁne a strange variable X as follows:
n−m+1 X= p − ts .
s =1 Notice that by our assumption, X can’t be zero.
Now suppose we have false match at shift s. Then p mod q = t s mod q, so p − t s is an integer multiple
of q, and this implies that X is also an integer multiple of q. In other words, if there is a false match,
then q must one of the prime divisors of X .
Since p < 10m and t s < 10m , we must have X < 10nm . Thus, by the second lemma, X has O(mn)
prime divisors. Since we chose q randomly from a set of π(nm2 log(nm2 )) = Ω(nm2 ) prime numbers,
the probability that q divides X is at most
O(nm)
Ω(nm2 ) =O 1
m . We have just proven the following amazing fact.
The probability of getting a false match is O(1/m).
Recall that the running time of KARPRABIN is O(n + mF ), where F is the number of false matches. By
using the really loose upper bound E[ F ] ≤ Pr[ F > 0] · n, we can conclude that the expected number of
false matches is O(n/m). Thus, the expected running time of the KARPRABIN algorithm is O(n). G.5 Random Prime Number? Actually choosing a random prime number is not particularly easy. The best method known is to
repeatedly generate a random integer and test to see if it’s prime. In practice, it’s enough to choose a
random probable prime. You can read about probable primes in the lecture notes on number-theoretic
algorithms, or in the textbook Randomized Algorithms by Rajeev Motwani and Prabhakar Raghavan
(Cambridge, 1995). Exercises
1. Describe and analyze a two-dimensional variant of KARPRABIN that searches for a given twodimensional pattern P [1 .. p][1 .. q] within a given two-dimensional ‘text’ T [1 .. m][1 .., n]. Your
algorithm should report all index pairs (i , j ) such that the subarray T [i .. i + p − 1][ j .. j + q − 1] is
identical to the given pattern, in O( pq + mn) expected time.
4 Algorithms Non-Lecture G: String Matching 2. Describe and analyze a variant of KARPRABIN that looks for strings inside labeled rooted trees. The
input consists of a pattern string P [1 .. m] and a rooted text tree T with n nodes, each labeled with
a single character. Nodes in T can have any number of children. Your algorithm should either
return a downward path in T whose labels match the string P , or report that there is no such path.
The expected running time of your algorithm should be O(m + n).
P
H E
I S A N
A Z
R E
E L M A
R C
W M Q S O K
H
F The string SEARCH appears on a downward path in the tree. 3. Describe and analyze a variant of KARPRABIN that searches for subtrees of ordered rooted binary
trees (every node has a left subtree and a right subtree, either or both of which may be empty).
The input consists of a pattern tree P with m nodes and a text tree T with n nodes. Your algorithm
should report all nodes v in T such that the subtree rooted at v is structurally identical to P . The
expected running time of your algorithm should be O(m + n). Ignore all search keys, labels, or
other data in the nodes; only the left/right pointer structure matters. The pattern tree (left) appears exactly twice in the text tree (right). 4. How important is the requirement that the ﬁngerprint modulus q is prime? Speciﬁcally, suppose q
is chosen uniformly at random in the range 1 .. N . If t s = p, what is the probability that ˜s = ˜ ?
t
p
What does this imply about the expected number of false matches? How large should N be to
guarantee expected running time O(m + n)? [Hint: This will require some additional number
theory.] 5 Algorithms Non-Lecture H: More String Matching
Philosophers gathered from far and near
To sit at his feat and hear and hear,
Though he never was heard
To utter a word
But “Abracadabra, abracadab,
Abracada, abracad,
Abraca, abrac, abra, ab!”
’Twas all he had,
’Twas all they wanted to hear, and each
Made copious notes of the mystical speech,
Which they published next –
A trickle of text
In the meadow of commentary.
Mighty big books were these,
In a number, as leaves of trees;
In learning, remarkably – very!
— Jamrach Holobom, quoted by Ambrose Bierce,
The Devil’s Dictionary (1911) H
H.1 More String Matching
Redundant Comparisons Let’s go back to the character-by-character method for string matching. Suppose we are looking for the
pattern ‘ABRACADABRA’ in some longer text using the (almost) brute force algorithm described in the
previous lecture. Suppose also that when s = 11, the substring comparison fails at the ﬁfth position;
the corresponding character in the text (just after the vertical line below) is not a C. At this point, our
algorithm would increment s and start the substring comparison from scratch.
HOCUSPOCUSABRA BRACADABRA...
ABRA /ADABRA
C
ABR ACADABRA If we look carefully at the text and the pattern, however, we should notice right away that there’s no
point in looking at s = 12. We already know that the next character is a B — after all, it matched P [2]
during the previous comparison — so why bother even looking there? Likewise, we already know that
the next two shifts s = 13 and s = 14 will also fail, so why bother looking there?
HOCUSPOCUSABRA BRACADABRA...
ABRA /ADABRA
C
A
/BR ACADABRA
A
/B RACADABRA
A BRACADABRA Finally, when we get to s = 15, we can’t immediately rule out a match based on earlier comparisons.
However, for precisely the same reason, we shouldn’t start the substring comparison over from scratch
— we already know that T [15] = P [4] = A. Instead, we should start the substring comparison at the
second character of the pattern, since we don’t yet know whether or not it matches the corresponding
text character.
If you play with this idea long enough, you’ll notice that the character comparisons should always
advance through the text. Once we’ve found a match for a text character, we never need to do
another comparison with that text character again. In other words, we should be able to optimize
the brute-force algorithm so that it always advances through the text.
1 Algorithms Non-Lecture H: More String Matching You’ll also eventually notice a good rule for ﬁnding the next ‘reasonable’ shift s. A preﬁx of a string
is a substring that includes the ﬁrst character; a sufﬁx is a substring that includes the last character. A
preﬁx or sufﬁx is proper if it is not the entire string. Suppose we have just discovered that T [i ] = P [ j ].
The next reasonable shift is the smallest value of s such that T [ s .. i − 1], which is a sufﬁx of the
previously-read text, is also a proper preﬁx of the pattern.
In this lecture, we’ll describe a string matching algorithm, published by Donald Knuth, James Morris,
and Vaughn Pratt in 1977, that implements both of these ideas. H.2 Finite State Machines If we have a string matching algorithm that follows our ﬁrst observation (that we always advance
through the text), we can interpret it as feeding the text through a special type of ﬁnite-state machine. A
ﬁnite state machine is a directed graph. Each node in the graph, or state, is labeled with a character
from the pattern, except for two special nodes labeled $ and ! . Each node has two outgoing edges, a
success edge and a failure edge. The success edges deﬁne a path through the characters of the pattern in
order, starting at $ and ending at ! . Failure edges always point to earlier characters in the pattern.
! $ A B A R R A
B C
A D A A ﬁnite state machine for the string ‘ABRADACABRA’.
Thick arrows are the success edges; thin arrows are the failure edges. We use the ﬁnite state machine to search for the pattern as follows. At all times, we have a current
text character T [i ] and a current node in the graph, which is usually labeled by some pattern character
P [ j ]. We iterate the following rules:
• If T [i ] = P [ j ], or if the current label is $ , follow the success edge to the next node and increment
i . (So there is no failure edge from the start node $ .)
• If T [i ] = P [ j ], follow the failure edge back to an earlier node, but do not change i .
For the moment, let’s simply assume that the failure edges are deﬁned correctly—we’ll come back to
this later. If we ever reach the node labeled ! , then we’ve found an instance of the pattern in the text,
and if we run out of text characters (i > n) before we reach ! , then there is no match.
The ﬁnite state machine is really just a (very!) convenient metaphor. In a real implementation, we
would not construct the entire graph. Since the success edges always go through the pattern characters
in order, we only have to remember where the failure edges go. We can encode this failure function in an
array fail[1 .. n], so that for each j there is a failure edge from node j to node fail[ j ]. Following a failure
edge back to an earlier state exactly corresponds, in our earlier formulation, to shifting the pattern
forward. The failure function fail[ j ] tells us how far to shift after a character mismatch T [i ] = P [ j ].
Here’s what the actual algorithm looks like: 2 Algorithms Non-Lecture H: More String Matching
KNUTHMORRISPRATT( T [1 .. n], P [1 .. m]):
j←1
for i ← 1 to n
while j > 0 and T [i ] = P [ j ]
j ← fail[ j ]
if j = m
〈〈Found it!〉〉
return i − m + 1
j ← j+1
return ‘none’ Before we discuss computing the failure function, let’s analyze the running time of KNUTHMORRISPRATT under the assumption that a correct failure function is already known. At each character
comparison, either we increase i and j by one, or we decrease j and leave i alone. We can increment i at
most n − 1 times before we run out of text, so there are at most n − 1 successful comparisons. Similarly,
there can be at most n − 1 failed comparisons, since the number of times we decrease j cannot exceed the
number of times we increment j . In other words, we can amortize character mismatches against earlier
character matches. Thus, the total number of character comparisons performed by KNUTHMORRISPRATT
in the worst case is O(n). H.3 Computing the Failure Function We can now rephrase our second intuitive rule about how to choose a reasonable shift after a character
mismatch T [i ] = P [ j ]:
P [1 .. fail[ j ] − 1] is the longest proper preﬁx of P [1 .. j − 1] that is also a sufﬁx of T [1 .. i − 1].
Notice, however, that if we are comparing T [i ] against P [ j ], then we must have already matched the
ﬁrst j − 1 characters of the pattern. In other words, we already know that P [1 .. j − 1] is a sufﬁx of
T [1 .. i − 1]. Thus, we can rephrase the preﬁx-sufﬁx rule as follows:
P [1 .. fail[ j ] − 1] is the longest proper preﬁx of P [1 .. j − 1] that is also a sufﬁx of P [1 .. j − 1].
This is the deﬁnition of the Knuth-Morris-Pratt failure function fail[ j ] for all j > 1.1 By convention we
set fail[1] = 0; this tells the KMP algorithm that if the ﬁrst pattern character doesn’t match, it should
just give up and try the next text character.
P [i ]
fail[i ] A B R A C A D A B R A 0 1 1 1 2 1 2 1 2 3 4 Failure function for the string ‘ABRACADABRA’
(Compare with the ﬁnite state machine on the previous page.) We could easily compute the failure function in O(m3 ) time by checking, for each j , whether every
preﬁx of P [1 .. j − 1] is also a sufﬁx of P [1 .. j − 1], but this is not the fastest method. The following
algorithm essentially uses the KMP search algorithm to look for the pattern inside itself!
1 Many algorithms textbooks, including CLRS, deﬁne a similar preﬁx function, denoted π[ j ], as follows:
P [1 .. π[ j ]] is the longest proper preﬁx of P [1 .. j ] that is also a sufﬁx of P [1 .. j ]. These two functions are not the same, but they are related by the simple equation π[ j ] = fail[ j + 1] − 1. The off-by-one
difference between the two functions adds a few extra +1s to the CLRS version of the algorithm. 3 Algorithms Non-Lecture H: More String Matching
COMPUTEFAILURE( P [1 .. m]):
j←0
for i ← 1 to m
fail[i ] ← j
(∗)
while j > 0 and P [i ] = P [ j ]
j ← fail[ j ]
j ← j+1 Here’s an example of this algorithm in action. In each line, the current values of i and j are indicated
by superscripts; $ represents the beginning of the string. (You should imagine pointing at P [ j ] with
your left hand and pointing at P [i ] with your right hand, and moving your ﬁngers according to the
algorithm’s directions.)
j ← 0, i ← 1
fail[i ] ← j
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← fail[ j ]
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← fail[ j ]
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← fail[ j ]
j ← fail[ j ]
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← fail[ j ]
j ← fail[ j ]
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← j + 1, i ← i + 1
fail[i ] ← j
j ← fail[ j ]
j ← fail[ j ] $j Ai B R A C A D A B R X ... R A C A D A B R X ... R A C A D A B R X ... A C A D A B R X ... A C A D A B R X ... C A D A B R X ... Ci A D A B R X ... A D A B R X ... 0 ... $ Aj Bi 0 1 $j A Bi ... $j Aj B Ri 0 $ 1 1 A B Ri ... $ Aj B R Ai 0 $ 1 1 1 A Bj R A ... 0 1 1 1 2 $ Aj B R A Ci $j A B R A Ci A D A B R X ... $ Aj B R A C Ai D A B R X ... 0 1 1 1 2 1 A Bj R A C A Di A B R X ... A B R X ... $ ... ... 0 1 1 1 2 1 2 $ Aj B R A C A Di $j A B R A C A Di A B R X ... $ Aj B R A C A D Ai B R X ... 0 1 1 1 2 1 2 1 A Bj R A C A D A Bi R X ... 0 1 1 1 2 1 2 1 2 A B Rj A C A D A B Ri X ... 0 1 1 1 2 1 2 1 2 3 ... A B R Aj C A D A B R Xi . . . 4 ... $
$
$ ... ...
... 0 1 1 1 2 1 2 1 2 3 $ Aj B R A C A D A B R Xi . . . $j A B R A C A D A B R Xi . . . ComputeFailure in action. Do this yourself by hand! Just as we did for KNUTHMORRISPRATT, we can analyze COMPUTEFAILURE by amortizing character mismatches against earlier character matches. Since there are at most m character matches, COMPUTEFAILURE
runs in O(m) time.
Let’s prove (by induction, of course) that COMPUTEFAILURE correctly computes the failure function.
The base case fail[1] = 0 is obvious. Assuming inductively that we correctly computed fail[1] through
fail[i ] in line (∗), we need to show that fail[i + 1] is also correct. Just after the i th iteration of line (∗),
we have j = fail[i ], so P [1 .. j − 1] is the longest proper preﬁx of P [1 .. i − 1] that is also a sufﬁx.
Let’s deﬁne the iterated failure functions failc [ j ] inductively as follows: fail0 [ j ] = j , and
c
c fail [ j ] = fail[ fail c −1 [ j ]] = fail[ fail[· · · [ fail[ j ]] · · · ]].
4 Algorithms Non-Lecture H: More String Matching In particular, if failc −1 [ j ] = 0, then failc [ j ] is undeﬁned. We can easily show by induction that every
string of the form P [1 .. failc [ j ] − 1] is both a proper preﬁx and a proper sufﬁx of P [1 .. i − 1], and in fact,
these are the only examples. Thus, the longest proper preﬁx/sufﬁx of P [1 .. i ] must be the longest string
of the form P [1 .. failc [ j ]] — that is, the one with smallest c — such that P [ failc [ j ]] = P [i ]. This is
exactly what the while loop in COMPUTEFAILURE computes; the (c + 1)th iteration compares P [ failc [ j ]] =
P [ failc +1 [i ]] against P [i ]. COMPUTEFAILURE is actually a dynamic programming implementation of the
following recursive deﬁnition of fail[i ]: fail[i ] = if i = 0, 0
c c max fail [i − 1] + 1 P [i − 1] = P [ fail [i − 1]]
c ≥1 H.4 otherwise. Optimizing the Failure Function We can speed up KNUTHMORRISPRATT slightly by making one small change to the failure function. Recall
that after comparing T [i ] against P [ j ] and ﬁnding a mismatch, the algorithm compares T [i ] against
P [ fail[ j ]]. With the current deﬁnition, however, it is possible that P [ j ] and P [ fail[ j ]] are actually the
same character, in which case the next character comparison will automatically fail. So why do the
comparison at all?
We can optimize the failure function by ‘short-circuiting’ these redundant comparisons with some
simple post-processing:
OPTIMIZEFAILURE( P [1 .. m], fail[1 .. m]):
for i ← 2 to m
if P [i ] = P [ fail[i ]]
fail[i ] ← fail[ fail[i ]] We can also compute the optimized failure function directly by adding three new lines (in bold) to the
COMPUTEFAILURE function.
COMPUTEOPTFAILURE( P [1 .. m]):
j←0
for i ← 1 to m
if P [i ] = P [ j ]
fail[i ] ← fail[ j ]
else
fail[i ] ← j
while j > 0 and P [i ] = P [ j ]
j ← fail[ j ]
j ← j+1 This optimization slows down the preprocessing slightly, but it may signiﬁcantly decrease the number
of comparisons at each text character. The worst-case running time is still O(n); however, the constant is
about half as big as for the unoptimized version, so this could be a signiﬁcant improvement in practice. 5 Algorithms Non-Lecture H: More String Matching ! $ A B A R R A
B C
A A D Optimized ﬁnite state machine for the string ‘ABRADACABRA’ P [i ]
fail[i ] A B R A C A D A B R A 0 1 1 0 2 0 2 0 1 1 0 Optimized failure function for ‘ABRACADABRA’, with changes in bold. Here are the unoptimized and optimized failure functions for a few more patterns:
P [i ] A N A N A B A N A N A N A unoptimized fail[i ] 0 1 1 2 3 4 1 2 3 4 5 6 5 optimized fail[i ] 0 1 0 1 0 4 0 1 0 1 0 6 0 Failure functions for ‘ANANABANANANA’. P [i ] A B A B C A B A B C A B C unoptimized fail[i ] 0 1 1 2 3 1 2 3 4 5 6 7 8 optimized fail[i ] 0 1 0 1 3 0 1 0 1 3 0 1 8 Failure functions for ‘ABABCABABCABC’. P [i ] A B B A B B A B A B B A B unoptimized fail[i ] 0 1 1 1 2 3 4 5 6 2 3 4 5 optimized fail[i ] 0 1 1 0 1 1 0 1 6 1 1 0 1 Failure functions for ‘ABBABBABABBAB’. Exercises
1. A palindrome is any string that is the same as its reversal, such as X, ABBA, or REDIVIDER. Describe
and analyze an algorithm that computes the longest palindrome that is a (not necessarily proper)
preﬁx of a given string T [1 .. n]. Your algorithm should run in O(n) time.
2. Describe a modiﬁcation of KNUTHMORRISPRATT in which the pattern can contain any number
of wildcard symbols * , each of which matches an arbitrary string. For example, the pattern
AIN
ABR* CAD* BRA appears in the text SCHABRAIN
AINCADBRANCH; in this case, the second * matches the
empty string. Your algorithm should run in O(m + n) time, where m is the length of the pattern
and n is the length of the text.
3. Describe a modiﬁcation of KNUTHMORRISPRATT in which the pattern can contain any number of
wildcard symbols ? , each of which matches an arbitrary single character. For example, the pattern
?
?
U
I
ABR? CAD? BRA appears in the text SCHABRU CADI BRANCH. Your algorithm should run in O(m + qn)
time, where m is the length of the pattern, n is the length of the text., and q is the number of ? s in
the pattern.
6 Algorithms Non-Lecture H: More String Matching 4. Describe another algorithm for the previous problem that runs in time O(m + kn), where k is the
number of runs of consecutive non-wildcard characters in the pattern. For example, the pattern
??? ??IS????
?
? FISH??? ?? ????
???B?? ????CUIT? has k = 4 runs. 5. Describe a modiﬁcation of KNUTHMORRISPRATT in which the pattern can contain any number of
wildcard symbols = , each of which matches the same arbitrary single character. For example, the
=
=
U
U
U
A
A
A
pattern = HOC= SPOC= S appears in the texts WHU HOCU SPOCU SOT and ABRA HOCA SPOCA SCADABRA, but
SHOCU SPOCE STIX. Your algorithm should run in O(m + n) time, where m is the
U
E
not in the text FRIS
length of the pattern and n is the length of the text. 6. Describe and analyze a variant of KNUTHMORRISPRATT that looks for strings inside labeled rooted
trees. The input consists of a pattern string P [1 .. m] and a rooted text tree T with n nodes, each
labeled with a single character. Nodes in T can have any number of children. Your algorithm
should either return a downward path in T whose labels match the string P , or report that there is
no such path. Your algorithm should run in O(m + n) time.
P
H E
I S A N
A Z
R E
E L M A
R C
W M Q S O K
H
F The string SEARCH appears on a downward path in the tree. 7. Describe and analyze a variant of KNUTHMORRISPRATT that searches for subtrees of ordered rooted
binary trees (every node has a left subtree and a right subtree, either or both of which may be
empty). The input consists of a pattern tree P with m nodes and a text tree T with n nodes. Your
algorithm should report all nodes v in T such that the subtree rooted at v is structurally identical
to P . Your algorithm should run in O(m + n) time. Ignore all search keys, labels, or other data in
the nodes; only the left/right pointer structure matters. The pattern tree (left) appears exactly twice in the text tree (right). 7 Algorithms Non-Lecture H: More String Matching 8. This problem considers the maximum length of a failure chain j → fail[ j ] → fail[ fail[ j ]] →
fail[ fail[ fail[ j ]]] → · · · → 0, or equivalently, the maximum number of iterations of the inner loop
of KNUTHMORRISPRATT. This clearly depends on which failure function we use: unoptimized or
optimized. Let m be an arbitrary positive integer.
(a) Describe a pattern A[1 .. m] whose longest unoptimized failure chain has length m.
(b) Describe a pattern B [1 .. m] whose longest optimized failure chain has length Θ(log m).
(c) Describe a pattern C [1 .. m] containing only two different characters, whose longest optimized
failure chain has length Θ(log m).
(d) Prove that for any pattern of length m, the longest optimized failure chain has length at most
O(log m). 8 Algorithms Lecture 11: Basic Graph Properties
Obie looked at the seein’ eye dog. Then at the twenty-seven 8 by 10 color glossy pictures with
the circles and arrows and a paragraph on the back of each one. . . and then he looked at the
seein’ eye dog. And then at the twenty-seven 8 by 10 color glossy pictures with the circles
and arrows and a paragraph on the back of each one and began to cry.
Because Obie came to the realization that it was a typical case of American blind justice,
and there wasn’t nothin’ he could do about it, and the judge wasn’t gonna look at the twentyseven 8 by 10 color glossy pictures with the circles and arrows and a paragraph on the back of
each one explainin’ what each one was, to be used as evidence against us.
And we was ﬁned ﬁfty dollars and had to pick up the garbage. In the snow.
But that’s not what I’m here to tell you about.
— Arlo Guthrie, “Alice’s Restaurant” (1966) I study my Bible as I gather apples.
First I shake the whole tree, that the ripest might fall.
Then I climb the tree and shake each limb,
and then each branch and then each twig,
and then I look under each leaf.
— Martin Luther 11
11.1 Basic Graph Properties
Deﬁnitions A graph G is a pair of sets (V, E ). V is a set of arbitrary objects that we call vertices1 or nodes. E is a set
of vertex pairs, which we call edges or occasionally arcs. In an undirected graph, the edges are unordered
pairs, or just sets of two vertices. In a directed graph, the edges are ordered pairs of vertices. We will
only be concerned with simple graphs, where there is no edge from a vertex to itself and there is at most
one edge from any vertex to any other.
Following standard (but admittedly confusing) practice, I’ll also use V to denote the number of
vertices in a graph, and E to denote the number of edges. Thus, in an undirected graph, we have
V
0 ≤ E ≤ 2 , and in a directed graph, 0 ≤ E ≤ V (V − 1).
We usually visualize graphs by looking at an embedding. An embedding of a graph maps each vertex
to a point in the plane and each edge to a curve or straight line segment between the two vertices.
A graph is planar if it has an embedding where no two edges cross. The same graph can have many
different embeddings, so it is important not to confuse a particular embedding with the graph itself. In
particular, planar graphs can have non-planar embeddings!
c
b d h a
a f
c e e d
f g h g
i b i A non-planar embedding of a planar graph with nine vertices, thirteen edges, and two connected components,
and a planar embedding of the same graph.
1 The singular of ‘vertices’ is vertex. The singular of ‘matrices’ is matrix. Unless you’re speaking Italian, there is no such
thing as a vertice, a matrice, an indice, an appendice, a helice, an apice, a vortice, a radice, a simplice, a codice, a directrice, a
dominatrice, a Unice, a Kleenice, an Asterice, an Obelice, a Dogmatice, a Getaﬁce, a Cacofonice, a Vitalstatistice, a Geriatrice,
or Jimi Hendrice! You will lose points for using any of these so-called words. 1 Algorithms Lecture 11: Basic Graph Properties There are other ways of visualizing and representing graphs that are sometimes also useful. For
example, the intersection graph of a collection of objects has a node for every object and an edge for
every intersecting pair. Whether a particular graph can be represented as an intersection graph depends
on what kind of object you want to use for the vertices. Different types of objects—line segments,
rectangles, circles, etc.—deﬁne different classes of graphs. One particularly useful type of intersection
graph is an interval graph, whose vertices are intervals on the real line, with an edge between any two
intervals that overlap.
f d e a g b h b d a c h fg i
c
(a) e i a (b) bc de i
gh f (c) The example graph is also the intersection graph of (a) a set of line segments, (b) a set of circles,
or (c) a set of intervals on the real line (stacked for visibility). If (u, v ) is an edge in an undirected graph, then u is a neighbor or v and vice versa. The degree of
a node is the number of neighbors. In directed graphs, we have two kinds of neighbors. If u → v is a
directed edge, then u is a predecessor of v and v is a successor of u. The in-degree of a node is the number
of predecessors, which is the same as the number of edges going into the node. The out-degree is the
number of successors, or the number of edges going out of the node.
A graph G = (V , E ) is a subgraph of G = (V, E ) if V ⊆ V and E ⊆ E .
A path is a sequence of edges, where each successive pair of edges shares a vertex, and all other edges
are disjoint. A graph is connected if there is a path from any vertex to any other vertex. A disconnected
graph consists of several connected components, which are maximal connected subgraphs. Two vertices
are in the same connected component if and only if there is a path between them.
A cycle is a path that starts and ends at the same vertex, and has at least one edge. A graph is acyclic
if no subgraph is a cycle; acyclic graphs are also called forests. Trees are special graphs that can be
deﬁned in several different ways. You can easily prove by induction (hint, hint, hint) that the following
deﬁnitions are equivalent.
• A tree is a connected acyclic graph.
• A tree is a connected component of a forest.
• A tree is a connected graph with at most V − 1 edges.
• A tree is a minimal connected graph; removing any edge makes the graph disconnected.
• A tree is an acyclic graph with at least V − 1 edges.
• A tree is a maximal acyclic graph; adding an edge between any two vertices creates a cycle.
A spanning tree of a graph G is a subgraph that is a tree and contains every vertex of G . Of course, a
graph can only have a spanning tree if it’s connected. A spanning forest of G is a collection of spanning
trees, one for each connected component of G . 2 Algorithms 11.2 Lecture 11: Basic Graph Properties Explicit Representations of Graphs There are two common data structures used to explicitly represent graphs: adjacency matrices2 and
adjacency lists.
The adjacency matrix of a graph G is a V × V matrix of indicator variables. Each entry in the matrix
indicates whether a particular edge is or is not in the graph:
A[i , j ] = (i , j ) ∈ E .
For undirected graphs, the adjacency matrix is always symmetric: A[i , j ] = A[ j , i ]. Since we don’t allow
edges from a vertex to itself, the diagonal elements A[i , i ] are all zeros.
Given an adjacency matrix, we can decide in Θ(1) time whether two vertices are connected by an
edge just by looking in the appropriate slot in the matrix. We can also list all the neighbors of a vertex
in Θ(V ) time by scanning the corresponding row (or column). This is optimal in the worst case, since
a vertex can have up to V − 1 neighbors; however, if a vertex has few neighbors, we may still have
to examine every entry in the row to see them all. Similarly, adjacency matrices require Θ(V 2 ) space,
regardless of how many edges the graph actually has, so it is only space-efﬁcient for very dense graphs.
a
b
c
d
e
f
g
h
i abcde f ghi
01100 0 000
10111 0 000
11011 0 000
01101 1 000
01110 1 000
00011 0 000
00000 0 010
00000 0 101
00000 0 110 a
b
c
d
e
f
g
h
i b
a
a
b
b
d
h
g
g c
c
b
c
c
e
i
i
h d
d
e
d e
e
f
f Adjacency matrix and adjacency list representations for the example graph. For sparse graphs—graphs with relatively few edges—we’re better off using adjacency lists. An
adjacency list is an array of linked lists, one list per vertex. Each linked list stores the neighbors of the
corresponding vertex.
For undirected graphs, each edge (u, v ) is stored twice, once in u’s neighbor list and once in v ’s
neighbor list; for directed graphs, each edge is stores only once. Either way, the overall space required
for an adjacency list is O(V + E ). Listing the neighbors of a node v takes O(1 + deg( v )) time; just scan the
neighbor list. Similarly, we can determine whether (u, v ) is an edge in O(1 + deg(u)) time by scanning
the neighbor list of u. For undirected graphs, we can speed up the search by simultaneously scanning
the neighbor lists of both u and v , stopping either we locate the edge or when we fall of the end of a list.
This takes O(1 + min{deg(u), deg( v )}) time.
The adjacency list structure should immediately remind you of hash tables with chaining. Just as
with hash tables, we can make adjacency list structure more efﬁcient by using something besides a
linked list to store the neighbors. For example, if we use a hash table with constant load factor, when
we can detect edges in O(1) expected time, just as with an adjacency list. In practice, this will only be
useful for vertices with large degree, since the constant overhead in both the space and search time is
larger for hash tables than for simple linked lists.
You might at this point ask why anyone would ever use an adjacency matrix. After all, if you use hash
tables to store the neighbors of each vertex, you can do everything as fast or faster with an adjacency list
as with an adjacency matrix, only using less space. The answer is that many graphs are only represented
2 See footnote 1. 3 Algorithms Lecture 11: Basic Graph Properties implicitly. For example, intersection graphs are usually represented implicitly by simply storing the list
of objects. As long as we can test whether two objects overlap in constant time, we can apply any graph
algorithm to an intersection graph by pretending that it is stored explicitly as an adjacency matrix. On
the other hand, any data structure build from records with pointers between them can be seen as a
directed graph. Algorithms for searching graphs can be applied to these data structures by pretending
that the graph is represented explicitly using an adjacency list.
To keep things simple, we’ll consider only undirected graphs for the rest of this lecture, although the
algorithms I’ll describe also work for directed graphs. 11.3 Traversing connected graphs Suppose we want to visit every node in a connected graph (represented either explicitly or implicitly).
The simplest method to do this is an algorithm called depth-ﬁrst search, which can be written either
recursively or iteratively. It’s exactly the same algorithm either way; the only difference is that we can
actually see the ‘recursion’ stack in the non-recursive version. Both versions are initially passed a source
vertex s.
ITERATIVEDFS(s):
PUSH(s)
while stack not empty
v ← POP
if v is unmarked
mark v
for each edge ( v, w )
PUSH(w ) RECURSIVEDFS( v ):
if v is unmarked
mark v
for each edge ( v, w )
RECURSIVEDFS(w ) Depth-ﬁrst search is one (perhaps the most common) instance of a general family of graph traversal
algorithms. The generic graph traversal algorithm stores a set of candidate edges in some data structure
that I’ll call a ‘bag’. The only important properties of a ‘bag’ are that we can put stuff into it and then
later take stuff back out. (In C++ terms, think of the ‘bag’ as a template for a real data structure.) Here’s
the algorithm:
TRAVERSE(s):
put (∅, s) in bag
while the bag is not empty
take ( p, v ) from the bag
()
if v is unmarked
mark v
parent( v ) ← p
for each edge ( v, w )
(†)
put ( v, w ) into the bag ( ) Notice that we’re keeping edges in the bag instead of vertices. This is because we want to remember,
whenever we visit a vertex v for the ﬁrst time, which previously-visited vertex p put v into the bag. The
vertex p is called the parent of v .
Lemma 1. TRAVERSE(s) marks every vertex in any connected graph exactly once, and the set of edges
( v, parent( v )) with parent( v ) = ∅ form a spanning tree of the graph.
Proof: ﬁrst, it should be obvious that no node is marked more than once.
Clearly, the algorithm marks s. Let v = s be a vertex, and let s → · · · → u → v be the path from s to v
with the minimum number of edges. Since the graph is connected, such a path always exists. (If s and v
4 Algorithms Lecture 11: Basic Graph Properties are neighbors, then u = s, and the path has just one edge.) If the algorithm marks u, then it must put
(u, v ) into the bag, so it must later take (u, v ) out of the bag, at which point v must be marked (if it isn’t
already). Thus, by induction on the shortest-path distance from s, the algorithm marks every vertex in
the graph.
Call an edge ( v, parent( v )) with parent( v ) = ∅ a parent edge. For any node v , the path of parent
edges v → parent( v ) → parent(parent( v )) → · · · eventually leads back to s, so the set of parent edges
form a connected graph. Clearly, both endpoints of every parent edge are marked, and the number of
parent edges is exactly one less than the number of vertices. Thus, the parent edges form a spanning
tree.
The exact running time of the traversal algorithm depends on how the graph is represented and
what data structure is used as the ‘bag’, but we can make a few general observations. Since each vertex
is visited at most once, the for loop (†) is executed at most V times. Each edge is put into the bag exactly
twice; once as (u, v ) and once as ( v, u), so line ( ) is executed at most 2 E times. ﬁnally, since we can’t
take more things out of the bag than we put in, line ( ) is executed at most 2 E + 1 times. 11.4 Examples Let’s ﬁrst assume that the graph is represented by an adjacency list, so that the overhead of the for
loop (†) is only a constant per edge.
• If we implement the ‘bag’ by using a stack, we have depth-ﬁrst search. Each execution of ( )
or ( ) takes constant time, so the overall running time is O(V + E ). Since the graph is connected,
V ≤ E + 1, so we can simplify the running time to O( E ). The spanning tree formed by the parent
edges is called a depth-ﬁrst spanning tree. The exact shape of the tree depends on the order in
which neighbor edges are pushed onto the stack, but the in general, depth-ﬁrst spanning trees are
long and skinny.
• If we use a queue instead of a stack, we have breadth-ﬁrst search. Again, each execution of ( )
or ( ) takes constant time, so the overall running time is still O( E ). In this case, the breadth-ﬁrst
spanning tree formed by the parent edges contains shortest paths from the start vertex s to every
other vertex in its connected component. The exact shape of the breadth-ﬁrst spanning tree
depends on the order in which neighbor edges are pushed onto the queue, but the in general,
shortest path trees are short and bushy. We’ll see shortest paths again in a future lecture.
b d b
f a
c d
f a e c e A depth-ﬁrst spanning tree and a breadth-ﬁrst spanning tree
of one component of the example graph, with start vertex a. • Suppose the edges of the graph are weighted. If we implement the ‘bag’ using a priority queue,
always extracting the minimum-weight edge in line ( ), then we we have what might be called
shortest-ﬁrst search. In this case, each execution of ( ) or ( ) takes O(log E ) time, so the overall
running time is O(V + E log E ), which simpliﬁes to O( E log E ) if the graph is connected. For this
algorithm, the set of parent edges form the minimum spanning tree of the connected component
of s. We’ll see minimum spanning trees again in the next lecture.
5 Algorithms Lecture 11: Basic Graph Properties If the graph is represented using an adjacency matrix instead of an adjacency list, ﬁnding all the
neighbors of each vertex in line (†) takes O(V ) time. Thus, depth- and breadth-ﬁrst search each take
O(V 2 ) time overall, and ‘shortest-ﬁrst search’ takes O(V 2 + E log E ) = O(V 2 log V ) time overall. 11.5 Searching disconnected graphs If the graph is disconnected, then TRAVERSE(s) only visits the nodes in the connected component of the
start vertex s. If we want to visit all the nodes in every component, we can use the following ‘wrapper’
around our generic traversal algorithm. Since TRAVERSE computes a spanning tree of one component,
TRAVERSEALL computes a spanning forest of the entire graph.
TRAVERSEALL(s):
for all vertices v
if v is unmarked
TRAVERSE( v ) Exercises
1. Prove that the following deﬁnitions are all equivalent.
• A tree is a connected acyclic graph.
• A tree is a connected component of a forest.
• A tree is a connected graph with at most V − 1 edges.
• A tree is a minimal connected graph; removing any edge makes the graph disconnected.
• A tree is an acyclic graph with at least V − 1 edges.
• A tree is a maximal acyclic graph; adding an edge between any two vertices creates a cycle.
2. Prove that any connected acyclic graph with n ≥ 2 vertices has at least two vertices with degree 1.
Do not use the words ‘tree’ of ‘leaf’, or any well-known properties of trees; your proof should
follow entirely from the deﬁnitions.
3. Let G be a connected graph, and let T be a depth-ﬁrst spanning tree of G rooted at some node v .
Prove that if T is also a breadth-ﬁrst spanning tree of G rooted at v , then G = T .
4. Whenever groups of pigeons gather, they instinctively establish a pecking order. For any pair
of pigeons, one pigeon always pecks the other, driving it away from food or potential mates.
The same pair of pigeons always chooses the same pecking order, even after years of separation,
no matter what other pigeons are around. Surprisingly, the overall pecking order can contain
cycles—for example, pigeon A pecks pigeon B , which pecks pigeon C , which pecks pigeon A.
(a) Prove that any ﬁnite set of pigeons can be arranged in a row from left to right so that every
pigeon pecks the pigeon immediately to its left. Pretty please.
(b) Suppose you are given a directed graph representing the pecking relationships among a set
of n pigeons. The graph contains one vertex per pigeon, and it contains an edge i → j if and
only if pigeon i pecks pigeon j . Describe and analyze an algorithm to compute a pecking
order for the pigeons, as guaranteed by part (a).
6 Algorithms Lecture 11: Basic Graph Properties 5. You are helping a group of ethnographers analyze some oral history data they have collected by
interviewing members of a village to learn about the lives of people lived there over the last two
hundred years. From the interviews, you have learned about a set of people, all now deceased,
whom we will denote P1 , P2 , . . . , Pn . The ethnographers have collected several facts about the
lifespans of these people. Speciﬁcally, for some pairs ( Pi , P j ), the ethnographers have learned one
of the following facts:
(a) Pi died before P j was born.
(b) Pi and P j were both alive at some moment.
Naturally, the ethnographers are not sure that their facts are correct; memories are not so good,
and all this information was passed down by word of mouth. So they’d like you to determine
whether the data they have collected is at least internally consistent, in the sense that there could
have existed a set of people for which all the facts they have learned simultaneously hold.
Describe and analyze and algorithm to answer the ethnographers’ problem. Your algorithm
should either output possible dates of birth and death that are consistent with all the stated facts,
or it should report correctly that no such dates exist.
6. Let G = (V, E ) be a given directed graph.
(a) The transitive closure G T is a directed graph with the same vertices as G , that contains any
edge u→ v if and only if there is a directed path from u to v in G . Describe an efﬁcient
algorithm to compute the transitive closure of G .
(b) The transitive reduction G T R is the smallest graph (meaning fewest edges) whose transitive
closure is G T . Describe an efﬁcient algorithm to compute the transitive reduction of G .
7. A graph (V, E ) is bipartite if the vertices V can be partitioned into two subsets L and R, such that
every edge has one vertex in L and the other in R.
(a) Prove that every tree is a bipartite graph.
(b) Describe and analyze an efﬁcient algorithm that determines whether a given undirected
graph is bipartite.
8. An Euler tour of a graph G is a closed walk through G that traverses every edge of G exactly once.
(a) Prove that a connected graph G has an Euler tour if and only if every vertex has even degree.
(b) Describe and analyze an algorithm to compute an Euler tour in a given graph, or correctly
report that no such graph exists.
9. The d -dimensional hypercube is the graph deﬁned as follows. There are 2d vertices, each labeled
with a different string of d bits. Two vertices are joined by an edge if their labels differ in exactly
one bit.
(a) A Hamiltonian cycle in a graph G is a cycle of edges in G that visits every vertex of G exactly
once. Prove that for all d ≥ 2, the d -dimensional hypercube has a Hamiltonian cycle.
(b) Which hypercubes have an Euler tour (a closed walk that traverses every edge exactly once)?
[Hint: This is very easy.]
7 Algorithms Lecture 11: Basic Graph Properties 10. Racetrack (also known as Graph Racers and Vector Rally) is a two-player paper-and-pencil racing
game that Jeff played on the bus in 5th grade.3 The game is played with a track drawn on a sheet
of graph paper. The players alternately choose a sequence of grid points that represent the motion
of a car around the track, subject to certain constraints explained below.
Each car has a position and a velocity, both with integer x - and y -coordinates. The initial
position is a point on the starting line, chosen by the player; the initial velocity is always (0, 0). At
each step, the player optionally increments or decrements either or both coordinates of the car’s
velocity; in other words, each component of the velocity can change by at most 1 in a single step.
The car’s new position is then determined by adding the new velocity to the car’s previous position.
The new position must be inside the track; otherwise, the car crashes and that player loses the
race. The race ends when the ﬁrst car reaches a position on the ﬁnish line.
Suppose the racetrack is represented by an n × n array of bits, where each 0 bit represents a
grid point inside the track, each 1 bit represents a grid point outside the track, the ‘starting line’ is
the ﬁrst column, and the ‘ﬁnish line’ is the last column.
Describe and analyze an algorithm to ﬁnd the minimum number of steps required to move a
car from the starting line to the ﬁnish line of a given racetrack. [Hint: Build a graph. What are the
vertices? What are the edges? What problem is this?]
position (1, 5)
(2, 5)
(4, 4)
(7, 4)
(9, 5)
(10, 7)
(10, 10)
(9, 14)
(9, 17)
(10, 19)
(12, 21)
(14, 22)
(16, 22)
(17, 21)
(19, 20)
(22, 20)
(25, 21) START (0, 0)
(1, 0)
(2, −1)
(3, 0)
(2, 1)
(1, 2)
(0, 3)
(−1, 4)
(0, 3)
(1, 2)
(2, 2)
(2, 1)
(2, 0)
(1, −1)
(2, −1)
(3, 0)
(3, 1) FINISH velocity A 16-step Racetrack run, on a 25 × 25 track. This is not the shortest run on this track. 11. Draughts/checkers is a game played on an m × m grid of squares, alternately colored light and
dark. (The game is usually played on an 8 × 8 or 10 × 10 board, but the rules easily generalize to
any board size.) Each dark square is occupied by at most one game piece (usually called a checker
in the U.S.), which is either black or white; light squares are always empty. One player (‘White’)
moves the white pieces; the other (‘Black’) moves the black pieces.
Consider the following simple version of the game, essentially American checkers or British
draughts, but where every piece is a king.4 Pieces can be moved in any of the four diagonal
3
The actual game is a bit more complicated than the version described here. In particular, in the actual game, the boundaries
of the track are a free-form curve, and (at least by default) the entire line segment between any two consecutive positions must
lie inside the track. In the version Jeff played, if a car does run off the track, the car starts its next turn with zero velocity, at
the legal grid point closest to where the car left the track.
4
Most other variants of draughts have ‘ﬂying kings’, which behave very differently than what’s described here. 8 Algorithms Lecture 11: Basic Graph Properties directions, either one or two steps at a time. On each turn, a player either moves one of her pieces
one step diagonally into an empty square, or makes a series of jumps with one of her checkers. In a
single jump, a piece moves to an empty square two steps away in any diagonal direction, but only
if the intermediate square is occupied by a piece of the opposite color; this enemy piece is captured
and immediately removed from the board. Multiple jumps are allowed in a single turn as long as
they are made by the same piece. A player wins if her opponent has no pieces left on the board.
Describe an algorithm that correctly determines whether White can capture every black piece,
thereby winning the game, in a single turn. The input consists of the width of the board (m), a list
of positions of white pieces, and a list of positions of black pieces. For full credit, your algorithm
should run in O(n) time, where n is the total number of pieces. 4 5 3 11 6 2 10 7 1 9 8 White wins in one turn. White cannot win in one turn from either of these positions. [Hint: The greedy strategy—make arbitrary jumps until you get stuck—does not always ﬁnd a
winning sequence of jumps even when one exists. See problem 8. Parity, parity, parity.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 9 Algorithms Lecture 12: Minimum Spanning Trees
We must all hang together, gentlemen, or else we shall most assuredly
hang separately.
— Benjamin Franklin, at the signing of the
Declaration of Independence (July 4, 1776)
It is a very sad thing that nowadays there is so little useless information.
— Oscar Wilde
A ship in port is safe, but that is not what ships are for.
— Rear Admiral Grace Murray Hopper 12
12.1 Minimum Spanning Trees
Introduction Suppose we are given a connected, undirected, weighted graph. This is a graph G = (V, E ) together with
a function w : E → IR that assigns a weight w (e) to each edge e. For this lecture, we’ll assume that the
weights are real numbers. Our task is to ﬁnd the minimum spanning tree of G , that is, the spanning
tree T minimizing the function
w(T ) =
w (e).
e∈ T To keep things simple, I’ll assume that all the edge weights are distinct: w (e) = w (e ) for any pair of
edges e and e . Distinct weights guarantee that the minimum spanning tree of the graph is unique.
Without this condition, there may be several different minimum spanning trees. For example, if all the
edges have weight 1, then every spanning tree is a minimum spanning tree with weight V − 1.
8 5
10
2 18 3 12 30 16 14
4 26 A weighted graph and its minimum spanning tree. If we have an algorithm that assumes the edge weights are unique, we can still use it on graphs
where multiple edges have the same weight, as long as we have a consistent method for breaking ties.
One way to break ties consistently is to use the following algorithm in place of a simple comparison.
SHORTEREDGE takes as input four integers i , j , k, l , and decides which of the two edges (i , j ) and (k, l )
has ‘smaller’ weight.
SHORTEREDGE(i , j , k, l )
if w (i , j ) < w (k, l ) return (i , j )
if w (i , j ) > w (k, l ) return (k, l )
if min(i , j ) < min(k, l ) return (i , j )
if min(i , j ) > min(k, l ) return (k, l )
if max(i , j ) < max(k, l ) return (i , j )
〈〈if max(i,j) < max(k,l) 〉〉 return (k, l ) 1 Algorithms 12.2 Lecture 12: Minimum Spanning Trees The Only Minimum Spanning Tree Algorithm There are several different methods for computing minimum spanning trees, but really they are all
instances of the following generic algorithm. The situation is similar to the previous lecture, where we
saw that depth-ﬁrst search and breadth-ﬁrst search were both instances of a single generic traversal
algorithm.
The generic minimum spanning tree algorithm maintains an acyclic subgraph F of the input graph G ,
which we will call an intermediate spanning forest. F is a subgraph of the minimum spanning tree of G ,
and every component of F is a minimum spanning tree of its vertices. Initially, F consists of n one-node
trees. The generic algorithm merges trees together by adding certain edges between them. When the
algorithm halts, F consists of a single n-node tree, which must be the minimum spanning tree. Obviously,
we have to be careful about which edges we add to the evolving forest, since not every edge is in the
minimum spanning tree.
The intermediate spanning forest F induces two special types of edges. An edge is useless if it is not
an edge of F , but both its endpoints are in the same component of F . For each component of F , we
associate a safe edge—the minimum-weight edge with exactly one endpoint in that component. Different
components might or might not have different safe edges. Some edges are neither safe nor useless—we
call these edges undecided.
All minimum spanning tree algorithms are based on two simple observations.
Lemma 1. The minimum spanning tree contains every safe edge and no useless edges.1
Proof: Let T be the minimum spanning tree. Suppose F has a ‘bad’ component whose safe edge
e = (u, v ) is not in T . Since T is connected, it contains a unique path from u to v , and at least one edge e
on this path has exactly one endpoint in the bad component. Removing e from the minimum spanning
tree and adding e gives us a new spanning tree. Since e is the bad component’s safe edge, we have
w (e ) > w (e), so the the new spanning tree has smaller total weight than T . But this is impossible— T is
the minimum spanning tree. So T must contain every safe edge.
Adding any useless edge to F would introduce a cycle. u e
v
e’ Proving that every safe edge is in the minimum spanning tree. The ‘bad’ component of F is highlighted. So our generic minimum spanning tree algorithm repeatedly adds one or more safe edges to the
evolving forest F . Whenever we add new edges to F , some undecided edges become safe, and others
become useless. To specify a particular algorithm, we must decide which safe edges to add, and how to
identify new safe and new useless edges, at each iteration of our generic template. 1 This lemma is actually a special case of two more general theorems. First, for any partition of the vertices of G into two
disjoint subsets, the minimum-weight edge with one endpoint in each subset is in the minimum spanning tree. Second, the
maximum-weight edge in any cycle in G is not in the minimum spanning tree. 2 Algorithms 12.3 Lecture 12: Minimum Spanning Trees Bor˙ vka’s Algorithm
u ˙
The oldest and arguably simplest minimum spanning tree algorithm was discovered by Boruvka in 1926,
2
long before computers even existed, and practically before the invention of graph theory! The algorithm
was rediscovered by Choquet in 1938; again by Florek, Łukaziewicz, Perkal, Stienhaus, and Zubrzycki in
1951; and again by Sollin some time in the early 1960s. Because Sollin was the only Western computer
scientist in this list—Choquet was a civil engineer; Florek and his co-authors were anthropologists—this
is often called ‘Sollin’s algorithm’, especially in the parallel computing literature.
˙
The Boruvka/Choquet/Florek/Łukaziewicz/Perkal/Stienhaus/Zubrzycki/Sollin algorithm can be
summarized in one line:
˙
B OR U VKA: Add all the safe edges and recurse. 8 5
10
2 18 3 12 30 16 18 12
14 14
4 26 26 Bor˙ vka’s algorithm run on the example graph. Thick edges are in F .
u
Arrows point along each component’s safe edge. Dashed edges are useless. ˙
At the beginning of each phase of the Boruvka algorithm, each component elects an arbitrary ‘leader’
node. The simplest way to hold these elections is a depth-ﬁrst search of F ; the ﬁrst node we visit in any
component is that component’s leader. Once the leaders are elected, we ﬁnd the safe edges for each
component, essentially by brute force. Finally, we add these safe edges to F .
FINDSAFEEDGES(V, E ):
for each leader v
safe( v ) ← ∞
for each edge (u, v ) ∈ E
u ← leader(u)
v ← leader( v )
if u = v
if w (u, v ) < w (safe(u))
safe(u) ← (u, v )
if w (u, v ) < w (safe( v ))
safe( v ) ← (u, v ) ˙
B OR U VKA(V, E ):
F = (V, ∅)
while F has more than one component
choose leaders using DFS
FINDSAFEEDGES(V, E )
for each leader v
add safe( v ) to F Each call to FINDSAFEEDGES takes O( E ) time, since it examines every edge. Since the graph is
˙
connected, it has at most E + 1 vertices. Thus, each iteration of the while loop in B OR U VKA takes O( E )
time, assuming the graph is represented by an adjacency list. Each iteration also reduces the number of
components of F by at least a factor of two—the worst case occurs when the components coalesce in
pairs. Since there are initially V components, the while loop iterates O(log V ) times. Thus, the overall
running time of Bor˙ vka’s algorithm is O (E log V ).
u
˙
Despite its relatively obscure origin, early algorithms researchers were aware of Boruvka’s algorithm,
˙
but dismissed it as being ‘too complicated’! As a result, despite its simplicity and efﬁciency, Boruvka’s
algorithm is rarely mentioned in algorithms and data structures textbooks. On the other hand, more
2 Leonard Euler published the ﬁrst graph theory result, his famous theorem about the bridges of Königsburg, in 1736.
However, the ﬁrst textbook on graph theory, written by Dénes König, was not published until 1936. 3 Algorithms Lecture 12: Minimum Spanning Trees ˙
recent algorithms to compute minimum spanning trees are all generalizations of Boruvka’s algorithm,
not the other two classical algorithms described next. 12.4 Jarník’s (‘Prim’s’) Algorithm The next oldest minimum spanning tree algorithm was ﬁrst described by the Polish mathematician
˙
Vojtˇch Jarník in a 1929 letter to Boruvka. The algorithm was independently rediscovered by Kruskal in
e
1956, by Prim in 1957, by Loberman and Weinberger in 1957, and ﬁnally by Dijkstra in 1958. Prim,
Loberman, Weinberger, and Dijkstra all (eventually) knew of and even cited Kruskal’s paper, but since
Kruskal also described two other minimum-spanning-tree algorithms in the same paper, this algorithm is
usually called ‘Prim’s algorithm’, or sometimes even ‘the Prim/Dijkstra algorithm’, even though by 1958
Dijkstra already had another algorithm (inappropriately) named after him.
In Jarník’s algorithm, the forest F contains only one nontrivial component T ; all the other components
are isolated vertices. Initially, T consists of an arbitrary vertex of the graph. The algorithm repeats the
following step until T spans the whole graph:
JARNÍK: Find T ’s safe edge and add it to T .
8 5 8 5 10
2
18 3 12 2
30 14
4 8 5 10
16 18 3 12 8 10
2
30 16 18 5
10 3 3
16 30 30 16 14
26 26 26 26 8 30 26 16 5 30 16 26 Jarník’s algorithm run on the example graph, starting with the bottom vertex.
At each stage, thick edges are in T , an arrow points along T ’s safe edge, and dashed edges are useless. To implement Jarník’s algorithm, we keep all the edges adjacent to T in a heap. When we pull the
minimum-weight edge off the heap, we ﬁrst check whether both of its endpoints are in T . If not, we add
the edge to T and then add the new neighboring edges to the heap. In other words, Jarník’s algorithm
is just another instance of the generic graph traversal algorithm we saw last time, using a heap as the
‘bag’! If we implement the algorithm this way, its running time is O( E log E ) = O( E log V ).
However, we can speed up the implementation by observing that the graph traversal algorithm visits
each vertex only once. Rather than keeping edges in the heap, we can keep a heap of vertices, where the
key of each vertex v is the length of the minimum-weight edge between v and T (or ∞ if there is no
such edge). Each time we add a new edge to T , we may need to decrease the key of some neighboring
vertices.
To make the description easier, we break the algorithm into two parts. JARNÍKINIT initializes the
vertex heap. JARNÍKLOOP is the main algorithm. The input consists of the vertices and edges of the graph,
plus the start vertex s. 4 Algorithms Lecture 12: Minimum Spanning Trees
JARNÍK(V, E , s):
JARNÍKINIT(V, E , s)
JARNÍKLOOP(V, E , s)
JARNÍKINIT(V, E , s):
for each vertex v ∈ V \ {s}
if ( v, s) ∈ E
edge( v ) ← ( v, s)
key( v ) ← w ( v, s)
else
edge( v ) ← NULL
key( v ) ← ∞
INSERT( v ) JARNÍKLOOP(V, E , s):
T ← ({s}, ∅)
for i ← 1 to |V | − 1
v ← EXTRACTMIN
add v and edge( v ) to T
for each edge (u, v ) ∈ E
if u ∈ T and key(u) > w (u, v )
/
edge(u) ← (u, v )
DECREASEKEY(u, w (u, v )) The running time of JARNÍK is dominated by the cost of the heap operations INSERT, EXTRACTMIN, and
DECREASEKEY. INSERT and EXTRACTMIN are each called O(V ) times once for each vertex except s, and
DECREASEKEY is called O( E ) times, at most twice for each edge. If we use a standard binary heap, each
of these operations requires O(log V ) time, so the overall running time of JARNÍK is O((V + E ) log V ) =
O (E log V ). The running time can be improved to O( E + V log V ) using a data structure called a Fibonacci
heap, which supports INSERT and DECREASEKEY in constant amortized time; this is faster than Bor˙ vka’s
u
algorithm unless E = O(V ). 12.5 Kruskal’s Algorithm The last minimum spanning tree algorithm I’ll discuss was ﬁrst described by Kruskal in 1956, in the same
paper where he rediscovered Jarnik’s algorithm. Kruskal was motivated by ‘a typewritten translation
˙
˙
(of obscure origin)’ of Boruvka’s original paper, claiming that Boruvka’s algorithm was ‘unnecessarily
3
elaborate’. This algorithm was also rediscovered in 1957 by Loberman and Weinberger, but somehow
avoided being renamed after them.
KRUSKAL: Scan all edges in increasing weight order; if an edge is safe, add it to F .
8 8 5 18 18 3 12 8 5 8 10 5 8 10 10 3 10
2 5
10 12 30 16 18 12 14 30 4 16 30 16 18 12 14
26 4 30 16 18 12 14
26 30 16 14
26 26 14
4 26 10
18 30 16 18 30 16 18 12 30 26 26 12 30 26 30 30 16 26 30 26 18 18 14 14 14 16 26 Kruskal’s algorithm run on the example graph. Thick edges are in F . Dashed edges are useless.
3 ˙
To be fair, Boruvka’s original paper was unnecessarily elaborate, but in his followup paper, also published in 1927, simpliﬁed
his algorithm to its current modern form. Kruskal was apparently unaware of Bor˙ vka’s second paper. Stupid Iron Curtain.
u 5 Algorithms Lecture 12: Minimum Spanning Trees Since we examine the edges in order from lightest to heaviest, any edge we examine is safe if and
only if its endpoints are in different components of the forest F . To prove this, suppose the edge e joins
two components A and B but is not safe. Then there would be a lighter edge e with exactly one endpoint
in A. But this is impossible, because (inductively) any previously examined edge has both endpoints in
the same component of F .
˙
Just as in Boruvka’s algorithm, each component of F has a ‘leader’ node. An edge joins two
˙
components of F if and only if the two endpoints have different leaders. But unlike Boruvka’s algorithm,
we do not recompute leaders from scratch every time we add an edge. Instead, when two components
are joined, the two leaders duke it out in a nationally-televised no-holds-barred steel-cage grudge
match.4 One of the two emerges victorious as the leader of the new larger component. More formally,
we will use our earlier algorithms for the UNION-FIND problem, where the vertices are the elements and
the components of F are the sets. Here’s a more formal description of the algorithm:
KRUSKAL(V, E ):
sort E by wieght
F ←∅
for each vertex v ∈ V
MAKESET( v )
for i ← 1 to | E |
(u, v ) ← i th lightest edge in E
if FIND(u) = FIND( v )
UNION(u, v )
add (u, v ) to F
return F In our case, the sets are components of F , and n = V . Kruskal’s algorithm performs O( E ) FIND
operations, two for each edge in the graph, and O(V ) UNION operations, one for each edge in the
minimum spanning tree. Using union-by-rank and path compression allows us to perform each UNION or
FIND in O(α( E , V )) time, where α is the not-quite-constant inverse-Ackerman function. So ignoring the
cost of sorting the edges, the running time of this algorithm is O( E α( E , V )).
We need O( E log E ) = O( E log V ) additional time just to sort the edges. Since this is bigger than the
time for the UNION-FIND data structure, the overall running time of Kruskal’s algorithm is O (E log V ),
exactly the same as Bor˙ vka’s algorithm, or Jarník’s algorithm with a normal (non-Fibonacci) heap.
u Exercises
1. Most classical minimum-spanning-tree algorithms use the notions of ‘safe’ and ‘useless’ edges
described in the lecture notes, but there is an alternate formulation. Let G be a weighted undirected
graph, where the edge weights are distinct. We say that an edge e is dangerous if it is the longest
edge in some cycle in G , and useful if it does not lie in any cycle in G .
(a) Prove that the minimum spanning tree of G contains every useful edge.
(b) Prove that the minimum spanning tree of G does not contain any dangerous edge.
(c) Describe and analyze an efﬁcient implementation of the “anti-Kruskal” MST algorithm:
Examine the edges of G in decreasing order; if an edge is dangerous, remove it from G . [Hint:
It won’t be as fast as Kruskal’s algorithm.]
4 Live at the Assembly Hall! Only $49.95 on Pay-Per-View! 6 Algorithms Lecture 12: Minimum Spanning Trees 2. Let G = (V, E ) be an arbitrary connected graph with weighted edges.
(a) Prove that for any partition of the vertices V into two subsets, the minimum-weight edge
with one endpoint in each subset is in the minimum spanning tree of G .
(b) Prove that the maximum-weight edge in any cycle of G is not in the minimum spanning tree
of G .
(c) Prove or disprove: The minimum spanning tree of G includes the minimum-weighted edge
in every cycle in G .
3. Throughout this lecture note, we assumed that no two edges in the input graph have equal weights,
which implies that the minimum spanning tree is unique. In fact, a weaker condition on the edge
weights implies MST uniqueness.
(a) Describe an edge-weighted graph that has a unique minimum spanning tree, even though
two edges have equal weights.
(b) Prove that an edge-weighted graph G has a unique minimum spanning tree if and only if the
following conditions hold:
• For any partition of the vertices of G into two subsets, the minimum-weight edge with
one endpoint in each subset is unique.
• The maximum-weight edge in any cycle of G is unique.
(c) Describe and analyze an algorithm to determine whether or not a graph has a unique
minimum spanning tree.
4. (a) Describe and analyze an algorithm to compute the maximum-weight spanning tree of a given
edge-weighted graph.
(b) A feedback edge set of a graph G is a subset F of the edges such that every cycle in G contains
at least one edge in F . In other words, removing every edge in F makes the graph G acyclic.
Describe and analyze a fast algorithm to compute the minimum weight feedback edge set of
of a given edge-weighted graph.
5. Consider a path between two vertices s and t in an undirected weighted graph G . The bottleneck
length of this path is the maximum weight of any edge in the path. The bottleneck distance
between s and t is the minimum bottleneck length of any path from s to t . (If there are no paths
from s to t , the bottleneck distance between s and t is ∞.) 11 1
6 t
3 4 7 5
2 10
s 9
8 12 The bottleneck distance between s and t is 5. 7 Algorithms Lecture 12: Minimum Spanning Trees Describe and analyze an algorithm to compute the bottleneck distance between every pair of
vertices in an arbitrary undirected weighted graph. Assume that no two edges have the same
weight.
6. Suppose you are given a graph G with weighted edges and a minimum spanning tree T of G .
(a) Describe an algorithm to update the minimum spanning tree when the weight of a single
edge e is decreased.
(b) Describe an algorithm to update the minimum spanning tree when the weight of a single
edge e is increased.
In both cases, the input to your algorithm is the edge e and its new weight; your algorithms should
modify T so that it is still a minimum spanning tree. [Hint: Consider the cases e ∈ T and e ∈ T
separately.]
7. (a) Describe and analyze and algorithm to ﬁnd the second smallest spanning tree of a given
graph G , that is, the spanning tree of G with smallest total weight except for the minimum
spanning tree.
(b) Describe and analyze an efﬁcient algorithm to compute, given a weighted undirected graph G
and an integer k, the k spanning trees of G with smallest weight.
8. We say that a graph G = (V, E ) is dense if E = Θ(V 2 ). Describe a modiﬁcation of Jarník’s minimumspanning tree algorithm that runs in O(V 2 ) time (independent of E ) when the input graph is
dense, using only simple data structures (and in particular, without using a Fibonacci heap).
˙
9. Consider an algorithm that ﬁrst performs k passes of Boruvka’s algorithm, and then runs Jarník’s
algorithm (with a Fibonacci heap) on the resulting contracted graph.
(a) What is the running time of this hybrid algorithm, as a function of V , E , and k?
(b) For which value of k is this running time minimized? What is the resulting running time?
10. Describe an algorithm to compute the minimum spanning tree of an n-vertex planar graph in O(n)
time. [Hint: Contracting an edge in a planar graph yields another planar graph. Any planar graph
with n vertices has at most 3n − 6 edges.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 8 Algorithms Lecture 13: Shortest Paths
Well, ya turn left by the ﬁre station in the village and take the old post road by
the reservoir and. . . no, that won’t do.
Best to continue straight on by the tar road until you reach the schoolhouse and
then turn left on the road to Bennett’s Lake until. . . no, that won’t work either.
East Millinocket, ya say? Come to think of it, you can’t get there from here.
— Robert Bryan and Marshall Dodge,
Bert and I and Other Stories from Down East (1961)
Hey farmer! Where does this road go?
Been livin’ here all my life, it ain’t gone nowhere yet.
Hey farmer! How do you get to Little Rock?
Listen stranger, you can’t get there from here.
Hey farmer! You don’t know very much do you?
No, but I ain’t lost.
— Michelle Shocked, “Arkansas Traveler" (1992) 13
13.1 Shortest Paths
Introduction Given a weighted directed graph G = (V, E , w ) with two special vertices, a source s and a target t , we
want to ﬁnd the shortest directed path from s to t . In other words, we want to ﬁnd the path p starting
at s and ending at t minimizing the function
w ( p) = w (e).
e∈ p For example, if I want to answer the question ‘What’s the fastest way to drive from my old apartment
in Champaign, Illinois to my wife’s old apartment in Columbus, Ohio?’, we might use a graph whose
vertices are cities, edges are roads, weights are driving times, s is Champaign, and t is Columbus.1 The
graph is directed since the driving times along the same road might be different in different directions.2
Perhaps counter to intuition, we will allow the weights on the edges to be negative. Negative edges
make our lives complicated, since the presence of a negative cycle might mean that there is no shortest
path. In general, a shortest path from s to t exists if and only if there is at least one path from s to t , but
there is no path from s to t that touches a negative cycle. If there is a negative cycle between s and t ,
then se can always ﬁnd a shorter path by going around the cycle one more time.
2
s −8 5 3
4 t 1 There is no shortest path from s to t . Every algorithm known for solving this problem actually solves (large portions of) the following
more general single source shortest path or SSSP problem: ﬁnd the shortest path from the source vertex s
1 West on Church, north on Prospect, east on I-74, south on I-465, east on Airport Expressway, north on I-65, east on I-70,
north on Grandview, east on 5th, north on Olentangy River, east on Dodridge, north on High, west on Kelso, south on Neil.
Depending on trafﬁc. We both live in Urbana now.
2
In 1999 and 2000, there was a speed trap on I-70 just east of the Indiana/Ohio border, but only for eastbound trafﬁc. 1 Algorithms Lecture 13: Shortest Paths to every other vertex in the graph. In fact, the problem is usually solved by ﬁnding a shortest path tree
rooted at s that contains all the desired shortest paths.
It’s not hard to see that if shortest paths are unique, then they form a tree. To prove this, it’s enough
to observe that any subpath of a shortest path is also a shortest path. If there are multiple shortest paths
to the same vertices, we can always choose one path to each vertex so that the union of the paths is a
tree. If there are shortest paths to two vertices u and v that diverge, then meet, then diverge again, we
can modify one of the paths so that the two paths only diverge once.
b
s c u a d x y v If s→a→ b→c →d → v and s→a→ x → y →d →u are both shortest paths,
then s→a→ b→c →d →u is also a shortest path. Shortest path trees and minimum spanning trees are usually very different. For one thing, there is
only one minimum spanning tree, but in general, there is a different shortest path tree for every source
vertex. Moreover, in general, all of these shortest path trees are different from the minimum spanning
tree.
8 5 8 5 10
2
18 10
3 12 2
30 16 18 12 14
4 3
30 16 14
26 4 26 A minimum spanning tree and a shortest path tree (rooted at the topmost vertex) of the same graph. All of the algorithms described in this lecture also work for undirected graphs, with some slight
modiﬁcations. Most importantly, we must speciﬁcally prohibit alternating back and forth across the same
undirected negative-weight edge. (Our unmodiﬁed algorithms would interpret any negative-weight
edge as a negative cycle of length 2.)
To emphasize the direction, I will consistently use the nonstandard notation u→ v to denote a directed
edge from u to v . 13.2 The Only SSSP Algorithm Just like graph traversal and minimum spanning trees, there are several different SSSP algorithms,
but they are all special cases of the a single generic algorithm, ﬁrst proposed by Ford in 1956, and
independently by Dantzig in 1957.3 Each vertex v in the graph stores two values, which (inductively)
describe a tentative shortest path from s to v .
• dist( v ) is the length of the tentative shortest s
3 v path, or ∞ if there is no such path. Speciﬁcally, Dantzig showed that the shortest path problem can be phrased as a linear programming problem, and then
described an interpretation of the simplex method (which Dantzig discovered) in terms of the original graph. His description
was equivalent to Ford’s relaxation strategy. 2 Algorithms Lecture 13: Shortest Paths • pred( v ) is the predecessor of v in the tentative shortest s
vertex. v path, or NULL if there is no such The predecessor pointers automatically deﬁne a tentative shortest path tree; they play the same role
as the ‘parent’ pointers in our generic graph traversal algorithm. We already know that dist(s) = 0 and
pred(s) = NULL. For every vertex v = s, we initially set dist( v ) = ∞ and pred( v ) = NULL to indicate that
we do not know of any path from s to v .
We call an edge u→ v tense if dist(u) + w (u→ v ) < dist( v ). If u→ v is tense, then the tentative shortest
path s v is incorrect, since the path s u→ v is shorter. Our generic algorithm repeatedly ﬁnds a tense
edge in the graph and relaxes it:
RELAX(u→ v ):
dist( v ) ← dist(u) + w (u→ v )
pred( v ) ← u If there are no tense edges, our algorithm is ﬁnished, and we have our desired shortest path tree.
The correctness of the relaxation algorithm follows directly from three simple claims:
1. If dist( v ) = ∞, then dist( v ) is the total weight of the predecessor chain ending at v :
s→ · · · →pred(pred( v ))→pred( v )→ v.
This is easy to prove by induction on the number of relaxation steps. (Hint, hint.)
2. If the algorithm halts, then dist( v ) ≤ w (s v ) for any path s
on the number of edges in the path s v . (Hint, hint.) v . This is easy to prove by induction 3. The algorithm halts if and only if there is no negative cycle reachable from s. The ‘only if’ direction
is easy—if there is a reachable negative cycle, then after the ﬁrst edge in the cycle is relaxed,
the cycle always has at least one tense edge. The ‘if’ direction follows from the fact that every
relaxation step reduces either the number of vertices with dist( v ) = ∞ by 1 or reduces the sum of
the ﬁnite shortest path lengths by the difference between two edge weights. Actually proving the ﬁrst two claims above is not as straightforward as I make it sound; there are
some unfortunate subtleties. It’s much easier to prove the following weaker claims, which also imply
correctness: • For every vertex v , dist( v ) is always greater than or equal to the shortest-path distance from s
to v . (Induction on the number of relaxation steps.)
• If no edge is tense, then for every vertex v , dist( v ) is the shortest-path distance from s to v .
(Induction on the number of edges in the true shortest path; see Bellman-Ford. If the edge
weights are non-negative, induction on the rank of the shortest-path distance also works; see
Dijkstra!) • If no edge is tense, then for every vertex v , the path s→ · · · →pred(pred( v ))→pred( v )→ v is a
shortest path from s to v . (Induction on the number of edges in the path; why does pred( v )
have its current value?) • The algorithm halts if there are no negative cycles reachable from s; see above. 3 Algorithms Lecture 13: Shortest Paths I haven’t said anything about how we detect which edges can be relaxed, or in what order we relax
them. In order to make this easier, we can reﬁne the relaxation algorithm slightly, into something closely
resembling the generic graph traversal algorithm. We maintain a ‘bag’ of vertices, initially containing
just the source vertex s. Whenever we take a vertex u out of the bag, we scan all of its outgoing edges,
looking for something to relax. Whenever we successfully relax an edge u→ v , we put v into the bag.
Unlike our generic graph traversal algorithm, the same vertex might be visited many times.
GENERIC SSSP(s):
INITSSSP(s)
put s in the bag
while the bag is not empty
take u from the bag
for all edges u→ v
if u→ v is tense
RELAX(u→ v )
put v in the bag INITSSSP(s):
dist(s) ← 0
pred(s) ← NULL
for all vertices v = s
dist( v ) ← ∞
pred( v ) ← NULL Just as with graph traversal, using different data structures for the ‘bag’ gives us different algorithms.
There are three obvious choices to try: a stack, a queue, and a heap. Unfortunately, if we use a stack, we
have to perform Θ(2V ) relaxation steps in the worst case! (Proving this is a good homework problem.)
The other two possibilities are much more efﬁcient. 13.3 Dijkstra’s Algorithm If we implement the bag as a heap, where the key of a vertex v is dist( v ), we obtain an algorithm
ﬁrst ‘published’4 by Leyzorek, Gray, Johnson, Ladew, Meaker, Petry, and Seitz in 1957, and then later
independently rediscovered by Edsger Dijkstra in 1959. A very similar algorithm was also described by
Dantzig in 1958.
Dijkstra’s algorithm, as it is universally known5 , is particularly well-behaved if the graph has no
negative-weight edges. In this case, it’s not hard to show (by induction, of course) that the vertices are
scanned in increasing order of their shortest-path distance from s. It follows that each vertex is scanned
at most once, and thus that each edge is relaxed at most once. Since the key of each vertex in the heap
is its tentative distance from s, the algorithm performs a DECREASEKEY operation every time an edge is
relaxed. Thus, the algorithm performs at most E DECREASEKEYs. Similarly, there are at most V INSERT
and EXTRACTMIN operations. Thus, if we store the vertices in a Fibonacci heap, the total running time of
Dijkstra’s algorithm is O (E + V log V ); if we use a regular binary heap, the running time is O( E log V ).
This analysis assumes that no edge has negative weight. Dijkstra’s algorithm (in the form I’m
presenting here) is still correct if there are negative edges6 , but the worst-case running time could be
exponential. (Proving this unfortunate fact is a good homework problem.)
4 in the ﬁrst annual report on a research project performed for the Combat Development Department of the Army Electronic
Proving Ground
5
I will follow this common convention, despite the historical inaccuracy, because I don’t think anybody wants to read about
the "Leyzorek-Gray-Johnson-Ladew-Meaker-Petry-Seitz algorithm".
6
Many textbooks present a version of Dijkstra’s algorithm that gives incorrect results for graphs with negative edges. 4 Algorithms Lecture 13: Shortest Paths
∞ ∞
3 2
1 ∞
0 1 5 0
12 8
∞ 2 ∞ ∞ ∞ 10 6 ∞ 12 3 6 8
∞ 3 3
0
s ∞
2 1 3 14 1 5 8 0
12 8
3 4 ∞ 3 12
7
3 4
6 0
s 9
5 4 10 7 6 2 4 9 4 10 3 4
6 7
3 0 12
7 0
s 4 12
5 4 10 3 4 0
s 1
0 7 ∞ 2 ∞ 5 8
∞ 3 4 10 7
4 ∞ 3 3
0
s Four phases of Dijkstra’s algorithm run on a graph with no negative edges.
At each phase, the shaded vertices are in the heap, and the bold vertex has just been scanned.
The bold edges describe the evolving shortest path tree. 13.4 The A∗ Heuristic A slight generalization of Dijkstra’s algorithm, commonly known as the A∗ algorithm, is frequently used
to ﬁnd a shortest path from a single source node s to a single target node t . A∗ uses a black-box function
GUESSDISTANCE( v, t ) that returns an estimate of the distance from v to t . The only difference between
Dijkstra and A∗ is that the key of a vertex v is dist( v ) + GUESSDISTANCE( v, t ).
The function GUESSDISTANCE is called admissible if GUESSDISTANCE( v, t ) never overestimates the
actual shortest path distance from v to t . If GUESSDISTANCE is admissible and the actual edge weights
are all non-negative, the A∗ algorithm computes the actual shortest path from s to t at least as quickly as
Dijkstra’s algorithm. The closer GUESSDISTANCE( v, t ) is to the real distance from v to t , the faster the
algorithm. However, in the worst case, the running time is still O( E + V log V ).
The heuristic is especially useful in situations where the actual graph is not known. For example, A∗
can be used to solve many puzzles (15-puzzle, Freecell, Shanghai, Sokoban, Atomix, Rush Hour, Rubik’s
Cube, . . . ) and other path planning problems where the starting and goal conﬁgurations are given, but
the graph of all possible conﬁgurations and their connections is not given explicitly. 13.5 Shimbel’s Algorithm (‘Bellman-Ford’) If we replace the heap in Dijkstra’s algorithm with a queue, we get an algorithm that was ﬁrst published
by Shimbel in 1955, then independently rediscovered by Moore in 1957, by Woodbury and Dantzig in
1957, and by Bellman in 1958. Since Bellman used the idea of relaxing edges, which was ﬁrst proposed
by Ford in 1956, this is usually called the ‘Bellman-Ford’ algorithm. Shimbel’s algorithm is efﬁcient even
if there are negative edges, and it can be used to quickly detect the presence of negative cycles. If there 5 Algorithms Lecture 13: Shortest Paths are no negative edges, however, Dijkstra’s algorithm is faster. (In fact, in practice, Dijkstra’s algorithm is
often faster even for graphs with negative edges.)
Is describing this algorithm with a queue really helpful, or just confusing? After all, it’s just a
straightforward dynamic programming algorithm. And the structure of the algorithm is very close to
the proof of correctness of the generic algorithm (induction on the number of edges in the shortest
path). The easiest way to analyze the algorithm is to break the execution into phases, by introducing an
imaginary token. Before we even begin, we insert the token into the queue. The current phase ends
when we take the token out of the queue; we begin the next phase by reinserting the token into the
queue. The 0th phase consists entirely of scanning the source vertex s. The algorithm ends when the
queue contains only the token. A simple inductive argument (hint, hint) implies the following invariant:
At the end of the i th phase, for every vertex v , dist( v ) is less than or equal to the length of
the shortest path s v consisting of i or fewer edges.
Since a shortest path can only pass through each vertex once, either the algorithm halts before the
V th phase, or the graph contains a negative cycle. In each phase, we scan each vertex at most once,
so we relax each edge at most once, so the running time of a single phase is O( E ). Thus, the overall
running time of Shimbel’s algorithm is O (V E ).
Once we understand how the phases of Shimbel’s algorithm behave, we can simplify the algorithm
considerably. Instead of using a queue to perform a partial breadth-ﬁrst search of the graph in each
phase, we can simply scan through the adjacency list directly and try to relax every edge in the graph.
SHIMBELSSSP(s)
INITSSSP(s)
repeat V times:
for every edge u→ v
if u→ v is tense
RELAX(u→ v )
for every edge u→ v
if u→ v is tense
return ‘Negative cycle!’ This is how most textbooks present the ‘Bellman-Ford’ algorithm.7 The O(V E ) running time of this
version of the algorithm should be obvious, but it may not be clear that the algorithm is still correct. To
prove correctness, we just have to show that our earlier invariant holds; as before, this can be proved by
induction on i .
7 In fact, this is closer to the description that Shimbel and Bellman used. Bob Tarjan recognized in the early 1980s that
Shimbel’s algorithm is equivalent to Dijkstra’s algorithm with a queue instead of a heap. 6 Algorithms Lecture 13: Shortest Paths
e
∞ −8
d
∞ −8 2
f
∞ 1
0 6 0 4 3 6 3
c 4
6
0
s e
9 e
9
2 −8 8 −3 6 5
b
2 −3 −1 8
3
c −3
−1 4 1
a 3 f
7 1
0 4 −2
a d
2 5
b
2 −3 2 −8
f
7 1
0 3
c
3 0
s d
1 −3
−1 6
a 3 0
s 5 8 −1 6
a f
7 b
2 −3 −3 8
∞
c 4 2
1 5
b
4 −1 d
4 f
∞ 1 −3 −3 −8 2 0 8
∞
a d
∞ 5
b
∞ −3 e
∞ e
∞ 6 0
s 3
c
3 0
s Four phases of Shimbel’s algorithm run on a directed graph with negative edges.
Nodes are taken from the queue in the order s a b c d f b a e d d a
, where is the token.
Shaded vertices are in the queue at the end of each phase. The bold edges describe the evolving shortest path tree. 13.6 Greedy Relaxation? Here’s another algorithm that ﬁts our generic framework, but which I’ve never seen analyzed.
Repeatedly relax the tensest edge.
Speciﬁcally, let’s deﬁne the ‘tension’ of an edge u→ v as follows:
tension(u→ v ) = max {0, dist( v ) − dist(u) − w (u→ v )}
(This is deﬁned even when dist( v ) = ∞ or dist(u) = ∞, as long as we treat ∞ just like some indescribably
large but ﬁnite number.) If an edge has zero tension, it’s not tense. If we relax an edge u→ v , then dist( v )
decreases tension(u→ v ) and tension(u→ v ) becomes zero.
Intuitively, we can keep the edges of the graph in some sort of heap, where the key of each edge is
its tension. Then we repeatedly pull out the tensest edge u→ v and relax it. Then we need to recompute
the tension of other edges adjacent to v . Edges leaving v possibly become more tense, and edges coming
into v possibly become less tense. So we need a heap that efﬁciently supports the operations INSERT,
EXTRACTMAX, INCREASEKEY, and DECREASEKEY.
If there are no negative cycles, this algorithm eventually halts with a shortest path tree, but how
quickly? Can the same edge be relaxed more than once, and if so, how many times? Is it faster if all the
edge weights are positive? Hmm.... This sounds like a good extra credit problem!8
8 I ﬁrst proposed this bizarre algorithm in 1998, the very ﬁrst time I taught an algorithms class. As far as I know, nobody has
even seriously attempted an analysis. Or maybe it has been analyzed, but it requires an exponential number of relaxation steps
in the worst case, so nobody’s ever bothered to publish it. 7 Algorithms Lecture 13: Shortest Paths Exercises
1. This question asks you to ﬁll in the remaining proof details for Ford’s generic shortest-path
algorithm: while at least one edge is tense, relax an arbitrary tense edge. You may assume that
the input graph does not contain a negative cycle, and that all shortest paths in the input graph
are unique.
(a) Prove that the generic shortest-path algorithm halts.
(b) Prove that after every call to RELAX, for every vertex v , either dist( v ) = ∞ or dist( v ) is the
total weight of some path from s to v .
(c) Prove that when the generic shortest-path algorithm halts, then for every vertex v , either
dist( v ) = ∞, or dist( v ) is the total weight of the predecessor chain ending at v :
s→ · · · →pred(pred( v ))→pred( v )→ v.
(d) Prove that when the generic shortest-path algorithm halts, then dist( v ) ≤ w (s
path s v . v ) for every 2. Prove the following statement for every integer i and every vertex v : At the end of the i th phase of
Shimbel’s algorithm, dist( v ) is less than or equal to the length of the shortest path s v consisting
of i or fewer edges.
3. Prove that Ford’s generic shortest-path algorithm (while the graph contains a tense edge, relax it)
can take exponential time in the worst case when implemented with a stack instead of a priority
queue (like Dijkstra) or a queue (like Shimbel’s algorithm). SpeciÞcally, for every positive integer n,
construct a weighted directed n-vertex graph Gn , such that the stack-based shortest-path algorithm
call RELAX Ω(2n ) times when Gn is the input graph. [Hint: Towers of Hanoi.]
4. Prove that Dijkstra’s shortest-path algorithm can require exponential time in the worst case when
edges are allowed to have negative weight. Speciﬁcally, for every positive integer n, construct
a weighted directed n-vertex graph Gn , such that Dijkstra’s algorithm calls RELAX Ω(2n ) times
when Gn is the input graph. [Hint: This should be easy if you’ve already solved the previous
problem.]
5. A looped tree is a weighted, directed graph built from a binary tree by adding an edge from every
leaf back to the root. Every edge has a non-negative weight. 5 4 17 0 23 8 9 16 A looped tree. 8 42 1 14 7 Algorithms Lecture 13: Shortest Paths (a) How much time would Dijkstra’s algorithm require to compute the shortest path between
two vertices u and v in a looped tree with n nodes?
(b) Describe and analyze a faster algorithm.
6. (a) Describe a modiﬁcation of Ford’s generic shortest-path algorithm that actually returns a
negative cycle if any such cycle is reachable from s, or a shortest-path tree if there is no such
cycle. You may assume that the unmodiﬁed algorithm halts in O(2V ) steps if there is no
negative cycle.
(b) Describe a modiﬁcation of Shimbel’s shortest-path algorithm that actually returns a negative
cycle if any such cycle is reachable from s, or a shortest-path tree if there is no such cycle.
The modiﬁed algorithm should still run in O(V E ) time.
7. For any edge e in any graph G , let G \ e denote the graph obtained by deleting e from G .
(a) Suppose we are given a directed graph G in which the shortest path σ from vertex s to
vertex t passes through every vertex of G . Describe an algorithm to compute the shortest-path
distance from s to t in G \ e, for every edge e of G , in O( E log V ) time. Your algorithm should
output a set of E shortest-path distances, one for each edge of the input graph. You may
assume that all edge weights are non-negative. [Hint: If we delete an edge of the original
shortest path, how do the old and new shortest paths overlap?]
(b) Let s and t be arbitrary vertices in an arbitrary directed graph G . Describe an algorithm to
compute the shortest-path distance from s to t in G \ e, for every edge e of G , in O( E log V )
time. Again, you may assume that all edge weights are non-negative.
8. Let G = (V, E ) be a connected directed graph with non-negative edge weights, let s and t be
vertices of G , and let H be a subgraph of G obtained by deleting some edges. Suppose we want to
reinsert exactly one edge from G back into H , so that the shortest path from s to t in the resulting
graph is as short as possible. Describe and analyze an algorithm that chooses the best edge to
reinsert, in O( E log V ) time.
9. Negative edges cause problems in shortest-path algorithms because of the possibility of negative
cycles. But what if the input graph has no cycles?
(a) Describe an efﬁcient algorithm to compute the shortest path between two nodes s and t
in a given directed acyclic graph with weighted edges. The edge weights could be positive,
negative, or zero.
(b) Describe an efﬁcient algorithm to compute the longest path between two nodes s and t in a
given directed acyclic graph with weighted edges.
10. After a grueling algorithms midterm, you decide to take the bus home. Since you planned ahead,
you have a schedule that lists the times and locations of every stop of every bus in ChampaignUrbana. Unfortunately, there isn’t a single bus that visits both your exam building and your home;
you must transfer between bus lines at least once.
Describe and analyze an algorithm to determine the sequence of bus rides that will get you
home as early as possible, assuming there are b different bus lines, and each bus stops n times per
9 Algorithms Lecture 13: Shortest Paths day. Your goal is to minimize your arrival time, not the time you actually spend traveling. Assume
that the buses run exactly on schedule, that you have an accurate watch, and that you are too
tired to walk between bus stops.
11. After graduating you accept a job with Aerophobes-R-Us, the leading traveling agency for people
who hate to ﬂy. Your job is to build a system to help customers plan airplane trips from one city to
another. All of your customers are afraid of ﬂying (and by extension, airports), so any trip you
plan needs to be as short as possible. You know all the departure and arrival times of all the ﬂights
on the planet.
Suppose one of your customers wants to ﬂy from city X to city Y . Describe an algorithm to
ﬁnd a sequence of ﬂights that minimizes the total time in transit—the length of time from the
initial departure to the ﬁnal arrival, including time at intermediate airports waiting for connecting
ﬂights. [Hint: Modify the input data and apply Dijkstra’s algorithm.]
12. Mulder and Scully have computed, for every road in the United States, the exact probability that
someone driving on that road won’t be abducted by aliens. Agent Mulder needs to drive from
Langley, Virginia to Area 51, Nevada. What route should he take so that he has the least chance of
being abducted?
More formally, you are given a directed graph G = (V, E ), where every edge e has an independent safety probability p(e). The safety of a path is the product of the safety probabilities of its
edges. Design and analyze an algorithm to determine the safest path from a given start vertex s to
a given target vertex t .
Las Vegas, NV 0.1
0.5 0.2 0.7
Memphis, TN
0.9 Langley, VA 0.5 Area 51, AZ For example, with the probabilities shown above, if Mulder tries to drive directly from Langley
to Area 51, he has a 50% chance of getting there without being abducted. If he stops in Memphis,
he has a 0.7 × 0.9 = 63% chance of arriving safely. If he stops ﬁrst in Memphis and then in Las
Vegas, he has a 1 − 0.7 × 0.1 × 0.5 = 96.5% chance of being abducted! (That’s how they got Elvis,
you know.)
13. On an overnight camping trip in Sunnydale National Park, you are woken from a restless sleep
by a scream. As you crawl out of your tent to investigate, a terriﬁed park ranger runs out of the
woods, covered in blood and clutching a crumpled piece of paper to his chest. As he reaches your
tent, he gasps, “Get out. . . while. . . you. . . ”, thrusts the paper into your hands, and falls to the
ground. Checking his pulse, you discover that the ranger is stone dead.
You look down at the paper and recognize a map of the park, drawn as an undirected
graph, where vertices represent landmarks in the park, and edges represent trails between those 10 Algorithms Lecture 13: Shortest Paths landmarks. (Trails start and end at landmarks and do not cross.) You recognize one of the vertices
as your current location; several vertices on the boundary of the map are labeled EXIT.
On closer examination, you notice that someone (perhaps the poor dead park ranger) has
written a real number between 0 and 1 next to each vertex and each edge. A scrawled note on
the back of the map indicates that a number next to an edge is the probability of encountering
a vampire along the corresponding trail, and a number next to a vertex is the probability of
encountering a vampire at the corresponding landmark. (Vampires can’t stand each other’s
company, so you’ll never see more than one vampire on the same trail or at the same landmark.)
The note warns you that stepping off the marked trails will result in a slow and painful death.
You glance down at the corpse at your feet. Yes, his death certainly looked painful. Wait, was
that a twitch? Are his teeth getting longer? After driving a tent stake through the undead ranger’s
heart, you wisely decide to leave the park immediately.
Describe and analyze an efﬁcient algorithm to ﬁnd a path from your current location to an
arbitrary EXIT node, such that the total expected number of vampires encountered along the path is
as small as possible. Be sure to account for both the vertex probabilities and the edge probabilities! — Randall Munroe, xkcd (http://xkcd.com/c69.html) c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 11 Algorithms Lecture 14: All-Pairs Shortest Paths
The tree which ﬁlls the arms grew from the tiniest sprout;
the tower of nine storeys rose from a (small) heap of earth;
the journey of a thousand li commenced with a single step.
— Lao-Tzu, Tao Te Ching, chapter 64 (6th century BC),
translated by J. Legge (1891)
And I would walk ﬁve hundred miles,
And I would walk ﬁve hundred more,
Just to be the man who walks a thousand miles
To fall down at your door.
— The Proclaimers, “Five Hundred Miles (I’m Gonna Be)”,
Sunshine on Leith (2001)
Almost there. . . Almost there. . .
— Red Leader [Drewe Henley], Star Wars (1977) 14 All-Pairs Shortest Paths 14.1 The Problem In the previous lecture, we saw algorithms to ﬁnd the shortest path from a source vertex s to a target
vertex t in a directed graph. As it turns out, the best algorithms for this problem actually ﬁnd the shortest
path from s to every possible target (or from every possible source to t ) by constructing a shortest path
tree. The shortest path tree speciﬁes two pieces of information for each node v in the graph:
• dist( v ) is the length of the shortest path (if any) from s to v ;
• pred( v ) is the second-to-last vertex (if any) the shortest path (if any) from s to v .
In this lecture, we want to generalize the shortest path problem even further. In the all pairs shortest
path problem, we want to ﬁnd the shortest path from every possible source to every possible destination.
Speciﬁcally, for every pair of vertices u and v , we need to compute the following information:
• dist(u, v ) is the length of the shortest path (if any) from u to v ;
• pred(u, v ) is the second-to-last vertex (if any) on the shortest path (if any) from u to v .
For example, for any vertex v , we have dist( v, v ) = 0 and pred( v, v ) = NULL. If the shortest path from u
to v is only one edge long, then dist(u, v ) = w (u→ v ) and pred(u, v ) = u. If there is no shortest path from
u to v —either because there’s no path at all, or because there’s a negative cycle—then dist(u, v ) = ∞
and pred( v, v ) = NULL.
The output of our shortest path algorithms will be a pair of V × V arrays encoding all V 2 distances
and predecessors. Many maps include a distance matrix—to ﬁnd the distance from (say) Champaign to
(say) Columbus, you would look in the row labeled ‘Champaign’ and the column labeled ‘Columbus’. In
these notes, I’ll focus almost exclusively on computing the distance array. The predecessor array, from
which you would compute the actual shortest paths, can be computed with only minor additions to the
algorithms I’ll describe (hint, hint). 14.2 Lots of Single Sources The obvious solution to the all-pairs shortest path problem is just to run a single-source shortest path
algorithm V times, once for every possible source vertex! Speciﬁcally, to ﬁll in the one-dimensional
subarray dist[s, ·], we invoke either Dijkstra’s or Shimbel’s algorithm starting at the source vertex s.
1 Algorithms Lecture 14: All-Pairs Shortest Paths
OBVIOUSAPSP(V, E , w ):
for every vertex s
dist[s, ·] ← SSSP(V, E , w, s) The running time of this algorithm depends on which single-source shortest path algorithm we use. If
we use Shimbel’s algorithm, the overall running time is Θ(V 2 E ) = O(V 4 ). If all the edge weights
are non-negative, we can use Dijkstra’s algorithm instead, which decreases the running time to
Θ(V E + V 2 log V ) = O(V 3 ). For graphs with negative edge weights, Dijkstra’s algorithm can take
exponential time, so we can’t get this improvement directly. 14.3 Reweighting One idea that occurs to most people is increasing the weights of all the edges by the same amount so
that all the weights become positive, and then applying Dijkstra’s algorithm. Unfortunately, this simple
idea doesn’t work. Different paths change by different amounts, which means the shortest paths in the
reweighted graph may not be the same as in the original graph.
3
s 2 2
4 t 4 Increasing all the edge weights by 2 changes the shortest path s to t . However, there is a more complicated method for reweighting the edges in a graph. Suppose each
vertex v has some associated cost c ( v ), which might be positive, negative, or zero. We can deﬁne a new
weight function w as follows:
w (u→ v ) = c (u) + w (u→ v ) − c ( v )
To give some intuition, imagine that when we leave vertex u, we have to pay an exit tax of c (u), and
when we enter v , we get c ( v ) as an entrance gift.
Now it’s not too hard to show that the shortest paths with the new weight function w are exactly
the same as the shortest paths with the original weight function w . In fact, for any path u
v from one
vertex u to another vertex v , we have
w (u v ) = c (u) + w (u v ) − c ( v ). We pay c (u) in exit fees, plus the original weight of of the path, minus the c ( v ) entrance gift. At every
intermediate vertex x on the path, we get c ( x ) as an entrance gift, but then immediately pay it back as
an exit tax! 14.4 Johnson’s Algorithm Johnson’s all-pairs shortest path algorithm ﬁnds a cost c ( v ) for each vertex, so that when the graph is
reweighted, every edge has non-negative weight.
Suppose the graph has a vertex s that has a path to every other vertex. Johnson’s algorithm computes
the shortest paths from s to every other vertex, using Shimbel’s algorithm (which doesn’t care if the edge
weights are negative), and then sets
c ( v ) = dist(s, v ), 2 Algorithms Lecture 14: All-Pairs Shortest Paths so the new weight of every edge is
w (u→ v ) = dist(s, u) + w (u→ v ) − dist(s, v ).
Why are all these new weights non-negative? Because otherwise, Shimbel’s algorithm wouldn’t be
ﬁnished! Recall that an edge u→ v is tense if dist(s, u) + w (u→ v ) < dist(s, v ), and that single-source
shortest path algorithms eliminate all tense edges. The only exception is if the graph has a negative
cycle, but then shortest paths aren’t deﬁned, and Johnson’s algorithm simply aborts.
But what if the graph doesn’t have a vertex s that can reach everything? No matter where we start
Shimbel’s algorithm, some of those vertex costs will be inﬁnite. Johnson’s algorithm avoids this problem
by adding a new vertex s to the graph, with zero-weight edges going from s to every other vertex, but
no edges going back into s. This addition doesn’t change the shortest paths between any other pair of
vertices, because there are no paths into s.
So here’s Johnson’s algorithm in all its glory.
JOHNSONAPSP(V, E , w ) :
create a new vertex s
for every vertex v ∈ V
w (s→ v ) ← 0
w ( v →s ) ← ∞
dist[s, ·] ← SHIMBEL(V, E , w, s)
abort if SHIMBEL found a negative cycle
for every edge (u, v ) ∈ E
w (u→ v ) ← dist[s, u] + w (u→ v ) − dist[s, v ]
for every vertex u ∈ V
dist[u, ·] ← DIJKSTRA(V, E , w , u)
for every vertex v ∈ V
dist[u, v ] ← dist[u, v ] − dist[s, u] + dist[s, v ] The algorithm spends Θ(V ) time adding the artiﬁcial start vertex s, Θ(V E ) time running SHIMBEL,
O( E ) time reweighting the graph, and then Θ(V E + V 2 log V ) running V passes of Dijkstra’s algorithm.
Thus, the overall running time is Θ(V E + V 2 log V ). 14.5 Dynamic Programming There’s a completely different solution to the all-pairs shortest path problem that uses dynamic programming instead of a single-source algorithm. For dense graphs where E = Ω(V 2 ), the dynamic programming
approach eventually leads to the same O(V 3 ) running time as Johnson’s algorithm, but with a much
simpler algorithm. In particular, the new algorithm avoids Dijkstra’s algorithm, which gets its efﬁciency
from Fibonacci heaps, which are rather easy to screw up in the implementation. In the rest of this
lecture, I will assume that the input graph contains no negative cycles.
As usual for dynamic programming algorithms, we ﬁrst need to come up with a recursive formulation
of the problem. Here is an “obvious" recursive deﬁnition for dist(u, v ):
dist(u, v ) = if u = v 0
min dist(u, x ) + w ( x → v )
x otherwise In other words, to ﬁnd the shortest path from u to v , try all possible predecessors x , compute the shortest
path from u to x , and then add the last edge u→ v . Unfortunately, this recurrence doesn’t work! In 3 Algorithms Lecture 14: All-Pairs Shortest Paths order to compute dist(u, v ), we ﬁrst have to compute dist(u, x ) for every other vertex x , but to compute
any dist(u, x ), we ﬁrst need to compute dist(u, v ). We’re stuck in an inﬁnite loop!
To avoid this circular dependency, we need an additional parameter that decreases at each recursion,
eventually reaching zero at the base case. One possibility is to include the number of edges in the
shortest path as this third magic parameter. So let’s deﬁne dist(u, v, k) to be the length of the shortest
path from u to v that uses at most k edges. Since we know that the shortest path between any two
vertices has at most V − 1 vertices, what we’re really trying to compute is dist(u, v, V − 1).
After a little thought, we get the following recurrence. 0
if u = v if k = 0 and u = v
dist(u, v, k) = ∞ min dist(u, x , k − 1) + w ( x → v )
otherwise
x Just like last time, the recurrence tries all possible predecessors of v in the shortest path, but now the
recursion actually bottoms out when k = 0.
Now it’s not difﬁcult to turn this recurrence into a dynamic programming algorithm. Even before
we write down the algorithm, though, we can tell that its running time will be Θ(V 4 ) simply because
recurrence has four variables—u, v , k, and x —each of which can take on V different values. Except for
the base cases, the algorithm itself is just four nested for loops. To make the algorithm a little shorter,
let’s assume that w ( v → v ) = 0 for every vertex v .
DYNAMICPROGRAMMINGAPSP(V, E , w ):
for all vertices u ∈ V
for all vertices v ∈ V
if u = v
dist[u, v, 0] ← 0
else
dist[u, v, 0] ← ∞
for k ← 1 to V − 1
for all vertices u ∈ V
for all vertices v ∈ V
dist[u, v, k] ← ∞
for all vertices x ∈ V
dist[u, v, k] ← min dist[u, v, k], dist[u, x , k − 1] + w ( x → v ) The last four lines actually evaluate the recurrence.
In fact, this algorithm is almost exactly the same as running Shimbel’s algorithm once for every
source vertex. The only difference is the innermost loop, which in Shimbel’s algorithm would read “for
all edges x → v ”. This simple change improves the running time to Θ(V 2 E ), assuming the graph is
stored in an adjacency list. 14.6 Divide and Conquer But we can make a more signiﬁcant improvement. The recurrence we just used broke the shortest path
into a slightly shorter path and a single edge, by considering all predecessors. Instead, let’s break it
into two shorter paths at the middle vertex on the path. This idea gives us a different recurrence for
dist(u, v, k). Once again, to simplify things, let’s assume w ( v → v ) = 0.
dist(u, v, k) = w (u→ v ) if k = 1 min dist(u, x , k/2) + dist( x , v, k/2) otherwise x 4 Algorithms Lecture 14: All-Pairs Shortest Paths This recurrence only works when k is a power of two, since otherwise we might try to ﬁnd the shortest
path with a fractional number of edges! But that’s not really a problem, since dist(u, v, 2 lg V ) gives us
the overall shortest distance from u to v . Notice that we use the base case k = 1 instead of k = 0, since
we can’t use half an edge.
Once again, a dynamic programming solution is straightforward. Even before we write down the
algorithm, we can tell the running time is Θ(V 3 log V )—we consider V possible values of u, v , and x ,
but only lg V possible values of k.
FASTDYNAMICPROGRAMMINGAPSP(V, E , w ):
for all vertices u ∈ V
for all vertices v ∈ V
dist[u, v, 0] ← w (u→ v )
for i ← 1 to lg V
〈〈k = 2i 〉〉
for all vertices u ∈ V
for all vertices v ∈ V
dist[u, v, i ] ← ∞
for all vertices x ∈ V
if dist[u, v, i ] > dist[u, x , i − 1] + dist[ x , v, i − 1]
dist[u, v, i ] ← dist[u, x , i − 1] + dist[ x , v, i − 1] 14.7 Aside: ‘Funny’ Matrix Multiplication There is a very close connection (ﬁrst observed by Shimbel, and later independently by Bellman) between
computing shortest paths in a directed graph and computing powers of a square matrix. Compare the
following algorithm for multiplying two n × n matrices A and B with the inner loop of our ﬁrst dynamic
programming algorithm. (I’ve changed the variable names in the second algorithm slightly to make the
similarity clearer.)
MATRIXMULTIPLY(A, B ):
for i ← 1 to n
for j ← 1 to n
C [i , j ] ← 0
for k ← 1 to n
C [i , j ] ← C [i , j ] + A[i , k] · B [k, j ]
APSPINNERLOOP:
for all vertices u
for all vertices v
D [u, v ] ← ∞
for all vertices x
D [u, v ] ← min D [u, v ], D[u, x ] + w [ x , v ] The only difference between these two algorithms is that we use addition instead of multiplication and
minimization instead of addition. For this reason, the shortest path inner loop is often referred to as
‘funny’ matrix multiplication.
DYNAMICPROGRAMMINGAPSP is the standard iterative algorithm for computing the (V − 1)th ‘funny
power’ of the weight matrix w . The ﬁrst set of for loops sets up the ‘funny identity matrix’, with zeros
on the main diagonal and inﬁnity everywhere else. Then each iteration of the second main for loop
computes the next ‘funny power’. FASTDYNAMICPROGRAMMINGAPSP replaces this iterative method for
computing powers with repeated squaring, exactly like we saw at the beginning of the semester. The
fast algorithm is simpliﬁed slightly by the fact that unless there are negative cycles, every ‘funny power’
after the V th is the same.
5 Algorithms Lecture 14: All-Pairs Shortest Paths There are faster methods for multiplying matrices, similar to Karatsuba’s divide-and-conquer algorithm for multiplying integers. (Google for ‘Strassen’s algorithm’.) Unfortunately, these algorithms us
subtraction, and there’s no ‘funny’ equivalent of subtraction. (What’s the inverse operation for min?)
So at least for general graphs, there seems to be no way to speed up the inner loop of our dynamic
programming algorithms.
Fortunately, this isn’t true. There is a beautiful randomized algorithm, due to Noga Alon, Zvi
Galil, Oded Margalit*, and Moni Noar,1 that computes all-pairs shortest paths in undirected graphs
in O( M (V ) log2 V ) expected time, where M (V ) is the time to multiply two V × V integer matrices. A
simpliﬁed version of this algorithm for unweighted graphs was discovered by Raimund Seidel.2 14.8 Floyd and Warshall’s Algorithm Our fast dynamic programming algorithm is still a factor of O(log V ) slower than Johnson’s algorithm. A
different formulation due to Floyd and Warshall removes this logarithmic factor. Their insight was to use
a different third parameter in the recurrence.
Number the vertices arbitrarily from 1 to V . For every pair of vertices u and v and every integer r ,
we deﬁne a path π(u, v, r ) as follows:
π(u, v, r ) := the shortest path from u to v where every intermediate vertex (that is, every
vertex except u and v ) is numbered at most r .
If r = 0, we aren’t allowed to use any intermediate vertices, so π(u, v, 0) is just the edge (if any) from
u to v . If r > 0, then either π(u, v, r ) goes through the vertex numbered r , or it doesn’t. If π(u, v, r ) does
contain vertex r , it splits into a subpath from u to r and a subpath from r to v , where every intermediate
vertex in these two subpaths is numbered at most r − 1. Moreover, the subpaths are as short as possible
with this restriction, so they must be π(u, r, r − 1) and π( r, v, r − 1). On the other hand, if π(u, v, r ) does
not go through vertex r , then every intermediate vertex in π(u, v, r ) is numbered at most r − 1; since
π(u, v, r ) must be the shortest such path, we have π(u, v, r ) = π(u, v, r − 1). u intermediate nodes ≤ r v = intermediate nodes ≤ r-1 u —
int
no erm
de ed
s ≤ iat
r-1 e or —
r v te
dia
me r-1
ter s ≤
in de
no Recursive structure of the restricted shortest path π(u, v, r ). This recursive structure implies the following recurrence for the length of π(u, v, r ), which we will
denote by dist(u, v, r ):
dist(u, v, r ) = w (u→ v ) if r = 0 min dist(u, v, r − 1), dist(u, r, r − 1) + dist( r, v, r − 1) otherwise We need to compute the shortest path distance from u to v with no restrictions, which is just dist(u, v, V ).
1
N. Alon, Z. Galil, O. Margalit*, and M. Naor. Witnesses for Boolean matrix multiplication and for shortest paths. Proc. 33rd
FOCS 417-426, 1992. See also N. Alon, Z. Galil, O. Margalit*. On the exponent of the all pairs shortest path problem. Journal
of Computer and System Sciences 54(2):255–262, 1997.
2
R. Seidel. On the all-pairs-shortest-path problem in unweighted undirected graphs. Journal of Computer and System
Sciences, 51(3):400-403, 1995. This is one of the few algorithms papers where (in the conference version at least) the
algorithm is completely described and analyzed in the abstract of the paper. 6 Algorithms Lecture 14: All-Pairs Shortest Paths Once again, we should immediately see that a dynamic programming algorithm that implements
this recurrence will run in Θ(V 3 ) time: three variables appear in the recurrence (u, v , and r ), each of
which can take on V possible values. Here’s one way to do it:
FLOYDWARSHALL(V, E , w ):
for u ← 1 to V
for v ← 1 to V
dist[u, v, 0] ← w (u→ v )
for r ← 1 to V
for u ← 1 to V
for v ← 1 to V
if dist[u, v, r − 1] < dist[u, r, r − 1] + dist[ r, v, r − 1]
dist[u, v, r ] ← dist[u, v, r − 1]
else
dist[u, v, r ] ← dist[u, r, r − 1] + dist[ r, v, r − 1] Exercises
1. Let G = (V, E ) be a directed graph with weighted edges; edge weights could be positive, negative,
or zero.
(a) How could we delete some node v from this graph, without changing the shortest-path
distance between any other pair of nodes? Describe an algorithm that constructs a directed
graph G = (V , E ) with weighted edges, where V = V \ { v }, and the shortest-path distance
between any two nodes in H is equal to the shortest-path distance between the same two
nodes in G , in O(V 2 ) time.
(b) Now suppose we have already computed all shortest-path distances in G . Describe an
algorithm to compute the shortest-path distances from v to every other node, and from every
other node to v , in the original graph G , in O(V 2 ) time.
(c) Combine parts (a) and (b) into another all-pairs shortest path algorithm that runs in O(V 3 )
time.
2. All of the algorithms discussed in this lecture fail if the graph contains a negative cycle. Johnson’s
algorithm detects the negative cycle in the initialization phase (via Shimbel’s algorithm) and
aborts; the dynamic programming algorithms just return incorrect results. However, all of these
algorithms can be modiﬁed to return correct shortest-path distances, even in the presence of
negative cycles. Speciﬁcally, if there is a path from vertex u to a negative cycle and a path from
that negative cycle to vertex v , the algorithm should report that dist[u, v ] = −∞. If there is no
directed path from u to v , the algorithm should return dist[u, v ] = ∞. Otherwise, dist[u, v ] should
equal the length of the shortest directed path from u to v .
(a) Describe how to modify Johnson’s algorithm to return the correct shortest-path distances,
even if the graph has negative cycles.
(b) Describe how to modify the Floyd-Warshall algorithm to return the correct shortest-path
distances, even if the graph has negative cycles.
3. Let G = (V, E ) be a directed graph with weighted edges; edge weights could be positive, negative,
or zero. Suppose the vertices of G are partitioned into k disjoint subsets V1 , V2 , . . . , Vk ; that is,
7 Algorithms Lecture 14: All-Pairs Shortest Paths every vertex of G belongs to exactly one subset Vi . For each i and j , let δ(i , j ) denote the minimum
shortest-path distance between vertices in Vi and vertices in Vj :
δ(i , j ) = min{dist(u, v ) | u ∈ Vi and v ∈ Vj }.
Describe an algorithm to compute δ(i , j ) for all i and j in time O(V 2 + kE log E ).
4. Recall3 that a deterministic ﬁnite automaton (DFA) is formally deﬁned as a tuple M = (Σ, Q, q0 , F, δ),
where the ﬁnite set Σ is the input alphabet, the ﬁnite set Q is the set of states, q0 ∈ Q is the start
state, F ⊆ Q is the set of ﬁnal (accepting) states, and δ : Q × Σ → Q is the transition function.
Equivalently, a DFA is a directed (multi-)graph with labeled edges, such that each symbol in Σ is
the label of exactly one edge leaving any vertex. There is a special ‘start’ vertex q0 , and a subset of
the vertices are marked as ‘accepting states’. Any string in Σ∗ describes a unique walk starting
at q0 ; a string in Σ∗ is accepted by M if this walk ends at a vertex in F .
Stephen Kleene4 proved that the language accepted by any DFA is identical to the language
described by some regular expression. This problem asks you to develop a variant of the FloydWarshall all-pairs shortest path algorithm that computes a regular expression that is equivalent to
the language accepted by a given DFA.
Suppose the input DFA M has n states, numbered from 1 to n, where (without loss of generality)
the start state is state 1. Let L (i , j , r ) denote the set of all words that describe walks in M from
state i to state j , where every intermediate state lies in the subset {1, 2, . . . , r }; thus, the language
accepted by the DFA is exactly
L (1, q, n).
q∈ F Let R(i , j , r ) be a regular expression that describes the language L (i , j , r ).
(a) What is the regular expression R(i , j , 0)?
(b) Write a recurrence for the regular expression R(i , j , r ) in terms of regular expressions of the
form R(i , j , r − 1).
(c) Describe a polynomial-time algorithm to compute R(i , j , n) for all states i and j . (Assume
that you can concatenate two regular expressions in O(1) time.)
5. Let G = (V, E ) be an undirected, unweighted, connected, n-vertex graph, represented by the
adjacency matrix A[1 .. n, 1 .. n]. In this problem, we will derive Seidel’s sub-cubic algorithm to
compute the n × n matrix D[1 .. n, 1 .. n] of shortest-path distances using fast matrix multiplication.
Assume that we have a subroutine MATRIXMULTIPLY that multiplies two n × n matrices in Θ(nω )
time, for some unknown constant ω ≥ 2.5
(a) Let G 2 denote the graph with the same vertices as G , where two vertices are connected by
a edge if and only if they are connected by a path of length at most 2 in G . Describe an
algorithm to compute the adjacency matrix of G 2 using a single call to MATRIXMULTIPLY and
O(n2 ) additional time.
3 Automata theory is a prerequisite for the algorithms class at UIUC.
Pronounced ‘clay knee’, not ‘clean’ or ‘clean-ee’ or ‘clay-nuh’ or ‘dimaggio’.
5
The matrix multiplication algorithm you already know runs in Θ(n3 ) time, but this is not the fastest algorithm known. The
current record is ω ≈ 2.376, due to Don Coppersmith and Shmuel Winograd. Determining the smallest possible value of ω is a
long-standing open problem; many people believe there is an undiscovered O(n2 )-time algorithm for matrix multiplication.
4 8 Algorithms Lecture 14: All-Pairs Shortest Paths (b) Suppose we discover that G 2 is a complete graph. Describe an algorithm to compute the
matrix D of shortest path distances in O(n2 ) additional time.
(c) Let D2 denote the (recursively computed) matrix of shortest-path distances in G 2 . Prove that
the shortest-path distance from node i to node j is either 2 · D2 [i , j ] or 2 · D2 [i , j ] − 1.
(d) Suppose G 2 is not a complete graph. Let X = D2 · A, and let deg(i ) denote the degree of
vertex i in the original graph G . Prove that the shortest-path distance from node i to node j
is 2 · D2 [i , j ] if and only if X [i , j ] ≥ D2 [i , j ] · deg(i ).
(e) Describe an algorithm to compute the matrix of shortest-path distances in G in O(nω log n)
time. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 9 Algorithms Lecture 15: Maximum Flows and Minimum Cuts
Col. Hogan: One of these wires disconnects the fuse,
the other one ﬁres the bomb.
Which one would you cut, Shultz?
Sgt. Schultz: Don’t ask me, this is a decision for an ofﬁcer.
Col. Hogan: All right. Which wire, Colonel Klink?
Col. Klink: This one. [points to the white wire]
Col. Hogan: You’re sure?
Col. Klink: Yes.
[Hogan cuts the black wire; the bomb stops ticking]
Col. Klink: If you knew which wire it was, why did you ask me?
Col. Hogan: I wasn’t sure which was the right one,
but I was certain you’d pick the wrong one.
— “A Klink, a Bomb, and a Short Fuse", Hogan’s Heroes (1966) 15 Maximum Flows and Minimum Cuts In the mid-1950s, Air Force researchers T. E. Harris and F. S. Ross published a classiﬁed report studying
the rail network that linked the Soviet Union to its satellite countries in Eastern Europe. The network
was modeled as a graph with 44 vertices, representing geographic regions, and 105 edges, representing
links between those regions in the rail network. Each edge was given a weight, representing the rate
at which material could be shipped from one region to the next. Essentially by trial and error, they
determined both the maximum amount of stuff that could be moved from Russia into Europe, as well as
the cheapest way to disrupt the network by removing links (or in less abstract terms, blowing up train
tracks), which they called ‘the bottleneck’. Their results (including the ﬁgure at the top of the page)
were only declassiﬁed in 1999.1 Harris and Ross’s map of the Warsaw Pact rail network This one of the ﬁrstFrom Harris and Ross [1955]: Schematicofagure 2ofmaximum oﬂowheanderminimum cut problems. For
recorded applications Fiig the the railway netw rk of t West n Sod ram
vie U o and Eastern E ropean co G s, ( h , E xi um ﬂow o value 1 ,000 tons fr vertices s and t called the
both problems, the inputtisnian directedugraphuntrie=witVa ma),malong f with63specialom
Russia to Eastern Europe, and a cut of capacity 163,000 tons indicated as ‘The bottleneck’.
source and target. As in the previous lectures, I will use u v to denote the directed edge from vertex u
to vertex v . Intuitively, the maximum ﬂow problem asks for the largest amount of material that can
The max-ﬂow min-cut theorem
be transported from one vertex to another; the minimum cut problem asks for the minimum damage
In the RAND Report of 19 November 1954, Ford and Fulkerson [1954] gave (next to deﬁning
needed to separate two maximum ﬂow problem and suggesting the simplex method for it) the max-ﬂow minthe vertices.
1 cut theorem for undirected graphs, saying that the maximum ﬂow value is equal to the Both the map and the storymwere itakencfromarating source and terminal. Their proof is nsurveyructive, history of combinatorial
minimu capac ty of a ut sep Alexander Schrijver’s fascinating ot const ‘On the
optimization (till 1960)’. but for planar graphs, with source and sink on the outer boundary, they give a polynomialtime, constructive method. In a report of 26 May 1955, Robacker [1955a] showed that the
max-ﬂow min-cut theorem can be derived also from the vertex-disjoint version of Menger’s
theorem.
1
As for the directed case, Ford and Fulkerson [1955] observed that the max-ﬂow min-cut
theorem holds also for directed graphs. Dantzig and Fulkerson [1955] showed, by extending
the results of Dantzig [1951a] on integer solutions for the transportation problem to the
25 Algorithms 15.1 Lecture 15: Maximum Flows and Minimum Cuts Flows An ( s , t )-ﬂow (or just a ﬂow if the source and target are clear from context) is a function f : E → IR≥0
that satisﬁes the following conservation constraint at every vertex v except possibly s and t :
f (u v ) = f ( v w ). u w In English, the total ﬂow into v is equal to the total ﬂow out of v . To keep the notation simple, we
assume here that f (u v ) = 0 if there is no edge u v in the graph.
The value of the ﬂow f , denoted | f |, is deﬁned as the excess ﬂow out of the source vertex s:
| f | := f (s w ) − f (u s) w u It’s not hard to show that the value | f | is also equal to the excess ﬂow into the target vertex t . First we
observe that
f (v w) −
v w f (u v ) = f (v w) − u v w f (u v ) = 0
v u because both summations count the total ﬂow across all edges. On the other hand, the conservation
constraint implies that
f (v w) −
v f (u v )
u w = f (t w) −
w f (u s) f (t w) − + u = |f | + w f (t w) −
w f (u t ) f (u t )
u . u It follows that
|f | = f (u t ) −
u f ( t w ).
w Now suppose we have another function c : E → IR≥0 that assigns a non-negative capacity c (e) to
each edge e. We say that a ﬂow f is feasible (with respect to c ) if f (e) ≤ c (e) for every edge e. Most of
the time we will consider only ﬂows that are feasible with respect to some ﬁxed capacity function c . We
say that a ﬂow f saturates edge e if f (e) = c (e), and avoids edge e if f (e) = 0. The maximum ﬂow
problem is to compute a feasible (s, t )-ﬂow in a given directed graph, with a given capacity function,
whose value is as large as possible.
0/5
5/15 10/20
0/15
s 5/10 10/10
0/10 t
5/20 10/10 An (s, t )-ﬂow with value 10. Each edge is labeled with its ﬂow/capacity. 2 Algorithms 15.2 Lecture 15: Maximum Flows and Minimum Cuts Cuts An ( s , t )-cut (or just cut if the source and target are clear from context) is a partition of the vertices
into disjoint subsets S and T —meaning S ∪ T = V and S ∩ T = ∅—where s ∈ S and t ∈ T .
If we have a capacity function c : E → IR≥0 , the capacity of a cut is the sum of the capacities of the
edges that start in S and end in T :
S, T = c ( v w ).
v ∈S w ∈ T (Again, if v w is not an edge in the graph, we assume c ( v w ) = 0.) Notice that the deﬁnition is
asymmetric; edges that start in T and end in S are unimportant. The minimum cut problem is to
compute an (s, t )-cut whose capacity is as large as possible.
5
15 20
10 s 15 10 10 t 20
10 An (s, t )-cut with capacity 15. Each edge is labeled with its capacity. Intuitively, the minimum cut is the cheapest way to disrupt all ﬂow from s to t . Indeed, it is not hard
to show that the value of any feasible ( s , t )-ﬂow is at most the capacity of any ( s , t )-cut. Choose
your favorite ﬂow f and your favorite cut (S , T ), and then follow the bouncing inequalities:
|f | = f (s w ) −
w f (u s) by deﬁnition u = f (v w) −
v ∈S w v ∈S w∈T f (u v ) by the convervation constraint f (u v ) removing duplicate edges u = f (v w) −
u∈ T f (v w) since f (u v ) ≥ 0 c(v w) ≤ since f (u v ) ≤ c ( v w ) v ∈S w ∈ T ≤
v ∈S w ∈ T = S, T by deﬁnition Our derivation actually implies the following stronger observation: | f | = S, T if and only if f saturates every edge from S to T and avoids every edge from T to S. Moreover, if we have a ﬂow f
and a cut (S , T ) that satisﬁes this equality condition, f must be a maximum ﬂow, and (S , T ) must be a
minimum cut. 3 Algorithms 15.3 Lecture 15: Maximum Flows and Minimum Cuts The Max-Flow Min-Cut Theorem Surprisingly, for any weighted directed graph, there is always a ﬂow f and a cut (S , T ) that satisfy the
equality condition. This is the famous max-ﬂow min-cut theorem:
The value of the maximum ﬂow is equal to the capacity of the minimum cut.
The rest of this section gives a proof of this theorem; we will eventually turn this proof into an algorithm.
Fix a graph G , vertices s and t , and a capacity function c : E → IR≥0 . The proof will be easier if we
assume that the capacity function is reduced: For any vertices u and v , either c (u v ) = 0 or c ( v u) = 0,
or equivalently, if an edge appears in G , then its reversal does not. This assumption is easy to enforce.
Whenever an edge u v and its reversal v u are both the graph, replace the edge u v with a path
u x v of length two, where x is a new vertex and c (u x ) = c ( x v ) = c (u v ). The modiﬁed graph
has the same maximum ﬂow value and minimum cut capacity as the original graph. Enforcing the one-direction assumption. Let f be a feasible ﬂow. We deﬁne a new capacity function c f : V × V → IR, called the residual
capacity, as follows: c (u v ) − f (u v ) if u v ∈ E
c f (u v ) = f ( v u) 0 if v u ∈ E . otherwise Since f ≥ 0 and f ≤ c , the residual capacities are always non-negative. It is possible to have c f (u v ) > 0
even if u v is not an edge in the original graph G . Thus, we deﬁne the residual graph G f = (V, E f ),
where E f is the set of edges whose residual capacity is positive. Notice that the residual capacities are
not necessarily reduced; it is quite possible to have both c f (u v ) > 0 and c f ( v u) > 0.
0/5 5 5/10 10/10
0/10 5 10 0/15
s 10 10 5/15 10/20 t s 10
15 5 5
5 10 5/20
10/10 t 15
10 A ﬂow f in a weighted graph G and the corresponding residual graph G f . Suppose there is no path from the source s to the target t in the residual graph G f . Let S be the set
of vertices that are reachable from s in G f , and let T = V \ S . The partition (S , T ) is clearly an (s, t )-cut.
For every vertex u ∈ S and v ∈ T , we have
c f (u v ) = (c (u v ) − f (u v )) + f ( v u) = 0,
which implies that c (u v ) − f (u v ) = 0 and f ( v u) = 0. In other words, our ﬂow f saturates every
edge from S to T and avoids every edge from T to S . It follows that | f | = S , T . Moreover, f is a
maximum ﬂow and (S , T ) is a minimum cut.
4 Algorithms Lecture 15: Maximum Flows and Minimum Cuts 5 5/5
10 10
s 5/15 10/20 5 10
10 15 5 0/15 5 t s 0/10 5/10 t 5 10 5/10 15
10 10/20
10/10 An augmenting path in G f with value F = 5 and the augmented ﬂow f . On the other hand, suppose there is a path s = v0 v1 · · · vr = t in G f . We refer to v0 v1 · · · vr
as an augmenting path. Let F = mini c f ( vi vi +1 ) denote the maximum amount of ﬂow that we can
push through the augmenting path in G f . We deﬁne a new ﬂow function f : E → IR as follows: f (u v ) + F
f (u v ) = f (u v ) − F f (u v ) if u v is in the augmenting path
if v u is in the augmenting path
otherwise To prove that the ﬂow f is feasible with respect to the original capacities c , we need to verify that
f ≥ 0 and f ≤ c . Consider an edge u v in G . If u v is in the augmenting path, then f (u v ) >
f (u v ) ≥ 0 and
f (u v ) = f (u v ) + F by deﬁnition of f ≤ f (u v ) + c f (u v ) by deﬁnition of F = f (u v ) + c (u v ) − f (u v ) by deﬁnition of c f = c (u v ) Duh. On the other hand, if the reversal v u is in the augmenting path, then f (u v ) < f (u v ) ≤ c (u v ),
which implies that
f (u v ) = f (u v ) − F by deﬁnition of f ≥ f (u v ) − c f ( v u) by deﬁnition of F = f (u v ) − f (u v ) by deﬁnition of c f =0 Duh. Finally, we observe that (without loss of generality) only the ﬁrst edge in the augmenting path leaves s,
so | f | = | f | + F > 0. In other words, f is not a maximum ﬂow.
This completes the proof! Exercises
1. Let (S , T ) and (S , T ) be minimum (s, t )-cuts in some ﬂow network G . Prove that (S ∩ S , T ∪ T )
and (S ∪ S , T ∩ T ) are also minimum (s, t )-cuts in G .
2. Suppose (S , T ) is a minimum (s, t )-cut in some ﬂow network. Prove that (S , T ) is also a minimum
( x , y )-cut for all vertices x ∈ S and y ∈ T .
5 Algorithms Lecture 15: Maximum Flows and Minimum Cuts 3. Cuts are sometimes deﬁned as subsets of the edges of the graph, instead of as partitions of its
vertices. In this problem, you will prove that these two deﬁnitions are almost equivalent.
We say that a subset X of (directed) edges separates s and t if every directed path from s to t
contains at least one (directed) edge in X . For any subset S of vertices, let δS denote the set of
directed edges leaving S ; that is, δS := {u v | u ∈ S , v ∈ S }.
(a) Prove that if (S , T ) is an (s, t )-cut, then δS separates s and t .
(b) Let X be an arbitrary subset of edges that separates s and t . Prove that there is an (s, t )-cut
(S , T ) such that δS ⊆ X .
(c) Let X be a minimal subset of edges that separates s and t . Prove that there is an (s, t )-cut
(S , T ) such that δS = X .
4. A ﬂow f is acyclic if the subgraph of directed edges with positive ﬂow contains no directed cycles.
(a) Prove that for any ﬂow f , there is an acyclic ﬂow with the same value as f . (In particular,
this implies that some maximum ﬂow is acyclic.)
(b) A path ﬂow assigns positive values only to the edges of one simple directed path from s to t .
Prove that every acyclic ﬂow can be written as the sum of a ﬁnite number of path ﬂows.
(c) Describe a ﬂow in a directed graph that cannot be written as the sum of path ﬂows.
(d) A cycle ﬂow assigns positive values only to the edges of one simple directed cycle. Prove that
every ﬂow can be written as the sum of a ﬁnite number of path ﬂows and cycle ﬂows.
(e) Prove that every ﬂow with value 0 can be written as the sum of a ﬁnite number of cycle ﬂows.
(Zero-value ﬂows are also called circulations.)
5. Suppose instead of capacities, we consider networks where each edge u v has a non-negative
demand d (u v ). Now an (s, t )-ﬂow f is feasible if and only if f (u v ) ≥ d (u v ) for every edge
u v . (Feasible ﬂow values can now be arbitrarily large.) A natural problem in this setting is to
ﬁnd a feasible (s, t )-ﬂow of minimum value.
(a) Describe an efﬁcient algorithm to compute a feasible (s, t )-ﬂow, given the graph, the demand
function, and the vertices s and t as input. [Hint: Find a ﬂow that is non-zero everywhere,
and then scale it up to make it feasible.]
(b) Suppose you have access to a subroutine MAXFLOW that computes maximum ﬂows in networks
with edge capacities. Describe an efﬁcient algorithm to compute a minimum ﬂow in a given
network with edge demands; your algorithm should call MAXFLOW exactly once.
(c) State and prove an analogue of the max-ﬂow min-cut theorem for this setting. (Do minimum
ﬂows correspond to maximum cuts?) c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 6 Algorithms Lecture 16: Max-Flow Algorithms
A process cannot be understood by stopping it. Understanding must move
with the ﬂow of the process, must join it and ﬂow with it.
— The First Law of Mentat, in Frank Herbert’s Dune (1965)
There’s a difference between knowing the path and walking the path.
— Morpheus [Laurence Fishburne], The Matrix (1999) 16
16.1 Max-Flow Algorithms
Recap Fix a directed graph G = (V, E ) that does not contain both an edge u v and its reversal v u, and ﬁx a
capacity function c : E → IR≥0 . For any ﬂow function f : E → IR≥0 , the residual capacity is deﬁned as c (u v ) − f (u v ) if u v ∈ E
c f (u v ) = f ( v u) 0 if v u ∈ E . otherwise The residual graph G f = (V, E f ), where E f is the set of edges whose non-zero residual capacity is positive.
0/5 5 5 10 0/15
s 10 10 5/15 10/20
5/10 10/10 t 0/10 s 10 15 5 t
5 10 5/20 5 15 10/10 10 A ﬂow f in a weighted graph G and its residual graph G f . In the last lecture, we proved the Max-ﬂow Min-cut Theorem: In any weighted directed graph network,
the value of the maximum (s, t )-ﬂow is equal to the capacity of the minimum (s, t )-cut. The proof of the
theorem is constructive. If the residual graph contains a path from s to t , then we can increase the ﬂow
by the minimum capacity of the edges on this path, so we must not have the maximum ﬂow. Otherwise,
we can deﬁne a cut (S , T ) whose capacity is the same as the ﬂow f , such that every edge from S to T is
saturated and every edge from T to S is empty, which implies that f is a maximum ﬂow and (S , T ) is a
minimum cut.
5 5/5
10 10
s 5/15 10/20 5 10
10 15 5 0/15 5 t s 0/10 5/10 t 5 10 5/10 15
10 10/20
10/10 An augmenting path in G f and the resulting (maximum) ﬂow f . 1 Algorithms 16.2 Lecture 16: Max-Flow Algorithms Ford-Fulkerson It’s not hard to realize that this proof translates almost immediately to an algorithm, ﬁrst developed by
Ford and Fulkerson in the 1950s: Starting with the zero ﬂow, repeatedly augment the ﬂow along any
path s
t in the residual graph, until there is no such path.
If every edge capacity is an integer, then every augmentation step increases the value of the ﬂow by
a positive integer. Thus, the algorithm halts after | f ∗ | iterations, where f ∗ is the actual maximum ﬂow.
Each iteration requires O( E ) time, to create the residual graph G f and perform a whatever-ﬁrst-search to
ﬁnd an augmenting path. Thus, in the words case, the Ford-Fulkerson algorithm runs in O (E | f ∗ |) time.
If we multiply all the capacities by the same (positive) constant, the maximum ﬂow increases
everywhere by the same constant factor. It follows that if all the edge capacities are rational, then the
Ford-Fulkerson algorithm eventually halts. However, if we allow irrational capacities, the algorithm can
loop forever, always ﬁnding smaller and smaller augmenting paths. Worse yet, this inﬁnite sequence
of augmentations may not even converge to the maximum ﬂow! Perhaps the simplest example of this
effect was discovered by Uri Zwick.
Consider the graph shown below, with six vertices and nine edges. Six of the edges have some large
integer capacity X , two have capacity 1, and one has capacity φ = ( 5 − 1)/2 ≈ 0.618034, chosen so
that 1 − φ = φ 2 . To prove that the Ford-Fulkerson algorithm can get stuck, we can watch the residual
capacities of the three horizontal edges as the algorithm progresses. (The residual capacities of the other
six edges will always be at least X − 3.)
s
X
1 X X
1 ϕ
X X X t A B C Uri Zwick’s non-terminating ﬂow example, and three augmenting paths. The Ford-Fulkerson algorithm starts by choosing the central augmenting path, shown in the large
ﬁgure above. The three horizontal edges„ in order from left to right, now have residual capacities 1, 0,
φ . Suppose inductively that the horizontal residual capacities are φ k−1 , 0, φ k for some non-negative
integer k.
1. Augment along B , adding φ k to the ﬂow; the residual capacities are now φ k+1 , φ k , 0.
2. Augment along C , adding φ k to the ﬂow; the residual capacities are now φ k+1 , 0, φ k .
3. Augment along B , adding φ k+1 to the ﬂow; the residual capacities are now 0, φ k+1 , φ k+2 .
4. Augment along A, adding φ k+1 to the ﬂow; the residual capacities are now φ k+1 , 0, φ k+2 . 2 Algorithms Lecture 16: Max-Flow Algorithms Thus, after 4n + 1 augmentation steps, the residual capacities are φ 2n−2 , 0, φ 2n−1 . As the number of
augmentation steps grows to inﬁnity, the value of the ﬂow converges to
∞ φi = 1 + 1+2
i =1 2
1−φ =4+ 5 < 7, even though the maximum ﬂow value is clearly 2X + 1.
Picky students might wonder at this point why we care about irrational capacities; after all, computers
can’t represent anything but (small) integers or (dyadic) rationals exactly. Good question! One reason is
that the integer restriction is literally artiﬁcial; it’s an artifact of actual computational hardware1 , not
an inherent feature of the abstract mathematical problem. Another reason, which is probably more
convincing to most practical computer scientists, is that the behavior of the algorithm with irrational
inputs tells us something about its worst-case behavior in practice given ﬂoating-point capacities—
terrible! Even with very reasonable capacities, a careless implementation of Ford-Fulkerson could enter
an inﬁnite loop simply because of round-off error! 16.3 Edmonds-Karp: Fat Pipes The Ford-Fulkerson algorithm does not specify which alternating path to use if there is more than one.
In 1972, Jack Edmonds and Richard Karp analyzed two natural heuristics for choosing the path. The
ﬁrst is essentially a greedy algorithm:
Choose the augmenting path with largest bottleneck value.
It’s a fairly easy to show that the maximum-bottleneck (s, t )-path in a directed graph can be computed
in O( E log V ) time using a variant of Jarník’s minimum-spanning-tree algorithm, or of Dijkstra’s shortest
path algorithm. Simply grow a directed spanning tree T , rooted at s. Repeatedly ﬁnd the highest-capacity
edge leaving T and add it to T , until T contains a path from s to t . Alternately, once could emulate
Kruskal’s algorithm—insert edges one at a time in decreasing capacity order until there is a path from s
to t —although this is less efﬁcient.
We can now analyze the algorithm in terms of the value of the maximum ﬂow f ∗ . Let f be any
ﬂow in G , and let f be the maximum ﬂow in the current residual graph G f . (At the beginning of the
algorithm, G f = G and f = f ∗ .) Let e be the bottleneck edge in the next augmenting path. Let S be the
set of vertices reachable from s through edges in G with capacity greater than c (e) and let T = V \ S .
By construction, T is non-empty, and every edge from S to T has capacity at most c (e). Thus, the
capacity of the cut (S , T ) is at most c (e) · E . On the other hand, the maxﬂow-mincut theorem implies
that S , T ≥ | f |. We conclude that c (e) ≥ | f |/ E .
The preceding argument implies that augmenting f along the maximum-bottleneck path in G f
multiplies the maximum ﬂow value in G f by a factor of at most 1 − 1/ E . In other words, the residual
ﬂow decays exponentially with the number of iterations. After E · ln| f ∗ | iterations, the maximum ﬂow
value in G f is at most
∗
∗
| f ∗ | · (1 − 1/ E ) E ·ln| f | < | f ∗ |e− ln| f | = 1.
(That’s Euler’s constant e, not the edge e. Sorry.) In particular, if all the capacities are integers, then after
E · ln| f ∗ | iterations, the maximum capacity of the residual graph is zero and f is a maximum ﬂow.
We conclude that for graphs with integer capacities, the Edmonds-Karp ‘fat pipe’ algorithm runs in
O (E 2 log E log| f ∗ |) time.
1 ...or perhaps the laws of physics. Yeah, right. Whatever. Like reality actually matters in this class. 3 Algorithms 16.4 Lecture 16: Max-Flow Algorithms Dinits/Edmonds-Karp: Short Pipes The second Edmonds-Karp heuristic was actually proposed by Ford and Fulkerson in their original
max-ﬂow paper, and ﬁrst analyzed by the Russian mathematician Dinits (sometimes transliterated Dinic)
in 1970. Edmonds and Karp published their independent and slightly weaker analysis in 1972. So
naturally, almost everyone refers to this algorithm as ‘Edmonds-Karp’.2
Choose the augmenting path with fewest edges.
The correct path can be found in O( E ) time by running breadth-ﬁrst search in the residual graph. More
surprisingly, the algorithm halts after a polynomial number of iterations, independent of the actual edge
capacities!
The proof of this upper bound relies on two observations about the evolution of the residual graph.
Let f i be the current ﬂow after i augmentation steps, let Gi be the corresponding residual graph. In
particular, f0 is zero everywhere and G0 = G . For each vertex v , let leveli ( v ) denote the unweighted
shortest path distance from s to v in Gi , or equivalently, the level of v in a breadth-ﬁrst search tree of Gi
rooted at s.
Our ﬁrst observation is that these levels can only increase over time.
Lemma 1. leveli +1 ( v ) ≥ leveli ( v ) for all vertices v and integers i .
Proof: The claim is trivial for v = s, since leveli (s) = 0 for all i . Choose an arbitrary vertex v = s, and
let s · · · u v be a shortest path from s to v in Gi +1 . (If there is no such path, then leveli +1 ( v ) = ∞,
and we’re done.) Because this is a shortest path, we have leveli +1 ( v ) = leveli +1 (u) + 1, and the inductive
hypothesis implies that leveli +1 (u) ≥ leveli (u).
We now have two cases to consider. If u v is an edge in Gi , then leveli ( v ) ≤ leveli (u) + 1, because
the levels are deﬁned by breadth-ﬁrst traversal.
On the other hand, if u v is not an edge in Gi , then v u must be an edge in the i th augmenting path.
Thus, v u must lie on the shortest path from s to t in Gi , which implies that leveli ( v ) = leveli (u) − 1 ≤
leveli (u) + 1.
In both cases, we have leveli +1 ( v ) = leveli +1 (u) + 1 ≥ leveli (u) + 1 ≥ leveli ( v ).
Whenever we augment the ﬂow, the bottleneck edge in the augmenting path disappears from the
residual graph, and some other edge in the reversal of the augmenting path may (re-)appear. Our second
observation is that an edge cannot appear or disappear too many times.
Lemma 2. During the execution of the Dinits/Edmonds-Karp algorithm, any edge u v disappears from
the residual graph G f at most V /2 times.
Proof: Suppose u v is in two residual graphs Gi and G j +1 , but not in any of the intermediate residual
graphs Gi +1 , . . . , G j , for some i < j . Then u v must be in the i th augmenting path, so leveli ( v ) =
leveli (u) + 1, and v u must be on the j th augmenting path, so level j ( v ) = level j (u) − 1. By the previous
lemma, we have
level j (u) = level j ( v ) + 1 ≥ leveli ( v ) + 1 = leveli (u) + 2.
2
To be fair, Edmonds and Karp discovered their algorithm a few years before publication—getting ideas into print takes time,
especially in the early 1970s—which is why some authors believe they deserve priority. I don’t buy it; Dinits also presumably
discovered his algorithm a few years before its publication. (In Soviet Union, result publish you.) On the gripping hand, Dinits’s
paper also described an improvement to the algorithm presented here that runs in O(V 2 E ) time instead of O(V E 2 ), so maybe
that ought to be called Dinits’s algorithm. 4 Algorithms Lecture 16: Max-Flow Algorithms In other words, the distance from s to u increased by at least 2 between the disappearance and
reappearance of u v . Since every level is either less than V or inﬁnite, the number of disappearances is
at most V /2.
Now we can derive an upper bound on the number of iterations. Since each edge can disappear at
most V /2 times, there are at most EV /2 edge disappearances overall. But at least one edge disappears
on each iteration, so the algorithm must halt after at most EV /2 iterations. Finally, since each iteration
requires O( E ) time, Dinits’ algorithm runs in O (V E 2 ) time overall. Exercises
1. A new assistant professor, teaching maximum ﬂows for the ﬁrst time, suggests the following greedy
modiﬁcation to the generic Ford-Fulkerson augmenting path algorithm. Instead of maintaining
a residual graph, just reduce the capacity of edges along the augmenting path! In particular,
whenever we saturate an edge, just remove it from the graph.
GREEDYFLOW(G , c , s, t ):
for every edge e in G
f (e) ← 0
while there is a path from s to t
π ← an arbitrary path from s to t
F ← minimum capacity of any edge in π
for every edge e in π
f (e) ← f (e) + F
if c (e) = F
remove e from G
else
c (e) ← c (e) − F
return f (a) Show that this algorithm does not always compute a maximum ﬂow.
(b) Prove that for any ﬂow network, if the Greedy Path Fairy tells you precisely which path π to
use at each iteration, then GREEDYFLOW does compute a maximum ﬂow. (Sadly, the Greedy
Path Fairy does not actually exist.)
2. Describe and analyze an algorithm to ﬁnd the maximum-bottleneck path from s to t in a ﬂow
network G in O( E log V ) time.
3. Describe a directed graph with irrational edge capacities, such that the Edmonds-Karp ‘fat pipe’
heuristic does not halt.
4. Describe an efﬁcient algorithm to check whether a given ﬂow network contains a unique maximum
ﬂow.
5. For any ﬂow network G and any vertices u and v , let bottleneckG (u, v ) denote the maximum, over
all paths π in G from u to v , of the minimum-capacity edge along π. Describe an algorithm
5 Algorithms Lecture 16: Max-Flow Algorithms to construct a spanning tree T of G such that bottleneck T (u, v ) = bottleneckG (u, v ). (Edges in T
inherit their capacities form G .)
One way to think about this problem is to imagine the vertices of the graph as islands, and the
edges as bridges. Each bridge has a maximum weight it can support. If a truck is carrying stuff
from u to v , how much can the truck carry? We don’t care what route the truck takes; the point is
that the smallest-weight edge on the route will determine the load.
6. We can speed up the Edmonds-Karp ‘fat pipe’ heuristic, at least for integer capacities, by relaxing
our requirements for the next augmenting path. Instead of ﬁnding the augmenting path with
maximum bottleneck capacity, we ﬁnd a path whose bottleneck capacity is at least half of maximum,
using the following capacity scaling algorithm.
The algorithm maintains a bottleneck threshold ∆; initially, ∆ is the maximum capacity among
all edges in the graph. In each phase, the algorithm augments along paths from s to t in which
every edge has residual capacity at least ∆. When there is no such path, the phase ends, we set
∆ ← ∆/2, and the next phase begins.
(a) How many phases will the algorithm execute in the worst case, if the edge capacities are
integers?
(b) Let f be the ﬂow at the end of a phase for a particular value of ∆. Let S be the nodes that are
reachable from s in the residual graph G f using only edges with residual capacity at least ∆,
and let T = V \ S . Prove that the capacity (with respect to G ’s original edge capacities) of the
cut (S , T ) is at most | f | + E · ∆.
(c) Prove that in each phase of the scaling algorithm, there are at most 2 E augmentations.
(d) What is the overall running time of the scaling algorithm, assuming all the edge capacities
are integers? c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 6 Algorithms Lecture 17: Applications of Maximum Flow
For a long time it puzzled me how something so expensive, so leading edge,
could be so useless, and then it occurred to me that a computer is a stupid
machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things.
They are, in short, a perfect match.
— Bill Bryson, Notes from a Big Country (1999) 17
17.1 Applications of Maximum Flow
Edge-Disjoint Paths One of the easiest applications of maximum ﬂows is computing the maximum number of edge-disjoint
paths between two speciﬁed vertices s and t in a directed graph G using maximum ﬂows. A set of paths
in G is edge-disjoint if each edge in G appears in at most one of the paths; several edge-disjoint paths
may pass through the same vertex, however.
If we give each edge capacity 1, then the maxﬂow from s to t assigns a ﬂow of either 0 or 1 to every
edge. Since any vertex of G lies on at most two saturated edges (one in and one out, or none at all),
the subgraph S of saturated edges is the union of several edge-disjoint paths and cycles. Moreover, the
number of paths is exactly equal to the value of the ﬂow. Extracting the actual paths from S is easy—just
follow any directed path in S from s to t , remove that path from S , and recurse.
Conversely, we can transform any collection of k edge-disjoint paths into a ﬂow by pushing one
unit of ﬂow along each path from s to t ; the value of the resulting ﬂow is exactly k. It follows that the
maxﬂow algorithm actually computes the largest possible set of edge-disjoint paths. The overall running
time is O(V E ), just like for maximum bipartite matchings.
The same algorithm can also be used to ﬁnd edge-disjoint paths in undirected graphs. We simply
replace every undirected edge in G with a pair of directed edges, each with unit capacity, and compute a
maximum ﬂow from s to t in the resulting directed graph G using the Ford-Fulkerson algorithm. For
any edge uv in G , if our max ﬂow saturates both directed edges u v and v u in G , we can remove
both edges from the ﬂow without changing its value. Thus, without loss of generality, the maximum ﬂow
assigns a direction to every saturated edge, and we can extract the edge-disjoint paths by searching the
graph of directed saturated edges. 17.2 Vertex Capacities and Vertex-Disjoint Paths Suppose we have capacities on the vertices as well as the edges. Here, in addition to our other constraints,
we require that for any vertex v other than s and t , the total ﬂow into v (and therefore the total ﬂow out
of v ) is at most some non-negative value c ( v ). How can we compute a maximum ﬂow with these new
constraints?
One possibility is to modify our existing algorithms to take these vertex capacities into account.
Given a ﬂow f , we can deﬁne the residual capacity of a vertex v to be its original capacity minus the
total ﬂow into v :
c f (v) = c(v) −
f (u v ).
u Since we cannot send any more ﬂow into a vertex with residual capacity 0 we remove from the
residual graph G f every edge u v that appears in G whose head vertex v is saturated. Otherwise, the
augmenting-path algorithm is unchanged.
But an even simpler method is to transform the input into a traditional ﬂow network, with only
edge capacities. Speciﬁcally, we replace every vertex v with two vertices vin and vout , connected by 1 Algorithms Lecture 17: Applications of Maximum Flow an edge vin vout with capacity c ( v ), and then replace every directed edge u v with the edge uout vin
(keeping the same capacity). Finally, we compute the maximum ﬂow from sout to t in in this modiﬁed
ﬂow network.
It is now easy to compute the maximum number of vertex-disjoint paths from s to t in any directed
graph. Simply give every vertex capacity 1, and compute a maximum ﬂow! 17.3 Maximum Matchings in Bipartite Graphs Another natural application of maximum ﬂows is ﬁnding large matchings in bipartite graphs. A matching
is a subgraph in which every vertex has degree at most one, or equivalently, a collection of edges such
that no two share a vertex. The problem is to ﬁnd the matching with the maximum number of edges in
a given bipartite graph.
We can solve this problem by reducing it to a maximum ﬂow problem as follows. Let G be the given
bipartite graph with vertex set U ∪ W , such that every edge joins a vertex in U to a vertex in W . We
create a new directed graph G by (1) orienting each edge from U to W , (2) adding two new vertices s
and t , (3) adding edges from s to every vertex in U , and (4) adding edges from each vertex in W to t .
Finally, we assign every edge in G a capacity of 1.
Any matching M in G can be transformed into a ﬂow f M in G as follows: For each edge uw in M ,
push one unit of ﬂow along the path s u w t . These paths are disjoint except at s and t , so the
resulting ﬂow satisﬁes the capacity constraints. Moreover, the value of the resulting ﬂow is equal to the
number of edges in M .
Conversely, consider any (s, t )-ﬂow f in G computed using the Ford-Fulkerson augmenting path
algorithm. Because the edge capacities are integers, the Ford-Fulkerson algorithm assigns an integer
ﬂow to every edge. (This is easy to verify by induction, hint, hint.) Moreover, since each edge has unit
capacity, the computed ﬂow either saturates ( f (e) = 1) or avoids ( f (e) = 0) every edge in G . Finally,
since at most one unit of ﬂow can enter any vertex in U or leave any vertex in W , the saturated edges
from U to W form a matching in G . The size of this matching is exactly | f |.
Thus, the size of the maximum matching in G is equal to the value of the maximum ﬂow in G , and
provided we compute the maxﬂow using augmenting paths, we can convert the actual maxﬂow into a
maximum matching. The maximum ﬂow has value at most min{|U |, |W |} = O(V ), so the Ford-Fulkerson
algorithm runs in O (V E ) time. s t A maximum matching in a bipartite graph G , and the corresponding maximum ﬂow in G . 17.4 Binary Assignment Problems Maximum-cardinality matchings are a special case of a general family of so-called assignment problems.1
An unweighted binary assignment problem involves two disjoint ﬁnite sets X and Y , which typically
represent two different kinds of resources, such as web pages and servers, jobs and machines, rows and
1 Most authors refer to ﬁnding a maximum-weight matching in a bipartite graph as the assignment problem. 2 Algorithms Lecture 17: Applications of Maximum Flow columns of a matrix, hospitals and interns, or customers and pints of ice cream. Our task is to choose
the largest possible collection of pairs ( x , y ) as possible, where x ∈ X and y ∈ Y , subject to several
constraints of the following form:
• Each element x ∈ X can appear in at most c ( x ) pairs.
• Each element y ∈ Y can appear in at most c ( y ) pairs.
• Each pair ( x , y ) ∈ X × Y can appear in the output at most c ( x , y ) times.
Each upper bound c ( x ), c ( y ), and c ( x , y ) is either a (typically small) non-negative integer or ∞.
Intuitively, we create each pair in our output by assigning an element of X to an element of y .
The maximum-matching problem is a special case, where c (z ) = 1 for all z ∈ X ∪ Y , and each c ( x , y )
is either 0 or 1, depending on whether the pair x y deﬁnes an edge in the underlying bipartite graph.
Here is a slightly more interesting example. A nearby school, famous for its onerous administrative
hurdles, decides to organize a dance. Every pair of students (one boy, one girl) who wants to dance
must register in advance. School regulations limit each boy-girl pair to at most three dances together,
and limits each student to at most ten dances overall. How can we maximize the number of dances?
This is a binary assignment problem for the set X of girls and the set Y of boys. For each girl x and boy
y , we have c ( x ) = 10, c ( y ) = 10, and either c ( x , y ) = 3 (if x and y registered to dance) or c ( x , y ) = 0
(if they didn’t).
This binary assignment problem can be reduces to a standard maximum ﬂow problem as follows.
We construct a ﬂow network G = (V, E ) with vertices X ∪ Y ∪ {s, t } and the following edges:
• an edge s x with capacity c ( x ) for each x ∈ X ,
• an edge y t with capacity c ( y ) for each y ∈ Y .
• an edge x y with capacity c ( x , y ) for each x ∈ X and y ∈ Y , and Because all the edges have integer capacities, the Ford-Fulkerson algorithm constructs an integer
maximum ﬂow f ∗ . This ﬂow can be decomposed into the sum of | f ∗ | paths of the form s x y t for
some x ∈ X and y ∈ Y . For each such path, we report the pair ( x , y ). (Equivalently, the pair ( x , y )
appears in our output collection f ( x y ) times.) It is easy to verify (hint, hint) that this collection of
pairs satisﬁes all the necessary constraints. Conversely, any legal collection of r pairs can be transformed
into a feasible integer ﬂow with value r in G . Thus, the largest legal collection of pairs corresponds to a
maximum ﬂow in G . So our algorithm is correct. 17.5 Baseball Elimination Every year millions of baseball fans eagerly watch their favorite team, hoping they will win a spot in the
playoffs, and ultimately the World Series. Sadly, most teams are “mathematically eliminated" days or
even weeks before the regular season ends. Often, it is easy to spot when a team is eliminated—they
can’t win enough games to catch up to the current leader in their division. But sometimes the situation
is more subtle.
For example, here are the actual standings from the American League East on August 30, 1996.
Team
New York Yankees
Baltimore Orioles
Boston Red Sox
Toronto Blue Jays
Detroit Lions Won–Lost Left NYY 75–59
71–63
69–66
63–72
49–86 28
28
27
27
27 3
8
7
3 3 BAL BOS TOR DET 3 8
2 7
7
0 3
4
0
0 2
7
4 0
0 0 Algorithms Lecture 17: Applications of Maximum Flow Detroit is clearly behind, but some die-hard Lions fans may hold out hope that their team can still win.
After all, if Detroit wins all 27 of their remaining games, they will end the season with 76 wins, more
than any other team has now. So as long as every other team loses every game. . . but that’s not possible,
because some of those other teams still have to play each other. Here is one complete argument:2
By winning all of their remaining games, Detroit can ﬁnish the season with a record of 76 and 86. If the Yankees
win just 2 more games, then they will ﬁnish the season with a 77 and 85 record which would put them ahead of
Detroit. So, let’s suppose the Tigers go undefeated for the rest of the season and the Yankees fail to win another
game.
The problem with this scenario is that New York still has 8 games left with Boston. If the Red Sox win all of
these games, they will end the season with at least 77 wins putting them ahead of the Tigers. Thus, the only way
for Detroit to even have a chance of ﬁnishing in ﬁrst place, is for New York to win exactly one of the 8 games
with Boston and lose all their other games. Meanwhile, the Sox must loss all the games they play agains teams
other than New York. This puts them in a 3-way tie for ﬁrst place. . . .
Now let’s look at what happens to the Orioles and Blue Jays in our scenario. Baltimore has 2 games left with
with Boston and 3 with New York. So, if everything happens as described above, the Orioles will ﬁnish with at
least 76 wins. So, Detroit can catch Baltimore only if the Orioles lose all their games to teams other than New
York and Boston. In particular, this means that Baltimore must lose all 7 of its remaining games with Toronto.
The Blue Jays also have 7 games left with the Yankees and we have already seen that for Detroit to ﬁnish in ﬁrst
place, Toronto must will all of these games. But if that happens, the Blue Jays will win at least 14 more games
giving them at ﬁnal record of 77 and 85 or better which means they will ﬁnish ahead of the Tigers. So, no matter
what happens from this point in the season on, Detroit can not ﬁnish in ﬁrst place in the American League East. There has to be a better way to ﬁgure this out!
Here is a more abstract formulation of the problem. Our input consists of two arrays W [1 .. n] and
G [1 .. n, 1 .. n], where W [i ] is the number of games team i has already won, and G [i , j ] is the number of
upcoming games between teams i and j . We want to determine whether team n can end the season
with the most wins (possibly tied with other teams).3
We model this question as an assignment problem: We want to assign a winner to each game, so that
team n comes in ﬁrst place. We have an assignment problem! Let R[i ] = j G [i , j ] denote the number
of remaining games for team i . We will assume that team n wins all R[n] of its remaining games. Then
team n can come in ﬁrst place if and only if every other team i wins at most W [n] + R[n] − W [i ] of its
R[i ] remaining games.
Since we want to assign winning teams to games, we start by building a bipartite graph, whose
n
nodes represent the games and the teams. We have 2 game nodes g i , j , one for each pair 1 ≤ i < j < n,
and n − 1 team nodes t i , one for each 1 ≤ i < n. For each pair i , j , we add edges g i , j t i and g i , j t j
with inﬁnite capacity. We add a source vertex s and edges s g i , j with capacity G [i , j ] for each pair i , j .
Finally, we add a target node t and edges t i t with capacity W [n] − W [i ] + R[n] for each team i .
Theorem: Team n can end the season in ﬁrst place if and only if there is a feasible ﬂow in this graph
that saturates every edge leaving s.
Proof: Suppose it is possible for team n to end the season in ﬁrst place. Then every team i < n wins at
most W [n]+ R[n] − W [i ] of the remaining games. For each game between team i and team j that team i
wins, add one unit of ﬂow along the path s g i , j t i t . Because there are exactly G [i , j ] games between
teams i and j , every edge leaving s is saturated. Because each team i wins at most W [n] + R[n] − W [i ]
games, the resulting ﬂow is feasible.
Conversely, Let f be a feasible ﬂow that saturates every edge out of s. Suppose team i wins exactly
f ( g i , j t i ) games against team j , for all i and j . Then teams i and j play f ( g i , j t i ) + f ( g i , j t j ) =
Both the example and this argument are taken from http://riot.ieor.berkeley.edu/~baseball/detroit.html.
We assume here that no games end in a tie (always true for Major League Baseball), and that every game is actually played
(not always true).
2 3 4 Algorithms Lecture 17: Applications of Maximum Flow f (s g i , j ) = G [i , j ] games, so every upcoming game is played. Moreover, each team i wins a total of
j f ( g i , j t i ) = f ( t i t ) ≤ W [ n] + R[ n] − W [ i ] upcoming games, and therefore at most W [ n] + R[ n]
games overall. Thus, if team n win all their upcoming games, they end the season in ﬁrst place.
So, to decide whether our favorite team can win, we construct the ﬂow network, compute a maximum
ﬂow, and report whether than maximum ﬂow saturates the edges leaving s. The ﬂow network has O(n2 )
vertices and O(n2 ) edges, and it can be constructed in O(n2 ) time. Using Dinitz’s algorithm, we can
compute the maximum ﬂow in O(V E 2 ) = O(n6 ) time.
The graph derived from the 1996 American League East standings is shown below. The total capacity
of the edges leaving s is 27 (there are 27 remaining games), but the total capacity of the edges entering t
is only 26. So the maximum ﬂow has value at most 26, which means that Detroit is mathematically
eliminated.
NYY
BAL NYY NYY
BOS 3
8
s 1 BAL 5 NYY
TOR 7 t
7 2 BOS BAL
BOS 7 BAL
TOR 13 TOR The ﬂow graph for the 1996 American League East standings. Unlabeled edges have inﬁnite capacity. Exercises
1. Given an undirected graph G = (V, E ), with three vertices u, v , and w , describe and analyze an
algorithm to determine whether there is a path from u to w that passes through v .
2. Let G = (V, E ) be a directed graph where for each vertex v , the in-degree and out-degree of v are
equal. Let u and v be two vertices G, and suppose G contains k edge-disjoint paths from u to v .
Under these conditions, must G also contain k edge-disjoint paths from v to u? Give a proof or a
counterexample with explanation.
3. A cycle cover of a given directed graph G = (V, E ) is a set of vertex-disjoint cycles that cover all the
vertices. Describe and analyze an efﬁcient algorithm to ﬁnd a cycle cover for a given graph, or
correctly report that no cycle cover exists. [Hint: Use bipartite matching!]
4. Consider a directed graph G = (V, E ) with multiple source vertices s1 , s2 , . . . , sσ and multiple target
vertices t 1 , t 1 , . . . , t τ , where no vertex is both a source and a target. A multiterminal ﬂow is a
function f : E → IR≥0 that satisﬁes the ﬂow conservation constraint at every vertex that is neither
a source nor a target. The value | f | of a multiterminal ﬂow is the total excess ﬂow out of all the
source vertices:
σ | f | := f (si w ) −
i =1 w f (u si )
u 5 Algorithms Lecture 17: Applications of Maximum Flow As usual, we are interested in ﬁnding ﬂows with maximum value, subject to capacity constraints
on the edges. (In particular, we don’t care how much ﬂow moves from any particular source to
any particular target.)
(a) Consider the following algorithm for computing multiterminal ﬂows. The variables f and f
represent ﬂow functions. The subroutine MAXFLOW(G , s, t ) solves the standard maximum
ﬂow problem with source s and target t .
MAXMULTIFLOW(G , s[1 .. σ], t [1 .. τ]):
f ←0
〈〈Initialize the ﬂow〉〉
for i ← 1 to σ
for j ← 1 to τ
f ← MAXFLOW(G f , s[i ], t [ j ])
f ←f +f
〈〈Update the ﬂow〉〉
return f Prove that this algorithm correctly computes a maximum multiterminal ﬂow in G .
(b) Describe a more efﬁcient algorithm to compute a maximum multiterminal ﬂow in G .
5. The Island of Sodor is home to a large number of towns and villages, connected by an extensive
rail network. Recently, several cases of a deadly contagious disease (either swine ﬂu or zombies;
reports are unclear) have been reported in the village of Ffarquhar. The controller of the Sodor
railway plans to close down certain railway stations to prevent the disease from spreading to
Tidmouth, his home town. No trains can pass through a closed station. To minimize expense
(and public notice), he wants to close down as few stations as possible. However, he cannot close
the Ffarquhar station, because that would expose him to the disease, and he cannot close the
Tidmouth station, because then he couldn’t visit his favorite pub.
Describe and analyze an algorithm to ﬁnd the minimum number of stations that must be closed
to block all rail travel from Ffarquhar to Tidmouth. The Sodor rail network is represented by an
undirected graph, with a vertex for each station and an edge for each rail connection between two
stations. Two special vertices f and t represent the stations in Ffarquhar and Tidmouth.
For example, given the following input graph, your algorithm should return the number 2. f t 6. Suppose we are given an array A[1 .. m][1 .. n] of non-negative real numbers. We want to round A
to an integer matrix, by replacing each entry x in A with either x or x , without changing the
sum of entries in any row or column of A. For example: 1.2 3.4 2.4
142 3.9 4.0 2.1 −→ 4 4 2
7.9 1.6 0.5
811
Describe an efﬁcient algorithm that either rounds A in this fashion, or reports correctly that no
such rounding is possible.
6 Algorithms Lecture 17: Applications of Maximum Flow 7. Ad-hoc networks are made up of low-powered wireless devices. In principle4 , these networks
can be used on battleﬁelds, in regions that have recently suffered from natural disasters, and in
other hard-to-reach areas. The idea is that a large collection of cheap, simple devices could be
distributed through the area of interest (for example, by dropping them from an airplane); the
devices would then automatically conﬁgure themselves into a functioning wireless network.
These devices can communication only within a limited range. We assume all the devices are
identical; there is a distance D such that two devices can communicate if and only if the distance
between them is at most D.
We would like our ad-hoc network to be reliable, but because the devices are cheap and
low-powered, they frequently fail. If a device detects that it is likely to fail, it should transmit
its information to some other backup device within its communication range. We require each
device x to have k potential backup devices, all within distance D of x ; we call these k devices
the backup set of x . Also, we do not want any device to be in the backup set of too many other
devices; otherwise, a single failure might affect a large fraction of the network.
So suppose we are given the communication radius D, parameters b and k, and an array
d [1 .. n, 1 .. n] of distances, where d [i , j ] is the distance between device i and device j . Describe
an algorithm that either computes a backup set of size k for each of the n devices, such that that
no device appears in more than b backup sets, or reports (correctly) that no good collection of
backup sets exists.
8. A rooted tree is a directed acyclic graph, in which every vertex has exactly one incoming edge,
except for the root, which has no incoming edges. Equivalently, a rooted tree consists of a root
vertex, which has edges pointing to the roots of zero or more smaller rooted trees. Describe a
polynomial-time algorithm to compute, given two rooted trees A and B , the largest common rooted
subtree of A and B .
[Hint: Let LC S (u, v ) denote the largest common subtree whose root in A is u and whose
root in B is v . Your algorithm should compute LC S (u, v ) for all vertices u and v using dynamic
programming. This would be easy if every vertex had O(1) children, and still straightforward if the
children of each node were ordered from left to right and the common subtree had to respect that
ordering. But for unordered trees with large degree, you need another trick to combine recursive
subproblems efﬁciently. Don’t waste your time trying to reduce the polynomial running time.] 4 but not really in practice c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 7 Algorithms Lecture 18: Extensions of Maximum Flow
“Who are you?" said Lunkwill, rising angrily from his seat. “What do you want?"
“I am Majikthise!" announced the older one.
“And I demand that I am Vroomfondel!" shouted the younger one.
Majikthise turned on Vroomfondel. “It’s alright," he explained angrily, “you don’t need to
demand that."
“Alright!" bawled Vroomfondel banging on an nearby desk. “I am Vroomfondel, and that is
not a demand, that is a solid fact! What we demand is solid facts!"
“No we don’t!" exclaimed Majikthise in irritation. “That is precisely what we don’t demand!"
Scarcely pausing for breath, Vroomfondel shouted, “We don’t demand solid facts! What we
demand is a total absence of solid facts. I demand that I may or may not be Vroomfondel!"
— Douglas Adams, The Hitchhiker’s Guide to the Galaxy (1979) 18 Extensions of Maximum Flow 18.1 Maximum Flows with Edge Demands Now suppose each directed edge e in has both a capacity c (e) and a demand d (e) ≤ c (e), and we want a
ﬂow f of maximum value that satisﬁes d (e) ≤ f (e) ≤ c (e) at every edge e. We call a ﬂow that satisﬁes
these constraints a feasible ﬂow. In our original setting, where d (e) = 0 for every edge e, the zero ﬂow
is feasible; however, in this more general setting, even determining whether a feasible ﬂow exists is a
nontrivial task.
Perhaps the easiest way to ﬁnd a feasible ﬂow (or determine that none exists) is to reduce the
problem to a standard maximum ﬂow problem, as follows. The input consists of a directed graph
G = (V, E ), nodes s and t , demand function d : E → IR, and capacity function c : E → IR. Let D denote
the sum of all edge demands in G :
D :=
d (u v ).
u v∈E We construct a new graph G = (V , E ) from G by adding new source and target vertices s and t ,
adding edges from s to each vertex in V , adding edges from each vertex in V to t , and ﬁnally adding
an edge from t to s. We also deﬁne a new capacity function c : E → IR as follows:
• For each vertex v ∈ V , we set c (s v) = u∈V d (u v ) and c ( v t ) = w ∈V d ( v w ). • For each edge u v ∈ E , we set c (u v ) = c (u v ) − d (u v ).
• Finally, we set c (s t ) = ∞.
Intuitively, we construct G by replacing any edge u v in G with three edges: an edge u v with
capacity c (u v ) − d (u v ), an edge s v with capacity d (u v ), and an edge u t with capacity d (u v ).
If this construction produces multiple edges from s to the same vertex v (or to t from the same vertex v ),
we merge them into a single edge with the same total capacity.
In G , the total capacity out of s and the total capacity into t are both equal to D. We call a ﬂow
with value exactly D a saturating ﬂow, since it saturates all the edges leaving s or entering t . If G has a
saturating ﬂow, it must be a maximum ﬂow, so we can ﬁnd it using any max-ﬂow algorithm.
Lemma 1. G has a feasible (s, t )-ﬂow if and only if G has a saturating (s , t )-ﬂow. 1 Algorithms Lecture 18: Extensions of Maximum Flow 8 11
4 5
5 7 3 13
0..5 s s' 6..20 2..15
5..10 5..10 4..10 5 14 6 0..15
s 13 t 10 7..20 t 6 t' 14
7 3 10
3 3..10 ∞ A ﬂow network G with demands and capacities (written d .. c ), and the transformed network G . Proof: Let f : E → IR be a feasible (s, t )-ﬂow in the original graph G . Consider the following function
f : E → IR:
f (u v ) = f (u v ) − d (u v )
f (s v) = for all u v ∈ E d (u v ) for all v ∈ V d (u → w ) for all v ∈ V u∈V f (v t ) =
w ∈V f ( t s) = | f |
We easily verify that f is a saturating (s , t )-ﬂow in G . The admissibility of f implies that f (e) ≥ d (e)
for every edge e ∈ E , so f (e) ≥ 0 everywhere. Admissibility also implies f (e) ≤ c (e) for every edge
e ∈ E , so f (e) ≤ c (e) everywhere. Tedious algebra implies that
f (u v ) =
u∈V f (v w)
w ∈V for every vertex v ∈ V (including s and t ). Thus, f is a legal (s , t )-ﬂow, and every edge out of s
or into t is clearly saturated. Intuitively, f diverts d (u v ) units of ﬂow from u directly to the new
target t , and injects the same amount of ﬂow into v directly from the new source s .
The same tedious algebra implies that for any saturating (s , t )-ﬂow f : E → IR for G , the function
f = f | E + d is a feasible (s, t )-ﬂow in G .
Thus, we can compute a feasible (s, t )-ﬂow for G , if one exists, by searching for a maximum (s , t )ﬂow in G and checking that it is saturating. Once we’ve found a feasible (s, t )-ﬂow in G , we can
transform it into a maximum ﬂow using an augmenting-path algorithm, but with one small change.
To ensure that every ﬂow we consider is feasible, we must redeﬁne the residual capacity of an edge as
follows: c (u v ) − f (u v ) if u v ∈ E ,
c f (u v ) = f ( v u) − d ( v u) if v u ∈ E , 0
otherwise. Otherwise, the algorithm is unchanged. If we use the Dinitz/Edmonds-Karp fat-pipe algorithm, we get
an overall running time of O (V E 2 ).
2 Algorithms Lecture 18: Extensions of Maximum Flow 8/8 11/11
4/4 5/5 7/7 2/5 3/3 0/13 s s' 3/13 0/5 0/14 t 0/6 0/6 t' 0/14
7/7 10/10 10/10 3/3 3/3
11/∞
3 2/0..5
7/7..20 2 5/2..15
1/1..15 s 5/5..10 4/4..10 4/4..10 3 13
s t 5 10 14 6 6/6..20 t 6 7 14 10/3..10 A saturating ﬂow f in G , the corresponding feasible ﬂow f in G , and the corresponding residual network G f . 18.2 Node Supplies and Demands Another useful variant to consider allows ﬂow to be injected or extracted from the ﬂow network at
vertices other than s or t . Let x : (V \ {s, t }) → IR be an excess function describing how much ﬂow is to
be injected (or extracted if the value is negative) at each vertex. We now want a maximum ‘ﬂow’ that
satisﬁes the variant balance condition
f (u v ) −
u∈V f (v w) = x (v)
w ∈V for every node v except s and t , or prove that no such ﬂow exists. As above, call such a function f a
feasible ﬂow.
As for ﬂows with edge demands, the only real difﬁculty in ﬁnding a maximum ﬂow under these
modiﬁed constraints is ﬁnding a feasible ﬂow (if one exists). We can reduce this problem to a standard
max-ﬂow problem, just as we did for edge demands.
To simplify the transformation, let us assume without loss of generality that the total excess in the
˜
network is zero:
v x ( v ) = 0. If the total excess is positive, we add an inﬁnite capacity edge t t ,
where ˜ is a new target node, and set x ( t ) = − v x ( v ). Similarly, if the total excess is negative, we add
t
an inﬁnite capacity edge ˜ s, where ˜ is a new source node, and set x (s) = − v x ( v ). In both cases,
s
s
every feasible ﬂow in the modiﬁed graph corresponds to a feasible ﬂow in the original graph.
As before, we modify G to obtain a new graph G by adding a new source s , a new target t , an
inﬁnite-capacity edge t s from the old target to the old source, and several edges from s and to t .
Speciﬁcally, for each vertex v , if x ( v ) > 0, we add a new edge s v with capacity x ( v ), and if x ( v ) < 0,
we add an edge v t with capacity − x ( v ). As before, we call an (s , t )-ﬂow in G saturating if every
edge leaving s or entering t is saturated; any saturating ﬂow is a maximum ﬂow. It is easy to check
that saturating ﬂows in G are in direct correspondence with feasible ﬂows in G ; we leave details as an
exercise (hint, hint).
3 Algorithms Lecture 18: Extensions of Maximum Flow Similar reductions allow us to solve several other variants of the maximum ﬂow problem using the
same path-augmentation algorithms. For example, we could associate capacities and lower bounds with
the vertices instead of (or in addition to) the edges. We could associate a range of excesses with every
node, instead of a single excess value. We can associate a cost c (e) with each edge, and then ask for the
maximum-value ﬂow f whose total cost e c (e) · f (e) is as small as possible. We could even apply all of
these extensions at once: upper bounds, lower bounds, and cost functions for the ﬂow through each
edge, into each vertex, and out of each vertex. 18.3 Minimum-Cost Flows — To be written —
18.4 Maximum-Weight Matchings Recall from the previous lecture that we can ﬁnd a maximum-cardinality matching in any bipartite graph
in O(V E ) time by reduction to the standard maximum ﬂow problem.
Now suppose the input graph has weighted edges, and we want to ﬁnd the matching with maximum
total weight. Given a bipartite graph G = (U × W, E ) and a non-negative weight function w : E → IR,
the goal is to compute a matching M whose total weight w ( M ) = uw ∈ M w (uw ) is as large as possible.
Max-weight matchings can’t be found directly using standard max-ﬂow algorithms1 , but we can modify
the algorithm for maximum-cardinality matchings described above.
It will be helpful to reinterpret the behavior of our earlier algorithm directly in terms of the original
bipartite graph instead of the derived ﬂow network. Our algorithm maintains a matching M , which is
initially empty. We say that a vertex is matched if it is an endpoint of an edge in M . At each iteration, we
ﬁnd an alternating path π that starts and ends at unmatched vertices and alternates between edges in
E \ M and edges in M . Equivalently, let G M be the directed graph obtained by orienting every edge in M
from W to U , and every edge in E \ M from U to W . An alternating path is just a directed path in G M
between two unmatched vertices. Any alternating path has odd length and has exactly one more edge in
E \ M than in M . The iteration ends by setting M ← M ⊕ π, thereby increasing the number of edges
in M by one. The max-ﬂow/min-cut theorem implies that when there are no more alternating paths, M
is a maximum matching. A matching M with 5 edges, an alternating path π, and the augmented matching M ⊕ π with 6 edges. If the edges of G are weighted, we only need to make two changes to the algorithm. First, instead of
looking for an arbitrary alternating path at each iteration, we look for the alternating path π such that
1
However, max-ﬂow algorithms can be modiﬁed to compute maximum weighted ﬂows, where every edge has both a capacity
and a weight, and the goal is to maximize u v w (u v ) f (u v ). 4 Algorithms Lecture 18: Extensions of Maximum Flow M ⊕ π has largest weight. Suppose we weight the edges in the residual graph G M as follows:
w (u w ) = −w (uw ) for all uw ∈ M w (w u) = w (uw ) for all uw ∈ M We now have w ( M ⊕ π) = w ( M ) − w (π). Thus, the correct augmenting path π must be the directed
path in G M with minimum total residual weight w (π). Second, because the matching with the
maximum weight may not be the matching with the maximum cardinality, we return the heaviest
matching considered in any iteration of the algorithm.
2 3 3 10 5 2
10 3 5 3 A maximum-weight matching is not necessarily a maximum-cardinality matching. Before we determine the running time of the algorithm, we need to check that it actually ﬁnds the
maximum-weight matching. After all, it’s a greedy algorithm, and greedy algorithms don’t work unless
you prove them into submission! Let Mi denote the maximum-weight matching in G with exactly i
edges. In particular, M0 = ∅, and the global maximum-weight matching is equal to Mi for some i .
(The ﬁgure above show M1 and M2 for the same graph.) Let Gi denote the directed residual graph
for Mi , let w i denote the residual weight function for Mi as deﬁned above, and let πi denote the directed
path in Gi such that w i (πi ) is minimized. To simplify the proof, I will assume that there is a unique
maximum-weight matching Mi of any particular size; this assumption can be enforced by applying a
consistent tie-breaking rule. With this assumption in place, the correctness of our algorithm follows
inductively from the following lemma.
Lemma 2. If G contains a matching with i + 1 edges, then Mi +1 = Mi ⊕ πi .
Proof: I will prove the equivalent statement Mi +1 ⊕ Mi = πi −1 . To simplify notation, call an edge in
Mi +1 ⊕ Mi red if it is an edge in Mi +1 , and blue if it is an edge in Mi .
The graph Mi +1 ⊕ Mi has maximum degree 2, and therefore consists of pairwise disjoint paths and
cycles, each of which alternates between red and blue edges. Since G is bipartite, every cycle must have
even length. The number of edges in Mi +1 ⊕ Mi is odd; speciﬁcally, Mi +1 ⊕ Mi has 2i + 1 − 2k edges,
where k is the number of edges that are in both matchings. Thus, Mi +1 ⊕ Mi contains an odd number of
paths of odd length, some number of paths of even length, and some number of cycles of even length.
Let γ be a cycle in Mi +1 ⊕ Mi . Because γ has an equal number of edges from each matching, Mi ⊕ γ
is another matching with i edges. The total weight of this matching is exactly w ( Mi ) − w i (γ), which
must be less than w ( Mi ), so w i (γ) must be positive. On the other hand, Mi +1 ⊕ γ is a matching with
i + 1 edges whose total weight is w ( Mi +1 ) + w i (γ) < w ( Mi +1 ), so w i (γ) must be negative! We conclude
that no such cycle γ exists; Mi +1 ⊕ Mi consists entirely of disjoint paths.
Exactly the same reasoning implies that no path in Mi +1 ⊕ Mi has an even number of edges.
Finally, since the number of red edges in Mi +1 ⊕ Mi is one more than the number of blue edges, the
number of paths that start with a red edge is exactly one more than the number of paths that start with
a blue edge. The same reasoning as above implies that Mi +1 ⊕ Mi does not contain a blue-ﬁrst path,
because we can pair it up with a red-ﬁrst path.
We conclude that Mi +1 ⊕ Mi consists of a single alternating path π whose ﬁrst edge is red. Since
w ( Mi +1 ) = w ( Mi ) − w i (π), the path π must be the one with minimum weight w i (π).
5 Algorithms Lecture 18: Extensions of Maximum Flow We can ﬁnd the alternating path πi using a single-source shortest path algorithm. Modify the
residual graph Gi by adding zero-weight edges from a new source vertex s to every unmatched node
in U , and from every unmatched node in W to a new target vertex t , exactly as in out unweighted
matching algorithm. Then πi is the shortest path from s to t in this modiﬁed graph. Since Mi is the
maximum-weight matching with i vertices, Gi has no negative cycles, so this shortest path is well-deﬁned.
We can compute the shortest path in Gi in O(V E ) time using Shimbel’s algorithm, so the overall running
time our algorithm is O (V 2 E ).
The residual graph Gi has negative-weight edges, so we can’t speed up the algorithm by replacing
Shimbel’s algorithm with Dijkstra’s. However, we can use a variant of Johnson’s all-pairs shortest path
algorithm to improve the running time to O (V E + V 2 log V ). Let di ( v ) denote the distance from s to v
˜
in the residual graph Gi , using the distance function w i . Let w i denote the modiﬁed distance function
˜
w i (u v ) = di −1 (u) + w i (u v ) − di −1 ( v ). As we argued in the discussion of Johnson’s algorithm, shortest
˜
˜
paths with respect to w i are still shortest paths with respect to w i . Moreover, w i (u v ) > 0 for every
edge u v in Gi :
• If u v is an edge in Gi −1 , then w i (u v ) = w i −1 (u v ) and di −1 ( v ) ≤ di −1 (u) + w i −1 (u v ).
• If u v is not in Gi −1 , then w i (u v ) = −w i −1 ( v u) and v u is an edge in the shortest path πi −1 ,
so di −1 (u) = di −1 ( v ) + w i −1 ( v u).
˜
˜
Let di ( v ) denote the shortest path distance from s to v with respect to the distance function w i .
˜
˜
Because w i is positive everywhere, we can quickly compute di ( v ) for all v using Dijkstra’s algorithm.
˜
This gives us both the shortest alternating path πi and the distances di ( v ) = di ( v ) + di −1 ( v ) needed for
the next iteration. Exercises
1. Suppose we are given a directed graph G = (V, E ), two vertices s an t , and a capacity function
c : V → IR+ . A ﬂow f is feasible if the total ﬂow into every vertex v is at most c ( v ):
f (u v ) ≤ c ( v ) for every vertex v .
u Describe and analyze an efﬁcient algorithm to compute a feasible ﬂow of maximum value.
2. Suppose we are given an n × n grid, some of whose cells are marked; the grid is represented by
an array M [1 .. n, 1 .. n] of booleans, where M [i , j ] = TRUE if and only if cell (i , j ) is marked. A
monotone path through the grid starts at the top-left cell, moves only right or down at each step,
and ends at the bottom-right cell. Our goal is to cover the marked cells with as few monotone
paths as possible. Greedily covering the marked cells in a grid with four monotone paths. 6 Algorithms Lecture 18: Extensions of Maximum Flow (a) Describe an algorithm to ﬁnd a monotone path that covers the largest number of marked
cells.
(b) There is a natural greedy heuristic to ﬁnd a small cover by monotone paths: If there are
any marked cells, ﬁnd a monotone path π that covers the largest number of marked cells,
unmark any cells covered by π those marked cells, and recurse. Show that this algorithm
does not always compute an optimal solution.
(c) Describe and analyze an efﬁcient algorithm to compute the smallest set of monotone paths
that covers every marked cell. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 7 Algorithms Non-Lecture I: Linear Programming
The greatest ﬂood has the soonest ebb;
the sorest tempest the most sudden calm;
the hottest love the coldest end; and
from the deepest desire oftentimes ensues the deadliest hate.
— Socrates
Th’ extremes of glory and of shame,
Like east and west, become the same.
— Samuel Butler, Hudibras Part II, Canto I (c. 1670)
Extremes meet, and there is no better example
than the haughtiness of humility.
— Ralph Waldo Emerson, “Greatness”,
in Letters and Social Aims (1876) I Linear Programming The maximum ﬂow/minimum cut problem is a special case of a very general class of problems called
linear programming. Many other optimization problems fall into this class, including minimum spanning
trees and shortest paths, as well as several common problems in scheduling, logistics, and economics.
Linear programming was used implicitly by Fourier in the early 1800s, but it was ﬁrst formalized
and applied to problems in economics in the 1930s by Leonid Kantorovich. Kantorivich’s work was
hidden behind the Iron Curtain (where it was largely ignored) and therefore unknown in the West.
Linear programming was rediscovered and applied to shipping problems in the early 1940s by Tjalling
Koopmans. The ﬁrst complete algorithm to solve linear programming problems, called the simplex method,
was published by George Dantzig in 1947. Koopmans ﬁrst proposed the name “linear programming" in a
discussion with Dantzig in 1948. Kantorovich and Koopmans shared the 1975 Nobel Prize in Economics
“for their contributions to the theory of optimum allocation of resources”. Dantzig did not; his work was
apparently too pure. Koopmans wrote to Kantorovich suggesting that they refuse the prize in protest
of Dantzig’s exclusion, but Kantorovich saw the prize as a vindication of his use of mathematics in
economics, which his Soviet colleagues had written off as “a means for apologists of capitalism”.
A linear programming problem asks for a vector x ∈ IRd that maximizes (or equivalently, minimizes)
a given linear function, among all vectors x that satisfy a given set of linear inequalities. The general
form of a linear programming problem is the following:
d maximize cj x j
j =1
d ai j x j ≤ bi for each i = 1 .. p ai j x j = bi for each i = p + 1 .. p + q ai j x j ≥ bi subject to for each i = p + q + 1 .. n j =1
d
j =1
d
j =1 Here, the input consists of a matrix A = (ai j ) ∈ IRn×d , a column vector b ∈ IRn , and a row vector c ∈ IRd .
Each coordinate of the vector x is called a variable. Each of the linear inequalities is called a constraint.
The function x → c · x is called the objective function. I will always use d to denote the number of
variables, also known as the dimension of the problem. The number of constraints is usually denoted n.
1 Algorithms Non-Lecture I: Linear Programming A linear programming problem is said to be in canonical form1 if it has the following structure:
d maximize cj x j
j =1
d ai j x j ≤ bi for each i = 1 .. n xj ≥ 0 subject to for each j = 1 .. d j =1 We can express this canonical form more compactly as follows. For two vectors x = ( x 1 , x 2 , . . . , x d ) and
y = ( y1 , y2 , . . . , yd ), the expression x ≥ y means that x i ≥ yi for every index i .
max c · x
s.t. Ax ≤ b
x≥0
Any linear programming problem can be converted into canonical form as follows:
• For each variable x j , add the equality constraint x j = x + − x − and the inequalities x + ≥ 0 and
j
j
j
x − ≥ 0.
j
• Replace any equality constraint
j ai j x j ≤ bi .
• Replace any upper bound j j ai j x j = bi with two inequality constraints ai j x j ≥ bi with the equivalent lower bound j j ai j x j ≥ bi and − ai j x j ≤ − bi . This conversion potentially double the number of variables and the number of constraints; fortunately, it
is rarely necessary in practice.
Another useful format for linear programming problems is slack form2 , in which every inequality is
of the form x j ≥ 0:
max c · x
s.t. Ax = b
x≥0
It’s fairly easy to convert any linear programming problem into slack form. Slack form is especially
useful in executing the simplex algorithm (which we’ll see in the next lecture). I.1 The Geometry of Linear Programming A point x ∈ IRd is feasible with respect to some linear programming problem if it satisﬁes all the linear
constraints. The set of all feasible points is called the feasible region for that linear program. The
feasible region has a particularly nice geometric structure that lends some useful intuition to the linear
programming algorithms we’ll see later.
Any linear equation in d variables deﬁnes a hyperplane in IRd ; think of a line when d = 2, or a plane
when d = 3. This hyperplane divides IRd into two halfspaces; each halfspace is the set of points that
satisfy some linear inequality. Thus, the set of feasible points is the intersection of several hyperplanes
1
2 Confusingly, some authors call this standard form.
Confusingly, some authors call this standard form. 2 Algorithms Non-Lecture I: Linear Programming (one for each equality constraint) and halfspaces (one for each inequality constraint). The intersection
of a ﬁnite number of hyperplanes and halfspaces is called a polyhedron. It’s not hard to verify that any
halfspace, and therefore any polyhedron, is convex—if a polyhedron contains two points x and y , then it
contains the entire line segment x y . A two-dimensional polyhedron (white) deﬁned by 10 linear inequalities. By rotating IRd (or choosing a coordinate frame) so that the objective function points downward, we
can express any linear programming problem in the following geometric form:
Find the lowest point in a given polyhedron.
With this geometry in hand, we can easily picture two pathological cases where a given linear
programming problem has no solution. The ﬁrst possibility is that there are no feasible points; in this
case the problem is called infeasible. For example, the following LP problem is infeasible:
maximize x − y
subject to 2 x + y ≤ 1
x+ y ≥2
x, y ≥ 0 An infeasible linear programming problem; arrows indicate the constraints. The second possibility is that there are feasible points at which the objective function is arbitrarily
large; in this case, we call the problem unbounded. The same polyhedron could be unbounded for some
objective functions but not others, or it could be unbounded for every objective function. A two-dimensional polyhedron (white) that is unbounded downward but bounded upward. 3 Algorithms I.2 Non-Lecture I: Linear Programming Example 1: Shortest Paths We can compute the length of the shortest path from s to t in a weighted directed graph by solving the
following very simple linear programming problem.
maximize dt subject to ds = 0
d v − du ≤ uv for every edge u v Here, u v is the length of the edge u v . Each variable d v represents a tentative shortest-path distance
from s to v . The constraints mirror the requirement that every edge in the graph must be relaxed.
These relaxation constraints imply that in any feasible solution, d v is at most the shortest path distance
from s to v . Thus, somewhat counterintuitively, we are correctly maximizing the objective function to
compute the shortest path! In the optimal solution, the objective function d t is the actual shortest-path
distance from s to t , but for any vertex v that is not on the shortest path from s to t , d v may be an
underestimate of the true distance from s to v . However, we can obtain the true distances from s to
every other vertex by modifying the objective function:
maximize dv
v subject to ds = 0 d v − du ≤ uv for every edge u v There is another formulation of shortest paths as an LP minimization problem using an indicator
variable x u v for each edge u v .
v xs w =1 xt w = −1 xv uv · xu w =0 for every vertex v = s, t xu minimize v ≥0 for every edge u v uv subject to xu s − u w xu t − u w xu
u v −
w Intuitively, x u v = 1 means u v lies on the shortest path from s to t , and x u v = 0 means u v does
not lie on this shortest path. The constraints merely state that the path should start at s, end at t , and
either pass through or avoid every other vertex v . Any path from s to t —in particular, the shortest
path—clearly implies a feasible point for this linear program.
However, there are other feasible solutions, possibly even optimal solutions, with non-integral values
that do not represent paths. Nevertheless, there is always an optimal solution in which every x e is either
0 or 1 and the edges e with x e = 1 comprise the shortest path. (This fact is by no means obvious, but a
proof is beyond the scope of these notes.) Moreover, in any optimal solution, even if not every x e is an
integer, the objective function gives the shortest path distance! I.3 Example 2: Maximum Flows and Minimum Cuts Recall that the input to the maximum (s, t )-ﬂow problem consists of a weighted directed graph G = (V, E ),
two special vertices s and t , and a function assigning a non-negative capacity ce to each edge e. Our task
4 Algorithms Non-Lecture I: Linear Programming is to choose the ﬂow f e across each edge e, as follows:
maximize fs − fu
fu v =0 v ≤ cu fu subject to s fu w w v ≥0 u fv w − w for every vertex v = s, t u
v for every edge u v
for every edge u v Similarly, the minimum cut problem can be formulated using ‘indicator’ variables similarly to the
shortest path problem. We have a variable S v for each vertex v , indicating whether v ∈ S or v ∈ T , and a
variable X u v for each edge u v , indicating whether u ∈ S and v ∈ T , where (S , T ) is some (s, t )-cut.3
minimize cu v · Xu v uv subject to Xu v + S v − Su ≥ 0 for every edge u v ≥0 for every edge u v Xu v Ss = 1
St = 0
Like the minimization LP for shortest paths, there can be optimal solutions that assign fractional values
to the variables. Nevertheless, the minimum value for the objective function is the cost of the minimum
cut, and there is an optimal solution for which every variable is either 0 or 1, representing an actual
minimum cut. No, this is not obvious; in particular, my claim is not a proof! I.4 Linear Programming Duality Each of these pairs of linear programming problems is related by a transformation called duality. For
any linear programming problem, there is a corresponding dual linear program that can be obtained by
a mechanical translation, essentially by swapping the constraints and the variables. The translation is
simplest when the LP is in canonical form:
Primal (Π)
max c · x
s.t. Ax ≤ b
x≥0 Dual ( ) ⇐⇒ min y · b
s.t. yA ≥ c
y≥0 We can also write the dual linear program in exactly the same canonical form as the primal, by swapping
the coefﬁcient vector c and the objective vector b, negating both vectors, and replacing the constraint
matrix A with its negative transpose.4
Dual ( ) Primal (Π)
max c · x
s.t. Ax ≤ b
x≥ 0 ⇐⇒ 3 max − b · y
s.t. −A y ≤ −c
y ≥0 These two linear programs are not quite syntactic duals; I’ve added two redundant variables Ss and S t to the min-cut
program to increase readability.
4
For the notational purists: In these formulations, x and b are column vectors, and y and c are row vectors. This is a
somewhat nonstandard choice. Yes, that means the dot in c · x is redundant. Sue me. 5 Algorithms Non-Lecture I: Linear Programming Written in this form, it should be immediately clear that duality is an involution: The dual of the dual
linear program is identical to the primal linear program Π. The choice of which LP to call the ‘primal’
and which to call the ‘dual’ is totally arbitrary.5
The Fundamental Theorem of Linear Programming. A linear program Π has an optimal solution x ∗
if and only if the dual linear program has an optimal solution y ∗ such that c · x ∗ = y ∗ Ax ∗ = y ∗ · b.
The weak form of this theorem is trivial to prove.
Weak Duality Theorem. If x is a feasible solution for a canonical linear program Π and y is a feasible
solution for its dual , then c · x ≤ yAx ≤ y · b.
Proof: Because x is feasible for Π, we have Ax ≤ b. Since y is positive, we can multiply both sides of
the inequality to obtain yAx ≤ y · b. Conversely, y is feasible for and x is positive, so yAx ≥ c · x .
It immediately follows that if c · x = y · b, then x and y are optimal solutions to their respective
linear programs. This is in fact a fairly common way to prove that we have the optimal value for a linear
program. I.5 Duality Example Before I prove the stronger duality theorem, let me ﬁrst provide some intuition about where this duality
thing comes from in the ﬁrst place.6 Consider the following linear programming problem:
maximize
subject to 4 x1 + x2 + 3 x3
x1 + 4 x2 ≤2 3 x1 − x2 + x3 ≤ 4
x1, x2, x3 ≥ 0
Let σ∗ denote the optimum objective value for this LP The feasible solution x = (1, 0, 0) gives us a lower
.
bound σ∗ ≥ 4. A different feasible solution x = (0, 0, 3) gives us a better lower bound σ∗ ≥ 9. We could
play this game all day, ﬁnding different feasible solutions and getting ever larger lower bounds. How do
we know when we’re done? Is there a way to prove an upper bound on σ∗ ?
In fact, there is. Let’s multiply each of the constraints in our LP by a new non-negative scalar value yi :
maximize
subject to 4 x1 + x2 + 3 x3 y1 ( x 1 + 4 x 2 ) ≤ 2 y1 y2 (3 x 1 − x 2 + x 3 ) ≤ 4 y2
x1, x2, x3 ≥ 0
Because each yi is non-negative, we do not reverse any of the inequalities. Any feasible solution
( x 1 , x 2 , x 3 ) must satisfy both of these inequalities, so it must also satisfy their sum:
( y1 + 3 y2 ) x 1 + (4 y1 − y2 ) x 2 + y2 x 3 ≤ 2 y1 + 4 y2 .
5
For historical reasons, maximization LPs tend to be called ‘primal’ and minimization LPs tend to be called ‘dual’. This is a
pointless religious tradition, nothing more. Duality is a relationship between LP problems, not a type of LP problem.
6
This example is taken from Robert Vanderbei’s excellent textbook Linear Programming: Foundations and Extensions
[Springer, 2001], but the idea appears earlier in Jens Clausen’s 1997 paper ‘Teaching Duality in Linear Programming: The
Multiplier Approach’. 6 Algorithms Non-Lecture I: Linear Programming Now suppose that each yi is larger than the i th coefﬁcient of the objective function:
y1 + 3 y2 ≥ 4, 4 y1 − y2 ≥ 1, y2 ≥ 3. This assumption lets us derive an upper bound on the objective value of any feasible solution:
4 x 1 + x 2 + 3 x 3 ≤ ( y1 + 3 y2 ) x 1 + (4 y1 − y2 ) x 2 + y2 x 3 ≤ 2 y1 + 4 y2 . (∗) ∗
∗
∗
In particular, by plugging in the optimal solution ( x 1 , x 2 , x 3 ) for the original LP we obtain the following
,
∗
upper bound on σ :
∗
∗
∗
σ∗ = 4 x 1 + x 2 + 3 x 3 ≤ 2 y1 + 4 y2 . Now it’s natural to ask how tight we can make this upper bound. How small can we make the
expression 2 y1 + 4 y2 without violating any of the inequalities we used to prove the upper bound? This
is just another linear programming problem.
minimize
subject to 2 y1 + 4 y2
y1 + 3 y2 ≥ 4
4 y1 − y2 ≥ 1
y2 ≥ 3
y1 , y2 ≥ 0 In fact, this is precisely the dual of our original linear program! Moreover, inequality (∗) is just an
instantiation of the Weak Duality Theorem. I.6 Strong Duality The Fundamental Theorem can be rephrased in the following form:
Strong Duality Theorem. If x ∗ is an optimal solution for a canonical linear program Π, then there is
an optimal solution y ∗ for its dual , such that c · x ∗ = y ∗ Ax ∗ = y ∗ · b.
Proof (Sketch): I’ll prove the theorem only for non-degenerate linear programs, in which (a) the
optimal solution (if one exists) is a unique vertex of the feasible region, and (b) at most d constraint
planes pass through any point. These non-degeneracy assumptions are relatively easy to enforce in
practice and can be removed from the proof at the expense of some technical detail. I will also prove the
theorem only for the case n ≥ d ; the argument for under-constrained LPs is similar (if not simpler).
Let x ∗ be the optimal solution for the linear program Π; non-degeneracy implies that this solution
is unique, and that exactly d of the n linear constraints are satisﬁed with equality. Without loss of
generality (by permuting the rows of A), we can assume that these are the ﬁrst d constraints.
So let A• be the d × d matrix containing the ﬁrst d rows of A, and let A◦ denote the other n − d rows.
Similarly, partition b into its ﬁrst d coordinates b• and everything else b◦ . Thus, we have partitioned the
inequality Ax ∗ ≤ b into a system of equations A• x ∗ = b• and a system of strict inequalities A◦ x ∗ < b◦ .
∗
∗
∗
∗
Now let y ∗ = ( y• , y◦ ) where y• = cA−1 and y◦ = 0. We easily verify that y ∗ · b = c · x ∗ :
•
∗
∗
∗
y ∗ · b = y• · b• + y◦ · b◦ = y• · b• = cA−1 b• = c · x ∗ .
• (The existence of the inverse matrix A−1 follows from our non-degeneracy assumption.) Similarly, it’s
•
easy to verify that y ∗ A ≥ c :
∗
∗
∗
y ∗ A = y• A∗ + y◦ A∗ = y• A∗ = c .
•
◦
•
7 Algorithms Non-Lecture I: Linear Programming Once we prove that y ∗ is non-negative, and therefore feasible, the Weak Duality Theorem implies
∗
the result. We chose y◦ = 0. As we will see below, the optimality of x ∗ implies the strict inequality
∗
y• > 0—we had to use optimality somewhere! This is the hardest part of the proof.
∗
The key insight is to give a geometric interpretation to the vector y• = cA−1 . Each row of the
•
∗
∗
d
linear system A• x = b• describes a hyperplane ai · x = bi in IR . The vector ai is normal to this
hyperplane and points out of the feasible region. The vectors a1 , . . . , ad are linearly independent (by
∗
non-degeneracy) and thus describe a coordinate frame for the vector space IRd . The deﬁnition of y• can
be rewritten as follows:
d c= ∗
y• A• =
i =1 yi∗ ai . ∗
words, y•
∗ In other
lists the coefﬁcients of the objective vector c in the coordinate frame a1 , . . . , ad .
The point x lies on exactly d constraint hyperplanes; any d − 1 of these hyperplanes determine a
line through x ∗ . For each 1 ≤ i ≤ d , let i denote the line that lies on all but the i th constraint plane,
and let vi denote a vector based at x ∗ that points into the halfspace ai x ≤ bi along the line i . This
vector lies along an edge of the feasible polytope. For all j = i , we have a j · vi = 0. Thus, we can write
A• vi = (0, . . . , 0, ai · vi , 0, . . . , 0)
where the scalar ai · vi appears in the i th coordinate. It follows that
∗
c · vi = y• A• vi = yi∗ (ai · vi ). The optimality of x ∗ implies that c · vi < 0, and because vi points into the feasible region while ai points
out, we have ai · vi < 0. We conclude that yi∗ > 0. We’re done! Exercises
1. (a) Describe precisely how to dualize a linear program written in slack form:
max c · x
s.t. Ax = b
x≥0
(b) Describe precisely how to dualize a linear program written in general form:
d maximize cj x j
j =1
d ai j x j ≤ bi for each i = 1 .. p ai j x j = bi for each i = p + 1 .. p + q ai j x j ≥ bi subject to for each i = p + q + 1 .. n j =1
d
j =1
d
j =1 In both cases, keep the number of dual variables as small as possible.
8 Algorithms Non-Lecture I: Linear Programming 2. A matrix A = (ai j ) is skew-symmetric if and only if a ji = −ai j for all indices i = j ; in particular,
every skew-symmetric matrix is square. A canonical linear program max{c · x | Ax ≤ b; x ≥ 0} is
self-dual if the matrix A is skew-symmetric and the objective vector c is equal to the constraint
vector b.
(a) Prove any self-dual linear program Π is equivalent to its dual program . (b) Show that any linear program Π with d variables and n constraints can be transformed into a
self-dual linear program with n + d variables and n + d constraints. The optimal solution to
the self-dual program should include both the optimal solution for Π (in d of the variables)
and the optimal solution for the dual program (in the other n variables).
3. (a) Model the maximum-cardinality bipartite matching problem as a linear programming problem. The input is a bipartite graph G = (U ∪ V ; E ), where E ⊆ U × V ; the output is the largest
matching in G . Your linear program should have one variable for each edge.
(b) Now dualize the linear program from part (a). What do the dual variables represent? What
does the objective function represent? What problem is this!?
4. An integer program is a linear program with the additional constraint that the variables must take
only integer values.
(a) Prove that deciding whether an integer program has a feasible solution is NP-complete.
(b) Prove that ﬁnding the optimal feasible solution to an integer program is NP-hard.
[Hint: Almost any NP-hard decision problem can be formulated as an integer program. Pick your
favorite.]
5. Helly’s theorem states that for any collection of convex bodies in IRd , if every d + 1 of them
intersect, then there is a point lying in the intersection of all of them. Prove Helly’s theorem for
the special case where the convex bodies are halfspaces. Equivalently, show that if a system of
linear inequalities Ax ≤ b does not have a solution, then we can select d + 1 of the inequalities
such that the resulting subsystem also does not have a solution. [Hint: Construct a dual LP from
the system by choosing a 0 cost vector.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 9 Algorithms Non-Lecture J: Linear Programming Algorithms
Simplicibus itaque verbis gaudet Mathematica Veritas, cum etiam per se
simplex sit Veritatis oratio. [And thus Mathematical Truth prefers simple
words, because the language of Truth is itself simple.]
— Tycho Brahe (quoting Seneca (quoting Euripides))
Epistolarum astronomicarum liber primus (1596)
When a jar is broken, the space that was inside
Merges into the space outside.
In the same way, my mind has merged in God;
To me, there appears no duality.
— Sankara, Viveka-Chudamani (c. 700), translator unknown J Linear Programming Algorithms In this lecture, we’ll see a few algorithms for actually solving linear programming problems. The most
famous of these, the simplex method, was proposed by George Dantzig in 1947. Although most variants
of the simplex algorithm performs well in practice, no simplex variant is known to run in sub-exponential
time in the worst case. However, if the dimension of the problem is considered a constant, there
are several linear programming algorithms that run in linear time. I’ll describe a particularly simple
randomized algorithm due to Raimund Seidel.
My approach to describing these algorithms will rely much more heavily on geometric intuition
than the usual linear-algebraic formalism. This works better for me, but your mileage may vary. For a
more traditional description of the simplex algorithm, see Robert Vanderbei’s excellent textbook Linear
Programming: Foundations and Extensions [Springer, 2001], which can be freely downloaded (but not
legally printed) from the author’s website. J.1 Bases, Feasibility, and Local Optimality Consider the canonical linear program max{c · x | Ax ≤ b, x ≥ 0}, where A is an n × d constraint
matrix, b is an n-dimensional coefﬁcient vector, and c is a d -dimensional objective vector. We will
interpret this linear program geometrically as looking for the lowest point in a convex polyhedron
in IRd , described as the intersection of n + d halfspaces. As in the last lecture, we will consider only
non-degenerate linear programs: Every subset of d constraint hyperplanes intersects in a single point; at
most d constraint hyperplanes pass through any point; and objective vector is linearly independent from
any d − 1 constraint vectors.
A basis is a subset of d constraints, which by our non-degeneracy assumption must be linearly
independent. The location of a basis is the unique point x that satisﬁes all d constraints with equality;
geometrically, x is the unique intersection point of the d hyperplanes. The value of a basis is c · x ,
n+d
where x is the location of the basis. There are precisely d bases. Geometrically, the set of constraint
hyperplanes deﬁnes a decomposition of IRd into convex polyhedra; this cell decomposition is called the
arrangement of the hyperplanes. Every subset of d hyperplanes (that is, every basis) deﬁnes a vertex of
this arrangement (the location of the basis). I will use the words ‘vertex’ and ‘basis’ interchangeably.
A basis is feasible if its location x satisﬁes all the linear constraints, or geometrically, if the point x is
a vertex of the polyhedron. If there are no feasible bases, the linear program is infeasible.
A basis is locally optimal if its location x is the optimal solution to the linear program with the
same objective function and only the constraints in the basis. Geometrically, a basis is locally optimal
if its location x is the lowest point in the intersection of those d halfspaces. A careful reading of the
proof of the Strong Duality Theorem reveals that local optimality is the dual equivalent of feasibility; a
basis is locally feasible for a linear program Π if and only if the same basis is feasible for the dual linear
1 Algorithms Non-Lecture J: Linear Programming Algorithms program . For this reason, locally optimal bases are sometimes also called dual feasible. If there are
no locally optimal bases, the linear program is unbounded.1
Two bases are neighbors if they have d − 1 constraints in common. Equivalently, in geometric terms,
two vertices are neighbors if they lie on a line determined by some d − 1 constraint hyperplanes. Every
basis is a neighbor of exactly d n other bases; to change a basis into one of its neighbors, there are d
choices for which constraint to remove and n choices for which constraint to add. The graph of vertices
and edges on the boundary of the feasible polyhedron is a subgraph of the basis graph.
The Weak Duality Theorem implies that the value of every feasible basis is less than or equal to
the value of every locally optimal basis; equivalently, every feasible vertex is higher than every locally
optimal vertex. The Strong Duality Theorem implies that (under our non-degeneracy assumption), if a
linear program has an optimal solution, it is the unique vertex that is both feasible and locally optimal.
Moreover, the optimal solution is both the lowest feasible vertex and the highest locally optimal vertex. J.2 The Primal Simplex Algorithm: Falling Marbles From a geometric standpoint, Dantzig’s simplex algorithm is very simple. The input is a set H of
halfspaces; we want the lowest vertex in the intersection of these halfspaces.
SIMPLEX1(H ):
if ∩H = ∅
return INFEASIBLE
x ← any feasible vertex
while x is not locally optimal
〈pivot downward, maintaining feasibility〉〉
if every feasible neighbor of x is higher than x
return UNBOUNDED
else
x ← any feasible neighbor of x that is lower than x
return x Let’s ignore the ﬁrst three lines for the moment. The algorithm maintains a feasible vertex x . At each
so-called pivot operation, the algorithm moves to a lower vertex, so the algorithm never visits the same
n+d
vertex more than once. Thus, the algorithm must halt after at most d pivots. When the algorithm
halts, either the feasible vertex x is locally optimal, and therefore the optimum vertex, or the feasible
vertex x is not locally optimal but has no lower feasible neighbor, in which case the feasible region must
be unbounded.
Notice that we have not speciﬁed which neighbor to choose at each pivot. Several different pivoting
rules have been proposed, but for almost every known pivot rule, there is an input polyhedron that
requires an exponential number of pivots under that rule. No pivoting rule is known that guarantees a
polynomial number of pivots in the worst case.2 J.3 The Dual Simplex Algorithm: Rising Bubbles We can also geometrically interpret the execution of the simplex algorithm on the dual linear program .
Again, the input is a set H of halfspaces, and we want the lowest vertex in the intersection of these
1 For non-degenerate linear programs, the feasible region is unbounded in the objective direction if and only if no basis is
locally optimal. However, there are degenerate linear programs with no locally optimal basis that are infeasible.
2
In 1957, Hirsch conjectured that for any linear programming instance with d variables and n + d constraints, starting at
any feasible basis, there is a sequence of at most n pivots that leads to the optimal basis. Hirsch’s conjecture is still open 50
years later; no counterexamples have ever been found, but no proof is known except in a few special cases. Truly our ignorance
is unbounded (or at least dual infeasible). 2 Algorithms Non-Lecture J: Linear Programming Algorithms halfspaces. By the Strong Duality Theorem, this is the same as the highest locally-optimal vertex in the
hyperplane arrangement.
SIMPLEX2(H ):
if there is no locally optimal vertex
return UNBOUNDED
x ← any locally optimal vertex
while x is not feasbile
〈〈pivot upward, maintaining local optimality〉〉
if every locally optimal neighbor of x is lower than x
return INFEASIBLE
else
x ← any locally-optimal neighbor of x that is higher than x
return x Let’s ignore the ﬁrst three lines for the moment. The algorithm maintains a locally optimal vertex x .
At each pivot operation, it moves to a higher vertex, so the algorithm never visits the same vertex more
n+d
than once. Thus, the algorithm must halt after at most d pivots. When the algorithm halts, either
the locally optimal vertex x is feasible, and therefore the optimum vertex, or the locally optimal vertex x
is not feasible but has no higher locally optimal neighbor, in which case the problem must be infeasible. The primal simplex (falling marble) algorithm in action. The dual simplex (rising bubble) algorithm in action. From the standpoint of linear algebra, there is absolutely no difference between running SIMPLEX1 on
any linear program Π and running SIMPLEX2 on the dual linear program . The actual code is identical.
The only difference between the two algorithms is how we interpret the linear algebra geometrically. J.4 Computing the Initial Basis To complete our description of the simplex algorithm, we need to describe how to ﬁnd the initial vertex x .
Our algorithm relies on the following simple observations.
First, the feasibility of a vertex does not depend at all on the choice of objective vector; a vertex is
either feasible for every objective function or for none. No matter how we rotate the polyhedron, every
feasible vertex stays feasible. Conversely (or by duality, equivalently), the local optimality of a vertex
does not depend on the exact location of the d hyperplanes, but only on their normal directions and
the objective function. No matter how we translate the hyperplanes, every locally optimal vertex stays 3 Algorithms Non-Lecture J: Linear Programming Algorithms locally optimal. In terms of the original matrix formulation, feasibility depends on A and b but not c ,
and local optimality depends on A and c but not b.
The second important observation is that every basis is locally optimal for some objective function.
Speciﬁcally, it sufﬁces to choose any vector that has a positive inner product with each of the normal
vectors of the d chosen hyperplanes. Equivalently, we can make any basis feasible by translating the
hyperplanes appropriately. Speciﬁcally, it sufﬁces to translate the chosen d hyperplanes so that they pass
through the origin, and then translate all the other halfspaces so that they strictly contain the origin.
Our strategy for ﬁnding our initial feasible vertex is to choose any vertex, choose a new objective
function that makes that vertex locally optimal, and then ﬁnd the optimal vertex for that objective
function by running the (dual) simplex algorithm. This vertex must be feasible, even after we restore
the original objective function! (a) (b) (c) (a) Choose any basis. (b) Rotate objective to make it locally optimal, and pivot ’upward’ to ﬁnd a feasible basis.
(c) Pivot downward to the optimum basis for the original objective. Equivalently, to ﬁnd an initial locally optimal vertex, we choose any vertex, translate the hyperplanes
so that that vertex becomes feasible, and then ﬁnd the optimal vertex for those translated constraints
using the (primal) simplex algorithm. This vertex must be locally optimal, even after we restore the
hyperplanes to their original locations! (a) (b) (c) (a) Choose any basis. (b) Translate constraints to make it feasible, and pivot downward to ﬁnd a locally optimal basis.
(c) Pivot upward to the optimum basis for the original constraints. 4 Algorithms Non-Lecture J: Linear Programming Algorithms Here are more complete descriptions of the simplex algorithm with this initialization rule, in both
primal and dual forms. As usual, the input is a set H of halfspaces, and the algorithms either return the
lowest vertex in the intersection of these halfspaces or report that no such vertex exists.
SIMPLEX1(H ):
x ← any vertex
˜
H ← any rotation of H that makes x locally optimal
while x is not feasible
˜
if every locally optimal neighbor of x is lower (wrt H ) than x
return INFEASIBLE
else
˜
x ← any locally optimal neighbor of x that is higher (wrt H ) than x
while x is not locally optimal
if every feasible neighbor of x is higher than x
return UNBOUNDED
else
x ← any feasible neighbor of x that is lower than x
return x
SIMPLEX2(H ):
x ← any vertex
˜
H ← any translation of H that makes x feasible
while x is not locally optimal
˜
if every feasible neighbor of x is higher (wrt H ) than x
return UNBOUNDED
else
˜
x ← any feasible neighbor of x that is lower (wrt H ) than x
while x is not feasible
if every locally optimal neighbor of x is lower than x
return INFEASIBLE
else
x ← any locally-optimal neighbor of x that is higher than x
return x J.5 Linear Expected Time for Fixed Dimensions In most geometric applications of linear programming, the number of variables is a small constant, but
the number of constraints may still be very large.
The input to the following algorithm is a set H of n halfspaces and a set B of b hyperplanes. (B stands
for basis.) The algorithm returns the lowest point in the intersection of the halfspaces in H and the
hyperplanes B . At the top level of recursion, B is empty. I will implicitly assume that the linear program
is both feasible and bounded. (If necessary, we can guarantee boundedness by adding a single halfspace
to H , and we can guarantee feasibility by adding a dimension.) A point x violates a constraint h if it is
not contained in the corresponding halfspace. 5 Algorithms Non-Lecture J: Linear Programming Algorithms
SEIDELLP(H , B ) :
if |B | = d
return B
if |H ∪ B | = d
return (H ∪ B )
h ← random element of H
x ← SEIDELLP(H \ h, B )
(∗)
if x violates h
return SEIDELLP(H \ h, B ∪ ∂ h)
else
return x The point x recursively computed in line (∗) is the optimal solution if and only if the random
halfspace h is not one of the d halfspaces that deﬁne the optimal solution. In other words, the probability
of calling SEIDELLP(H , B ∪ h) is exactly (d − b)/n. Thus, we have the following recurrence for the
expected number of recursive calls for this algorithm: if b = d or n + b = d
1
T ( n, b ) =
d−b · T (n − 1, b + 1) otherwise
T (n − 1, b) +
n
The recurrence is somewhat simpler if we write δ = d − b: if δ = 0 or n = δ
1
T ( n, δ ) =
δ T (n − 1, δ) + · T (n − 1, δ − 1) otherwise
n
It’s easy to prove by induction that T (n, δ) = O(δ! n):
T (n, δ) = T (n − 1, δ) +
≤ δ ! ( n − 1) + δ
n δ
n · T (n − 1, δ − 1) (δ − 1)! · (n − 1) = δ ! ( n − 1) + δ ! [induction hypothesis] n−1
n ≤ δ! n
At the top level of recursion, we perform one violation test in O(d ) time. In each of the base cases,
we spend O(d 3 ) time computing the intersection point of d hyperplanes, and in the ﬁrst base case, we
spend O(d n) additional time testing for violations. More careful analysis implies that the algorithm runs
in O (d ! · n ) expected time. Exercises
1. Fix a non-degenerate linear program in canonical form with d variables and n + d constraints.
(a) Prove that every feasible basis has exactly d feasible neighbors.
(b) Prove that every locally optimal basis has exactly n locally optimal neighbors.
6 Algorithms Non-Lecture J: Linear Programming Algorithms 2. Suppose you have a subroutine that can solve linear programs in polynomial time, but only if they
are both feasible and bounded. Describe an algorithm that solves arbitrary linear programs in
polynomial time. Your algorithm should return an optimal solution if one exists; if no optimum
exists, your algorithm should report that the input instance is UNBOUNDED or INFEASIBLE, whichever
is appropriate. [Hint: Add one variable and one constraint.]
3. (a) Give an example of a non-empty polyhedron Ax ≤ b that is unbounded for every objective
vector c .
(b) Give an example of an infeasible linear program whose dual is also infeasible.
In both cases, your linear program will be degenerate.
4. Describe and analyze an algorithm that solves the following problem in O(n) time: Given n red
points and n blue points in the plane, either ﬁnd a line that separates every red point from every
blue point, or prove that no such line exists.
5. The single-source shortest path problem can be formulated as a linear programming problem, with
one variable d v for each vertex v = s in the input graph, as follows:
maximize dv
v dv ≤ s→ v for every edge s → v d v − du ≤ subject to u→ v for every edge u → v with u = s
for every vertex v = s dv ≥ 0 This problem asks you to describe the behavior of the simplex algorithm on this linear program in
terms of distances. Assume that the edge weights u→ v are all non-negative and that there is a
unique shortest path between any two vertices in the graph.
(a) What is a basis for this linear program? What is a feasible basis? What is a locally optimal
basis?
(b) Show that in the optimal basis, every variable d v is equal to the shortest-path distance from s
to v .
(c) Describe the primal simplex algorithm for the shortest-path linear program directly in terms
of vertex distances. In particular, what does it mean to pivot from a feasible basis to a
neighboring feasible basis, and how can we execute such a pivot quickly?
(d) Describe the dual simplex algorithm for the shortest-path linear program directly in terms of
vertex distances. In particular, what does it mean to pivot from a locally optimal basis to a
neighboring locally optimal basis, and how can we execute such a pivot quickly?
(e) Is Dijkstra’s algorithm an instance of the simplex method? Justify your answer.
(f) Is Shimbel’s algorithm an instance of the simplex method? Justify your answer.
6. The maximum (s, t )-ﬂow problem can be formulated as a linear programming problem, with one 7 Algorithms Non-Lecture J: Linear Programming Algorithms variable fu→ v for each edge u → v in the input graph:
f s→w − maximize
w fu→s
u fu→ v = 0 f v →w − subject to
w for every vertex v = s, t u fu→ v ≤ cu→ v for every edge u → v
fu→ v ≥ 0 for every edge u → v This problem asks you to describe the behavior of the simplex algorithm on this linear program in
terms of ﬂows.
(a) What is a basis for this linear program? What is a feasible basis? What is a locally optimal
basis?
(b) Show that the optimal basis represents a maximum ﬂow.
(c) Describe the primal simplex algorithm for the ﬂow linear program directly in terms of ﬂows.
In particular, what does it mean to pivot from a feasible basis to a neighboring feasible basis,
and how can we execute such a pivot quickly?
(d) Describe the dual simplex algorithm for the ﬂow linear program directly in terms of ﬂows. In
particular, what does it mean to pivot from a locally optimal basis to a neighboring locally
optimal basis, and how can we execute such a pivot quickly?
(e) Is the Ford-Fulkerson augmenting path algorithm an instance of the simplex method? Justify
your answer. [Hint: There is a one-line argument.]
7. (a) Formulate the minimum spanning tree problem as an instance of linear programming. Try to
minimize the number of variables and constraints.
(b) In your MST linear program, what is a basis? What is a feasible basis? What is a locally
optimal basis?
(c) Describe the primal simplex algorithm for your MST linear program directly in terms of the
input graph. In particular, what does it mean to pivot from a feasible basis to a neighboring
feasible basis, and how can we execute such a pivot quickly?
(d) Describe the dual simplex algorithm for your MST linear program directly in terms of the
input graph. In particular, what does it mean to pivot from a locally optimal basis to a
neighboring locally optimal basis, and how can we execute such a pivot quickly?
˙
(e) Which of the classical MST algorithms (Boruvka, Jarník, Kruskal, reverse greedy), if any, are
instances of the simplex method? Justify your answer. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 8 Algorithms Lecture 20: Adversary Arguments
An adversary means opposition and competition,
but not having an adversary means grief and loneliness.
— Zhuangzi (Chuang-tsu) c. 300 BC
It is possible that the operator could be hit by an asteroid and your $20 could
fall off his cardboard box and land on the ground, and while you were picking
it up, $5 could blow into your hand. You therefore could win $5 by a simple
twist of fate.
— Penn Jillette, explaining how to win at Three-Card Monte (1999) 20
20.1 Adversary Arguments
Three-Card Monte Until Times Square was sanitized into TimesSquareLandTM by Mayor Rudy Guiliani, you could often ﬁnd
dealers stealing tourists’ money using a game called ‘Three Card Monte’ or ‘Spot the Lady’. The dealer
has three cards, say the Queen of Hearts and the two and three of clubs. The dealer shufﬂes the cards
face down on a table (usually slowly enough that you can follow the Queen), and then asks the tourist
to bet on which card is the Queen. In principle, the tourist’s odds of winning are at least one in three.
In practice, however, the tourist never1 wins, because the dealer cheats. There are actually four cards;
before he even starts shufﬂing the cards, the dealer palms the queen or sticks it up his sleeve. No matter
what card the tourist bets on, the dealer turns over a black card. If the tourist gives up, the dealer slides
the queen under one of the cards and turns it over, showing the tourist ‘where the queen was all along’.
If the dealer is really good, the tourist won’t see the dealer changing the cards and will think maybe
the queen was there all along and he just wasn’t smart enough to ﬁgure that out. As long as the dealer
doesn’t reveal all the black cards at once, the tourist has no way to prove that the dealer cheated!2 20.2 n -Card Monte
Now let’s consider a similar game, but with an algorithm acting as the tourist and with bits instead of
cards. Suppose we have an array of n bits and we want to determine if any of them is a 1. Obviously we
can ﬁgure this out by just looking at every bit, but can we do better? Is there maybe some complicated
tricky algorithm to answer the question “Any ones?” without looking at every bit? Well, of course not,
but how do we prove it?
The simplest proof technique is called an adversary argument. The idea is that an all-powerful
malicious adversary (the dealer) pretends to choose an input for the algorithm (the tourist). When the
algorithm wants looks at a bit (a card), the adversary sets that bit to whatever value will make the
algorithm do the most work. If the algorithm does not look at enough bits before terminating, then
there will be several different inputs, each consistent with the bits already seen, the should result in
different outputs. Whatever the algorithm outputs, the adversary can ‘reveal’ an input that is has all the
examined bits but contradicts the algorithm’s output, and then claim that that was the input that he was
using all along. Since the only information the algorithm has is the set of bits it examined, the algorithm
cannot distinguish between a malicious adversary and an honest user who actually chooses an input in
advance and answers all queries truthfully.
For the n-card monte problem, the adversary originally pretends that the input array is all zeros—
whenever the algorithm looks at a bit, it sees a 0. Now suppose the algorithms stops before looking at all
1 What, never? No, NEVER. Anyone you see winning at Three Card Monte is a shill.
Even if the dealer isn’t very good, he cheats anyway. The shills will protect him from any angry tourists who realize they’ve
been ripped off, and shake down any tourist who refuses to pay. You cannot win this game.
2 1 Algorithms Lecture 20: Adversary Arguments three bits. If the algorithm says ‘No, there’s no 1,’ the adversary changes one of the unexamined bits to a
1 and shows the algorithm that it’s wrong. If the algorithm says ‘Yes, there’s a 1,’ the adversary reveals
the array of zeros and again proves the algorithm wrong. Either way, the algorithm cannot tell that the
adversary has cheated.
One absolutely crucial feature of this argument is that the adversary makes absolutely no assumptions
about the algorithm. The adversary strategy can’t depend on some predetermined order of examining
bits, and it doesn’t care about anything the algorithm might or might not do when it’s not looking at bits.
Any algorithm that doesn’t examine every bit falls victim to the adversary. 20.3 Finding Patterns in Bit Strings Let’s make the problem a little more complicated. Suppose we’re given an array of n bits and we want to
know if it contains the substring 01, a zero followed immediately by a one. Can we answer this question
without looking at every bit?
It turns out that if n is odd, we don’t have to look at all the bits. First we look the bits in every even
position: B [2], B [4], . . . , B [n − 1]. If we see B [i ] = 0 and B [ j ] = 1 for any i < j , then we know the
pattern 01 is in there somewhere—starting at the last 0 before B [ j ]—so we can stop without looking at
any more bits. If we see only 1s followed by 0s, we don’t have to look at the bit between the last 0 and
the ﬁrst 1. If every even bit is a 0, we don’t have to look at B [1], and if every even bit is a 1, we don’t
have to look at B [n]. In the worst case, our algorithm looks at only n − 1 of the n bits.
But what if n is even? In that case, we can use the following adversary strategy to show that any
algorithm does have to look at every bit. The adversary will attempt to produce an ‘input’ string B
without the substring 01; all such strings have the form 11 . . . 100 . . . 0. The adversary maintains two
indices and r and pretends that the preﬁx B [1 .. ] contains only 1s and the sufﬁx B [ r .. n] contains
only 0s. Initially = 0 and r = n + 1. 111111
↑ 0000
↑
r What the adversary is thinking; represents an unknown bit. The adversary maintains the invariant that r − , the length of the undecided portion of the ‘input’
string, is even. When the algorithm looks at a bit between and r , the adversary chooses whichever value
preserves the parity of the intermediate chunk of the array, and then moves either or r . Speciﬁcally,
here’s what the adversary does when the algorithm examines bit B [i ]. (Note that I’m specifying the
adversary strategy as an algorithm!)
HIDE01(i ):
if i ≤
B [i ] ← 1
else if i ≥ r
B [i ] ← 0
else if i − is even
B [i ] ← 0
r←i
else
B [i ] ← 1
←i It’s fairly easy to prove that this strategy forces the algorithm to examine every bit. If the algorithm
doesn’t look at every bit to the right of r , the adversary could replace some unexamined bit with a 1.
2 Algorithms Lecture 20: Adversary Arguments Similarly, if the algorithm doesn’t look at every bit to the left of , the adversary could replace some
unexamined bit with a zero. Finally, if there are any unexamined bits between and r , there must be at
least two such bits (since r − is always even) and the adversary can put a 01 in the gap.
In general, we say that a bit pattern is evasive if we have to look at every bit to decide if a string of n
bits contains the pattern. So the pattern 1 is evasive for all n, and the pattern 01 is evasive if and only if
n is even. It turns out that the only patterns that are evasive for all values of n are the one-bit patterns 0
and 1. 20.4 Evasive Graph Properties Another class of problems for which adversary arguments give good lower bounds is graph problems
where the graph is represented by an adjacency matrix, rather than an adjacency list. Recall that the
adjacency matrix of an undirected n-vertex graph G = (V, E ) is an n × n matrix A, where A[i , j ] =
(i , j ) ∈ E . We are interested in deciding whether an undirected graph has or does not have a certain
property. For example, is the input graph connected? Acyclic? Planar? Complete? A tree? We call
n
a graph property evasive if we have to look look at all 2 entries in the adjacency matrix to decide
whether a graph has that property.
An obvious example of an evasive graph property is emptiness: Does the graph have any edges at
all? We can show that emptiness is evasive using the following simple adversary strategy. The adversary
maintains two graphs E and G . E is just the empty graph with n vertices. Initially G is the complete
graph on n vertices. Whenever the algorithm asks about an edge, the adversary removes that edge from
G (unless it’s already gone) and answers ‘no’. If the algorithm terminates without examining every edge,
then G is not empty. Since both G and E are consistent with all the adversary’s answers, the algorithm
must give the wrong answer for one of the two graphs. 20.5 Connectedness Is Evasive Now let me give a more complicated example, connectedness. Once again, the adversary maintains two
graphs, Y and M (‘yes’ and ‘maybe’). Y contains all the edges that the algorithm knows are deﬁnitely in
the input graph. M contains all the edges that the algorithm thinks might be in the input graph, or in
other words, all the edges of Y plus all the unexamined edges. Initially, Y is empty and M is complete.
Here’s the strategy that adversary follows when the adversary asks whether the input graph contains
the edge e. I’ll assume that whenever an algorithm examines an edge, it’s in M but not in Y ; in other
words, algorithms never ask about the same edge more than once.
HIDECONNECTEDNESS(e):
if M \ {e} is connected
remove (i , j ) from M
return 0
else
add e to Y
return 1 Notice that the graphs Y and M are both consistent with the adversary’s answers at all times. The
adversary strategy maintains a few other simple invariants.
• Y is a subgraph of M . This is obvious.
• M is connected. This is also obvious. 3 Algorithms Lecture 20: Adversary Arguments • If M has a cycle, none of its edges are in Y . If M has a cycle, then deleting any edge in that
cycle leaves M connected.
• Y is acyclic. This follows directly from the previous invariant.
• If Y = M , then Y is disconnected. The only connected acyclic graph is a tree. Suppose Y is a
tree and some edge e is in M but not in Y . Then there is a cycle in M that contains e, all of whose
other edges are in Y . This violated our third invariant.
We can also think about the adversary strategy in terms of minimum spanning trees. Recall the
anti-Kruskal algorithm for computing the maximum spanning tree of a graph: Consider the edges one
at a time in increasing order of length. If removing an edge would disconnect the graph, declare it
part of the spanning tree (by adding it to Y ); otherwise, throw it away (by removing it from M ). If the
n
algorithm examines all 2 possible edges, then Y and M are both equal to the maximum spanning tree
of the complete n-vertex graph, where the weight of an edge is the time when the algorithm asked about
it.
n
Now, if an algorithm terminates before examining all 2 edges, then there is at least one edge in M
that is not in Y . Since the algorithm cannot distinguish between M and Y , even though M is connected
and Y is not, the algorithm cannot possibly give the correct output for both graphs. Thus, in order to be
correct, any algorithm must examine every edge—Connectedness is evasive! 20.6 An Evasive Conjecture A graph property is nontrivial is there is at least one graph with the property and at least one graph
without the property. (The only trivial properties are ‘Yes’ and ‘No’.) A graph property is monotone if it is
closed under taking subgraphs — if G has the property, then any subgraph of G has the property. For
example, emptiness, planarity, acyclicity, and non-connectedness are monotone. The properties of being
a tree and of having a vertex of degree 3 are not monotone.
Conjecture 1 (Aanderraa, Karp, and Rosenberg). Every nontrivial monotone property of n-vertex
graphs is evasive.
The Aanderraa-Karp-Rosenberg conjecture has been proven when n = p e for some prime p and
positive integer exponent e—the proof uses some interesting results from algebraic topology3 —but it is
still open for other values of n.’
There are non-trivial non-evasive graph properties, but all known examples are non-monotone. One
such property—‘scorpionhood’—is described in an exercise at the end of this lecture note. 20.7 Finding the Minimum and Maximum Last time, we saw an adversary argument that ﬁnding the largest element of an unsorted set of n
numbers requires at least n − 1 comparisons. Let’s consider the complexity of ﬁnding the largest and
smallest elements. More formally:
Given a sequence X = 〈 x 1 , x 2 , . . . , x n 〉 of n distinct numbers, ﬁnd indices i and j such that
x i = min X and x j = max X .
Let ∆ be a contractible simplicial complex whose automorphism group Aut(∆) is vertex-transitive, and let Γ be a vertextransitive subgroup of Aut(∆). If there are normal subgroups Γ1 Γ2 Γ such that |Γ1 | = pα for some prime p and integer α,
|Γ/Γ2 | = qβ for some prime q and integer β , and Γ2 /Γ1 is cyclic, then ∆ is a simplex.
3 No, this will not be on the ﬁnal exam. 4 Algorithms Lecture 20: Adversary Arguments How many comparisons do we need to solve this problem? An upper bound of 2n − 3 is easy: ﬁnd the
minimum in n − 1 comparisons, and then ﬁnd the maximum of everything else in n − 2 comparisons.
Similarly, a lower bound of n − 1 is easy, since any algorithm that ﬁnds the min and the max certainly
ﬁnds the max.
We can improve both the upper and the lower bound to 3n/2 − 2. The upper bound is established
by the following algorithm. Compare all n/2 consecutive pairs of elements x 2i −1 and x 2i , and put the
smaller element into a set S and the larger element into a set L . if n is odd, put x n into both L and S .
Then ﬁnd the smallest element of S and the largest element of L . The total number of comparisons is at
most
n
n
n
3n
−1 +
−1 =
+
− 2.
2
2
2
2
build S and L compute min S compute max L For the lower bound, we use an adversary argument. The adversary marks each element + if it might
be the maximum element, and − if it might be the minimum element. Initially, the adversary puts both
marks + and − on every element. If the algorithm compares two double-marked elements, then the
adversary declares one smaller, removes the + mark from the smaller element, and removes the − mark
from the larger one. In every other case, the adversary can answer so that at most one mark needs to be
removed. For example, if the algorithm compares a double marked element against one labeled −, the
adversary says the one labeled − is smaller and removes the − mark from the other. If the algorithm
compares to +’s, the adversary must unmark one of the two.
Initially, there are 2n marks. At the end, in order to be correct, exactly one item must be marked +
and exactly one other must be marked −, since the adversary can make any + the maximum and any −
the minimum. Thus, the algorithm must force the adversary to remove 2n − 2 marks. At most n/2
comparisons remove two marks; every other comparison removes at most one mark. Thus, the adversary
strategy forces any algorithm to perform at least 2n − 2 − n/2 = 3n/2 − 2 comparisons. 20.8 Finding the Median Finally, let’s consider the median problem: Given an unsorted array X of n numbers, ﬁnd its n/2th largest
entry. (I’ll assume that n is even to eliminate pesky ﬂoors and ceilings.) More formally:
Given a sequence 〈 x 1 , x 2 , . . . , x n 〉 of n distinct numbers, ﬁnd the index m such that x m is the
n/2th largest element in the sequence.
To prove a lower bound for this problem, we can use a combination of information theory and two
adversary arguments. We use one adversary argument to prove the following simple lemma:
Lemma 1. Any comparison tree that correctly ﬁnds the median element also identiﬁes the elements
smaller than the median and larger than the median.
Proof: Suppose we reach a leaf of a decision tree that chooses the median element x m , and there is still
some element x i that isn’t known to be larger or smaller than x m . In other words, we cannot decide
based on the comparisons that we’ve already performed whether x i < x m or x i > x m . Then in particular
no element is known to lie between x i and x m . This means that there must be an input that is consistent
with the comparisons we’ve performed, in which x i and x m are adjacent in sorted order. But then we
can swap x i and x m , without changing the result of any comparison, and obtain a different consistent
input in which x i is the median, not x m . Our decision tree gives the wrong answer for this ‘swapped’
input. 5 Algorithms Lecture 20: Adversary Arguments This lemma lets us rephrase the median-ﬁnding problem yet again.
Given a sequence X = 〈 x 1 , x 2 , . . . , x n 〉 of n distinct numbers, ﬁnd the indices of its n/2 − 1
largest elements L and its n/2th largest element x m .
Now suppose a ‘little birdie’ tells us the set L of elements larger than the median. This information
ﬁxes the outcomes of certain comparisons—any item in L is bigger than any element not in L —so we
can ‘prune’ those comparisons from the comparison tree. The pruned tree ﬁnds the largest element of
X \ L (the median of X ), and thus must have depth at least n/2 − 1. In fact, the adversary argument in
the last lecture implies that every leaf in the pruned tree must have depth at least n/2 − 1, so the pruned
tree has at least 2n/2−1 leaves.
n
There are n/2−1 ≈ 2n / n/2 possible choices for the set L . Every leaf in the original comparison
tree is also a leaf in exactly one of the
least n/2−1 2n/2−1 ≈ 23n/2 /
depth at least
n n
n/2−1 pruned trees, so the original comparison tree must have at n/2 leaves. Thus, any comparison tree that ﬁnds the median must have
n
2 − 1 + lg n
n/ 2 − 1 = 3n
2 − O(log n). A more complicated adversary argument (also involving pruning the comparison tree with little birdies)
improves this lower bound to 2n − o(n).
n
A similar argument implies that at least n − k + lg k−1 = Ω((n − k) + k log(n/k)) comparisons
are required to ﬁnd the kth largest element in an n-element set. This bound is tight up to constant
factors for all k ≤ n/2; there is an algorithm that uses at most O(n + k log(n/k)) comparisons. Moreover,
this lower bound is exactly tight when k = 1 or k = 2. In fact, these are the only values of k ≤ n/2 for
which the exact complexity of the selection problem is known. Even the case k = 3 is still open! Exercises
1. (a) Let X be a set containing an odd number of n-bit strings. Prove that any algorithm that
decides whether a given n-bit string is an element of X must examine every bit of the input
string in the worst case.
(b) Give a one-line proof that the bit pattern 01 is evasive for all even n.
(c) Prove that the bit pattern 11 is evasive if and only if n mod 3 = 1.
(d) Prove that the bit pattern 111 is evasive if and only if n mod 4 = 0 or 3. 2. Suppose we are given the adjacency matrix of a directed graph G with n vertices. Describe an
algorithm that determines whether G has a sink by probing only O(n) bits in the input matrix. A
sink is a vertex that has an incoming edge from every other vertex, but no outgoing edges. 3. A scorpion is an undirected graph with three special vertices: the sting, the tail, and the body. The
sting is connected only to the tail; the tail is connected only to the sting and the body; and the
body is connected to every vertex except the sting. The rest of the vertices (the head, eyes, legs,
antennae, teeth, gills, ﬂippers, wheels, etc.) can be connected arbitrarily. Describe an algorithm
that determines whether a given n-vertex graph is a scorpion by probing only O(n) entries in the
adjacency matrix.
6 Algorithms Lecture 20: Adversary Arguments 4. Prove using an advesary argument that acyclicity is an evasive graph property. [Hint: Kruskal.] 5. Prove that ﬁnding the second largest element in an n-element array requires exactly n − 2 + lg n
comparisons in the worst case. Prove the upper bound by describing and analyzing an algorithm;
prove the lower bound using an adversary argument. 6. Let T be a perfect ternary tree where every leaf has depth . Suppose each of the 3 leaves of T is
labeled with a bit, either 0 or 1, and each internal node is labeled with a bit that agrees with the
majority of its children.
(a) Prove that any deterministic algorithm that determines the label of the root must examine all
3 leaf bits in the worst case.
(b) Describe and analyze a randomized algorithm that determines the root label, such that
the expected number of leaves examined is o(3 ). (You may want to review the notes on
randomized algorithms.) 7. UIUC has just ﬁnished constructing the new Reingold Building, the tallest dormitory on campus. In
order to determine how much insurance to buy, the university administration needs to determine
the highest safe ﬂoor in the building. A ﬂoor is consdered safe if a drunk student an egg can fall
from a window on that ﬂoor and land without breaking; if the egg breaks, the ﬂoor is considered
unsafe. Any ﬂoor that is higher than an unsafe ﬂoor is also considered unsafe. The only way to
determine whether a ﬂoor is safe is to drop an egg from a window on that ﬂoor.
You would like to ﬁnd the lowest unsafe ﬂoor L by performing as few tests as possible;
unfortunately, you have only a very limited supply of eggs.
(a) Prove that if you have only one egg, you can ﬁnd the lowest unsafe ﬂoor with L tests. [Hint:
Yes, this is trivial.]
(b) Prove that if you have only one egg, you must perform at least L tests in the worst case. In
other words, prove that your algorithm from part (a) is optimal. [Hint: Use an adversary
argument.]
(c) Describe an algorithm to ﬁnd the lowest unsafe ﬂoor using two eggs and only O( L ) tests.
[Hint: Ideally, each egg should be dropped the same number of times. How many ﬂoors can
you test with n drops?]
(d) Prove that if you start with two eggs, you must perform at least Ω( L ) tests in the worst
case. In other words, prove that your algorithm from part (c) is optimal.
(e) Describe an algorithm to ﬁnd the lowest unsafe ﬂoor using k eggs, using as few tests as
possible, and prove your algorithm is optimal for all values of k. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision. 7 Algorithms Lecture 21: NP-Hard Problems
The wonderful thing about standards is that
there are so many of them to choose from.
— Real Admiral Grace Murray Hopper
If a problem has no solution, it may not be a problem, but a fact —
not to be solved, but to be coped with over time.
— Shimon Peres 21
21.1 NP-Hard Problems
‘Efﬁcient’ Problems A generally-accepted minimum requirement for an algorithm to be considered ‘efﬁcient’ is that its
running time is polynomial: O(nc ) for some constant c , where n is the size of the input.1 Researchers
recognized early on that not all problems can be solved this quickly, but we had a hard time ﬁguring
out exactly which ones could and which ones couldn’t. there are several so-called NP-hard problems,
which most people believe cannot be solved in polynomial time, even though nobody can prove a
super-polynomial lower bound.
Circuit satisﬁability is a good example of a problem that we don’t know how to solve in polynomial
time. In this problem, the input is a boolean circuit: a collection of AND, OR, and NOT gates connected by
wires. We will assume that there are no loops in the circuit (so no delay lines or ﬂip-ﬂops). The input to
the circuit is a set of m boolean (TRUE/FALSE) values x 1 , . . . , x m . The output is a single boolean value.
Given speciﬁc input values, we can calculate the output of the circuit in polynomial (actually, linear)
time using depth-ﬁrst-search, since we can compute the output of a k-input gate in O(k) time. x
y x∧y x
y x∨y x x An And gate, an Or gate, and a Not gate. x1
x2
x3
x4
x5
A boolean circuit. inputs enter from the left, and the output leaves to the right. The circuit satisﬁability problem asks, given a circuit, whether there is an input that makes the
circuit output TRUE, or conversely, whether the circuit always outputs FALSE. Nobody knows how to solve
this problem faster than just trying all 2m possible inputs to the circuit, but this requires exponential
time. On the other hand, nobody has ever proved that this is the best we can do; maybe there’s a clever
algorithm that nobody has discovered yet!
1 This notion of efﬁciency was independently formalized by Alan Cobham (The intrinsic computational difﬁculty of functions.
Logic, Methodology, and Philosophy of Science (Proc. Int. Congress), 24–30, 1965), Jack Edmonds (Paths, trees, and ﬂowers.
Canadian Journal of Mathematics 17:449–467, 1965), and Michael Rabin (Mathematical theory of automata. Proceedings of
the 19th ACM Symposium in Applied Mathematics, 153–175, 1966), although similar notions were considered more than a
decade earlier by Kurt Gödel and John von Neumann. 1 Algorithms 21.2 Lecture 21: NP-Hard Problems P, NP, and co-NP A decision problem is a problem whose output is a single boolean value: YES or NO.2 Let me deﬁne three
classes of decision problems:
• P is the set of decision problems that can be solved in polynomial time.3 Intuitively, P is the set of
problems that can be solved quickly.
• NP is the set of decision problems with the following property: If the answer is YES, then there is
a proof of this fact that can be checked in polynomial time. Intuitively, NP is the set of decision
problems where we can verify a YES answer quickly if we have the solution in front of us.
• co-NP is the opposite of NP. If the answer to a problem in co-NP is NO, then there is a proof of this
fact that can be checked in polynomial time.
For example, the circuit satisﬁability problem is in NP. If the answer is YES, then any set of m input
values that produces TRUE output is a proof of this fact; we can check the proof by evaluating the circuit
in polynomial time. It is widely believed that circuit satisﬁability is not in P or in co-NP, but nobody
actually knows.
Every decision problem in P is also in NP. If a problem is in P, we can verify YES answers in polynomial
time recomputing the answer from scratch! Similarly, any problem in P is also in co-NP.
One of the most important open questions in theoretical computer science is whether or not P = NP.
Nobody knows. Intuitively, it should be obvious that P = NP; the homeworks and exams in this class
and others have (I hope) convinced you that problems can be incredibly hard to solve, even when the
solutions are obvious in retrospect. But nobody knows how to prove it.
A more subtle but still open question is whether NP and co-NP are different. Even if we can verify
every YES answer quickly, there’s no reason to think that we can also verify NO answers quickly. For
example, as far as we know, there is no short proof that a boolean circuit is not satisﬁable. It is generally
believed that NP = co-NP, but nobody knows how to prove it. co−NP NP
P What we think the world looks like. 21.3 NP-hard, NP-easy, and NP-complete A problem Π is NP-hard if a polynomial-time algorithm for Π would imply a polynomial-time algorithm
for every problem in NP. In other words:
Π is NP-hard ⇐⇒ If Π can be solved in polynomial time, then P=NP
2
Technically, I should be talking about languages, which are just sets of bit strings. The language associated with a decision
problem is the set of bit strings for which the answer is YES. For example, for the problem is ‘Is the input graph connected?’,
the corresponding language is the set of connected graphs, where each graph is represented as a bit string (for example, its
adjacency matrix).
3
More formally, P is the set of languages that can be recognized in polynomial time by a single-tape Turing machine. If you
want to learn more about Turing machines, take CS 579. 2 Algorithms Lecture 21: NP-Hard Problems Intuitively, if we could solve one particular NP-hard problem quickly, then we could quickly solve any
problem whose solution is easy to understand, using the solution to that one special problem as a
subroutine. NP-hard problems are at least as hard as any problem in NP.4
Saying that a problem is NP-hard is like saying ‘If I own a dog, then it can speak ﬂuent English.’ You
probably don’t know whether or not I own a dog, but you’re probably pretty sure that I don’t own a
talking dog. Nobody has a mathematical proof that dogs can’t speak English—the fact that no one has
ever heard a dog speak English is evidence, as are the hundreds of examinations of dogs that lacked the
proper mouth shape and braiNPower, but mere evidence is not a mathematical proof. Nevertheless, no
sane person would believe me if I said I owned a dog that spoke ﬂuent English. So the statement ‘If I
own a dog, then it can speak ﬂuent English’ has a natural corollary: No one in their right mind should
believe that I own a dog! Likewise, if a problem is NP-hard, no one in their right mind should believe it
can be solved in polynomial time.
Finally, a problem is NP-complete if it is both NP-hard and an element of NP (or ‘NP-easy’). NPcomplete problems are the hardest problems in NP. If anyone ﬁnds a polynomial-time algorithm for even
one NP-complete problem, then that would imply a polynomial-time algorithm for every NP-complete
problem. Literally thousands of problems have been shown to be NP-complete, so a polynomial-time
algorithm for one (that is, all) of them seems incredibly unlikely.
NP−hard
co−NP
NP
NP−complete P More of what we think the world looks like. It is not immediately clear that any decision problems are NP-hard or NP-complete. NP-hardness
is already a lot to demand of a problem; insisting that the problem also have a nondeterministic
polynomial-time algorithm seems almost completely unreasonable. The following remarkable theorem
was ﬁrst published by Steve Cook in 1971 and independently by Leonid Levin in 1973.5 I won’t even
sketch the proof, since I’ve been (deliberately) vague about the deﬁnitions.
The Cook-Levin Theorem. Circuit satisﬁability is NP-complete.
More formally, a problem Π is NP-hard if and only if, for any problem Π in NP, there is a polynomial-time Turing reduction
from Π to Π—a Turing reduction just means a reduction that can be executed on a Turing machine. Polynomial-time Turing
reductions are also called Cook reductions.
Complexity theorists prefer to deﬁne NP-hardness in terms of polynomial-time many-one reductions, which are also called
Karp reductions. A many-one reduction from one language Π to another language Π is an function f : Σ∗ → Σ∗ such that
x ∈ Π if and only if f ( x ) ∈ Π. Every Karp reduction is a Cook reduction, but not vice versa. Every reduction (between
decision problems) in these notes is a Karp reduction. This deﬁnition is preferred primarily because NP is closed under Karp
reductions, but believed not to be closed under Cook reductions. Moreover, the two deﬁnitions of NP-hardness are equivalent
only if NP=co-NP, which is considered unlikely. In fact, there are natural problems that are (1) NP-hard with respect to Cook
reductions, but (2) NP-hard with respect to Karp reductions only if P=NP! On the other hand, the Karp deﬁnition only applies
to decision problems, or more formally, sets of bit-strings.
To make things even more confusing, both Cook and Karp originally deﬁned NP-hardness in terms of logarithmic-space
reductions. Every logarithmic-space reduction is a polynomial-time reduction, but (we think) not vice versa. It is an open
question whether relaxing the set of allowed (Cook or Karp) reductions from logarithmic-space to polynomial-time changes the
set of NP-hard problems.
5
Levin submitted his results, and discussed them in talks, several years before they were published. So naturally, in
accordance with Stigler’s Law, this result is often called ‘Cook’s Theorem’. Cook won the Turing award for his proof; Levin did
not.
4 3 Algorithms 21.4 Lecture 21: NP-Hard Problems Reductions and SAT To prove that a problem is NP-hard, we use a reduction argument. Reducing problem A to another
problem B means describing an algorithm to solve problem A under the assumption that an algorithm for
problem B already exists. You’re already used to doing reductions, only you probably call it something
else, like writing subroutines or utility functions, or modular programming. To prove something is
NP-hard, we describe a similar transformation between problems, but not in the direction that most
people expect.
You should tattoo the following rule of onto the back of your hand.
To prove that problem A is NP-hard, reduce a known NP-hard problem to A.
In other words, to prove that your problem is hard, you need to describe an algorithm to solve a
different problem, which you already know is hard, using a mythical algorithm for your problem as a
subroutine. The essential logic is a proof by contradiction. Your reduction shows implies that if your
problem were easy, then the other problem would be easy, too. Equivalently, since you know the other
problem is hard, your problem must also be hard.
For example, consider the formula satisﬁability problem, usually just called SAT . The input to SAT is
a boolean formula like
¯
¯
(a ∨ b ∨ c ∨ d ) ⇔ (( b ∧ ¯) ∨ (a ⇒ d ) ∨ (c = a ∧ b)),
c
and the question is whether it is possible to assign boolean values to the variables a, b, c , . . . so that the
formula evaluates to TRUE.
To show that SAT is NP-hard, we need to give a reduction from a known NP-hard problem. The only
problem we know is NP-hard so far is circuit satisﬁability, so let’s start there. Given a boolean circuit, we
can transform it into a boolean formula by creating new output variables for each gate, and then just
writing down the list of gates separated by and. For example, we could transform the example circuit
into a formula as follows:
x1 y1 y4
y5 x2
x3
x4
x5 y7
y2 y8 y3
y6 ( y1 = x 1 ∧ x 4 ) ∧ ( y2 = x 4 ) ∧ ( y3 = x 3 ∧ y2 ) ∧ ( y4 = y1 ∨ x 2 ) ∧
( y5 = x 2 ) ∧ ( y6 = x 5 ) ∧ ( y7 = y3 ∨ y5 ) ∧ ( y8 = y4 ∧ y7 ∧ y6 ) ∧ y8
A boolean circuit with gate variables added, and an equivalent boolean formula. Now the original circuit is satisﬁable if and only if the resulting formula is satisﬁable. Given a
satisfying input to the circuit, we can get a satisfying assignment for the formula by computing the
output of every gate. Given a satisfying assignment for the formula, we can get a satisfying input the the
circuit by just ignoring the gate variables yi .
We can transform any boolean circuit into a formula in linear time using depth-ﬁrst search, and the
size of the resulting formula is only a constant factor larger than the size of the circuit. Thus, we have a
polynomial-time reduction from circuit satisﬁability to SAT:
4 Algorithms Lecture 21: NP-Hard Problems
O(n) −−→ boolean formula
boolean circuit −
SAT
trivial TRUE or FALSE ←−− TRUE or FALSE
−
TCSAT (n) ≤ O(n) + TSAT (O(n)) =⇒ TSAT (n) ≥ TCSAT (Ω(n)) − O(n) The reduction implies that if we had a polynomial-time algorithm for SAT, then we’d have a polynomialtime algorithm for circuit satisﬁability, which would imply that P=NP. So SAT is NP-hard.
To prove that a boolean formula is satisﬁable, we only have to specify an assignment to the variables
that makes the formula TRUE. We can check the proof in linear time just by reading the formula from
left to right, evaluating as we go. So SAT is also in NP, and thus is actually NP-complete. 21.5 3SAT (from SAT) A special case of SAT that is particularly useful in proving NP-hardness results is called 3SAT.
A boolean formula is in conjunctive normal form (CNF) if it is a conjunction (AND) of several clauses,
each of which is the disjunction (OR) of several literals, each of which is either a variable or its negation.
For example:
clause ¯
(a ∨ b ∨ c ∨ d ) ∧ ( b ∨ ¯ ∨ d ) ∧ (a ∨ c ∨ d ) ∧ (a ∨ ¯ )
c¯
b
A 3CNF formula is a CNF formula with exactly three literals per clause; the previous example is not a
3CNF formula, since its ﬁrst clause has four literals and its last clause has only two. 3SAT is just SAT
restricted to 3CNF formulas: Given a 3CNF formula, is there an assignment to the variables that makes
the formula evaluate to TRUE?
We could prove that 3SAT is NP-hard by a reduction from the more general SAT problem, but it’s
easier just to start over from scratch, with a boolean circuit. We perform the reduction in several stages.
1. Make sure every AND and OR gate has only two inputs. If any gate has k > 2 inputs, replace it with a
binary tree of k − 1 two-input gates.
2. Write down the circuit as a formula, with one clause per gate. This is just the previous reduction.
3. Change every gate clause into a CNF formula. There are only three types of clauses, one for each
type of gate:
a= b∧c ¯
¯
−→ (a ∨ ¯ ∨ ¯) ∧ (a ∨ b) ∧ (a ∨ c )
bc
¯
−→ (a ∨ b ∨ c ) ∧ (a ∨ ¯ ) ∧ (a ∨ ¯)
b
c a= b∨c
¯b
a = ¯ −→ (a ∨ b) ∧ (a ∨ ¯ )
b 4. Make sure every clause has exactly three literals. Introduce new variables into each one- and
two-literal clause, and expand it into two clauses as follows:
¯
¯y
a −→ (a ∨ x ∨ y ) ∧ (a ∨ x ∨ y ) ∧ (a ∨ x ∨ ¯ ) ∧ (a ∨ x ∨ ¯ )
y
¯
a ∨ b −→ (a ∨ b ∨ x ) ∧ (a ∨ b ∨ x ) 5 Algorithms Lecture 21: NP-Hard Problems For example, if we start with the same example circuit we used earlier, we obtain the following 3CNF
formula. Although this may look a lot more ugly and complicated than the original circuit at ﬁrst glance,
it’s actually only a constant factor larger—every binary gate in the original circuit has been transformed
into at most ﬁve clauses. Even if the formula size were a large polynomial function (like n573 ) of the
circuit size, we would still have a valid reduction.
( y1 ∨ x 1 ∨ x 4 ) ∧ ( y1 ∨ x 1 ∨ z1 ) ∧ ( y1 ∨ x 1 ∨ z1 ) ∧ ( y1 ∨ x 4 ∨ z2 ) ∧ ( y1 ∨ x 4 ∨ z2 )
∧ ( y2 ∨ x 4 ∨ z3 ) ∧ ( y2 ∨ x 4 ∨ z3 ) ∧ ( y2 ∨ x 4 ∨ z4 ) ∧ ( y2 ∨ x 4 ∨ z4 )
∧ ( y3 ∨ x 3 ∨ y2 ) ∧ ( y3 ∨ x 3 ∨ z5 ) ∧ ( y3 ∨ x 3 ∨ z5 ) ∧ ( y3 ∨ y2 ∨ z6 ) ∧ ( y3 ∨ y2 ∨ z6 )
∧ ( y4 ∨ y1 ∨ x 2 ) ∧ ( y4 ∨ x 2 ∨ z7 ) ∧ ( y4 ∨ x 2 ∨ z7 ) ∧ ( y4 ∨ y1 ∨ z8 ) ∧ ( y4 ∨ y1 ∨ z8 )
∧ ( y5 ∨ x 2 ∨ z9 ) ∧ ( y5 ∨ x 2 ∨ z9 ) ∧ ( y5 ∨ x 2 ∨ z10 ) ∧ ( y5 ∨ x 2 ∨ z10 )
∧ ( y6 ∨ x 5 ∨ z11 ) ∧ ( y6 ∨ x 5 ∨ z11 ) ∧ ( y6 ∨ x 5 ∨ z12 ) ∧ ( y6 ∨ x 5 ∨ z12 )
∧ ( y7 ∨ y3 ∨ y5 ) ∧ ( y7 ∨ y3 ∨ z13 ) ∧ ( y7 ∨ y3 ∨ z13 ) ∧ ( y7 ∨ y5 ∨ z14 ) ∧ ( y7 ∨ y5 ∨ z14 )
∧ ( y8 ∨ y4 ∨ y7 ) ∧ ( y8 ∨ y4 ∨ z15 ) ∧ ( y8 ∨ y4 ∨ z15 ) ∧ ( y8 ∨ y7 ∨ z16 ) ∧ ( y8 ∨ y7 ∨ z16 )
∧ ( y9 ∨ y8 ∨ y6 ) ∧ ( y9 ∨ y8 ∨ z17 ) ∧ ( y9 ∨ y8 ∨ z17 ) ∧ ( y9 ∨ y6 ∨ z18 ) ∧ ( y9 ∨ y6 ∨ z18 )
∧ ( y9 ∨ z19 ∨ z20 ) ∧ ( y9 ∨ z19 ∨ z20 ) ∧ ( y9 ∨ z19 ∨ z20 ) ∧ ( y9 ∨ z19 ∨ z20 )
This process transforms the circuit into an equivalent 3CNF formula; the output formula is satisﬁable
if and only if the input circuit is satisﬁable. As with the more general SAT problem, the formula is only a
constant factor larger than any reasonable description of the original circuit, and the reduction can be
carried out in polynomial time. Thus, we have a polynomial-time reduction from circuit satisﬁability to
3SAT:
O(n) boolean circuit −
−−→ 3CNF formula
3SAT
trivial TRUE or FALSE ←−− TRUE or FALSE
−
TCSAT (n) ≤ O(n) + T3SAT (O(n)) =⇒ T3SAT (n) ≥ TCSAT (Ω(n)) − O(n) We conclude 3SAT is NP-hard. And because 3SAT is a special case of SAT, it is also in NP. Therefore,
3SAT is NP-complete. 21.6 Maximum Independent (from 3SAT) For the next few problems we consider, the input is a simple, unweighted graph, and the problem asks
for the size of the largest or smallest subgraph satisfying some structural property.
Let G be an arbitrary graph. An independent set in G is a subset of the vertices of G with no edges
between them. The maximum independent set problem, or simply MAXINDSET, asks for the size of the
largest independent set in a given graph.
I’ll prove that MAXINDSET is NP-hard (but not NP-complete, since it isn’t a decision problem) using a
reduction from 3SAT. I’ll describe a reduction from a 3CNF formula into a graph that has an independent
set of a certain size if and only if the formula is satisﬁable. The graph has one node for each instance
of each literal in the formula. Two nodes are connected by an edge if (1) they correspond to literals
in the same clause, or (2) they correspond to a variable and its inverse. For example, the formula
¯
(a ∨ b ∨ c ) ∧ ( b ∨ ¯ ∨ d ) ∧ (a ∨ c ∨ d ) ∧ (a ∨ ¯ ∨ d ) is transformed into the following graph.
c¯
b¯ 6 Algorithms Lecture 21: NP-Hard Problems a c
b ‾
b b
a c
‾ ‾
d ‾
d
d
a
‾ c A graph derived from a 3CNF formula, and an independent set of size 4.
Black edges join literals from the same clause; red (heavier) edges join contradictory literals. Now suppose the original formula had k clauses. Then I claim that the formula is satisﬁable if and
only if the graph has an independent set of size k.
1. independent set =⇒ satisfying assignment: If the graph has an independent set of k vertices,
then each vertex must come from a different clause. To obtain a satisfying assignment, we assign
the value TRUE to each literal in the independent set. Since contradictory literals are connected by
edges, this assignment is consistent. There may be variables that have no literal in the independent
set; we can set these to any value we like. The resulting assignment satisﬁes the original 3CNF
formula.
2. satisfying assignment =⇒ independent set: If we have a satisfying assignment, then we can
choose one literal in each clause that is TRUE. Those literals form an independent set in the graph.
Thus, the reduction is correct. Since the reduction from 3CNF formula to graph takes polynomial time,
we conclude that MAXINDSET is NP-hard. Here’s a diagram of the reduction:
O(n) 3CNF formula with k clauses −
−−→ graph with 3k nodes
MAXINDSET
O(1) TRUE or FALSE ←−− maximum independent set size
−
T3SAT (n) ≤ O(n) + TMAXINDSET (O(n)) 21.7 =⇒ TMAXINDSET (n) ≥ T3SAT (Ω(n)) − O(n) Clique (from Independent Set) A clique is another name for a complete graph, that is, a graph where every pair of vertices is connected
by an edge. The maximum clique size problem, or simply MAXCLIQUE, is to compute, given a graph, the
number of nodes in its largest complete subgraph. A graph with maximum clique size 4. 7 Algorithms Lecture 21: NP-Hard Problems There is an easy proof that MAXCLIQUE is NP-hard, using a reduction from MAXINDSET. Any graph G
has an edge-complement G with the same vertices, but with exactly the opposite set of edges—(u, v ) is an
edge in G if and only if it is not an edge in G . A set of vertices is independent in G if and only if the
same vertices deﬁne a clique in G . Thus, we can compute the largest independent in a graph simply by
computing the largest clique in the complement of the graph.
O(n) −−→ complement graph G
graph G −
MAXCLIQUE
trivial largest independent set ←−− largest clique
− 21.8 Vertex Cover (from Independent Set) A vertex cover of a graph is a set of vertices that touches every edge in the graph. The MINVERTEXCOVER
problem is to ﬁnd the smallest vertex cover in a given graph.
Again, the proof of NP-hardness is simple, and relies on just one fact: If I is an independent set in a
graph G = (V, E ), then V \ I is a vertex cover. Thus, to ﬁnd the largest independent set, we just need to
ﬁnd the vertices that aren’t in the smallest vertex cover of the same graph.
trivial −−→ graph G = (V, E )
graph G = (V, E ) −
MINVERTEXCOVER
O(n) largest independent set V \ S ←−− smallest vertex cover S
− 21.9 Graph Coloring (from 3SAT) A k-coloring of a graph is a map C : V → {1, 2, . . . , k} that assigns one of k ‘colors’ to each vertex, so
that every edge has two different colors at its endpoints. The graph coloring problem is to ﬁnd the
smallest possible number of colors in a legal coloring. To show that this problem is NP-hard, it’s enough
to consider the special case 3COLORABLE: Given a graph, does it have a 3-coloring?
To prove that 3COLORABLE is NP-hard, we use a reduction from 3SAT. Given a 3CNF formula, we
produce a graph as follows. The graph consists of a truth gadget, one variable gadget for each variable
in the formula, and one clause gadget for each clause in the formula.
The truth gadget is just a triangle with three vertices T , F , and X , which intuitively stand for
TRUE, FALSE, and OTHER. Since these vertices are all connected, they must have different colors in any
3-coloring. For the sake of convenience, we will name those colors TRUE, FALSE, and OTHER. Thus, when
we say that a node is colored TRUE, all we mean is that it must be colored the same as the node T .
The variable gadget for a variable a is also a triangle joining two new nodes labeled a and a to
node X in the truth gadget. Node a must be colored either TRUE or FALSE, and so node a must be colored
either FALSE or TRUE, respectively.
Finally, each clause gadget joins three literal nodes to node T in the truth gadget using ﬁve new
unlabeled nodes and ten edges; see the ﬁgure below. If all three literal nodes in the clause gadget are
colored FALSE, then the rightmost vertex in the gadget cannot have one of the three colors. Since the
variable gadgets force each literal node to be colored either TRUE or FALSE, in any valid 3-coloring, at
least one of the three literal nodes is colored TRUE. I need to emphasize here that the ﬁnal graph contains
¯
only one node T , only one node F , and only two nodes a and a for each variable. 8 Algorithms Lecture 21: NP-Hard Problems a X X
T F b a T c a Gadgets for the reduction from 3SAT to 3-Colorability:
The truth gadget, a variable gadget for a, and a clause gadget for (a ∨ b ∨ ¯).
c The proof of correctness is just brute force. If the graph is 3-colorable, then we can extract a satisfying
assignment from any 3-coloring—at least one of the three literal nodes in every clause gadget is colored
TRUE. Conversely, if the formula is satisﬁable, then we can color the graph according to any satisfying
assignment.
O(n) 3CNF formula −
−−→ graph
3COLORABLE
trivial TRUE or FALSE ←−− TRUE or FALSE
−
¯
For example, the formula (a ∨ b ∨ c ) ∧ ( b ∨ ¯ ∨ d ) ∧ (a ∨ c ∨ d ) ∧ (a ∨ ¯ ∨ d ) that I used to illustrate the
c¯
b¯
MAXCLIQUE reduction would be transformed into the following graph. The 3-coloring is one of several
that correspond to the satisfying assignment a = c = TRUE, b = d = FALSE.
T F
X a a b b c c d d A 3-colorable graph derived from a satisﬁable 3CNF formula. We can easily verify that a graph has been correctly 3-colored in linear time: just compare the
endpoints of every edge. Thus, 3COLORING is in NP, and therefore NP-complete. Moreover, since
3COLORING is a special case of the more general graph coloring problem—What is the minimum number
of colors?—the more problem is also NP-hard, but not NP-complete, because it’s not a decision problem. 9 Algorithms 21.10 Lecture 21: NP-Hard Problems Hamiltonian Cycle (from Vertex Cover) A Hamiltonian cycle in a graph is a cycle that visits every vertex exactly once. This is very different from
an Eulerian cycle, which is actually a closed walk that traverses every edge exactly once. Eulerian cycles
are easy to ﬁnd and construct in linear time using a variant of depth-ﬁrst search. Finding Hamiltonian
cycles, on the other hand, is NP-hard.
To prove this, we use a reduction from the vertex cover problem. Given a graph G and an integer k,
we need to transform it into another graph G , such that G has a Hamiltonian cycle if and only if G has
a vertex cover of size k. As usual, our transformation uses several gadgets.
• For each edge (u, v ) in G , we have an edge gadget in G consisting of twelve vertices and fourteen
edges, as shown below. The four corner vertices (u, v, 1), (u, v, 6), ( v, u, 1), and ( v, u, 6) each have
an edge leaving the gadget. A Hamiltonian cycle can only pass through an edge gadget in one of
three ways. Eventually, these will correspond to one or both of the vertices u and v being in the
vertex cover.
(u,v,1) (u,v,2) (u,v,3) (u,v,4) (u,v,5) (u,v,6) (v,u,1) (v,u,2) (v,u,3) (v,u,4) (v,u,5) (v,u,6)
An edge gadget for (u, v ) and the only possible Hamiltonian paths through it. • G also contains k cover vertices, simply numbered 1 through k.
• Finally, for each vertex u in G , we string together all the edge gadgets for edges (u, v ) into a single
vertex chain, and then connect the ends of the chain to all the cover vertices. Speciﬁcally, suppose
u has d neighbors v1 , v2 , . . . , vd . Then G has d − 1 edges between (u, vi , 6) and (u, vi +1 , 1), plus k
edges between the cover vertices and (u, v1 , 1), and ﬁnally k edges between the cover vertices and
(u, vd , 6).
1
2
3 (v,x) (v,y) (v,z) (w,v) (x,v) (y,v) (z,v) ... (v,w) k The vertex chain for v : all edge gadgets involving v are strung together and joined to the k cover vertices. It’s not hard to prove that if { v1 , v2 , . . . , vk } is a vertex cover of G , then G has a Hamiltonian cycle—
start at cover vertex 1, through traverse the vertex chain for v1 , then visit cover vertex 2, then traverse
the vertex chain for v2 , and so forth, eventually returning to cover vertex 1. Conversely, any Hamiltonian
cycle in G alternates between cover vertices and vertex chains, and the vertex chains correspond to the
k vertices in a vertex cover of G . (This is a little harder to prove.) Thus, G has a vertex cover of size k if
and only if G has a Hamiltonian cycle.
10 Algorithms Lecture 21: NP-Hard Problems (u,v)
1
(v,u)
u v
(u,w) (v,x) (w,u) w (v,w) (w,v) (x,v) x (w,x)
2
(x,w) The original graph G with vertex cover { v, w }, and the transformed graph G with a corresponding Hamiltonian cycle.
Vertex chains are colored to match their corresponding vertices. The transformation from G to G takes at most O(n2 ) time, so the Hamiltonian cycle problem is
NP-hard. Moreover, since we can easily verify a Hamiltonian cycle in linear time, the Hamiltonian cycle
problem is in NP, and therefore is NP-complete.
A closely related problem to Hamiltonian cycles is the famous traveling salesman problem—Given a
weighted graph G , ﬁnd the shortest cycle that visits every vertex. Finding the shortest cycle is obviously
harder than determining if a cycle exists at all, so the traveling salesman problem is also NP-hard. 21.11 Subset Sum (from Vertex Cover) The last problem that we will prove NP-hard is the SUBSETSUM problem considered in the very ﬁrst
lecture on recursion: Given a set X of integers and an integer t , determine whether X has a subset
whose elements sum to t .
To prove this problem is NP-hard, we apply a reduction from the vertex cover problem. Given a
graph G and an integer k, we need to transform it into set of integers X and an integer t , such that X
has a subset that sums to t if and only if G has an vertex cover of size k. Our transformation uses just
two ‘gadgets’; these are integers representing vertices and edges in G .
Number the edges of G arbitrarily from 0 to m − 1. Our set X contains the integer bi := 4i for each
edge i , and the integer
a v := 4 m +
4i
i ∈∆( v ) for each vertex v , where ∆( v ) is the set of edges that have v has an endpoint. Alternately, we can think
of each integer in X as an (m + 1)-digit number written in base 4. The mth digit is 1 if the integer
represents a vertex, and 0 otherwise. For each i < m, the i th digit is 1 if the integer represents edge i or
one of its endpoints, and 0 otherwise. Finally, we set the target sum
m−1 t := k · 4 m + 2 · 4i .
i =0 11 Algorithms Lecture 21: NP-Hard Problems Now let’s prove that the reduction is correct. First, suppose there is a vertex cover of size k in the
original graph G . Consider the subset X C ⊆ X that includes a v for every vertex v in the vertex cover, and
bi for every edge i that has exactly one vertex in the cover. The sum of these integers, written in base 4,
has a 2 in each of the ﬁrst m digits; in the most signiﬁcant digit, we are summing exactly k 1’s. Thus,
the sum of the elements of X C is exactly t .
On the other hand, suppose there is a subset X ⊆ X that sums to t . Speciﬁcally, we must have
av +
v ∈V bi = t
i∈E for some subsets V ⊆ V and E ⊆ E . Again, if we sum these base-4 numbers, there are no carries in
the ﬁrst m digits, because for each i there are only three numbers in X whose i th digit is 1. Each edge
number bi contributes only one 1 to the i th digit of the sum, but the i th digit of t is 2. Thus, for each
edge in G , at least one of its endpoints must be in V . In other words, V is a vertex cover. On the other
hand, only vertex numbers are larger than 4m , and t /4m = k, so V has at most k elements. (In fact,
it’s not hard to see that V has exactly k elements.)
For example, given the four-vertex graph used on the previous page to illustrate the reduction to
Hamiltonian cycle, our set X might contain the following base-4 integers:
buv
buw
b vw
bv x
bw x := 0100004 = 256
:= 0010004 = 64
:= 0001004 = 16
:= 0000104 =
4
:= 0000014 =
1 au := 1110004 = 1344
a v := 1101104 = 1300
aw := 1011014 = 1105
a x := 1000114 = 1029
If we are looking for a vertex cover of size 2, our target sum would be t := 2222224 = 2730. Indeed, the
vertex cover { v, w } corresponds to the subset {a v , aw , buv , buw , b v x , bw x }, whose sum is 1300 + 1105 +
256 + 64 + 4 + 1 = 2730.
The reduction can clearly be performed in polynomial time. Since VERTEXCOVER is NP-hard, it follows
that SUBSETSUM is NP-hard.
There is one subtle point that needs to be emphasized here. Way back at the beginning of the
semester, we developed a dynamic programming algorithm to solve SUBSETSUM in time O(nt ). Isn’t this
a polynomial-time algorithm? Nope. True, the running time is polynomial in n and t , but in order to
qualify as a true polynomial-time algorithm, the running time must be a polynomial function of the size
of the input. The values of the elements of X and the target sum t could be exponentially larger than the
number of input bits. Indeed, the reduction we just described produces exponentially large integers,
which would force our dynamic programming algorithm to run in exponential time. Algorithms like this
are called pseudo-polynomial-time, and any NP-hard problem with such an algorithm is called weakly
NP-hard. 21.12 Other Useful NP-hard Problems Literally thousands of different problems have been proved to be NP-hard. I want to close this note by
listing a few NP-hard problems that are useful in deriving reductions. I won’t describe the NP-hardness 12 Algorithms Lecture 21: NP-Hard Problems for these problems, but you can ﬁnd most of them in Garey and Johnson’s classic Scary Black Book of
NP-Completeness.6
• PLANARCIRCUITSAT: Given a boolean circuit that can be embedded in the plane so that no two
wires cross, is there an input that makes the circuit output TRUE? This can be proved NP-hard by
reduction from the general circuit satisﬁability problem, by replacing each crossing with a small
series of gates. (This is an easy7 exercise.)
• NOTALLEQUAL3SAT: Given a 3CNF formula, is there an assignment of values to the variables so that
every clause contains at least one TRUE literal and at least one FALSE literal? This can be proved
NP-hard by reduction from the usual 3SAT.
• PLANAR3SAT: Given a 3CNF boolean formula, consider a bipartite graph whose vertices are the
clauses and variables, where an edge indicates that a variable (or its negation) appears in a clause.
If this graph is planar, the 3CNF formula is also called planar. The PLANAR3SAT problem asks,
given a planar 3CNF formula, whether it has a satisfying assignment. This can be proved NP-hard
by reduction from PLANARCIRCUITSAT.8
• EXACT3DIMENSIONALMATCHING or X3M: Given a set S and a collection of three-element subsets
of S , called triples, is there a sub-collection of disjoint triples that exactly cover S ? This can be
proved NP-hard by a reduction from 3SAT.
• PARTITION: Given a set S of n integers, are there subsets A and B such that A ∪ B = S , A ∩ B = ∅,
and
a=
b?
a ∈A b∈B This can be proved NP-hard by a simple reduction from SUBSETSUM. Like SUBSETSUM, the PARTITION
problem is only weakly NP-hard.
• 3PARTITION: Given a set S of 3n integers, can it be partitioned into n disjoint subsets, each with 3
elements, such that every subset has exactly the same sum? Note that this is very different from
the PARTITION problem; I didn’t make up the names. This can be proved NP-hard by reduction from
X3M. Unlike PARTITION, the 3PARTITION problem is strongly NP-hard, that is, it remains NP-hard
even if the input numbers are less than some polynomial in n. The similar problem of dividing a
set of 2n integers into n equal-weight two-element sets can be solved in O(n log n) time.
• SETCOVER: Given a collection of sets
= {S1 , S2 , . . . , Sm }, ﬁnd the smallest sub-collection of Si ’s
that contains all the elements of i Si . This is a generalization of both VERTEXCOVER and X3M.
• HITTINGSET: Given a collection of sets
= {S1 , S2 , . . . , Sm }, ﬁnd the minimum number of elements
of i Si that hit every set in . This is also a generalization of VERTEXCOVER.
• LONGESTPATH: Given a non-negatively weighted graph G and two vertices u and v , what is the
longest simple path from u to v in the graph? A path is simple if it visits each vertex at most once.
This is a generalization of the HAMILTONIANPATH problem. Of course, the corresponding shortest
path problem is in P.
6
Michael Garey and David Johnson. Computers and Intractability: A Guide to the Theory of NP-Completeness. W. H. Freeman
and Co., 1979.
7
or at least nondeterministically easy
8
Surprisingly, PLANARNOTALLEQUAL3SAT is solvable in polynomial time! 13 Algorithms Lecture 21: NP-Hard Problems • STEINERTREE: Given a weighted, undirected graph G with some of the vertices marked, what is
the minimum-weight subtree of G that contains every marked vertex? If every vertex is marked,
the minimum Steiner tree is just the minimum spanning tree; if exactly two vertices are marked,
the minimum Steiner tree is just the shortest path between them. This can be proved NP-hard by
reduction to HAMILTONIANPATH.
Most interesting puzzles and solitaire games have been shown to be NP-hard, or to have NP-hard
generalizations. (Arguably, if a game or puzzle isn’t at least NP-hard, it isn’t interesting!) Some
familiar examples include Minesweeper (by reduction from CIRCUITSAT)9 , Tetris (by reduction from
3PARTITION)10 , and Sudoku (by reduction from 3SAT)11 . Most two-player games12 like tic-tac-toe, reversi,
checkers, chess, go, mancala—or more accurately, appropriate generalizations of these constant-size
games to arbitrary board sizes—are not just NP-hard, but PSPACE-hard or even EXP-hard.13 Exercises
1. Consider the following problem, called B OXDEPTH: Given a set of n axis-aligned rectangles in the
plane, how big is the largest subset of these rectangles that contain a common point?
(a) Describe a polynomial-time reduction from B OXDEPTH to MAXCLIQUE.
(b) Describe and analyze a polynomial-time algorithm for B OXDEPTH. [Hint: O(n3 ) time should
be easy, but O(n log n) time is possible.]
(c) Why don’t these two results imply that P=NP?
2. (a) Describe a polynomial-time reduction from PARTITION to SUBSETSUM.
(b) Describe a polynomial-time reduction from SUBSETSUM to PARTITION.
3. (a) Describe and analyze and algorithm to solve PARTITION in time O(nM ), where n is the size of
the input set and M is the sum of the absolute values of its elements.
Richard Kaye. Minesweeper is NP-complete. Mathematical Intelligencer 22(2):9–15, 2000. http://www.mat.bham.ac.
uk/R.W.Kaye/minesw/minesw.pdf
10
Ron Breukelaar*, Erik D. Demaine, Susan Hohenberger*, Hendrik J. Hoogeboom, Walter A. Kosters, and David LibenNowell*. Tetris is Hard, Even to Approximate. International Journal of Computational Geometry and Applications 14:41–68,
2004. The reduction was considerably simpliﬁed between its discovery in 2002 and its publication in 2004.
11
Takayuki Yato and Takahiro Seta. Complexity and Completeness of Finding Another Solution and Its Application to Puzzles.
IEICE Transactions on Fundamentals of Electronics, Communications and Computer Sciences E86-A(5):1052–1060, 2003.
http://www-imai.is.s.u-tokyo.ac.jp/~yato/data2/MasterThesis.pdf.
12
For a good (but now slightly dated) overview of known results on the computational complexity of games and puzzles, see
Erik D. Demaine’s survey “Playing Games with Algorithms: Algorithmic Combinatorial Game Theory” at http://arxiv.org/abs/
cs.CC/0106019.
13
PSPACE and EXP are the next two big steps above NP in the complexity hierarchy.
PSPACE is the set of decision problems that can be solved using polynomial space. Every problem in NP (and therefore in P)
is also in PSPACE. It is generally believed that NP = PSPACE, but nobody can even prove that P = PSPACE. A problem Π is
PSPACE-hard if, for any problem Π that can be solved using polynomial space, there is a polynomial-time many-one reduction
from Π to Π. If any PSPACE-hard problem is in NP, then PSPACE=NP.
c
EXP (also called EXPTIME) is the set of decision problems that can be solved in exponential time: at most 2n for some c > 0.
Every problem in PSPACE (and therefore in NP (and therefore in P)) is also in EXP. It is generally believed that PSPACE EXP,
but nobody can even prove that NP = EXP. We do know that P EXP, but we do not know of a single natural decision problem
in P \ EXP. A problem Π is EXP-hard if, for any problem Π that can be solved in exponential time, there is a polynomial-time
many-one reduction from Π to Π. If any EXP-hard problem is in PSPACE, then EXP=PSPACE.
Then there’s NEXP then EXPSPACE, then EEXP then NEEXP then EEXPSPACE, and so on ad inﬁnitum. Whee!
,
,
,
9 14 Algorithms Lecture 21: NP-Hard Problems (b) Why doesn’t this algorithm imply that P=NP?
4. A boolean formula is in disjunctive normal form (or DNF ) if it consists of a disjunction (OR) or
several terms, each of which is the conjunction (AND) of one or more literals. For example, the
formula
(a ∧ b ∧ c ) ∨ ( b ∧ c ) ∨ (a ∧ b ∧ c )
is in disjunctive normal form. DNF-SAT asks, given a boolean formula in disjunctive normal form,
whether that formula is satisﬁable.
(a) Show that DNF-SAT is in P.
(b) What is the error in the following argument that P=NP?
Suppose we are given a boolean formula in conjunctive normal form with at most three
literals per clause, and we want to know if it is satisﬁable. We can use the distributive law
to construct an equivalent formula in disjunctive normal form. For example, (a ∨ b ∨ c ) ∧ (a ∨ b) ⇐⇒ (a ∧ b) ∨ ( b ∧ a) ∨ (c ∧ a) ∨ (c ∧ b)
Now we can use the algorithms from part (a) to determine, in polynomial time, whether the
resulting DNF formula is satisﬁable. We have just solved 3SAT in polynomial time! Since
3SAT is NP-hard, we must conclude that P=NP. 5. (a) Describe and analyze a polynomial-time algorithm for 2PARTITION. Given a set S of 2n positive
integers, your algorithm will determine in polynomial time whether the elements of S can be
split into n disjoint pairs whose sums are all equal.
(b) Describe and analyze a polynomial-time algorithm for 2COLOR. Given an undirected graph G ,
your algorithm will determine in polynomial time whether G has a proper coloring that uses
only two colors.
(c) Describe and analyze a polynomial-time algorithm for 2SAT. Given a boolean formula Φ in
conjunctive normal form, with exactly two literals per clause, your algorithm will determine
in polynomial time whether Φ has a satisfying assignment.
6. (a) Prove that PLANARCIRCUITSAT is NP-complete.
(b) Prove that NOTALLEQUAL3SAT is NP-complete.
(c) Prove that the following variant of 3SAT is NP-complete: Given a formula φ in conjunctive
normal form where each clause contains at most 3 literals and each variable appears in at
most 3 clauses, is φ satisﬁable?
7. The problem 12COLOR is deﬁned as follows: Given a graph, can we color each vertex with one of
twelve colors, so that every edge touches two different colors? Prove that 12COLOR is NP-hard.
8. Jeff tries to make his students happy. At the beginning of class, he passes out a questionnaire that
lists a number of possible course policies in areas where he is ﬂexible. Every student is asked to
respond to each possible course policy with one of “strongly favor”, “mostly neutral”, or “strongly
oppose”. Each student may respond with “strongly favor” or “strongly oppose” to at most ﬁve
questions. Because Jeff’s students are very understanding, each student is happy if (but only if) he
15 Algorithms Lecture 21: NP-Hard Problems or she prevails in just one of his or her strong policy preferences. Either describe a polynomial-time
algorithm for setting course policy to maximize the number of happy students, or show that the
problem is NP-hard.
9. (a) Using the gadget on the right below, prove that deciding whether a given planar graph is
3-colorable is NP-complete. [Hint: Show that the gadget can be 3-colored, and then replace
any crossings in a planar embedding with the gadget appropriately.]
(b) Using part (a) and the middle gadget below, prove that deciding whether a planar graph
with maximum degree 4 is 3-colorable is NP-complete. [Hint: Replace any vertex with degree
greater than 4 with a collection of gadgets connected so that no degree is greater than four.] 3
1
3 0
0 (a) Gadget for planar 3-colorability. (b) Gadget for degree-4 planar 3-colorability. 4 2 2
4 (c) A careful 5-coloring. 10. Recall that a 5-coloring of a graph G is a function that assigns each vertex of G an ‘color’ from
the set {0, 1, 2, 3, 4}, such that for any edge uv , vertices u and v are assigned different ’colors’.
A 5-coloring is careful if the colors assigned to adjacent vertices are not only distinct, but differ
by more than 1 (mod 5). Prove that deciding whether a given graph has a careful 5-coloring is
NP-complete. [Hint: Reduce from the standard 5COLOR problem.]
11. Prove that the following problems are NP-complete.
(a) Given two undirected graphs G and H , is G isomorphic to a subgraph of H ?
(b) Given an undirected graph G , does G have a spanning tree in which every node has degree
at most 17?
(c) Given an undirected graph G , does G have a spanning tree with at most 42 leaves?
12. The RECTANGLETILING problem asks, given a ‘large’ rectangle R and several ‘small’ rectangles
r1 , r2 , . . . , rn , whether the small rectangles can be placed inside the larger rectangle with no gaps
or overlaps. Prove that RECTANGLETILING is NP-complete.
13. (a) A tonian path in a graph G is a path that goes through at least half of the vertices of G . Show
that determining whether a graph has a tonian path is NP-complete.
(b) A tonian cycle in a graph G is a cycle that goes through at least half of the vertices of G . Show
that determining whether a graph has a tonian cycle is NP-complete. [Hint: Use part (a).]
14. For each problem below, either describe a polynomial-time algorithm or prove that the problem is
NP-complete.
16 Algorithms Lecture 21: NP-Hard Problems (a) A double-Eulerian circuit in an undirected graph G is a closed walk that traverses every edge
in G exactly twice. Given a graph G , does G have a double-Eulerian circuit?
(b) A double-Hamiltonian circuit in an undirected graph G is a closed walk that visits every vertex
in G exactly twice. Given a graph G , does G have a double-Hamiltonian circuit?
15. A boolean formula in exclusive-or conjunctive normal form (XCNF) is a conjunction (AND) of several
clauses, each of which is the exclusive-or of several literals; that is, a clause is true if and only if
it contains an odd number of true literals. The XCNF-SAT problem asks whether a given XCNF
formula is satisﬁable. Either describe a polynomial-time algorithm for XCNF-SAT or prove that it is
NP-hard.
16. Let G be an undirected graph with weighted edges. A heavy Hamiltonian cycle is a cycle C that
passes through each vertex of G exactly once, such that the total weight of the edges in C is at
least half of the total weight of all edges in G . Prove that deciding whether a graph has a heavy
Hamiltonian cycle is NP-complete.
5 8
1 6 2 4
5 12 7 8 3 9 5 A heavy Hamiltonian cycle. The cycle has total weight 34; the graph has total weight 67. 17. Pebbling is a solitaire game played on an undirected graph G , where each vertex has zero or more
pebbles. A single pebbling move consists of removing two pebbles from a vertex v and adding one
pebble to an arbitrary neighbor of v . (Obviously, the vertex v must have at least two pebbles
before the move.) The PEBBLEDESTRUCTION problem asks, given a graph G = (V, E ) and a pebble
count p( v ) for each vertex v , whether is there a sequence of pebbling moves that removes all but
one pebble. Prove that PEBBLEDESTRUCTION is NP-complete.
18. The next time you are at a party, one of the guests will suggest everyone play a round of Three-Way
Mumbletypeg, a game of skill and dexterity that requires three teams and a knife. The ofﬁcial
Rules of Three-Way Mumbletypeg (ﬁxed during the Holy Roman Three-Way Mumbletypeg Council
in 1625) require that (1) each team must have at least one person, (2) any two people on the
same team must know each other, and (3) everyone watching the game must be on one of the
three teams. Of course, it will be a really fun party; nobody will want to leave. There will be
several pairs of people at the party who don’t know each other. The host of the party, having
heard thrilling tales of your prowess in all things algorithmic, will hand you a list of which pairs of
party-goers know each other and ask you to choose the teams, while he sharpens the knife.
Either describe and analyze a polynomial time algorithm to determine whether the party-goers
can be split into three legal Three-Way Mumbletypeg teams, or prove that the problem is NP-hard. 17 Algorithms Lecture 21: NP-Hard Problems 19. (a) Suppose you are given a magic black box that can determine in polynomial time, given an
arbitrary weighted graph G , the length of the shortest Hamiltonian cycle in G . Describe and
analyze a polynomial-time algorithm that computes, given an arbitrary weighted graph G ,
the shortest Hamiltonian cycle in G , using this magic black box as a subroutine.
(b) Suppose you are given a magic black box that can determine in polynomial time, given an
arbitrary graph G , the number of vertices in the largest complete subgraph of G . Describe and
analyze a polynomial-time algorithm that computes, given an arbitrary graph G , a complete
subgraph of G of maximum size, using this magic black box as a subroutine.
(c) Suppose you are given a magic black box that can determine in polynomial time, given an
arbitrary weighted graph G , whether G is 3-colorable. Describe and analyze a polynomialtime algorithm that either computes a proper 3-coloring of a given graph or correctly reports
that no such coloring exists, using the magic black box as a subroutine. [Hint: The input to
the magic black box is a graph. Just a graph. Vertices and edges. Nothing else.]
(d) Suppose you are given a magic black box that can determine in polynomial time, given an
arbitrary boolean formula Φ, whether Φ is satisﬁable. Describe and analyze a polynomialtime algorithm that either computes a satisfying assignment for a given boolean formula or
correctly reports that no such assignment exists, using the magic black box as a subroutine.
(e) Suppose you are given a magic black box that can determine in polynomial time, given an
initial Tetris conﬁguration and a ﬁnite sequence of Tetris pieces, whether a perfect player
can play every piece. (This problem is NP-hard.) Describe and analyze a polynomialtime algorithm that computes the shortest Hamiltonian cycle in a given weighted graph in
polynomial time, using this magic black box as a subroutine. [Hint: Use part (a). You do not
need to know anything about Tetris to solve this problem.] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 18 Algorithms Non-Lecture K: Approximation Algorithms
Le mieux est l’ennemi du bien. [The best is the enemy of the good.]
— Voltaire, La Bégueule (1772)
Who shall forbid a wise skepticism, seeing that there is no practical question
on which any thing more than an approximate solution can be had?
— Ralph Waldo Emerson, Representative Men (1850)
Now, distrust of corporations threatens our still-tentative economic recovery;
it turns out greed is bad, after all.
— Paul Krugman, “Greed is Bad”, The New York Times, June 4, 2002. K
K.1 Approximation Algorithms
Load Balancing On the future smash hit reality-TV game show Grunt Work, scheduled to air Thursday nights at 3am
(2am Central) on ESPNπ, the contestants are given a series of utterly pointless tasks to perform. Each
task has a predetermined time limit; for example, “Sharpen this pencil for 17 seconds”, or “Pour pig’s
blood on your head and sing The Star-Spangled Banner for two minutes”, or “Listen to this 75-minute
algorithms lecture”. The directors of the show want you to assign each task to one of the contestants,
so that the last task is completed as early as possible. When your predecessor correctly informed the
directors that their problem is NP-hard, he was immediately ﬁred. “Time is money!” they screamed at
him. “We don’t need perfection. Wake up, dude, this is television!”
Less facetiously, suppose we have a set of n jobs, which we want to assign to m machines. We are
given an array T [1 .. n] of non-negative numbers, where T [ j ] is the running time of job j . We can
describe an assignment by an array A[1 .. n], where A[ j ] = i means that job j is assigned to machine i .
The makespan of an assignment is the maximum time that any machine is busy:
makespan(A) = max
i T [ j]
A[ j ]=i The load balancing problem is to compute the assignment with the smallest possible makespan.
It’s not hard to prove that the load balancing problem is NP-hard by reduction from PARTITION: The
array T [1 .. n] can be evenly partitioned if and only if there is an assignment to two machines with
makespan exactly i T [i ]/2. A slightly more complicated reduction from 3PARTITION implies that the
load balancing problem is strongly NP-hard. If we really need the optimal solution, there is a dynamic
programming algorithm that runs in time O(nM m ), where M is the minimum makespan, but that’s just
horrible.
There is a fairly natural and efﬁcient greedy heuristic for load balancing: consider the jobs one at a
time, and assign each job to the machine with the earliest ﬁnishing time.
GREEDYLOADBALANCE( T [1 .. n], m):
for i ← 1 to m
Total[i ] ← 0
for j ← 1 to n
min ← 1
for i ← 2 to n
if Total[i ] < Total[min]
min ← i
A[ j ] ← min
Total[min] ← Total[min] + T [ j ]
return A[1 .. m] 1 Algorithms Non-Lecture K: Approximation Algorithms Theorem 1. The makespan of the assignment computed by GREEDYLOADBALANCE is at most twice the
makespan of the optimal assignment.
Proof: Fix an arbitrary input, and let OP T denote the makespan of its optimal assignment. The
approximation bound follows from two trivial observations. First, the makespan of any assignment (and
therefore of the optimal assignment) is at least the duration of the longest job. Second, the makespan of
any assignment is at least the total duration of all the jobs divided by the number of machines.
OP T ≥ max T [ j ] and j OP T ≥ 1
m n T [ j]
j =1 Now consider the assignment computed by GREEDYLOADBALANCE. Suppose machine i has the largest
total running time, and let j be the last job assigned to job i . Our ﬁrst trivial observation implies that
T [ j ] ≤ OP T . To ﬁnish the proof, we must show that Total[i ] − T [ j ] ≤ OP T . Job j was assigned to
machine i because it had the smallest ﬁnishing time, so Total[i ] − T [ j ] ≤ Total[k] for all k. (Some
values Total[k] may have increased since job j was assigned, but that only helps us.) In particular,
Total[i ] − T [ j ] is less than or equal to the average ﬁnishing time over all machines. Thus,
Total[i ] − T [ j ] ≤ 1
m m Total[i ] =
i =1 1
m n T [ j ] ≤ OP T
j =1 by our second trivial observation. We conclude that the makespan Total[i ] is at most 2 · OP T . makespan j ≤ OPT ≤ OPT
i Proof that GreedyLoadBalance is a 2-approximation algorithm GREEDYLOADBALANCE is an online algorithm: It assigns jobs to machines in the order that the jobs
appear in the input array. Online approximation algorithms are useful in settings where inputs arrive
in a stream of unknown length—for example, real jobs arriving at a real scheduling algorithm. In this
online setting, it may be impossible to compute an optimum solution, even in cases where the ofﬂine
problem (where all inputs are known in advance) can be solved in polynomial time. The study of online
algorithms could easily ﬁll an entire one-semester course (alas, not this one).
In our original ofﬂine setting, we can improve the approximation factor by sorting the jobs before
piping them through the greedy algorithm.
SORTEDGREEDYLOADBALANCE( T [1 .. n], m):
sort T in decreasing order
return GREEDYLOADBALANCE( T, m) Theorem 2. The makespan of the assignment computed by SORTEDGREEDYLOADBALANCE is at most 3/2
times the makespan of the optimal assignment. 2 Algorithms Non-Lecture K: Approximation Algorithms Proof: Let i be the busiest machine in the schedule computed by SORTEDGREEDYLOADBALANCE. If only
one job is assigned to machine i , then the greedy schedule is actually optimal, and the theorem is
trivially true. Otherwise, let j be the last job assigned to machine i . Since each of the ﬁrst m jobs
is assigned to a unique processor, we must have j ≥ m + 1. As in the previous proof, we know that
Total[i ] − T [ j ] ≤ OP T .
In the optimal assignment, at least two of the ﬁrst m + 1 jobs, say jobs k and , must be scheduled
on the same processor. Thus, T [k] + T [ ] ≤ OP T . Since max{k, } ≤ m + 1 ≤ j , and the jobs are sorted
in decreasing order by direction, we have
T [ j ] ≤ T [m + 1] ≤ T [max{k, }] = min { T [k], T [ ]} ≤ OP T /2.
We conclude that the makespan Total[i ] is at most 3 · OP T /2, as claimed. K.2 Generalities Consider an arbitrary optimization problem. Let OP T (X ) denote the value of the optimal solution for a
given input X , and let A(X ) denote the value of the solution computed by algorithm A given the same
input X . We say that A is an α(n )-approximation algorithm if and only if
OP T (X )
A(X ) ≤ α(n) and A(X )
OP T (X ) ≤ α(n) for all inputs X of size n. The function α(n) is called the approximation factor for algorithm A. For
any given algorithm, only one of these two inequalities will be important. For maximization problems,
where we want to compute a solution whose cost is as small as possible, the ﬁrst inequality is trivial.
For maximization problems, where we want a solution whose value is as large as possible, the second
inequality is trivial. A 1-approximation algorithm always returns the exact optimal solution.
Especially for problems where exact optimization is NP-hard, we have little hope of completely
characterizing the optimal solution. The secret to proving that an algorithm satisﬁes some approximation
ratio is to ﬁnd a useful function of the input that provides both lower bounds on the cost of the optimal
solution and upper bounds on the cost of the approximate solution. For example, if OP T (X ) ≥ f (X )/2
and A(X ) ≤ 5 f (X ), for any function f , then A is a 10-approximation algorithm. Finding the right
intermediate function can be a delicate balancing act. K.3 Greedy Vertex Cover Recall that the vertex color problem asks, given a graph G , for the smallest set of vertices of G that cover
every edge. This is one of the ﬁrst NP-hard problems introduced in the ﬁrst week of class. There is a
natural and efﬁcient greedy heuristic1 for computing a small vertex cover: mark the vertex with the
largest degree, remove all the edges incident to that vertex, and recurse.
GREEDYVERTEXCOVER(G ):
C ←∅
while G has at least one edge
v ← vertex in G with maximum degree
G←G\v
C ←C∪v
return C
1 A heuristic is an algorithm that doesn’t work. 3 Algorithms Non-Lecture K: Approximation Algorithms Obviously this algorithm doesn’t compute the optimal vertex cover—that would imply P=NP!—but
it does compute a reasonably close approximation.
Theorem 3. GREEDYVERTEXCOVER is an O(log n)-approximation algorithm.
Proof: For all i , let Gi denote the graph G after i iterations of the main loop, and let di denote the
maximum degree of any node in Gi −1 . We can deﬁne these variables more directly by adding a few extra
lines to our algorithm:
GREEDYVERTEXCOVER(G ):
C ←∅
G0 ← G
i←0
while Gi has at least one edge
i ← i+1
vi ← vertex in Gi −1 with maximum degree
di ← degGi−1 ( vi )
Gi ← Gi −1 \ vi
C ← C ∪ vi
return C Let |Gi −1 | denote the number of edges in the graph Gi −1 . Let C ∗ denote the optimal vertex cover of
G , which consists of OP T vertices. Since C ∗ is also a vertex cover for Gi −1 , we have
v ∈C ∗ degGi−1 ( v ) ≥ |Gi −1 |. In other words, the average degree in Gi of any node in C ∗ is at least |Gi −1 |/OP T . It follows that Gi −1
has at least one node with degree at least |Gi −1 |/OP T . Since di is the maximum degree of any node
in Gi −1 , we have
| G i −1 |
di ≥
OP T
Moreover, for any j ≥ i − 1, the subgraph G j has no more edges than Gi −1 , so di ≥ |G j |/OP T . This
observation implies that
OP T OP T di ≥
i =1 i =1 | G i −1 |
OP T OP T ≥
i =1 |GOP T |
OP T OP T = |GOP T | = |G | − di .
i =1 In other words, the ﬁrst OP T iterations of GREEDYVERTEXCOVER remove at least half the edges of G . Thus,
after at most OP T lg|G | ≤ 2 OP T lg n iterations, all the edges of G have been removed, and the algorithm
terminates. We conclude that GREEDYVERTEXCOVER computes a vertex cover of size O(OP T log n).
So far we’ve only proved an upper bound on the approximation factor of GREEDYVERTEXCOVER;
perhaps a more careful analysis would imply that the approximation factor is only O(log log n), or even
O(1). Alas, no such improvement is possible. For any integer n, a simple recursive construction gives us
an n-vertex graph for which the greedy algorithm returns a vertex cover of size Ω(OP T · log n). Details
are left as an exercise for the reader. 4 Algorithms K.4 Non-Lecture K: Approximation Algorithms Set Cover and Hitting Set The greedy algorithm for vertex cover can be applied almost immediately to two more general problems:
set cover and hitting set. The input for both of these problems is a set system (X , F), where X is a ﬁnite
ground set, and F is a family of subsets of X .2 A set cover of a set system (X , F) is a subfamily of sets in
F whose union is the entire ground set X . A hitting set for (X , F) is a subset of the ground set X that
intersects every set in F.
An undirected graph can be cast as a set system in two different ways. In one formulation, the
ground set X contains the vertices, and each edge deﬁnes a set of two vertices in F. In this formulation,
a vertex cover is a hitting set. In the other formulation, the edges are the ground set, the vertices deﬁne
the family of subsets, and a vertex cover is a set cover.
Here are the natural greedy algorithms for ﬁnding a small set cover and ﬁnding a small hitting set.
GREEDYSETCOVER ﬁnds a set cover whose size is at most O(log |F|) times the size of smallest set cover.
GREEDYHITTINGSET ﬁnds a hitting set whose size is at most O(log |X |) times the size of the smallest
hitting set.
GREEDYSETCOVER(X , F):
C←∅
while X = ∅
S ← arg max |S ∩ X | GREEDYHITTINGSET(X , F):
H ←∅
while F = ∅
x ← arg max |{S ∈ F | x ∈ S }| S ∈F x ∈X F ← F \ {S ∈ F | x ∈ S }
H ← H ∪ {x}
return H X ← X \S
C ← C ∪ {S }
return C The similarity between these two algorithms is no coincidence. For any set system (X , F), there is a
dual set system (F, X ∗ ) deﬁned as follows. For any element x ∈ X in the ground set, let x ∗ denote the
subfamily of sets in F that contain x :
x ∗ = {S ∈ F | x ∈ S } .
Finally, let X ∗ denote the collection of all subsets of the form x ∗ :
X ∗ = x∗ x ∈S . As an example, suppose X is the set of letters of alphabet and F is the set of last names of student taking
CS 573 this semester. Then X ∗ has 26 elements, each containing the subset of CS 573 students whose
last name contains a particular letter of the alphabet. For example, m∗ is the set of students whose last
names contain the letter m.
There is a direct one-to-one correspondence between the ground set X and the dual set family X ∗ . It
is a tedious but instructive exercise to prove that the dual of the dual of any set system is isomorphic to
the original set system—(X ∗ , F∗ ) is essentially the same as (X , F). It is also easy to prove that a set cover
for any set system (X , F) is also a hitting set for the dual set system (F, X ∗ ), and therefore a hitting set
for any set system (X , F) is isomorphic to a set cover for the dual set system (F, X ∗ ). K.5 Vertex Cover, Again The greedy approach doesn’t always lead to the best approximation algorithms. Consider the following
alternate heuristic for vertex cover:
2 A matroid (see the lecture note on greedy algorithms) is a special type of set system. 5 Algorithms Non-Lecture K: Approximation Algorithms
DUMBVERTEXCOVER(G ):
C ←∅
while G has at least one edge
(u, v ) ← any edge in G
G ← G \ {u, v }
C ← C ∪ {u, v }
return C The minimum vertex cover—in fact, every vertex cover—contains at least one of the two vertices u
and v chosen inside the while loop. It follows immediately that DUMBVERTEXCOVER is a 2-approximation
algorithm!
The same idea can be extended to approximate the minimum hitting set for any set system (X , F),
where the approximation factor is the size of the largest set in F. K.6 Traveling Salesman: The Bad News The traveling salesman problem3 problem asks for the shortest Hamiltonian cycle in a weighted undirected
graph. To keep the problem simple, we can assume without loss of generality that the underlying graph
is always the complete graph Kn for some integer n; thus, the input to the traveling salesman problem is
n
just a list of the 2 edge lengths.
Not surprisingly, given its similarity to the Hamiltonian cycle problem, it’s quite easy to prove that
the traveling salesman problem is NP-hard. Let G be an arbitrary undirected graph with n vertices. We
can construct a length function for Kn as follows:
(e) = 1 if e is an edge in G , 2 otherwise. Now it should be obvious that if G has a Hamiltonian cycle, then there is a Hamiltonian cycle in Kn whose
length is exactly n; otherwise every Hamiltonian cycle in Kn has length at least n + 1. We can clearly
compute the lengths in polynomial time, so we have a polynomial time reduction from Hamiltonian
cycle to traveling salesman. Thus, the traveling salesman problem is NP-hard, even if all the edge lengths
are 1 and 2.
There’s nothing special about the values 1 and 2 in this reduction; we can replace them with any
values we like. By choosing values that are sufﬁciently far apart, we can show that even approximating
the shortest traveling salesman tour is NP-hard. For example, suppose we set the length of the ‘absent’
edges to n + 1 instead of 2. Then the shortest traveling salesman tour in the resulting weighted graph
either has length exactly n (if G has a Hamiltonian cycle) or has length at least 2n (if G does not have a
Hamiltonian cycle). Thus, if we could approximate the shortest traveling salesman tour within a factor
of 2 in polynomial time, we would have a polynomial-time algorithm for the Hamiltonian cycle problem.
Pushing this idea to its limits us the following negative result.
Theorem 4. For any function f (n) that can be computed in time polynomial in n, there is no polynomialtime f (n)-approximation algorithm for the traveling salesman problem on general weighted graphs,
unless P=NP.
3 This is sometimes bowdlerized into the traveling salesperson problem. That’s just silly. Who ever heard of a traveling
salesperson sleeping with the farmer’s child? 6 Algorithms K.7 Non-Lecture K: Approximation Algorithms Traveling Salesman: The Good News Even though the general traveling salesman problem can’t be approximated, a common special case can
be approximated fairly easily. The special case requires the edge lengths to satisfy the so-called triangle
inequality:
(u, w ) ≤ (u, v ) + ( v, w ) for any vertices u, v, w.
This inequality is satisﬁed for geometric graphs, where the vertices are points in the plane (or some
higher-dimensional space), edges are straight line segments, and lengths are measured in the usual
Euclidean metric. Notice that the length functions we used above to show that the general TSP is hard
to approximate do not (always) satisfy the triangle inequality.
With the triangle inequality in place, we can quickly compute a 2-approximation for the traveling
salesman tour as follows. First, we compute the minimum spanning tree T of the weighted input graph;
this can be done in O(n2 log n) time (where n is the number of vertices of the graph) using any of several
classical algorithms. Second, we perform a depth-ﬁrst traversal of T , numbering the vertices in the order
that we ﬁrst encounter them. Because T is a spanning tree, every vertex is numbered. Finally, we return
the cycle obtained by visiting the vertices according to this numbering.
6
7 6
5 7 1
2 1
4 3 5 2 3
4 A minimum spanning tree T , a depth-ﬁrst traversal of T , and the resulting approximate traveling salesman tour. Let OP T denote the cost of the optimal TSP tour, let M S T denote the total length of the minimum
spanning tree, and let A be the length of the tour computed by our approximation algorithm. Consider
the ‘tour’ obtained by walking through the minimum spanning tree in depth-ﬁrst order. Since this tour
traverses every edge in the tree exactly twice, its length is 2 · M S T . The ﬁnal tour can be obtained from
this one by removing duplicate vertices, moving directly from each node to the next unvisited node.; the
triangle inequality implies that taking these shortcuts cannot make the tour longer. Thus, A ≤ 2 · M S T .
On the other hand, if we remove any edge from the optimal tour, we obtain a spanning tree (in fact a
spanning path) of the graph; thus, M S T ≥ OP T . We conclude that A ≤ 2 · OP T ; our algorithm computes
a 2-approximation of the optimal tour.
We can improve this approximation factor using the following algorithm discovered by Nicos
Christoﬁdes in 1976. As in the previous algorithm, we start by constructing the minimum spanning
tree T . Then let O be the set of vertices with odd degree in T ; it is an easy exercise (hint, hint) to show
that the number of vertices in O is even.
In the next stage of the algorithm, we compute a minimum-cost perfect matching M of these odddegree vertices. A prefect matching is a collection of edges, where each edge has both endpoints in O
and each vertex in O is adjacent to exactly one edge; we want the perfect matching of minimum total
length. Later in the semester, we will see an algorithm to compute M in polynomial time.
Now consider the multigraph T ∪ M ; any edge in both T and M appears twice in this multigraph.
This graph is connected, and every vertex has even degree. Thus, it contains an Eulerian circuit: a
closed walk that uses every edge exactly once. We can compute such a walk in O(n) time with a simple
modiﬁcation of depth-ﬁrst search. To obtain the ﬁnal approximate TSP tour, we number the vertices in
7 Algorithms Non-Lecture K: Approximation Algorithms the order they ﬁrst appear in some Eulerian circuit of T ∪ M , and return the cycle obtained by visiting
the vertices according to that numbering.
7
6 7
5 6 1
2 5
1 4 2 3 4
3 A minimum spanning tree T , a minimum-cost perfect matching M of the odd vertices in T ,
an Eulerian circuit of T ∪ M , and the resulting approximate traveling salesman tour. Theorem 5. Given a weighted graph that obeys the triangle inequality, the Christoﬁdes heuristic
computes a (3/2)-approximation of the minimum traveling salesman tour.
Proof: Let A denote the length of the tour computed by the Christoﬁdes heuristic; let OP T denote the
length of the optimal tour; let M S T denote the total length of the minimum spanning tree; let M OM
denote the total length of the minimum odd-vertex matching.
The graph T ∪ M , and therefore any Euler tour of T ∪ M , has total length M S T + M OM . By the
triangle inequality, taking a shortcut past a previously visited vertex can only shorten the tour. Thus,
A ≤ M S T + M OM .
By the triangle inequality, the optimal tour of the odd-degree vertices of T cannot be longer than
OP T . Any cycle passing through of the odd vertices can be partitioned into two perfect matchings, by
alternately coloring the edges of the cycle red and green. One of these two matchings has length at most
OP T /2. On the other hand, both matchings have length at least M OM . Thus, M OM ≤ OP T /2.
Finally, recall our earlier observation that M S T ≤ OP T .
Putting these three inequalities together, we conclude that A ≤ 3 · OP T /2, as claimed. K.8 k -center Clustering The k-center clustering problem is deﬁned as follows. We are given a set P = { p1 , p2 , . . . , pn } of n points
in the plane4 and an integer k. Our goal to ﬁnd a collection of k circles that collectively enclose all the
input points, such that the radius of the largest circle is as large as possible. More formally, we want to
compute a set C = {c1 , c2 , . . . , ck } of k center points, such that the following cost function is minimized:
cost(C ) := max min | pi c j |.
i j Here, | pi c j | denotes the Euclidean distance between input point pi and center point c j . Intuitively, each
input point is assigned to its closest center point; the points assigned to a given center c j comprise a
cluster. The distance from c j to the farthest point in its cluster is the radius of that cluster; the cluster is
contained in a circle of this radius centered at c j . The k-center clustering cost cost(C ) is precisely the
maximum cluster radius.
This problem turns out to be NP-hard, even to approximate within a factor of roughly 1.8. However,
there is a natural greedy strategy, ﬁrst analyzed in 1985 by Teoﬁlo Gonzalez5 , that is guaranteed to
4 The k-center problem can be deﬁned over any metric space, and the approximation analysis in this section holds in any
metric space as well. The analysis in the next section, however, does require that the points come from the Euclidean plane.
5
Teoﬁlo F. Gonzalez. Clustering to minimize the maximum inter-cluster distance. Theoretical Computer Science 38:293-306,
1985. 8 Algorithms Non-Lecture K: Approximation Algorithms produce a clustering whose cost is at most twice optimal. Choose the k center points one at a time,
starting with an arbitrary input point as the ﬁrst center. In each iteration, choose the input point that is
farthest from any earlier center point to be the next center point. The ﬁrst ﬁve iterations of Gonzalez’s k-center clustering algorithm. In the pseudocode below, di denotes the current distance from point pi to its nearest center, and
r j denotes the maximum of all di (or in other words, the cluster radius) after the ﬁrst j centers have
been chosen. The algorithm includes an extra iteration to computer the ﬁnal clustering cost rk .
GONZALEZKCENTER( P, k):
for i ← 1 to n
di ← ∞
c1 ← p1
for j ← 1 to k
rj ← 0
for i ← 1 to n
di ← min{di , | pi c j |}
if r j < di
r j ← d i ; c j +1 ← p i
return {c1 , c2 , . . . , ck } GONZALEZKCENTER clearly runs in O(nk) time. Using more advanced data structures, Tomas Feder
and Daniel Greene6 described an algorithm to compute exactly the same clustering in only O(n log k)
time.
Theorem 6. GONZALEZKCENTER computes a 2-approximation to the optimal k-center clustering.
Proof: Let r ∗ be the optimal k-center clustering radius for some point set P , and let r be the clustering
radius computed by GONZALEZKCENTER. Suppose for purposes of proving a contradiction that r > 2 r ∗ .
Then among the ﬁrst k + 1 center points selected by GONZALEZKCENTER, every pair has distance at least
r ≥ 2 r ∗ . Thus, by the triangle inequality, any ball of radius r ∗ contains at most one of these k + 1 center
points. But this implies that at least k + 1 balls of radius r ∗ are required to cover P , which contradicts
the deﬁnition of r ∗ . K.9 Approximation Schemes With just a little more work, we can compute an arbitrarily close approximation of the optimal kclustering, using a so-called approximation scheme. An approximation scheme accepts both an instance
of the problem and a value > 0 as input, and it computes a (1 + )-approximation of the optimal
6
Tomas Feder* and Daniel H. Greene. Optimal algorithms for approximate clustering. Proc. 20th STOC, 1988. Unlike
Gonzalez’s algorithm, Feder and Greene’s faster algorithm does not work over arbitrary metric spaces; it requires that the
input points come from some IRd and that distances are measured in some L p metric. The time analysis also assumes that the
distance between any two points can be computed in O(1) time. 9 Algorithms Non-Lecture K: Approximation Algorithms output for that instance. As I mentioned earlier, computing even a 1.8-approximation is NP-hard, so
we cannot expect our approximation scheme to run in polynomial time; nevertheless, at least for small
values of k, the approximation scheme will be considerably more efﬁcient than any exact algorithm.
Our approximation scheme works in three phases:
1. Compute a 2-approximate clustering of the input set P using GONZALEZKCENTER. Let r be the cost
of this clustering.
2. Create a regular grid of squares of width δ = r /2 2. Let Q be a subset of P containing one point
from each non-empty cell of this grid.
3. Compute an optimal set of k centers for Q. Return these k centers as the approximate k-center
clustering for P .
The ﬁrst phase requires O(nk) time. By our earlier analysis, we have r ∗ ≤ r ≤ 2 r ∗ , where r ∗ is the
optimal k-center clustering cost for P .
The second phase can be implemented in O(n) time using a hash table, or in O(n log n) time by
standard sorting, by associating approximate coordinates ( x /δ , y /δ ) to each point ( x , y ) ∈ P and
removing duplicates. The key observation is that the resulting point set Q is signiﬁcantly smaller than P .
We know P can be covered by k balls of radius r ∗ , each of which touches O( r ∗ /δ2 ) = O(1/ 2 ) grid cells.
It follows that |Q| = O(k/ 2 ).
Let T (n, k) be the running time of an exact k-center clustering algorithm, given n points as input.
If this were a computational geometry class, we might see a “brute force” algorithm that runs in time
T (n, k) = O(nk+2 ); the fastest algorithm currently known7 runs in time T (n, k) = nO( k) . If we use this
algorithm, our third phase requires (k/ 2 )O( k) time.
It remains to show that the optimal clustering for Q implies a (1 + )-approximation of the optimal
clustering for P . Suppose the optimal clustering of Q consists of k balls B1 , B2 , . . . , Bk , each of radius
˜. Clearly ˜ ≤ r ∗ , since any set of k balls that cover P also cover any subset of P . Each point in P \ Q
r
r
shares a grid cell with some point in Q, and therefore is within distance δ 2 of some point in Q.
Thus, if we increase the radius of each ball Bi by δ 2, the expanded balls must contain every point
in P . We conclude that the optimal centers for Q gives us a k-center clustering for P of cost at most
r ∗ + δ 2 ≤ r ∗ + r /2 ≤ r ∗ + r ∗ = (1 + ) r ∗ .
The total running time of the approximation scheme is O(nk + (k/ 2 )O( k) ). This is still exponential
in the input size if k is large (say n or n/100), but if k and are ﬁxed constants, the running time is
linear in the number of input points. K.10 An FPTAS for Subset Sum An approximation scheme whose running time, for any ﬁxed , is polynomial in n is called a polynomialtime approximation scheme or PTAS (usually pronounced “pee taz"). If in addition the running time
depends only polynomially on , the algorithm is called a fully polynomial-time approximation scheme
or FPTAS (usually pronounced “eff pee taz"). For example, an approximation scheme with running time
6
O(n2 / 2 ) is an FPTAS; an approximation scheme with running time O(n1/ ) is a PTAS but not an FPTAS;
and our approximation scheme for k-center clustering is not a PTAS.
The last problem we’ll consider is the SUBSETSUM problem: Given a set X containing n positive
integers and a target integer t , determine whether X has a subset whose elements sum to t . The lecture
notes on NP-completeness include a proof that SUBSETSUM is NP-hard. As stated, this problem doesn’t
7 R. Z. Hwang, R. C. T. Lee, and R. C. Chan. The slab dividing approach to solve the Euclidean p-center problem. Algorithmica
9(1):1–22, 1993. 10 Algorithms Non-Lecture K: Approximation Algorithms allow any sort of approximation—the answer is either TRUE or FALSE.8 So we will consider a related
optimization problem instead: Given set X and integer t , ﬁnd the subset of X whose sum is as large as
possible but no larger than t .
We have already seen a dynamic programming algorithm to solve the decision version SUBSETSUM
in time O(nt ); a similar algorithm solves the optimization version in the same time bound. Here is a
different algorithm, whose running time does not depend on t :
SUBSETSUM(X [1 .. n], t ):
S0 ← {0}
for i ← 1 to n
Si ← Si −1 ∪ (Si −1 + X [i ])
remove all elements of Si bigger than t
return max Sn Here Si −1 + X [i ] denotes the set {s + X [i ] | s ∈ Si −1 }. If we store each Si in a sorted array, the i th
iteration of the for-loop requires time O(|Si −1 |). Each set Si contains all possible subset sums for the
ﬁrst i elements of X ; thus, Si has at most 2i elements. On the other hand, since every element of Si
is an integer between 0 and t , we also have |Si | ≤ t + 1. It follows that the total running time of this
n
algorithm is i =1 O(|Si −1 |) = O(min{2n , nt }).
Of course, this is only an estimate of worst-case behavior. If several subsets of X have the same
sum, the sets Si will have fewer elements, and the algorithm will be faster. The key idea for ﬁnding an
approximate solution quickly is to ‘merge’ nearby elements of Si —if two subset sums are nearly equal,
ignore one of them. On the one hand, merging similar subset sums will introduce some error into the
output, but hopefully not too much. On the other hand, by reducing the size of the set of sums we need
to maintain, we will make the algorithm faster, hopefully signiﬁcantly so.
Here is our approximation algorithm. We make only two changes to the exact algorithm: an initial
sorting phase and an extra FILTERing step inside the main loop.
FILTER( Z [1 .. k], δ):
SORT( Z )
j←1
Y [ j ] ← Z [i ]
for i ← 2 to k
if Z [i ] > (1 + δ) · Y [ j ]
j ← j+1
Y [ j ] ← Z [i ]
return Y [1 .. j ] APPROXSUBSETSUM(X [1 .. n], k, ):
SORT(X )
R0 ← {0}
for i ← 1 to n
R i ← R i −1 ∪ (R i −1 + X [i ])
Ri ← FILTER(Ri , /2n )
remove all elements of R i bigger than t
return max R n Theorem 7. APPROXSUBSETSUM returns a (1 + )-approximation of the optimal subset sum, given any
such that 0 < ≤ 1.
Proof: The theorem follows from the following claim, which we prove by induction:
For any element s ∈ Si , there is an element r ∈ R i such that r ≤ s ≤ r · (1 + n/2)i .
The claim is trivial for i = 0. Let s be an arbitrary element of Si , for some i > 0. There are two cases to
consider: either x ∈ Si −1 , or x ∈ Si −1 + x i .
8 Do, or do not. There is no ‘try’. (Are old one thousand when years you, alphabetical also in order talk will you.) 11 Algorithms Non-Lecture K: Approximation Algorithms (1) Suppose s ∈ Si −1 . By the inductive hypothesis, there is an element r ∈ R i −1 such that r ≤ s ≤
r · (1 + n/2)i −1 . If r ∈ R i , the claim obviously holds. On the other hand, if r ∈ R i , there must
be an element r ∈ R i such that r < r ≤ r (1 + n/2), which implies that
r < r ≤ s ≤ r · (1 + n/2)i −1 ≤ r · (1 + n/2)i ,
so the claim holds.
(2) Suppose s ∈ Si −1 + x i . By the inductive hypothesis, there is an element r ∈ R i −1 such that
r ≤ s − x i ≤ r · (1 + n/2)i −1 . If r + x i ∈ R i , the claim obviously holds. On the other hand, if
r + x i ∈ R i , there must be an element r ∈ R i such that r < r + x i ≤ r (1 + n/2), which implies
that
r < r + x i ≤ s ≤ r · (1 + n/2)i −1 + x i
≤ ( r − x i ) · (1 + n/2)i + x i
≤ r · (1 + n/2)i − x i · ((1 + n/2)i − 1)
≤ r · (1 + n/2)i .
so the claim holds.
Now let s∗ = max Sn and r ∗ = max R n . Clearly r ∗ ≤ s∗ , since R n ⊆ Sn . Our claim implies that there
is some r ∈ R n such that s∗ ≤ r · (1 + /2n)n . But r cannot be bigger than r ∗ , so s∗ ≤ r ∗ · (1 + /2n)n .
The inequalities e x ≥ 1 + x for all x , and e x ≤ 2 x + 1 for all 0 ≤ x ≤ 1, imply that (1 + /2n)n ≤ e /2 ≤
1+ .
Theorem 8. APPROXSUBSETSUM runs in O((n3 log n)/ ) time.
Proof: Assuming we keep each set R i in a sorted array, we can merge the two sorted arrays R i −1 and
R i −1 + x i in O(|R i −1 |) time. FILTERin R i and removing elements larger than t also requires only O(|R i −1 |)
time. Thus, the overall running time of our algorithm is O( i |R i |); to express this in terms of n and ,
we need to prove an upper bound on the size of each set R i .
Let δ = /2n. Because we consider the elements of X in increasing order, every element of R i is
between 0 and i · x i . In particular, every element of R i −1 + x i is between x i and i · x i . After FILTERing, at
most one element r ∈ R i lies in the range (1 + δ)k ≤ r < (1 + δ)k+1 , for any k. Thus, at most log1+δ i
elements of R i −1 + x i survive the call to FILTER. It follows that
|R i | = |R i −1 | +
≤ |R i −1 | +
≤ |R i −1 | +
≤ |R i −1 | + log i
log(1 + δ)
log n [ i ≤ n] log(1 + δ)
2 ln n [e x ≤ 1 + 2 x for all 0 ≤ x ≤ 1] δ
n ln n [δ = /2n] Unrolling this recurrence into a summation gives us the upper bound |R i | ≤ i · (n ln n)/ = O((n2 log n)/ ).
We conclude that the overall running time of APPROXSUBSETSUM is O((n3 log n)/ ), as claimed. 12 Algorithms Non-Lecture K: Approximation Algorithms Exercises
1. (a) Prove that for any set of jobs, the makespan of the greedy assignment is at most (2 − 1/m)
times the makespan of the optimal assignment, where m is the number of machines.
(b) Describe a set of jobs such that the makespan of the greedy assignment is exactly (2 − 1/m)
times the makespan of the optimal assignment, where m is the number of machines.
(c) Describe an efﬁcient algorithm to solve the minimum makespan scheduling problem exactly
if every processing time T [i ] is a power of two.
2. (a) Find the smallest graph (minimum number of edges) for which GREEDYVERTEXCOVER does
not return the smallest vertex cover.
(b) For any integer n, describe an n-vertex graph for which GREEDYVERTEXCOVER returns a vertex
cover of size OP T · Ω(log n).
3. (a) Find the smallest graph (minimum number of edges) for which DUMBVERTEXCOVER does not
return the smallest vertex cover.
(b) Describe an inﬁnite family of graphs for which DUMBVERTEXCOVER returns a vertex cover of
size 2 · OP T .
4. Consider the following heuristic for constructing a vertex cover of a connected graph G : return
the set of non-leaf nodes in any depth-ﬁrst spanning tree of G .
(a) Prove that this heuristic returns a vertex cover of G .
(b) Prove that this heuristic returns a 2-approximation to the minimum vertex cover of G .
(c) Describe an inﬁnite family of graphs for which this heuristic returns a vertex cover of size
2 · OP T .
5. Consider the following optimization version of the PARTITION problem. Given a set X of positive
integers, our task is to partition X into disjoint subsets A and B such that max{ A, B } is as
small as possible. This problem is clearly NP-hard. Determine the approximation ratio of the
following polynomial-time approximation algorithm. Prove your answer is correct.
PARTITION(X [1 .. n]):
Sort X in increasing order
a ← 0; b ← 0
for i ← 1 to n
if a < b
a ← a + X [i ]
else
b ← b + X [i ]
return max{a, b} 6. The chromatic number χ (G ) of a graph G is the minimum number of colors required to color
the vertices of the graph, so that every edge has endpoints with different colors. Computing the
chromatic number exactly is NP-hard.
13 Algorithms Non-Lecture K: Approximation Algorithms Prove that the following problem is also NP-hard: Given an n-vertex graph G , return any
integer between χ (G ) and χ (G ) + 573. [Note: This does not contradict the possibility of a constant
factor approximation algorithm.]
7. Let G = (V, E ) be an undirected graph, each of whose vertices is colored either red, green, or blue.
An edge in G is boring if its endpoints have the same color, and interesting if its endpoints have
different colors. The most interesting 3-coloring is the 3-coloring with the maximum number of
interesting edges, or equivalently, with the fewest boring edges. Computing the most interesting
3-coloring is NP-hard, because the standard 3-coloring problem is a special case.
(a) Let zzz(G ) denote the number of boring edges in the most interesting 3-coloring of a graph G .
100
Prove that it is NP-hard to approximate zzz(G ) within a factor of 1010 .
(b) Let wow(G ) denote the number of interesting edges in the most interesting 3-coloring of G .
Suppose we assign each vertex in G a random color from the set {red, green, blue}. Prove
that the expected number of interesting edges is at least 2 wow(G ).
3
8. Consider the following algorithm for coloring a graph G .
TREECOLOR(G ):
T ← any spanning tree of G
Color the tree T with two colors
c←2
for each edge (u, v ) ∈ G \ T
T ← T ∪ {(u, v )}
if color(u) = color( v ) 〈〈Try recoloring u with an existing color 〉〉
for i ← 1 to c
if no neighbor of u in T has color i
color(u) ← i
if color(u) = color( v ) 〈〈Try recoloring v with an existing color 〉〉
for i ← 1 to c
if no neighbor of v in T has color i
color( v ) ← i
if color(u) = color( v )
c ← c+1
color(u) ← c 〈〈Give up and create a new color 〉〉 (a) Prove that this algorithm colors any bipartite graph with just two colors.
(b) Let ∆(G ) denote the maximum degree of any vertex in G . Prove that this algorithm colors
any graph G with at most ∆(G ) colors. This trivially implies that TREECOLOR is a ∆(G )approximation algorithm.
(c) Prove that TREECOLOR is not a constant-factor approximation algorithm.
9. The KNAPSACK problem can be deﬁned as follows. We are given a ﬁnite set of elements X where
each element x ∈ X has a non-negative size and a non-negative value, along with an integer
capacity c . Our task is to determine the maximum total value among all subsets of X whose total
size is at most c . This problem is NP-hard. Speciﬁcally, the optimization version of SUBSETSUM is a
special case, where each element’s value is equal to its size.
14 Algorithms Non-Lecture K: Approximation Algorithms Determine the approximation ratio of the following polynomial-time approximation algorithm.
Prove your answer is correct.
APPROXKNAPSACK(X , c ):
return max{GREEDYKNAPSACK(X , c ), PICKBESTONE(X , c )}
GREEDYKNAPSACK(X , c ):
Sort X in decreasing order by the ratio value/size
S ← 0; V ← 0
for i ← 1 to n
if S + size( x i ) > c
return V
S ← S + size( x i )
V ← V + value( x i )
return V PICKBESTONE(X , c ):
Sort X in increasing order by size
V ←0
for i ← 1 to n
if size( x i ) > c
return V
if value( x i ) > V
V ← value( x i )
return V 10. In the bin packing problem, we are given a set of n items, each with weight between 0 and 1, and
we are asked to load the items into as few bins as possible, such that the total weight in each bin
is at most 1. It’s not hard to show that this problem is NP-Hard; this question asks you to analyze
a few common approximation algorithms. In each case, the input is an array W [1 .. n] of weights,
and the output is the number of bins used.
FIRSTFIT(W [1 .. n]):
b←0 NEXTFIT(W [1 .. n]):
b←0
Total[0] ← ∞ for i ← 1 to n
j ← 1; f ound ← FALSE
while j ≤ b and f ound = FALSE
if Total[ j ] + W [i ] ≤ 1
Total[ j ] ← Total[ j ] + W [i ]
f ound ← TRUE
j ← j+1 for i ← 1 to n
if Total[ b] + W [i ] > 1
b ← b+1
Total[ b] ← W [i ]
else
Total[ b] ← Total[ b] + W [i ] if f ound = FALSE
b ← b+1
Total[ b] = W [i ] return b return b (a) Prove that NEXTFIT uses at most twice the optimal number of bins.
(b) Prove that FIRSTFIT uses at most twice the optimal number of bins.
(c) Prove that if the weight array W is initially sorted in decreasing order, then FIRSTFIT uses at
most (4 · OP T + 1)/3 bins, where OP T is the optimal number of bins. The following facts
may be useful (but you need to prove them if your proof uses them):
• In the packing computed by FIRSTFIT, every item with weight more than 1/3 is placed in
one of the ﬁrst OP T bins.
• FIRSTFIT places at most OP T − 1 items outside the ﬁrst OP T bins.
11. Given a graph G with edge weights and an integer k, suppose we wish to partition the the vertices
of G into k subsets S1 , S2 , . . . , Sk so that the sum of the weights of the edges that cross the partition
(that is, have endpoints in different subsets) is as large as possible.
15 Algorithms Non-Lecture K: Approximation Algorithms (a) Describe an efﬁcient (1 − 1/k)-approximation algorithm for this problem.
(b) Now suppose we wish to minimize the sum of the weights of edges that do not cross the
partition. What approximation ratio does your algorithm from part (a) achieve for the new
problem? Justify your answer.
12. The lecture notes describe a (3/2)-approximation algorithm for the metric traveling salesman
problem. Here, we consider computing minimum-cost Hamiltonian paths. Our input consists of
a graph G whose edges have weights that satisfy the triangle inequality. Depending upon the
problem, we are also given zero, one, or two endpoints.
(a) If our input includes zero endpoints, describe a (3/2)-approximation to the problem of
computing a minimum cost Hamiltonian path.
(b) If our input includes one endpoint u, describe a (3/2)-approximation to the problem of
computing a minimum cost Hamiltonian path that starts at u.
(c) If our input includes two endpoints u and v , describe a (5/3)-approximation to the problem
of computing a minimum cost Hamiltonian path that starts at u and ends at v .
13. Suppose we are given a collection of n jobs to execute on a machine containing a row of m
processors. When the i th job is executed, it occupies a contiguous set of prox[i ] processors for
time[i ] seconds. A schedule for a set of jobs assigns each job an interval of processors and a starting
time, so that no processor works on more than one job at any time. The makespan of a schedule is
the time from the start to the ﬁnish of all jobs. Finally, the parallel scheduling problem asks us to
compute the schedule with the smallest possible makespan.
(a) Prove that the parallel scheduling problem is NP-hard.
(b) Give an algorithm that computes a 3-approximation of the minimum makespan of a set of
jobs in O(m log m) time. That is, if the minimum makespan is M , your algorithm should
compute a schedule with make-span at most 3 M . You can assume that n is a power of 2.
14. Consider the greedy algorithm for metric TSP: start at an arbitrary vertex u, and at each step,
travel to the closest unvisited vertex.
(a) Show that the greedy algorithm for metric TSP is an O(log n)-approximation, where n is the
number of vertices. [Hint: Argue that the kth least expensive edge in the tour output by the
greedy algorithm has weight at most OPT/(n − k + 1); try k = 1 and k = 2 ﬁrst.]
(b) Show that the greedy algorithm for metric TSP is no better than an O(log n)-approximation.
That is, describe an inﬁnite family of weighted graphs such that the greedy algorithm returns
a cycle whose weight is Ω(log n) times the optimal TSP tour. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 16 Algorithms Appendix: Solving Recurrences
“. . . O Zarathustra, who you are and must become” behold you are the
teacher of the eternal recurrence – that is your destiny! That you as the
ﬁrst must teach this doctrine – how could this great destiny not be your
greatest danger and sickness too?
— Friedrich Nietzsche, Also sprach Zarathustra (1885)
[translated by Walter Kaufmann] Solving Recurrences
1 Introduction A recurrence is a recursive description of a function, usually of the form F : IN → IR, or a description
of such a function in terms of itself. Like all recursive structures, a recurrence consists of one or more
base cases and one or more recursive cases. Each of these cases is an equation or inequality, with some
function value f (n) on the left side. The base cases give explicit values for a (typically ﬁnite, typically
small) subset of the possible values of n. The recursive cases relate the function value f (n) to function
value f (k) for one or more integers k < n; typically, each recursive case applies to an inﬁnite number of
possible values of n.
For example, the following recurrence (written in two different but standard ways) describes the
identity function f (n) = n:
f ( n) = 0 if n = 0 f (0) = 0 f (n − 1) + 1 otherwise f (n) = f (n − 1) + 1 for all n > 0 In both presentations, the ﬁrst line is the only base case, and the second line is the only recursive case.
The same function can satisfy many different recurrences; for example, both of the following recurrences
also describe the identity function: 0
if n = 0
if n = 0 0
f ( n) = f ( n) = 1
if n = 1 f ( n/2 ) + f ( n/2 ) otherwise 2 · f ( n/ 2) f ( n − 1) + 1 if n is even and n > 0 if n is odd We say that a particular function satisﬁes a recurrence, or is the solution to a recurrence, if each
of the statements in the recurrence is true. Most recurrences—at least, those that we will encounter
in this class—have a solution; moreover, if every case of the recurrence is an equation, that solution is
unique. Speciﬁcally, if we transform the recursive formula into a recursive algorithm, the solution to the
recurrence is the function computed by that algorithm!
Recurrences arise naturally in the analysis of algorithms, especially recursive algorithms. In many
cases, we can express the running time of an algorithm as a recurrence, where the recursive cases of the
recurrence correspond exactly to the recursive cases of the algorithm. Recurrences are also useful tools
for solving counting problems—How many objects of a particular kind exist?
By itself, a recurrence is not a satisfying description of the running time of an algorithm or a bound on
the number of widgets. Instead, we need a closed-form solution to the recurrence; this is a non-recursive
description of a function that satisﬁes the recurrence. For recurrence equations, we sometimes prefer
an exact closed-form solution, but such a solution may not exist, or may be too complex to be useful.
Thus, for most recurrences, especially those arising in algorithm analysis, we can be satisﬁed with an
asymptotic solution of the form Θ( f (n)), for some explicit (non-recursive) function g (n).
1 Algorithms Appendix: Solving Recurrences For recursive inequalities, we prefer a tight solution; this is a function that would still satisfy the
recurrence if all the inequalities were replaced with the corresponding equations. Again, exactly tight
solutions may not exist, or may be too complex to be useful, so we may have to settle for a looser
solution and/or an asymptotic solution of the form O( g (n)) or Ω( g (n)). 2 The Ultimate Method: Guess and Conﬁrm Ultimately, there is only one fail-safe method to solve any recurrence: Guess the answer, and then prove it correct by induction.
Later sections of these notes describe techniques to generate guesses that are guaranteed to be correct,
provided you use them correctly. But if you’re faced with a recurrence that doesn’t seem to ﬁt any of these
methods, or if you’ve forgotten how those techniques work, don’t despair! If you guess a closed-form
solution and then try to verify your guess inductively, usually either the proof will succeed, in which
case you’re done, or the proof will fail, in which case the failure will help you reﬁne your guess. Where
you get your initial guess is utterly irrelevant1 —from a classmate, from a textbook, on the web, from the
answer to a different problem, scrawled on a bathroom wall in Siebel, included in a care package from
your mom, dictated by the machine elves, whatever. If you can prove that the answer is correct, then it’s
correct! 2.1 Tower of Hanoi The classical Tower of Hanoi problem gives us the recurrence T (n ) = 2 T (n − 1) + 1 with base case
T (0) = 0. Just looking at the recurrence we can guess that T (n) is something like 2n . If we write out
the ﬁrst few values of T (n), we discover that they are each one less than a power of two.
T (0) = 0, T (1) = 1, T (2) = 3, T (3) = 7, T (4) = 15, T (5) = 31, T (6) = 63, ..., It looks like T (n ) = 2n − 1 might be the right answer. Let’s check.
T (0) = 0 = 20 − 1
T ( n) = 2 T ( n − 1) + 1
= 2(2n−1 − 1) + 1 [induction hypothesis] n =2 −1 [algebra] We were right! Hooray, we’re done!
Another way we can guess the solution is by unrolling the recurrence, by substituting it into itself:
T ( n) = 2 T ( n − 1) + 1
= 2 (2 T (n − 2) + 1) + 1
= 4 T ( n − 2) + 3
= 4 (2 T (n − 3) + 1) + 3
= 8 T ( n − 2) + 7
= ···
1 . . . except of course during exams, where you aren’t supposed to use any outside sources 2 Algorithms Appendix: Solving Recurrences It looks like unrolling the initial Hanoi recurrence k times, for any non-negative integer k, will give us
the new recurrence T (n) = 2k T (n − k) + (2k − 1). Let’s prove this by induction:
T (n) = 2 T (n − 1) + 1 [k = 0, by deﬁnition] T (n) = 2k−1 T (n − (k − 1)) + (2k−1 − 1)
=2 k −1 2 T (n − k) + 1 + (2 k k −1 [inductive hypothesis] − 1) [initial recurrence for T (n − (k − 1))] k = 2 T (n − k) + (2 − 1) [algebra] Our guess was correct! In particular, unrolling the recurrence n times give us the recurrence T (n) =
2n T (0) + (2n − 1). Plugging in the base case T (0) = 0 give us the closed-form solution T (n) = 2n − 1. 2.2 Fibonacci numbers Let’s try a less trivial example: the Fibonacci numbers Fn = Fn −1 + Fn −2 with base cases F0 = 0 and
F1 = 1. There is no obvious pattern in the ﬁrst several values (aside from the recurrence itself), but we
can reasonably guess that Fn is exponential in n. Let’s try to prove inductively that Fn ≤ α · c n for some
constants a > 0 and c > 1 and see how far we get.
Fn = Fn−1 + Fn−2
≤ α · c n−1 + α · c n−2 [“induction hypothesis”] n ≤ α · c ???
The last inequality is satisﬁed if c n ≥ c n−1 + c n−2 , or more simply, if c 2 − c − 1 ≥ 0. The smallest value
of c that works is φ = (1 + 5)/2 ≈ 1.618034; the other root of the quadratic equation has smaller
absolute value, so we can ignore it.
So we have most of an inductive proof that Fn ≤ α · φ n for some constant α. All that we’re missing
are the base cases, which (we can easily guess) must determine the value of the coefﬁcient a. We quickly
compute
F0
F1
0
1
= = 0 and
= ≈ 0.618034 > 0,
0
1
1
φ
φ
φ
so the base cases of our induction proof are correct as long as α ≥ 1/φ . It follows that Fn ≤ φ n −1 for all
n ≥ 0.
What about a matching lower bound? Essentially the same inductive proof implies that Fn ≥ β · φ n
for some constant β , but the only value of β that works for all n is the trivial β = 0! We could try to
ﬁnd some lower-order term that makes the base case non-trivial, but an easier approach is to recall that
asymptotic Ω( ) bounds only have to work for sufﬁciently large n. So let’s ignore the trivial base case
F0 = 0 and assume that F2 = 1 is a base case instead. Some more easy calculation gives us
F2
φ2 = 1
φ2 ≈ 0.381966 < 1
φ . Thus, the new base cases of our induction proof are correct as long as β ≤ 1/φ 2 , which implies that
Fn ≥ φ n −2 for all n ≥ 1.
Putting the upper and lower bounds together, we obtain the tight asymptotic bound Fn = Θ(φ n ). It
is possible to get a more exact solution by speculatively reﬁning and conforming our current bounds, but
it’s not easy. Fortunately, if we really need it, we can get an exact solution using the annihilator method,
which we’ll see later in these notes.
3 Algorithms 2.3 Appendix: Solving Recurrences Mergesort Mergesort is a classical recursive divide-and-conquer algorithm for sorting an array. The algorithm splits
the array in half, recursively sorts the two halves, and then merges the two sorted subarrays into the
ﬁnal sorted array.
MERGE(A[1 .. n], m):
i ← 1; j ← m + 1
for k ← 1 to n
if j > n
B [k] ← A[i ];
else if i > m
B [k] ← A[ j ];
else if A[i ] < A[ j ]
B [k] ← A[i ];
else
B [k] ← A[ j ]; MERGESORT(A[1 .. n]):
if (n > 1)
m ← n/ 2
MERGESORT(A[1 .. m])
MERGESORT(A[m + 1 .. n])
MERGE(A[1 .. n], m) i ← i+1
j ← j+1
i ← i+1
j ← j+1 for k ← 1 to n
A[k] ← B [k] Let T (n) denote the worst-case running time of MERGESORT when the input array has size n. The
MERGE subroutine clearly runs in Θ(n) time, so the function T (n) satisﬁes the following recurrence:
T ( n) = Θ(1)
T if n = 1, n/ 2 +T n/ 2 + Θ(n) otherwise. For now, let’s consider the special case where n is a power of 2; this assumption allows us to take the
ﬂoors and ceilings out of the recurrence. (We’ll see how to deal with the ﬂoors and ceilings later; the
short version is that they don’t matter.)
Because the recurrence itself is given only asymptotically—in terms of Θ( ) expressions—we can’t
hope for anything but an asymptotic solution. So we can safely simplify the recurrence further by
removing the Θ’s; any asymptotic solution to the simpliﬁed recurrence will also satisfy the original
recurrence. (This simpliﬁcation is actually important for another reason; if we kept the asymptotic
expressions, we might be tempted to simplify them inappropriately.)
Our simpliﬁed recurrence now looks like this:
T ( n) = 1 if n = 1, 2 T (n/2) + n otherwise. To guess at a solution, let’s try unrolling the recurrence.
T (n) = 2 T (n/2) + n
= 2 2 T (n/4) + n/2 + n
= 4 T ( n/ 4) + 2 n
= 8 T ( n/ 8) + 3 n = · · ·
It looks like T (n) satisﬁes the recurrence T (n) = 2k T (n/2k ) + kn for any positive integer k. Let’s verify
this by induction. 4 Algorithms Appendix: Solving Recurrences T (n) = 2 T (n/2) + n = 21 T (n/21 ) + 1 · n [k = 1, given recurrence] T ( n) = 2 k − 1 T ( n/ 2 k −1 ) + ( k − 1) n
=2 k −1 k 2 T ( n/ 2 ) + n/ 2 k −1 [inductive hypothesis]
+ ( k − 1) n [substitution] = 2k T (n/2k ) + kn [algebra] Our guess was right! The recurrence becomes trivial when n/2k = 1, or equivalently, when k = log2 n:
T (n) = nT (1) + n log2 n = n log2 n + n.
Finally, we have to put back the Θ’s we stripped off; our ﬁnal closed-form solution is T (n ) = Θ(n log n ). 2.4 An uglier divide-and-conquer example Consider the divide-and-conquer recurrence T (n ) = n · T ( n ) + n . This doesn’t ﬁt into the form
required by the Master Theorem (which we’ll see below), but it still sort of resembles the Mergesort
recurrence—the total size of the subproblems at the ﬁrst level of recursion is n—so let’s guess that
T (n) = O(n log n), and then try to prove that our guess is correct. (We could also attack this recurrence
by unrolling, but let’s see how far just guessing will take us.)
Let’s start by trying to prove an upper bound T (n) ≤ a n lg n for all sufﬁciently large n and some
constant a to be determined later:
T ( n) =
≤ n · T ( n) + n
n·a n lg n+n [induction hypothesis] = (a/2)n lg n + n [algebra] ≤ an lg n [algebra] The last inequality assumes only that 1 ≤ (a/2) log n,or equivalently, that n ≥ 22/a . In other words, the
induction proof is correct if n is sufﬁciently large. So we were right!
But before you break out the champagne, what about the multiplicative constant a? The proof worked
for any constant a, no matter how small. This strongly suggests that our upper bound T (n) = O(n log n)
is not tight. Indeed, if we try to prove a matching lower bound T (n) ≥ b n log n for sufﬁciently large n,
we run into trouble.
T ( n) =
≥ n · T ( n) + n
n· b n log n+n [induction hypothesis] = ( b/2)n log n + n
≥ bn log n
The last inequality would be correct only if 1 > ( b/2) log n, but that inequality is false for large values
of n, no matter which constant b we choose.
Okay, so Θ(n log n) is too big. How about Θ(n)? The lower bound is easy to prove directly:
T ( n) = n · T ( n) + n ≥ n But an inductive proof of the upper bound fails.
T ( n) =
≤ n · T ( n) + n
n·a n+n [induction hypothesis] = ( a + 1) n [algebra] ≤ an
5 Algorithms Appendix: Solving Recurrences Hmmm. So what’s bigger than n and smaller than n lg n? How about n
T ( n) = n · T ( n) + n ≤ n·a lg n n+n = ( a / 2) n [induction hypothesis] lg n + n ≤ an lg n? for large enough n lg n [algebra] Okay, the upper bound checks out; how about the lower bound?
T ( n) = n · T ( n) + n ≥ n· b n = ( b / 2) n
≥ bn lg n+n [induction hypothesis] lg n + n [algebra] lg n No, the last step doesn’t work. So Θ(n lg n) doesn’t work.
Okay. . . what else is between n and n lg n? How about n lg lg n?
T ( n) = n · T ( n) + n ≤ n·a n+n n lg lg [induction hypothesis] = a n lg lg n − a n + n
≤ a n lg lg n [algebra] if a ≥ 1 Hey look at that! For once, our upper bound proof requires a constraint on the hidden constant a. This
is an good indication that we’ve found the right answer. Let’s try the lower bound:
T ( n) = n · T ( n) + n ≥ n· b n+n n lg lg [induction hypothesis] = b n lg lg n − b n + n
≥ b n lg lg n [algebra] if b ≤ 1 Hey, it worked! We have most of an inductive proof that T (n) ≤ an lg lg n for any a ≥ 1 and most of
an inductive proof that T (n) ≥ bn lg lg n for any b ≤ 1. Technically, we’re still missing the base cases in
both proofs, but we can be fairly conﬁdent at this point that T (n ) = Θ(n log log n ). 3 Divide and Conquer Recurrences (Recursion Trees) Many divide and conquer algorithms give us running-time recurrences of the form
T ( n) = a T ( n/ b ) + f ( n) (1) where a and b are constants and f (n) is some other function. There is a simple and general technique
for solving many recurrences in this and similar forms, using a recursion tree. The root of the recursion
tree is a box containing the value f (n); the root has a children, each of which is the root of a (recursively
deﬁned) recursion tree for the function T (n/ b).
Equivalently, a recursion tree is a complete a-ary tree where each node at depth i contains the
value f (n/ b i ). The recursion stops when we get to the base case(s) of the recurrence. Because we’re
only looking for asymptotic bounds, the exact base case doesn’t matter; we can safely assume that
T (1) = Θ(1), or even that T (n) = Θ(1) for all n ≤ 10100 . I’ll also assume for simplicity that n is an
integral power of b; we’ll see how to avoid this assumption later (but to summarize: it doesn’t matter). 6 Algorithms Appendix: Solving Recurrences Now T (n) is just the sum of all values stored in the recursion tree. For each i , the i th level of the
tree contains a i nodes, each with value f (n/ b i ). Thus,
L a i f ( n/ b i ) T ( n) = (Σ) i =0 where L is the depth of the recursion tree. We easily see that L = log b n , because n/ b L = 1. The base
case f (1) = Θ(1) implies that the last non-zero term in the summation is Θ(a L ) = Θ(alog b n ) = Θ(nlog b a ).
For most divide-and-conquer recurrences, the level-by-level sum (Σ) is a geometric series—each term
is a constant factor larger or smaller than the previous term. In this case, only the largest term in the
geometric series matters; all of the other terms are swallowed up by the Θ(·) notation.
f(n)
a
f(n/b) f(n)
+ f(n/b)
a f(n/b)
a f(n/b²)
f(n/b²) ²)
f(n/b
f(n/b²) a⋅f(n/b) f(n/b)
a a f(n/b²)
f(n/b²)
f(n/b²) ²)
f(n/b²) ²)
f(n/b
f(n/b
f(n/b²)
f(n/b²) +
a²⋅ f(n/b²) f(n/b²)
f(n/b²) ²)
f(n/b
f(n/b²) + +
f(n/b ) ) L f(n/b) ) L f(n/b) ) L f(n/b) ) L f(n/b) ) L f(n/b) ) L f(n/b) ) L f(n/b) ) L
f(n/b L)f(n/b L)f(n/b L)f(n/b L)f(n/b L)f(n/b L)f(n/b L)f(n/b L) )
f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b L)
f(n/b L L L f(n/bL L L f(n/bL L L f(n/bL L L f(n/bL L L f(n/bL L L f(n/bL L L f(n/b L L
f(n/b )f(n/b )f(n/b )f(n/b )f(n/b )f(n/b )f(n/b )f(n/b ) )
f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b ) )f(n/b L)
f(n/b L f(n/b L f(n/b L f(n/b L f(n/b L f(n/b L f(n/b L f(n/b
f(n/b
f(n/b
f(n/b
f(n/b
f(n/b
f(n/b
f(n/b
f(n/b
LL LL LL LL LL LL LL LL aL⋅f(n/bL) A recursion tree for the recurrence T (n) = a T (n/ b) + f (n) Here are several examples of the recursion-tree technique in action:
• Mergesort (simpliﬁed): T (n ) = 2 T (n /2) + n
There are 2i nodes at level i , each with value n/2i , so every term in the level-by-level sum (Σ)
is the same:
L T (n) = n.
i =0 The recursion tree has L = log2 n levels, so T (n ) = Θ(n log n ).
• Randomized selection: T (n ) = T (3n /4) + n
The recursion tree is a single path. The node at depth i has value (3/4)i n, so the level-by-level
sum (Σ) is a decreasing geometric series:
L (3/4)i n. T ( n) =
i =0 This geometric series is dominated by its initial term n, so T (n ) = Θ(n ). The recursion tree has
L = log4/3 n levels, but so what? 7 Algorithms Appendix: Solving Recurrences • Karatsuba’s multiplication algorithm: T (n ) = 3 T (n /2) + n
There are 3i nodes at depth i , each with value n/2i , so the level-by-level sum (Σ) is an
increasing geometric series:
L (3/2)i n. T ( n) =
i =0 This geometric series is dominated by its ﬁnal term (3/2) L n. Each leaf contributes 1 to this term;
thus, the ﬁnal term is equal to the number of leaves in the tree! The recursion tree has L = log2 n
levels, and therefore 3log2 n = nlog2 3 leaves, so T (n ) = Θ(n log2 3 ).
• T (n ) = 2 T (n /2) + n / lg n
The sum of all the nodes in the i th level is n/(lg n − i ). This implies that the depth of the tree
is at most lg n − 1. The level sums are neither constant nor a geometric series, so we just have to
evaluate the overall sum directly.
Recall (or if you’re seeing this for the ﬁrst time: Behold!) that the nth harmonic number H n is
the sum of the reciprocals of the ﬁrst n positive integers:
n 1 H n :=
i =1 i It’s nat hard to show that H n = Θ(log n); in fact, we have the stronger inequalities ln(n + 1) ≤
H n ≤ ln n + 1.
lg n−1 T ( n) =
i =0 n
lg n − i lg n n =
j =1 j = nHlg n = Θ(n lg lg n ) • T (n ) = 4 T (n /2) + n lg n
There are 4i nodes at each level i , each with value (n/2i ) lg(n/2i ) = (n/2i )(lg n − i ); again,
the depth of the tree is at most lg n − 1. We have the following summation:
lg n−1 n2i (lg n − i ) T ( n) =
i =0 We can simplify this sum by substituting j = lg n − i :
lg n lg n n2 j /2 j = Θ( j 2 ) n2lg n− j j = T ( n) =
j =i j =i ∞ The last step uses the fact that i =1 j /2 j = 2. Although this is not quite a geometric series, it is
still dominated by its largest term.
• Ugly divide and conquer: T (n ) = n · T ( n) + n We solved this recurrence earlier by guessing the right answer and verifying, but we can use
recursion trees to get the correct answer directly. The degree of the nodes in the recursion tree is
8 Algorithms Appendix: Solving Recurrences no longer constant, so we have to be a bit more careful, but the same basic technique still applies.
−L
It’s not hard to see that the nodes in any level sum to n. The depth L satisﬁes the identity n2 = 2
(we can’t get all the way down to 1 by taking square roots), so L = lg lg n and T (n ) = Θ(n lg lg n ).
• Randomized quicksort: T (n ) = T (3n /4) + T (n /4) + n
This recurrence isn’t in the standard form described earlier, but we can still solve it using
recursion trees. Now modes in the same level of the recursion tree have different values, and
different leaves are at different levels. However, the nodes in any complete level (that is, above
any of the leaves) sum to n. Moreover, every leaf in the recursion tree has depth between log4 n
and log4/3 n. To derive an upper bound, we overestimate T (n) by ignoring the base cases and
extending the tree downward to the level of the deepest leaf. Similarly, to derive a lower bound,
we overestimate T (n) by counting only nodes in the tree up to the level of the shallowest leaf.
These observations give us the upper and lower bounds n log4 n ≤ T (n) ≤ n log4/3 n. Since these
bounds differ by only a constant factor, we have T (n ) = Θ(n log n ).
• Deterministic selection: T (n ) = T (n /5) + T (7n /10) + n
Again, we have a lopsided recursion tree. If we look only at complete levels of the tree, we ﬁnd
that the level sums form a descending geometric series T (n) = n + 9n/10 + 81n/100 + · · · . We can
get an upper bound by ignoring the base cases entirely and growing the tree out to inﬁnity, and
we can get a lower bound by only counting nodes in complete levels. Either way, the geometric
series is dominated by its largest term, so T (n ) = Θ(n ).
3 1 • Randomized search trees: T (n ) = 4 T (n /4) + 4 T (3n /4) + 1
This looks like a divide-and-conquer recurrence, but what does it mean to have a quarter of
a child? The right approach is to imagine that each node in the recursion tree has a weight in
addition to its value. Alternately, we get a standard recursion tree again if we add a second real
parameter to the recurrence, deﬁning T (n) = T (n, 1), where
T (n, α) = T (n/4, α/4) + T (3n/4, 3α/4) + α.
In each complete level of the tree, the (weighted) node values sum to exactly 1. The leaves of the
recursion tree are at different levels, but all between log4 n and log4/3 n. So we have upper and
lower bounds log4 n ≤ T (n) ≤ log4/3 n, which differ by only a constant factor, so T (n ) = Θ(log n ).
• Ham-sandwich trees: T (n ) = T (n /2) + T (n /4) + 1
Again, we have a lopsided recursion tree. If we only look at complete levels, we ﬁnd that the
level sums form an ascending geometric series T (n) = 1 + 2 + 4 + · · · , so the solution is dominated
by the number of leaves. The recursion tree has log4 n complete levels, so there are more than
2log4 n = nlog4 2 = n; on the other hand, every leaf has depth at most log2 n, so the total number
of leaves is at most 2log2 n = n. Unfortunately, the crude bounds n
T (n ) n are the best we
can derive using the techniques we know so far!
The following theorem completely describes the solution for any divide-and-conquer recurrence in
the ‘standard form’ T (n) = aT (n/ b) + f (n), where a and b are constants and f (n) is a polynomial. This
theorem allows us to bypass recursion trees for ‘standard’ recurrences, but many people (including Jeff)
ﬁnd it harder to remember than the more general recursion-tree technique. Your mileage may vary.
9 Algorithms Appendix: Solving Recurrences The Master Theorem. The recurrence T (n) = aT (n/ b) + f (n) can be solved as follows.
• If a f (n/ b) = κ f (n) for some constant κ < 1, then T (n) = Θ( f (n)).
• If a f (n/ b) = K f (n) for some constant K > 1, then T (n) = Θ(nlog b a ).
• If a f (n/ b) = f (n), then T (n) = Θ( f (n) log b n).
• If none of these three cases apply, you’re on your own.
Proof: If f (n) is a constant factor larger than a f ( b/n), then by induction, the sum is a descending
geometric series. The sum of any geometric series is a constant times its largest term. In this case, the
largest term is the ﬁrst term f (n).
If f (n) is a constant factor smaller than a f ( b/n), then by induction, the sum is an ascending
geometric series. The sum of any geometric series is a constant times its largest term. In this case, this is
the last term, which by our earlier argument is Θ(nlog b a ).
Finally, if a f ( b/n) = f (n), then by induction, each of the L + 1 terms in the sum is equal to f (n). 4 Linear Recurrences (Annihilators) Another common class of recurrences, called linear recurrences, arises in the context of recursive
backtracking algorithms and counting problems. These recurrences express each function value f (n) as
a linear combination of a small number of nearby values f (n − 1), f (n − 2), f (n − 3), . . . . The Fibonacci
recurrence is a typical example: if n = 0
0
F ( n) = 1
if n = 1 F (n − 1) + F (n − 2) otherwise It turns out that the solution to any linear recurrence is a simple combination of polynomial and
exponential functions in n. For example, we can verify by induction that the linear recurrence if n = 0
1
T (n) = 0
if n = 1 or n = 2 3 T (n − 1) − 8 T (n − 2) + 4 T (n − 3) otherwise has the closed-form solution T (n ) = (n − 3)2n + 4. First we check the base cases:
T (0) = (0 − 3)20 + 4 = 1
T (1) = (1 − 3)21 + 4 = 0
T (2) = (2 − 3)22 + 4 = 0 And now the recursive case:
T ( n) = 3 T ( n − 1) − 8 T ( n − 2) + 4 T ( n − 3)
= 3((n − 4)2n−1 + 4) − 8((n − 5)2n−2 + 4) + 4((n − 6)2n−3 + 4)
384
12 40 24
=
−+
n · 2n −
−
+
2n + (2 − 8 + 4) · 4
248
2
4
8
= ( n − 3) · 2 n + 4
10 Algorithms Appendix: Solving Recurrences But how could we have possibly come up with that solution? In this section, I’ll describe a general
method for solving linear recurrences that’s arguably easier than the induction proof! 4.1 Operators Our technique for solving linear recurrences relies on the theory of operators. Operators are higher-order
functions, which take one or more functions as input and produce different functions as output. For
example, your ﬁrst two semesters of calculus focus almost exclusively on the differential and integral
operators ddx and d x . All the operators we will need are combinations of three elementary building
blocks:
• Sum: ( f + g )(n) := f (n) + g (n)
• Scale: (α · f )(n) := α · ( f (n))
• Shift: (E f )(n) := f (n + 1)
The shift and scale operators are linear, which means they can be distributed over sums; for example,
for any functions f , g , and h, we have E ( f − 3( g − h)) = E f + (−3)E g + 3E h.
We can combine these building blocks to obtain more complex compound operators. For example, the
compound operator E − 2 is deﬁned by setting (E − 2) f := E f + (−2) f for any function f . We can also
apply the shift operator twice: (E (E f ))(n) = f (n + 2); we write usually E 2 f as a synonym for E (E f ).
More generally, for any positive integer k, the operator E k shifts its argument k times: E k f (n) = f (n + k).
Similarly, (E − 2)2 is shorthand for the operator (E − 2)(E − 2), which applies (E − 2) twice.
For example, here are the results of applying different operators to the function f (n) = 2n :
2 f (n) = 2 · 2n = 2n+1
3 f ( n) = 3 · 2 n
E f ( n) = 2 n +1
E 2 f ( n) = 2 n +2
( E − 2) f ( n) = E f ( n) − 2 f ( n) = 2 n +1 − 2 n +1 = 0
(E 2 − 1) f (n) = E 2 f (n) − f (n) = 2n+2 − 2n = 3 · 2n
These compound operators can be manipulated exactly as though they were polynomials over the
‘variable’ E . In particular, we can ‘factor’ compound operators into ‘products’ of simpler operators,
and the order of the factors is unimportant. For example, the compound operators E 2 − 3E + 2 and
(E − 1)(E − 2) are equivalent:
Let
Then g (n) := (E − 2) f (n) = f (n + 1) − 2 f (n). (E − 1)(E − 2) f (n) = (E − 1) g (n)
= g ( n + 1) − g ( n)
= ( f (n + 2) − 2 f (n − 1)) − ( f (n + 1) − 2 f (n))
= f (n + 2) − 3 f (n + 1) + 2 f (n)
= (E 2 − 3E + 2) f (n). It is an easy exercise to conﬁrm that E 2 − 3E + 2 is also equivalent to the operator (E − 2)(E − 1). 11 Algorithms Appendix: Solving Recurrences The following table summarizes everything we need to remember about operators.
Operator
addition
subtraction
multiplication
shift
k-fold shift
composition distribution 4.2 Deﬁnition
( f + g )(n) := f (n) + g (n)
( f − g )(n) := f (n) − g (n)
(α · f )(n) := α · ( f (n))
E f ( n ) := f ( n + 1 )
E k f (n) := f (n + k)
( X + Y ) f := X f + Y f
( X − Y ) f := X f − Y f
X Y f := X ( Y f ) = Y ( X f )
X ( f + g) = X f + X g Annihilators An annihilator of a function f is any nontrivial operator that transforms f into the zero function. (We
can trivially annihilate any function by multiplying it by zero, so as a technical matter, we do not consider
the zero operator to be an annihilator.) Every compound operator we consider annihilates a speciﬁc
class of functions; conversely, every function composed of polynomial and exponential functions has a
unique (minimal) annihilator.
We have already seen that the operator (E − 2) annihilates the function 2n . It’s not hard to see
that the operator (E − c ) annihilates the function α · c n , for any constants c and α. More generally, the
operator (E − c ) annihilates the function a n if and only if c = a:
( E − c ) a n = E a n − c · a n = a n +1 − c · a n = ( a − c ) a n .
Thus, (E − 2) is essentially the only annihilator of the function 2n .
What about the function 2n + 3n ? The operator (E − 2) annihilates the function 2n , but leaves the
function 3n unchanged. Similarly, (E − 3) annihilates 3n while negating the function 2n . But if we apply
both operators, we annihilate both terms:
(E − 2)(2n + 3n ) = E (2n + 3n ) − 2(2n + 3n )
= (2n+1 + 3n+1 ) − (2n+1 + 2 · 3n ) = 3n
=⇒ (E − 3)(E − 2)(2n + 3n ) = (E − 3)3n = 0
In general, for any integers a = b, the operator (E − a)(E − b) = (E − b)(E − a) = (E 2 − (a + b)E + a b)
annihilates any function of the form αa n + β b n , but nothing else.
What about the operator (E − a)(E − a) = (E − a)2 ? It turns out that this operator annihilates all
functions of the form (αn + β )a n :
(E − a)((αn + β )a n ) = (α(n + 1) + β )a n+1 − a(αn + β )a n
= αa n+1
=⇒ (E − a)2 ((αn + β )a n ) = (E − a)(αa n+1 ) = 0
More generally, the operator (E − a)d annihilates all functions of the form p(n) · a n , where p(n) is a
polynomial of degree at most d − 1. For example, (E − 1)3 annihilates any polynomial of degree at
most 2. 12 Algorithms Appendix: Solving Recurrences The following table summarizes everything we need to remember about annihilators.
Operator
E −1
E−a
(E − a)(E − b)
(E − a0 )(E − a1 ) · · · (E − ak )
(E − 1)2
(E − a)2
(E − a)2 (E − b)
(E − a) Functions annihilated
α
αa n
n
αa + β b n
[if a = b]
k
n
i =0 α i a i [if ai distinct] αn + β
(αn + β )a n
(αn + β )a b + γ b n
d −1
i
i =0 α i n d a [if a = b] n If X annihilates f , then X also annihilates E f .
If X annihilates both f and g , then X also annihilates f ± g .
If X annihilates f , then X also annihilates α f , for any constant α.
If X annihilates f and Y annihilates g , then X Y annihilates f ± g . 4.3 Annihilating Recurrences Given a linear recurrence for a function, it’s easy to extract an annihilator for that function. For many
recurrences, we only need to rewrite the recurrence in operator notation. Once we have an annihilator,
we can factor it into operators of the form (E − c ); the table on the previous page then gives us a generic
solution with some unknown coefﬁcients. If we are given explicit base cases, we can determine the
coefﬁcients by examining a few small cases; in general, this involves solving a small system of linear
equations. If the base cases are not speciﬁed, the generic solution almost always gives us an asymptotic
solution. Here is the technique step by step:
1.
2.
3.
4.
5. Write the recurrence in operator form
Extract an annihilator for the recurrence
Factor the annihilator (if necessary)
Extract the generic solution from the annihilator
Solve for coefﬁcients using base cases (if known) Here are several examples of the technique in action:
• r (n ) = 5 r (n − 1), where r (0) = 3.
1. We can write the recurrence in operator form as follows:
r (n) = 5 r (n − 1) =⇒ r (n + 1) − 5 r (n) = 0 =⇒ (E − 5) r (n) = 0.
2. We immediately see that (E − 5) annihilates the function r (n).
3. The annihilator (E − 5) is already factored.
4. Consulting the annihilator table on the previous page, we ﬁnd the generic solution r (n) = α5n
for some constant α.
5. The base case r (0) = 3 implies that α = 3. 13 Algorithms Appendix: Solving Recurrences We conclude that r (n ) = 3 · 5n . We can easily verify this closed-form solution by induction:
r (0) = 3 · 50 = 3 [deﬁnition] r (n) = 5 r (n − 1) [deﬁnition] = 5 · (3 · 5 n−1 ) [induction hypothesis] n =5 ·3 [algebra] • Fibonacci numbers: F (n ) = F (n − 1) + F (n − 2), where F (0) = 0 and F (1) = 1.
1. We can rewrite the recurrence as (E 2 − E − 1) F (n) = 0.
2. The operator E 2 − E − 1 clearly annihilates F (n).
ˆ
3. The quadratic formula implies that the annihilator E 2 − E − 1 factors into (E − φ )(E − φ ),
ˆ = (1 − 5)/2 = 1 − φ = −1/φ .
where φ = (1 + 5)/2 ≈ 1.618034 is the golden ratio and φ
ˆˆ
ˆ
4. The annihilator implies that F (n ) = αφ n + αφ n for some unknown constants α and α.
5. The base cases give us two equations in two unknowns:
ˆ
F (0) = 0 = α + α
ˆˆ
F (1) = 1 = αφ + αφ
ˆ
Solving this system of equations gives us α = 1/(2φ − 1) = 1/ 5 and α = −1/ 5.
We conclude with the following exact closed form for the nth Fibonacci number:
F (n ) = ˆ
φn − φn
5 = 1 1+ 5 2 5 n − 1
5 1− 5 n 2 With all the square roots in this formula, it’s quite amazing that Fibonacci numbers are integers.
However, if we do all the math correctly, all the square roots cancel out when i is an integer. (In
fact, this is pretty easy to prove using the binomial theorem.)
• Towers of Hanoi: T (n ) = 2 T (n − 1) + 1, where T (0) = 0. This is our ﬁrst example of a
non-homogeneous recurrence, which means the recurrence has one or more non-recursive terms.
1. We can rewrite the recurrence as (E − 2) T (n) = 1.
2. The operator (E − 2) doesn’t quite annihilate the function; it leaves a residue of 1. But we
can annihilate the residue by applying the operator (E − 1). Thus, the compound operator
(E − 1)(E − 2) annihilates the function.
3. The annihilator is already factored.
4. The annihilator table gives us the generic solution T (n) = α2n + β for some unknown
constants α and β .
5. The base cases give us T (0) = 0 = α20 + β and T (1) = 1 = α21 + β . Solving this system of
equations, we ﬁnd that α = 1 and β = −1.
We conclude that T (n ) = 2n − 1.
14 Algorithms Appendix: Solving Recurrences For the remaining examples, I won’t explicitly enumerate the steps in the solution.
• Height-balanced trees: H (n ) = H (n − 1) + H (n − 2) + 1, where H (−1) = 0 and H (0) = 1. (Yes,
we’re starting at −1 instead of 0. So what?)
We can rewrite the recurrence as (E 2 − E − 1)H = 1. The residue 1 is annihilated by (E − 1),
so the compound operator (E − 1)(E 2 − E − 1) annihilates the recurrence. This operator factors
ˆ
ˆ
into (E − 1)(E − φ )(E − φ ), where φ = (1 + 5)/2 and φ = (1 − 5)/2. Thus, we get the generic
n
n
ˆ
solution H (n) = α · φ + β + γ · φ , for some unknown constants α, β , γ that satisfy the following
system of equations:
ˆ
ˆ
H (−1) = 0 = αφ −1 + β + γφ −1 = α/φ + β − γ/φ
ˆ
H (0) = 1 = αφ 0 + β + γφ 0 = α + β + γ
ˆ
H (1) = 2 = αφ 1 + β + γφ 1 ˆ
= αφ + β + γφ Solving this system (using Cramer’s rule or Gaussian elimination), we ﬁnd that α = ( 5 + 2)/ 5,
β = −1, and γ = ( 5 − 2)/ 5. We conclude that
H (n ) = 5+2 1+ 5 5 2 n −1+ 5−2
5 1− 5 2 n . • T (n ) = 3 T (n − 1) − 8 T (n − 2) + 4 T (n − 3), where T (0) = 1, T (1) = 0, and T (2) = 0. This was
our original example of a linear recurrence.
We can rewrite the recurrence as ( E 3 − 3 E 2 + 8 E − 4) T = 0, so we immediately have an
annihilator E 3 − 3 E 2 + 8 E − 4. Using high-school algebra, we can factor the annihilator into
(E − 2)2 (E − 1), which implies the generic solution T (n) = αn2n + β 2n + γ. The constants α, β ,
and γ are determined by the base cases:
T (0) = 1 = α · 0 · 20 + β 20 + γ = β +γ T (1) = 0 = α · 1 · 21 + β 21 + γ = 2α + 2β + γ
T (2) = 0 = α · 2 · 22 + β 22 + γ = 8α + 4β + γ
Solving this system of equations, we ﬁnd that α = 1, β = −3, and γ = 4, so T (n ) = (n − 3)2n + 4.
• T (n ) = T (n − 1) + 2 T (n − 2) + 2n − n 2
We can rewrite the recurrence as (E 2 − E − 2) T (n) = E 2 (2n − n2 ). Notice that we had to shift
up the non-recursive parts of the recurrence when we expressed it in this form. The operator
(E − 2)(E − 1)3 annihilates the residue 2n − n2 , and therefore also annihilates the shifted residue
E 2 (2n + n2 ). Thus, the operator (E − 2)(E − 1)3 (E 2 − E − 2) annihilates the entire recurrence. We can
factor the quadratic factor into (E − 2)( E + 1), so the annihilator factors into (E − 2)2 (E − 1)3 (E + 1).
So the generic solution is T (n ) = αn 2n + β 2n + γn 2 + δn + + η(−1)n . The coefﬁcients α, β , γ,
δ, , η satisfy a system of six equations determined by the ﬁrst six function values T (0) through
T (5). For almost2 every set of base cases, we have α = 0, which implies that T (n ) = Θ(n 2n ).
For a more detailed explanation of the annihilator method, see George Lueker, Some techniques for
solving recurrences, ACM Computing Surveys 12(4):419-436, 1980.
2 In fact, the only possible solutions with α = 0 have the form −2n−1 − n2 /2 − 5n/2 + η(−1)n for some constant η. 15 Algorithms 5 Appendix: Solving Recurrences Transformations Sometimes we encounter recurrences that don’t ﬁt the structures required for recursion trees or annihilators. In many of those cases, we can transform the recurrence into a more familiar form, by deﬁning a
new function in terms of the one we want to solve. There are many different kinds of transformations,
but these three are probably the most useful:
• Domain transformation: Deﬁne a new function S (n) = T ( f (n)) with a simpler recurrence, for
some simple function f .
• Range transformation: Deﬁne a new function S (n) = f ( T (n)) with a simpler recurrence, for some
simple function f .
• Difference transformation: Simplify the recurrence for T (n) by considering the difference T (n) −
T (n − 1).
Here are some examples of these transformations in action.
• Unsimpliﬁed Mergesort: T (n ) = T ( n /2 ) + T ( n /2 ) + Θ(n )
When n is a power of 2, we can simplify the mergesort recurrence to T (n) = 2 T (n/2) + Θ(n),
which has the solution T (n) = Θ(n log n). Unfortunately, for other values values of n, this simpliﬁed
recurrence is incorrect. When n is odd, then the recurrence calls for us to sort a fractional number
of elements! Worse yet, if n is not a power of 2, we will never reach the base case T (1) = 1.
So we really need to solve the original recurrence. We have no hope of getting an exact solution,
even if we ignore the Θ( ) in the recurrence; the ﬂoors and ceilings will eventually kill us. But
we can derive a tight asymptotic solution using a domain transformation—we can rewrite the
function T (n) as a nested function S ( f (n)), where f (n) is a simple function and the function S ( )
has an simpler recurrence.
First let’s overestimate the time bound, once by pretending that the two subproblem sizes are
equal, and again to eliminate the ceiling:
T ( n) ≤ 2 T n/ 2 + n ≤ 2 T (n/2 + 1) + n. Now we deﬁne a new function S (n) = T (n + α), where α is a unknown constant, chosen so that
S (n) satisﬁes the Master-Theorem-ready recurrence S (n) ≤ 2S (n/2) + O(n). To ﬁgure out the
correct value of α, we compare two versions of the recurrence for the function T (n + α):
S (n) ≤ 2S (n/2) + O(n) =⇒ T (n + α) ≤ 2 T (n/2 + α) + O(n) T ( n) ≤ 2 T ( n/ 2 + 1) + n =⇒ T (n + α) ≤ 2 T ((n + α)/2 + 1) + n + α For these two recurrences to be equal, we need n/2 + α = (n + α)/2 + 1, which implies that α = 2.
The Master Theorem now tells us that S (n) = O(n log n), so
T (n) = S (n − 2) = O((n − 2) log(n − 2)) = O(n log n).
A similar argument implies the matching lower bound T (n) = Ω(n log n). So T (n ) = Θ(n log n )
after all, just as though we had ignored the ﬂoors and ceilings from the beginning!
Domain transformations are useful for removing ﬂoors, ceilings, and lower order terms from
the arguments of any recurrence that otherwise looks like it ought to ﬁt either the Master Theorem
or the recursion tree method. But now that we know this, we don’t need to bother grinding
through the actual gory details!
16 Algorithms Appendix: Solving Recurrences • Ham-Sandwich Trees: T (n ) = T (n /2) + T (n /4) + 1
As we saw earlier, the recursion tree method only gives us the uselessly loose bounds n
T ( n)
n for this recurrence, and the recurrence is in the wrong form for annihilators. The authors
who discovered ham-sandwich trees (yes, this is a real data structure) solved this recurrence by
guessing the solution and giving a complicated induction proof.
But a simple transformation allows us to solve the recurrence in just a few lines. We deﬁne
a new function t (k) = T (2k ), which satisﬁes the simpler linear recurrence t (k) = t (k − 1) +
t (k − 2) + 1. This recurrence should immediately remind you of Fibonacci numbers. Sure enough,
the annihilator method implies the solution t (k) = Θ(φ k ), where φ = (1 + 5)/2 is the golden
ratio. We conclude that
T (n) = t (lg n) = Θ(φ lg n ) = Θ(n lg φ ) ≈ Θ(n0.69424 ).
Many other divide-and-conquer recurrences can be similarly transformed into linear recurrences
and then solved with annihilators. Consider once more the simpliﬁed mergesort recurrence
T (n) = 2 T (n/2) + n. The function t (k) = T (2k ) satisﬁes the recurrence t (k) = 2 t (k − 1) + 2k .
The annihilator method gives us the generic solution t (k) = Θ(k · 2k ), which implies that T (n) =
t (lg n) = Θ(n log n), just as we expected.
On the other hand, for some recurrences like T (n) = T (n/3) + T (2n/3) + n, the recursion tree
method gives an easy solution, but there’s no way to transform the recurrence into a form where
we can apply the annihilator method directly.3 • Random Binary Search Trees: T (n ) = 1 3 T (3n /4) + 1
4
4
This looks like a divide-and-conquer recurrence, so we might be tempted to apply recursion
trees, but what does it mean to have a quarter of a child? If we’re not comfortable with weighted
recursion trees, we can instead consider a new function U (n) = n · T (n), which satisﬁes the
recurrence U (n) = U (n/4) + U (3n/4) + n. As we’ve already seen, recursion trees imply that
U (n) = Θ(n log n), which immediately implies that T (n ) = Θ(log n ). • Randomized Quicksort: T (n ) = 2 n −1
n T (n /4) + T (k ) + n k =0 This is our ﬁrst example of a full history recurrence; each function value T (n) is deﬁned in
terms of all previous function values T (k) with k < n. Before we can apply any of our existing
techniques, we need to convert this recurrence into an equivalent limited history form by shifting
and subtracting away common terms. To make this step slightly easier, we ﬁrst multiply both sides
of the recurrence by n to get rid of the fractions. However, we can still get a solution via functional transformations as follows. The function t (k) = T ((3/2)k ) satisﬁes
the recurrence t (n) = t (n − 1) + t (n − λ) + (3/2)k , where λ = log3/2 3 = 2.709511 . . . . The characteristic function for this
recurrence is ( r λ − r λ−1 − 1)( r − 3/2), which has a double root at r = 3/2 and nowhere else. Thus, t (k) = Θ(k(3/2)k ), which
implies that T (n) = t (log3/2 n) = Θ(n log n).
3 17 Algorithms Appendix: Solving Recurrences
n−1 T ( j ) + n2 n · T ( n) = 2 [multiply both sides by n] k =0
n−2 T ( j ) + (n − 1)2 ( n − 1) · T ( n − 1) = 2 [shift] k =0 nT (n) − (n − 1) T (n − 1) = 2 T (n − 1) + 2n − 1
n+1
1
T ( n) =
T (n − 1) + 2 −
n
n [subtract]
[simplify] We can solve this limited-history recurrence using another functional transformation. We
deﬁne a new function t (n) = T (n)/(n + 1), which satisﬁes the simpler recurrence
t ( n) = t ( n − 1) + 2
n+1 − 1
n( n + 1) , which we can easily unroll into a summation. If we only want an asymptotic solution, we can
simplify the ﬁnal recurrence to t (n) = t (n − 1) + Θ(1/n), which unrolls into a very familiar
summation:
n t ( n) = Θ(1/i ) = Θ(H n ) = Θ(log n).
i =1 Finally, substituting T (n) = (n + 1) t (n) gives us a solution to the original recurrence: T (n ) =
Θ(n log n ). 18 Algorithms Appendix: Solving Recurrences Exercises
1. For each of the following recurrences, ﬁrst guess an exact closed-form solution, and then prove
your guess is correct. You are free to use any method you want to make your guess—unrolling
the recurrence, writing out the ﬁrst several values, induction proof template, recursion trees,
annihilators, transformations, ‘It looks like that other one’, whatever—but please describe your
method. All functions are from the non-negative integers to the reals. If it simpliﬁes your solutions,
n
express them in terms of Fibonacci numbers Fn , harmonic numbers H n , binomial coefﬁcients k ,
factorials n!, and/or the ﬂoor and ceiling functions x and x .
(a) A(n) = A(n − 1) + 1, where A(0) = 0.
(b) B (n) = 0 if n < 5 B ( n − 5) + 2 otherwise (c) C (n) = C (n − 1) + 2n − 1, where C (0) = 0.
(d) D(n) = D(n − 1) + n
2
n , where D(0) = 0. (e) E (n) = E (n − 1) + 2 , where E (0) = 0.
(f) F (n) = 3 · F (n − 1), where F (0) = 1.
(g) G (n) = G (n−1)
,
G (n−2) where G (0) = 1 and G (1) = 2. [Hint: This is easier than it looks.] (h) H (n) = H (n − 1) + 1/n, where H (0) = 0.
(i) I (n) = I (n − 2) + 3/n, where I (0) = I (1) = 0. [Hint: Consider even and odd n separately.]
(j) J (n) = J (n − 1)2 , where J (0) = 2.
(k) K (n) = K ( n/2 ) + 1, where K (0) = 0.
(l) L (n) = L (n − 1) + L (n − 2), where L (0) = 2 and L (1) = 1.
[Hint: Write the solution in terms of Fibonacci numbers.]
(m) M (n) = M (n − 1) · M (n − 2), where M (0) = 2 and M (1) = 1.
[Hint: Write the solution in terms of Fibonacci numbers.]
(n) N (n) = 1 +
(p) P (n) =
(q) Q(n) = n (N (k − 1) + N (n − k)), where N (0) = 1. k =1 n−1
k =0 (k · P (k − 1)), where P (0) = 1. 1
,
2−Q(n−1) where Q(0) = 0. (r) R(n) = max {R(k − 1) + R(n − k) + n}
1≤k≤n (s) S (n) = max {S (k − 1) + S (n − k) + 1}
1≤k≤n (t) T (n) = min { T (k − 1) + T (n − k) + n}
1≤k≤n (u) U (n) = min {U (k − 1) + U (n − k) + 1}
1≤k≤n (v) V (n) = max n/3≤k≤2n/3 { V ( k − 1) + V ( n − k ) + n}
19 Algorithms Appendix: Solving Recurrences 2. Use recursion trees to solve each of the following recurrences.
(a) A(n) = 2A(n/4) + n (b) B (n) = 2B (n/4) + n
(c) C (n) = 2C (n/4) + n2
(d) D(n) = 3 D(n/3) + n (e) E (n) = 3 E (n/3) + n
(f) F (n) = 3 F (n/3) + n2
(g) G (n) = 4G (n/2) + n (h) H (n) = 4H (n/2) + n
(i) I (n) = 4 I (n/2) + n2
(j) J (n) = J (n/2) + J (n/3) + J (n/6) + n
(k) K (n) = K (n/2) + K (n/3) + K (n/6) + n2
(l) L (n) = L (n/15) + L (n/10) + 2 L (n/6) +
(m) M (n) = 2 n M ( 2 n) + (n) N (n) = 2 n N ( 2 n) + n (p) P (n) = n 2 n P ( 2 n) + n2 n (q) Q(n) = Q(n − 3) + 8n — Don’t use annihilators!
(r) R(n) = 2R(n − 2) + 4n — Don’t use annihilators!
(s) S (n) = 4S (n − 1) + 2n — Don’t use annihilators!
3. Make up a bunch of linear recurrences and then solve them using annihilators.
4. Solve the following recurrences, using any tricks at your disposal.
lg n T ( n/ 2 i ) + n (a) T (n) = [Hint: Assume n is a power of 2.] i =1 (b) More to come. . . 20 Algorithms Non-Lecture L: Fibonacci Heaps
A little and a little, collected together, become a great deal; the heap in the barn
consists of single grains, and drop and drop makes an inundation.
— Saadi (1184–1291)
The trees that are slow to grow bear the best fruit.
— Molière (1622–1673)
Promote yourself but do not demote another.
— Rabbi Israel Salanter (1810–1883)
Fall is my favorite season in Los Angeles, watching the birds change color and fall
from the trees.
— David Letterman L Fibonacci Heaps L.1 Mergeable Heaps A mergeable heap is a data structure that stores a collection of keys1 and supports the following operations.
• INSERT: Insert a new key into a heap. This operation can also be used to create a new heap
containing just one key.
• FINDMIN: Return the smallest key in a heap.
• DELETEMIN: Remove the smallest key from a heap.
• MERGE: Merge two heaps into one. The new heap contains all the keys that used to be in the old
heaps, and the old heaps are (possibly) destroyed.
If we never had to use DELETEMIN, mergeable heaps would be completely trivial. Each “heap” just
stores to maintain the single record (if any) with the smallest key. INSERTs and MERGEs require only
one comparison to decide which record to keep, so they take constant time. FINDMIN obviously takes
constant time as well.
If we need DELETEMIN, but we don’t care how long it takes, we can still implement mergeable
heaps so that INSERTs, MERGEs, and FINDMINs take constant time. We store the records in a circular
doubly-linked list, and keep a pointer to the minimum key. Now deleting the minimum key takes Θ(n)
time, since we have to scan the linked list to ﬁnd the new smallest key.
In this lecture, I’ll describe a data structure called a Fibonacci heap that supports INSERTs, MERGEs,
and FINDMINs in constant time, even in the worst case, and also handles DELETEMIN in O(log n) amortized
time. That means that any sequence of n INSERTs, m MERGEs, f FINDMINs, and d DELETEMINs takes
O(n + m + f + d log n) time. L.2 Binomial Trees and Fibonacci Heaps A Fibonacci heap is a circular doubly linked list, with a pointer to the minimum key, but the elements of the
list are not single keys. Instead, we collect keys together into structures called binomial heaps. Binomial
heaps are trees that satisfy the heap property—every node has a smaller key than its children—and have
the following special recursive structure.
A kth order binomial tree, which I’ll abbreviate Bk , is deﬁned recursively. B0 is a single node. For all
k > 0, Bk consists of two copies of Bk−1 that have been linked together, meaning that the root of one
Bk−1 has become a new child of the other root.
1 In the earlier lecture on treaps, I called these keys priorities to distinguish them from search keys. 1 Algorithms Non-Lecture L: Fibonacci Heaps B4
B4
B5
Binomial trees of order 0 through 5. Binomial trees have several useful properties, which are easy to prove by induction (hint, hint).
• The root of Bk has degree k.
• The children of the root of Bk are the roots of B0 , B1 , . . . , Bk−1 .
• Bk has height k.
• Bk has 2k nodes.
• Bk can be obtained from Bk−1 by adding a new child to every node.
• Bk has k
d nodes at depth d , for all 0 ≤ d ≤ k. • Bk has 2k−h−1 nodes with height h, for all 0 ≤ h < k, and one node (the root) with height k.
Although we normally don’t care in this class about the low-level details of data structures, we need
to be speciﬁc about how Fibonacci heaps are actually implemented, so that we can be sure that certain
operations can be performed quickly. Every node in a Fibonacci heap points to four other nodes: its
parent, its ‘next’ sibling, its ‘previous’ sibling, and one of its children. The sibling pointers are used to
join the roots together into a circular doubly-linked root list. In each binomial tree, the children of each
node are also joined into a circular doubly-linked list using the sibling pointers.
min min A high-level view and a detailed view of the same Fibonacci heap. Null pointers are omitted for clarity. With this representation, we can add or remove nodes from the root list, merge two root lists together,
link one two binomial tree to another, or merge a node’s list of children with the root list, in constant
time, and we can visit every node in the root list in constant time per node. Having established that
these primitive operations can be performed quickly, we never again need to think about the low-level
representation details. 2 Algorithms L.3 Non-Lecture L: Fibonacci Heaps Operations on Fibonacci Heaps The INSERT, MERGE, and FINDMIN algorithms for Fibonacci heaps are exactly like the corresponding
algorithms for linked lists. Since we maintain a pointer to the minimum key, FINDMIN is trivial. To insert
a new key, we add a single node (which we should think of as a B0 ) to the root list and (if necessary)
update the pointer to the minimum key. To merge two Fibonacci heaps, we just merge the two root lists
and keep the pointer to the smaller of the two minimum keys. Clearly, all three operations take O(1)
time.
Deleting the minimum key is a little more complicated. First, we remove the minimum key from the
root list and splice its children into the root list. Except for updating the parent pointers, this takes O(1)
time. Then we scan through the root list to ﬁnd the new smallest key and update the parent pointers of
the new roots. This scan could take Θ(n) time in the worst case. To bring down the amortized deletion
time, we apply a CLEANUP algorithm, which links pairs of equal-size binomial heaps until there is only
one binomial heap of any particular size.
Let me describe the CLEANUP algorithm in more detail, so we can analyze its running time. The
following algorithm maintains a global array B [1 .. lg n ], where B [i ] is a pointer to some previouslyvisited binomial heap of order i , or NULL if there is no such binomial heap. Notice that CLEANUP
simultaneously resets the parent pointers of all the new roots and updates the pointer to the minimum
key. I’ve split off the part of the algorithm that merges binomial heaps of the same order into a separate
subroutine MERGEDUPES.
MERGEDUPES( v ):
w ← B [deg( v )]
while w = NULL
B [deg( v )] ← NULL
if key( v ) ≤ key(w )
swap v
w
remove w from the root list
link w to v
w ← B [deg( v )]
B [deg( v )] ← v CLEANUP:
newmin ← some node in the root list
for i ← 0 to lg n
B [i ] ← NULL
for all nodes v in the root list
parent( v ) ← NULL ( )
if key(newmin) > key( v )
newmin ← v
MERGEDUPES( v )
0 B 1 2 0 3 v 1 2 3 0 B v 1 2 ( ) 3 B v MergeDupes( v ), ensuring that no earlier root has the same degree as v . Notices that MERGEDUPES is careful to merge heaps so that the heap property is maintained—the
heap whose root has the larger key becomes a new child of the heap whose root has the smaller key.
This is handled by swapping v and w if their keys are in the wrong order.
The running time of CLEANUP is O( r ), where r is the length of the root list just before CLEANUP is
called. The easiest way to see this is to count the number of times the two starred lines can be executed:
line ( ) is executed once for every node v on the root list, and line ( ) is executed at most once for
every node w on the root list. Since DELETEMIN does only a constant amount of work before calling
CLEANUP, the running time of DELETEMIN is O( r ) = O( r + deg(min)) where r is the number of roots
before DELETEMIN begins, and min is the node deleted. 3 Algorithms Non-Lecture L: Fibonacci Heaps Although deg(min) is at most lg n, we can still have r = Θ(n) (for example, if nothing has been
deleted yet), so the worst-case time for a DELETEMIN is Θ(n). After a DELETEMIN, the root list has length
O(log n), since all the binomial heaps have unique orders and the largest has order at most lg n . L.4 Amortized Analysis of DELETEMIN To bound the amortized cost, observe that each insertion increments r . If we charge a constant ‘cleanup
tax’ for each insertion, and use the collected tax to pay for the CLEANUP algorithm, the unpaid cost of a
DELETEMIN is only O(deg(min)) = O(log n).
More formally, deﬁne the potential of the Fibonacci heap to be the number of roots. Recall that
the amortized time of an operation can be deﬁned as its actual running time plus the increase in
potential, provided the potential is initially zero (it is) and we never have negative potential (we
never do). Let r be the number of roots before a DELETEMIN, and let r denote the number of roots
afterwards. The actual cost of DELETEMIN is r + deg(min), and the number of roots increases by r − r ,
so the amortized cost is r + deg(min). Since r = O(log n) and the degree of any node is O(log n),
the amortized cost of DELETEMIN is O(log n).
Each INSERT adds only one root, so its amortized cost is still constant. A MERGE actually doesn’t
change the number of roots, since the new Fibonacci heap has all the roots from its constituents and no
others, so its amortized cost is also constant. L.5 Decreasing Keys In some applications of heaps, we also need the ability to delete an arbitrary node. The usual way to do
this is to decrease the node’s key to −∞, and then use DELETEMIN. Here I’ll describe how to decrease
the key of a node in a Fibonacci heap; the algorithm will take O(log n) time in the worst case, but the
amortized time will be only O(1).
Our algorithm for decreasing the key at a node v follows two simple rules.
1. Promote v up to the root list. (This moves the whole subtree rooted at v .)
2. As soon as two children of any node w have been promoted, immediately promote w .
In order to enforce the second rule, we now mark certain nodes in the Fibonacci heap. Speciﬁcally, a
node is marked if exactly one of its children has been promoted. If some child of a marked node is
promoted, we promote (and unmark) that node as well. Whenever we promote a marked node, we
unmark it; this is theonly way to unmark a node. (Speciﬁcally, splicing nodes into the root list during a
DELETEMIN is not considered a promotion.)
Here’s a more formal description of the algorithm. The input is a pointer to a node v and the new
value k for its key.
PROMOTE( v ):
unmark v
if parent( v ) = NULL
remove v from parent( v )’s list of children
insert v into the root list
if parent( v ) is marked
PROMOTE(parent( v ))
else
mark parent( v ) DECREASEKEY( v, k):
key( v ) ← k
update the pointer to the smallest key
PROMOTE( v ) 4 Algorithms Non-Lecture L: Fibonacci Heaps The PROMOTE algorithm calls itself recursively, resulting in a ‘cascading promotion’. Each consecutive
marked ancestor of v is promoted to the root list and unmarked, otherwise unchanged. The lowest
unmarked ancestor is then marked, since one of its children has been promoted.
a
b
f
l g m n h c
i d j f e k l a
b m
g p o c h j i n d f e k l a
b m
g p o h n c
i d e k j o p
f
l
p b m g
n h a
e c j d
k i a f
l m p o b
g
n h c e j d
k i
o Decreasing the keys of four nodes: ﬁrst f , then d , then j , and ﬁnally h. Dark nodes are marked.
DecreaseKey(h) causes nodes b and a to be recursively promoted. The time to decrease the key of a node v is O(1 + #consecutive marked ancestors of v ). Binomial
heaps have logarithmic depth, so if we still had only full binomial heaps, the running time would be
O(log n). Unfortunately, promoting nodes destroys the nice binomial tree structure; our trees no longer
have logarithmic depth! In fact, DECREASEKEY runs in Θ(n) time in the worst case.
To compute the amortized cost of DECREASEKEY, we’ll use the potential method, just as we did for
DELETEMIN. We need to ﬁnd a potential function Φ that goes up a little whenever we do a little work,
and goes down a lot whenever we do a lot of work. DECREASEKEY unmarks several marked ancestors and
possibly also marks one node. So the number of marked nodes might be an appropriate potential function
here. Whenever we do a little bit of work, the number of marks goes up by at most one; whenever we
do a lot of work, the number of marks goes down a lot.
More precisely, let m and m be the number of marked nodes before and after a DECREASEKEY
operation. The actual time (ignoring constant factors) is
t = 1 + #consecutive marked ancestors of v
and if we set Φ = m, the increase in potential is
m − m ≤ 1 − #consecutive marked ancestors of v .
Since t + ∆Φ ≤ 2, the amortized cost of DECREASEKEY is O(1) . L.6 Bounding the Degree But now we have a problem with our earlier analysis of DELETEMIN. The amortized time for a DELETEMIN
is still O( r + deg(min)). To show that this equaled O(log n), we used the fact that the maximum degree
of any node is O(log n), which implies that after a CLEANUP the number of roots is O(log n). But now
that we don’t have complete binomial heaps, this ‘fact’ is no longer obvious!
So let’s prove it. For any node v , let | v | denote the number of nodes in the subtree of v , including v
itself. Our proof uses the following lemma, which ﬁnally tells us why these things are called Fibonacci
heaps.
Lemma 1. For any node v in a Fibonacci heap, | v | ≥ Fdeg( v )+2 .
5 Algorithms Non-Lecture L: Fibonacci Heaps Proof: Label the children of v in the chronological order in which they were linked to v . Consider the
situation just before the i th oldest child w i was linked to v . At that time, v had at least i − 1 children
(possibly more). Since CLEANUP only links trees with the same degree, we had deg(w i ) = deg( v ) ≥ i − 1.
Since that time, at most one child of w i has been promoted away; otherwise, w i would have been
promoted to the root list by now. So currently we have deg(w i ) ≥ i − 2.
We also quickly observe that deg(w i ) ≥ 0. (Duh.)
Let sd be the minimum possible size of a tree with degree d in any Fibonacci heap. Clearly s0 = 1;
for notational convenience, let s−1 = 1 also. By our earlier argument, the i th oldest child of the root has
degree at least max{0, i − 2}, and thus has size at least max{1, si −2 } = si −2 . Thus, we have the following
recurrence:
d sd ≥ 1 + s i −2
i =1 If we assume inductively that si ≥ Fi +2 for all −1 ≤ i < d (with the easy base cases s−1 = F1 and
s0 = F2 ), we have
d sd ≥ 1 + F i = F d +2 .
i =1 (The last step was a practice problem in Homework 0.) By deﬁnition, | v | ≥ sdeg( v ) .
You can easily show (using either induction or the annihilator method) that Fk+2 > φ k where
φ= 1+ 5
2 ≈ 1.618 is the golden ratio. Thus, Lemma 1 implies that
deg( v ) ≤ logφ | v | = O(log| v |). Thus, since the size of any subtree in an n-node Fibonacci heap is obviously at most n, the degree of any
node is O(log n), which is exactly what we wanted. Our earlier analysis is still good. L.7 Analyzing Everything Together Unfortunately, our analyses of DELETEMIN and DECREASEKEY used two different potential functions.
Unless we can ﬁnd a single potential function that works for both operations, we can’t claim both
amortized time bounds simultaneously. So we need to ﬁnd a potential function Φ that goes up a little
during a cheap DELETEMIN or a cheap DECREASEKEY, and goes down a lot during an expensive DELETEMIN
or an expensive DECREASEKEY.
Let’s look a little more carefully at the cost of each Fibonacci heap operation, and its effect on
both the number of roots and the number of marked nodes, the things we used as out earlier potential
functions. Let r and m be the numbers of roots and marks before each operation, and let r and m be
the numbers of roots and marks after the operation.
operation
INSERT
MERGE
DELETEMIN
DECREASEKEY actual cost
1
1
r + deg(min)
1+m−m r −r
1
0
r −r
1+m−m m −m
0
0
0
m −m In particular, notice that promoting a node in DECREASEKEY requires constant time and increases the
number of roots by one, and that we promote (at most) one unmarked node. 6 Algorithms Non-Lecture L: Fibonacci Heaps If we guess that the correct potential function is a linear combination of our old potential functions r
and m and play around with various possibilities for the coefﬁcients, we will eventually stumble across
the correct answer:
Φ = r + 2m
To see that this potential function gives us good amortized bounds for every Fibonacci heap operation,
let’s add two more columns to our table.
operation
INSERT
MERGE
DELETEMIN
DECREASEKEY actual cost
1
1
r + deg(min)
1+m−m r −r
1
0
r −r
1+m−m Φ −Φ
1
0
r −r
1+m −m m −m
0
0
0
m −m amortized cost
2
1
r + deg(min)
2 Since Lemma 1 implies that r + deg(min) = O(log n), we’re ﬁnally done! (Whew!) L.8 Fibonacci Trees To give you a little more intuition about how Fibonacci heaps behave, let’s look at a worst-case
construction for Lemma 1. Suppose we want to remove as many nodes as possible from a binomial heap
of order k, by promoting various nodes to the root list, but without causing any cascading promotions.
The most damage we can do is to promote the largest subtree of every node. Call the result a Fibonacci
tree of order k + 1, and denote it f k+1 . As a base case, let f1 be the tree with one (unmarked) node, that
is, f1 = B0 . The reason for shifting the index should be obvious after a few seconds. Fibonacci trees of order 1 through 6. Light nodes have been promoted away; dark nodes are marked. Recall that the root of a binomial tree Bk has k children, which are roots of B0 , B1 , . . . , Bk−1 . To
convert Bk to f k+1 , we promote the root of Bk−1 , and recursively convert each of the other subtrees Bi
to f i +1 . The root of the resulting tree f k+1 has degree k − 1, and the children are the roots of smaller
Fibonacci trees f1 , f2 , . . . , f k−1 . We can also consider Bk as two copies of Bk−1 linked together. It’s quite
easy to show that an order-k Fibonacci tree consists of an order k − 2 Fibonacci tree linked to an order
k − 1 Fibonacci tree. (See the picture below.)
B6 B5 B4 B3 B2 f7 B6
B 1B 0 B5
f5 B5 f4 f7 f3 f2 f1
f5 f6 Comparing the recursive structures of B6 and f7 . Since f1 and f2 both have exactly one node, the number of nodes in an order-k Fibonacci tree is
exactly the kth Fibonacci number! (That’s why we changed in the index.) Like binomial trees, Fibonacci
trees have lots of other nice properties that easy to prove by induction (hint, hint):
7 Algorithms Non-Lecture L: Fibonacci Heaps • The root of f k has degree k − 2.
• f k can be obtained from f k−1 by adding a new unmarked child to every marked node and then
marking all the old unmarked nodes.
• f k has height k/2 − 1.
• f k has Fk−2 unmarked nodes, Fk−1 marked nodes, and thus Fk nodes altogether.
k − d −2 • f k has d −1 unmarked nodes,
all 0 ≤ d ≤ k/2 − 1. k − d −2
d marked nodes, and k − d −1
d total nodes at depth d , for • f k has Fk−2h−1 nodes with height h, for all 0 ≤ h ≤ k/2 − 1, and one node (the root) with height
k/2 − 1. I stopped covering Fibonacci heaps in my undergraduate algorithms class a few years ago, even
though they are a great illustration of data structure design principles and amortized analysis. My
main reason for skipping them is that the algorithms are relatively complicated, which both hinders
understanding and limits any practical advantages over regular binary heaps. (The popular algorithms textbook CLRS dismisses Fibonacci heaps as “predominantly of theoretical interest” because of
programming complexity and large constant factors in its running time.)
Nevertheless, students interested in data structures are strongly advised to become familiar with
binomial and Fibonacci heaps, since they share key structural properties with other data structures (such
as union-ﬁnd trees and Bentley-Saxe dynamization) and illustrate important data structure techniques
(like lazy rebuilding). Also, Fibonacci heaps (and more recent extensions like Chazelle’s soft heaps) are
key ingredients in the fastest algorithms known for some problems. 8 Algorithms Non-Lecture M: Number-Theoretic Algorithms
And it’s one, two, three,
What are we ﬁghting for?
Don’t tell me, I don’t give a damn,
Next stop is Vietnam; [or: This time we’ll kill Saddam]
And it’s ﬁve, six, seven,
Open up the pearly gates,
Well there ain’t no time to wonder why,
Whoopee! We’re all going to die.
— Country Joe and the Fish
“I-Feel-Like-I’m-Fixin’-to-Die Rag" (1967)
There are 0 kinds of mathematicians:
Those who can count modulo 2 and those who can’t.
— Anonymous
God created the integers; all the rest is the work of man.
— Kronecker M Number Theoretic Algorithms M.1 Greatest Common Divisors Before we get to any actual algorithms, we need some deﬁnitions and preliminary results. Unless
speciﬁcally indicated otherwise, all variables in this lecture are integers.
The symbol Z (from the German word “Zahlen", meaning ‘numbers’ or ‘to count’) to denote the set
Z
of integers. We say that one integer d divides another integer n, or that d is a divisor of n, if the quotient
n/d is also an integer. Symbolically, we can write this deﬁnition as follows:
d | n ⇐⇒ n
d = n
d In particular, zero is not a divisor of any integer—∞ is not an integer—but every other integer is a
divisor of zero. If d and n are positive, then d | n immediately implies that d ≤ n.
Any integer n can be written in the form n = qd + r for some non-negative integer 0 ≤ r ≤ |d − 1|.
Moreover, the choices for the quotient q and remainder r are unique:
q= n
d and r = n mod d = n − d n
d . Note that the remainder n mod d is always non-negative, even if n < 0 or d < 0 or both.1
If d divides two integers m and n, we say that d is a common divisor of m and n. It’s trivial to
prove (by deﬁnition crunching) that any common divisor of m and n also divides any integer linear
combination of m and n:
(d | m) and (d | n) =⇒ d | (am + bn)
The greatest common divisor of m and n, written gcd(m, n),2 is the largest integer that divides both
m and n. Sometimes this is also called the greater common denominator. The greatest common divisor
has another useful characterization as the smallest element of another set.
Lemma 1. g cd (m, n) is the smallest positive integer of the form am + bn.
1 The sign rules for the C/C++/Java % operator are just plain stupid. I can’t count the number of times I’ve had to write x = (x+n)%n; instead of x %= n;. Frickin’ idiots. Gah! Do not use the notation (m, n) for greatest common divisor, unless you want Notation Kitty to visit you in the middle of the
night and remove your eyes while you sleep. Nice kitty.
2 1 Algorithms Non-Lecture M: Number-Theoretic Algorithms Proof: Let s be the smallest positive integer of the form am + bn. Any common divisor of m and n is also
a divisor of s = am + bn. In particular, gcd(m, n) is a divisor of s, which implies that gcd(m, n) ≤ s .
To prove the other inequality, let’s show that s | m by calculating m mod s.
m mod s = m − s m
s m = m − (am + bn) s = m 1−a m
s + n −b m
s We observe that m mod s is an integer linear combination of m and n. Since m mod s < s, and s is the
smallest positive integer linear combination, m mod s cannot be positive. So it must be zero, which
implies that s | m, as we claimed. By a symmetric argument, s | n. Thus, s is a common divisor of m and
n. A common divisor can’t be greater than the greatest common divisor, so s ≤ gcd(m, n) .
These two inequalities imply that s = gcd(m, n), completing the proof. M.2 Euclid’s GCD Algorithm We can compute the greatest common divisor of two given integers by recursively applying two simple
observations:
gcd(m, n) = gcd(m, n − m)
and
gcd(n, 0) = n
The following algorithm uses the ﬁrst observation to reduce the input and recurse; the second observation
provides the base case.
SLOWGCD(m, n):
m ← |m|; n ← |n|
if m < n
swap m ↔ n
while n > 0
m← m−n
if m < n
swap m ↔ n
return m The ﬁrst few lines just ensure that m ≥ n ≥ 0. Each iteration of the main loop decreases one of the
numbers by at least 1, so the running time is O(m + n). This bound is tight in the worst case; consider
the case n = 1. Unfortunately, this is terrible. The input consists of just log m + log n bits; as a function
of the input size, this algorithm runs in exponential time.
Let’s think for a moment about what the main loop computes between swaps. We start with two
numbers m and n and repeatedly subtract n from m until we can’t any more. This is just a (slow) recipe
for computing m mod n! That means we can speed up the algorithm by using mod instead of subtraction.
EUCLIDGCD(m, n):
m ← |m|; n ← |n|
if m < n
swap m ↔ n
while n > 0
m ← m mod n
swap m ↔ n
return m 2 () Algorithms Non-Lecture M: Number-Theoretic Algorithms This algorithm swaps m and n at every iteration, because m mod n is always less than n. This is almost
universally called Euclid’s algorithm, because the main idea is included in Euclid’s Elements.3
The easiest way to analyze this algorithm is to work backward. First, let’s consider the number of
iterations of the main loop, or equivalently, the number times line ( ) is executed. To keep things simple,
let’s assume that m > n > 0, so the ﬁrst three lines are redundant, and the algorithm performs at least
one iteration. Recall that the Fibonacci numbers(!) are deﬁned as F0 = 0, F1 = 1, and Fk = Fk−1 + Fk−2
for all k > 1.
Lemma 2. If the algorithm performs k iterations, then m ≥ Fk+2 and n ≥ Fk+1 .
Proof (by induction on k ): If k = 1, we have the trivial bounds n ≥ 1 = F2 and m ≥ 2 = F3 .
Suppose k > 1. The ﬁrst iteration of the loop replaces (m, n) with (n, m mod n). The algorithm
performs k − 1 more iterations, so the inductive hypothesis implies that n ≥ Fk+1 and m mod n ≥ Fk .
We’ve assumed that m > n, so m ≥ m + n(1 − m/n ) = n +(m mod n). We conclude that m ≥ Fk+1 + Fk =
F k +1 .
Theorem 1. EUCLIDGCD(m, n) runs in O(log m) iterations.
Proof: Let k be the number of iterations. Lemma 2 implies that m ≥ Fk+2 ≥ φ k+2 / 5 − 1, where
φ = (1 + 5)/2 (by the annihilator method). Thus, k ≤ logφ ( 5(m + 1)) − 2 = O(log m).
What about the actual running time? Every number used by the algorithm has O(log m) bits.
Computing the remainder of one b-bit integer by another using the grade-school long division algorithm
requires O( b2 ) time. So crudely, the running time is O( b2 log m) = O(log3 m). More careful analysis
reduces the time bound to O(log2 m) . We can make the algorithm even faster by using a fast integer
division algorithm (based on FFTs, for example). M.3 Modular Arithmetic and Algebraic Groups Modular arithmetic is familiar to anyone who’s ever wondered how many minutes are left in an exam
that ends at 9:15 when the clock says 8:54.
When we do arithmetic ‘modulo n’, what we’re really doing is a funny kind of arithmetic on the
elements of following set: Z n = {0, 1, 2, . . . , n − 1}
Z
Modular addition and subtraction satisﬁes all the axioms that we expect implicitly:
• Z n is closed under addition mod n: For any a, b ∈ Z n , their sum a + b mod n is also in Z n
Z
Z
Z
3
However, Euclid’s exposition was a little, erm, informal by current standards, primarily because the Greeks didn’t know
about induction. He basically said “Try one iteration. If that doesn’t work, try three iterations." In modern language, Euclid’s
algorithm would be written as follows, assuming m ≥ n > 0. ACTUALEUCLIDGCD(m, n):
if n | m
return n
else
return n mod (m mod n) This algorithm is obviously incorrect; consider the input m = 3, n = 2. Nevertheless, mathematics and algorithms students have
applied ‘Euclidean induction’ to a vast number of problems, only to scratch their heads in dismay when they don’t get any
credit. 3 Algorithms Non-Lecture M: Number-Theoretic Algorithms • Addition is associative: (a + b mod n) + c mod n = a + ( b + c mod n) mod n.
• Zero is an additive identity element: 0 + a mod n = a + 0 mod n = a mod n. Z
Z
• Every element a ∈ Z n has an inverse b ∈ Z n such that a + b mod n = 0. Speciﬁcally, if a = 0, then
b = 0; otherwise, b = n − a.
Any set with a binary operator that satisﬁes the closure, associativity, identity, and inverse axioms is
called a group. Since Z n is a group under an ‘addition’ operator, we call it an additive group. Moreover,
Z
because addition is commutative (a + b mod n = b + a mod n), we can call (Z n , + mod n) is an abelian
Z
4
additive group.
What about multiplication? Z n is closed under multiplication mod n, multiplication mod n is
Z
associative (and commutative), and 1 is a multiplicative identity, but some elements do not have
multiplicative inverses. Formally, we say that Z n is a ring under addition and multiplication modulo n.
Z
If n is composite, then the following theorem shows that we can factor the ring Z n into two smaller
Z
rings. The Chinese Remainder Theorem is named a third-century Chinese mathematician and algorismist
Sun Tzu (or Sun Zi).5
The Chinese Remainder Theorem. If p ⊥ q, then Z pq ∼ Z p × Z q .
Z =Z
Z
Okay, okay, before we prove this, let’s deﬁne all the notation. The product Z p × Z q is the set of
Z
Z
ordered pairs {(a, b) | a ∈ Z p , b ∈ Z q }, where addition, subtraction, and multiplication are deﬁned as
Z
Z
follows:
(a, b) + (c , d ) = (a + c mod p, b + d mod q)
(a, b) − (c , d ) = (a − c mod p, b − d mod q)
(a, b) · (c , d ) = (ac mod p, bd mod q)
It’s not hard to check that Z p × Z q is a ring under these operations, where (0, 0) is the additive identity
Z
Z
and (1, 1) is the multiplicative identity. The funky equal sign ∼ means that these two rings are isomorphic:
=
there is a bijection between the two sets that is consistent with the arithmetic operations.
As an example, the following table describes the bijection between Z 15 and Z 3 × Z 5 :
Z
Z
Z
0
1
2
3
4
00
6 12 3
9
1 10 1
7 13 4
2 5 11 2
8 14
For instance, we have 8 = (2, 3) and 13 = (1, 3), and
(2, 3) + (1, 3) = (2 + 1 mod 3, 3 + 3 mod 5) = (0, 1) = 6 = 21 mod 15 = (8 + 13) mod 15.
(2, 3) · (1, 3) = (2 · 1 mod 3, 3 · 3 mod 5) = (2, 4) = 14 = 104 mod 15 = (8 · 13) mod 15.
Proof: The functions n → (n mod p, n mod q) and (a, b) → aq(q mod p) + bp( p mod q) are inverses of
each other, and each map preserves the ring structure.
We can extend the Chinese remainder theorem inductively as follows:
4 after the Norwegian mathematical prodigy Niels Henrik Abel, who (among many other things) proved the insolubility of
quintic equations at the ripe old age of 22.
5
The author of The Art of War, who had the same name, lived more than 500 years earlier. 4 Algorithms Non-Lecture M: Number-Theoretic Algorithms
r
i =1 The Real Chinese Remainder Theorem. Suppose n =
r
Z n ∼ i =1 Z pi = Z p1 × Z p2 × · · · × Z pr .
Z=
Z
Z
Z
Z pi , where pi ⊥ p j for all i and j . Then Thus, if we want to perform modular arithmetic where the modulus n is very large, we can improve
the performance of our algorithms by breaking n into several relatively prime factors, and performing
modular arithmetic separately modulo each factor.
So we can do modular addition, subtraction, and multiplication; what about division? As I said
earlier, not every element of Z n has a multiplicative inverse. The most obvious example is 0, but there
Z
can be others. For example, 3 has no multiplicative inverse in Z 15 ; there is no integer x such that
Z
3 x mod 15 = 1. On the other hand, 0 is the only element of Z 7 without a multiplicative inverse:
Z
1 · 1 ≡ 2 · 4 ≡ 3 · 5 ≡ 6 · 6 ≡ 1 (mod 7)
These examples suggest (I hope) that x has a multiplicative inverse in Z n if and only if a and x are
Z
relatively prime. This is easy to prove as follows. If x y mod n = 1, then x y + kn = 1 for some integer
k. Thus, 1 is an integer linear combination of x and n, so Lemma 1 implies that gcd( x , n) = 1. On the
other hand, if x ⊥ n, then a x + bn = 1 for some integers a and b, which implies that a x mod n = 1.
Let’s deﬁne the set Z ∗ to be the set of elements if Z n that have multiplicative inverses.
Zn
Z Z ∗ = { a ∈ Z n | a ⊥ n}
Zn
Z
It is a tedious exercise to show that Z ∗ is an abelian group under multiplication modulo n. As long as
Zn
we stick to elements of this group, we can reasonably talk about ‘division mod n’.
We denote the number of elements in Z ∗ by φ (n); this is called Euler’s totient function. This
Zn
function is remarkably badly-behaved, but there is a relatively simple formula for φ (n) (not surprisingly)
involving prime numbers and division:
p−1 φ ( n) = n
p|n p I won’t prove this formula, but the following intuition is helpful. If we start with Z n and throw out
Z
all n/2 multiples of 2, all n/3 multiples of 3, all n/5 multiples of 5, and so on. Whenever we throw
out multiples of p, we multiply the size of the set by ( p − 1)/ p. At the end of this process, we’re left
with precisely the elements of Z ∗ . This is not a proof! On the one hand, this argument throws out some
Zn
numbers (like 6) more than once, so our estimate seems too low. On the other hand, there are actually
n/ p multiples of p in Z n , so our estimate seems too high. Surprisingly, these two errors exactly cancel
Z
each other out. M.4 Toward Primality Testing In this last section, we discuss algorithms for detecting whether a number is prime. Large prime numbers
are used primarily (but not exclusively) in cryptography algorithms.
A positive integer is prime if it has exactly two positive divisors, and composite if it has more than
two positive divisors. The integer 1 is neither prime nor composite. Equivalently, an integer n ≥ 2 is
prime if n is relatively prime with every positive integer smaller than n. We can rephrase this deﬁnition
yet again: n is prime if and only if φ (n) = n − 1.
The obvious algorithm for testing whether a number is prime is trial division: simply try every
possible nontrivial divisor between 2 and n.
5 Algorithms Non-Lecture M: Number-Theoretic Algorithms
TRIALDIVPRIME(n) :
for d ← 1 to
n
if n mod d = 0
return COMPOSITE
return PRIME Unfortunately, this algorithm is horribly slow. Even if we could do the remainder computation in constant
time, the overall running time of this algorithm would be Ω( n), which is exponential in the number of
input bits.
This might seem completely hopeless, but fortunately most composite numbers are quite easy to
detect as composite. Consider, for example, the related problem of deciding whether a given integer n,
whether n = me for any integers m > 1 and e > 1. We can solve this problem in polynomial time with the
following straightforward algorithm. The subroutine ROOT(n, i ) computes n1/i essentially by binary
search. (I’ll leave the analysis as a simple exercise.)
ROOT(n, i ):
r ←0
for ← (lg n)/i down to 1
if ( r + 2 )i ≤ n
r ← r +2
return r EXACTPOWER?(n):
for i ← 2 to lg n
if (ROOT(n, i ))i = n
return TRUE
return FALSE To distinguish between arbitrary prime and composite numbers, we need to exploit some results
about Z ∗ from group theory and number theory. First, we deﬁne the order of an element x ∈ Z ∗ as the
Zn
Zn
smallest positive integer k such that x k ≡ 1 (mod n). For example, in the group Z ∗ = {1, 2, 4, 7, 8, 11, 13, 14},
Z15
the number 2 has order 4, and the number 11 has order 2. For any x ∈ Z ∗ , we can partition the elements
Zn
∗
k
of Z n into equivalence classes, by declaring a ∼ x b if a ≡ b · x for some integer k. The size of every
Z
equivalence class is exactly the order of x . Because the equivalence classes must be disjoint, we can
conclude that the order of any element divides the size of the group . We can express this observation
more succinctly as follows:
Euler’s Theorem. aφ (n) ≡ 1 (mod n).6
The most interesting special case of this theorem is when n is prime.
Fermat’s Little Theorem. If p is prime, then a p ≡ a (mod p).7
This theorem leads to the following efﬁcient pseudo-primality test.
6 This is not Euler’s only theorem; he had thousands. It’s not even his most famous theorem. His second most famous
theorem is the formula v + e − f = 2 relating the vertices, edges and faces of any planar map. His most famous theorem is the
magic formula eπi + 1 = 0. Naming something after a mathematician or physicist (as in ‘Euler tour’ or ‘Gaussian elimination’ or
‘Avogadro’s number’) is considered a high compliment. Using a lower case letter (‘abelian group’) is even better; abbreviating
(‘volt’, ‘amp’) is better still. The number e was named after Euler.
7
This is not Fermat’s only theorem; he had hundreds, most of them stated without proof. Fermat’s Last Theorem wasn’t the
last one he published, but the last one proved. Amazingly, despite his dislike of writing proofs, Fermat was almost always right.
In that respect, he was very different from you and me. 6 Algorithms Non-Lecture M: Number-Theoretic Algorithms
FERMATPSEUDOPRIME(n) :
choose an integer a between 1 and n − 1
if a n mod n = a
return COMPOSITE!
else
return PRIME? In practice, this algorithm is both fast and effective. The (empirical) probability that a random
100-digit composite number will return PRIME? is roughly 10−30 , even if we always choose a = 2.
Unfortunately, there are composite numbers that always pass this test, no matter which value of a
we use. A Carmichael number is a composite integer n such that a n ≡ a (mod n) for every integer a.
Thus, Fermat’s Little Theorem can be used to distinguish between two types of numbers: (primes and
Carmichael numbers) and everything else. Carmichael numbers are extremely rare; in fact, it was proved
only in the 1990s that there are an inﬁnite number of them.
To deal with Carmichael numbers effectively, we need to look more closely at the structure of the
group Z ∗ . We say that Z ∗ is cyclic if it contains an element of order φ (n); such an element is called a
Zn
Zn
generator. Successive powers of any generator cycle through every element of the group in some order.
For example, the group Z ∗ = {1, 2, 4, 5, 7, 8} is cyclic, with two generators: 2 and 5, but Z ∗ is not
Z9
Z15
∗
cyclic. The following theorem completely characterizes which groups Z n are cyclic.
Z
The Cycle Theorem. Z ∗ is cyclic if and only if n = 2, 4, p e , or 2 p e for some odd prime p and positive
Zn
integer e.
This theorem has two relatively simple corollaries.
The Discrete Log Theorem. Suppose Z ∗ is cyclic and g is a generator. Then g x ≡ g y (mod n) if and
Zn
only if x ≡ y (mod φ (n)).
Proof: Suppose g x ≡ g y (mod n). By deﬁnition of ‘generator’, the sequence 〈1, g , g 2 , . . .〉 has period
φ (n). Thus, x ≡ y (mod φ (n)). On the other hand, if x ≡ y (mod φ (n)), then x = y + kφ (n) for
some integer k, so g x = g y +kφ (n) = g y · ( g φ (n) )k . Euler’s Theorem now implies that ( g φ (n) )k ≡ 1k ≡ 1
(mod n), so g x ≡ g y (mod n).
The 1 Theorem. Suppose n = p e for some odd prime p and positive integer e. The only elements
x ∈ Z ∗ that satisfy the equation x 2 ≡ 1 (mod n) are x = 1 and x = n − 1.
Zn
Proof: Obviously 12 ≡ 1 (mod n) and (n − 1)2 = n2 − 2n + 1 ≡ 1 (mod n).
Suppose x 2 ≡ 1 (mod n) where n = p e . By the Cycle Theorem, Z ∗ is cyclic. Let g be a generator of
Zn
Z ∗ , and suppose x = g k . Then we immediately have x 2 = g 2k ≡ g 0 = 1 (mod p e ). The Discrete Log
Zn
Theorem implies that 2k ≡ 0 (mod φ ( p e )). Because p is and odd prime, we have φ ( p e ) = ( p − 1) p e−1 ,
which is even. Thus, the equation 2k ≡ 0 (mod φ ( p e )) has just two solutions: k = 0 and k = φ ( p e )/2.
By the Cycle Theorem, either x = 1 or x = g φ (n)/2 . Because x = n − 1 is also a solution to the original
equation, we must have g φ (n)/2 ≡ n − 1 (mod n).
This theorem leads to a different pseudo-primality algorithm:
SQRT1PSEUDOPRIME(n):
choose a number a between 2 and n − 2
if a2 mod n = 1
return COMPOSITE!
else
return PRIME? 7 Algorithms Non-Lecture M: Number-Theoretic Algorithms As with the previous pseudo-primality test, there are composite numbers that this algorithm cannot
identify as composite: powers of primes, for instance. Fortunately, however, the set of composites that
always pass the 1 test is disjoint from the set of numbers that always pass the Fermat test. In particular,
Carmichael numbers never have the form p e . M.5 The Miller-Rabin Primality Test The following randomized algorithm, adapted by Michael Rabin from an earlier deterministic algorithm
of Gary Miller∗ , combines the Fermat test and the 1 test. The algorithm repeats the same two tests s
times, where s is some user-chosen parameter, each time with a random value of a.
MILLERRABIN(n):
write n − 1 = 2 t u where u is odd
for i ← 1 to s
a ← RANDOM(2, n − 2)
if EUCLIDGCD(a, n) = 1
return COMPOSITE! 〈〈obviously!〉〉 u x 0 ← a mod n
for j ← 1 to t
x j ← x 2−1 mod n
j
if x j = 1 and x j −1 = 1 and x j −1 = n − 1
return COMPOSITE! 〈〈by the 1 Theorem〉〉
if x t = 1
return COMPOSITE! 〈〈 x t = a n−1 mod n〉〉
〈〈by Fermat’s Little Theorem〉〉 return PRIME? First let’s consider the running time; for simplicity, we assume that all integer arithmetic is done using
the quadratic-time grade school algorithms. We can compute u and t in O(log n) time by scanning the bits
in the binary representation of n. Euclid’s algorithm takes O(log2 n) time. Computing au mod n requires
O(log u) = O(log n) multiplications, each of which takes O(log2 n) time. Squaring x j takes O(log2 n)
time. Overall, the running time for one iteration of the outer loop is O(log3 n + t log2 n) = O(log3 n),
because t ≤ lg n. Thus, the total running time of this algorithm is O(s log3 n) . If we set s = O(log n),
this running time is polynomial in the size of the input.
Fine, so it’s fast, but is it correct? Like the earlier pseudoprime testing algorithms, a prime input
will always cause MILLERRABIN to return PRIME?. Composite numbers, however, may not always
return COMPOSITE!; because we choose the number a at random, there is a small probability of error.8
Fortunately, the error probability can be made ridiculously small—in practice, less than the probability
that random quantum ﬂuctuations will instantly transform your computer into a kitten—by setting
s ≈ 1000.
Theorem 2. If n is composite, MILLERRABIN(n) returns COMPOSITE! with probability at least 1 − 2−s .
Proof: First, suppose n is not a Carmichael number. Let F be the set of elements of Z ∗ that pass the
Zn
Fermat test:
F = {a ∈ Z ∗ | a n−1 ≡ 1 (mod n)}.
Zn
8
If instead, we try all possible values of a, we obtain an exact primality testing algorithm, but it runs in exponential time.
Miller’s original deterministic algorithm examined every value of a in a carefully-chosen subset of Z ∗ . If the Extended Riemann
Zn
Hypothesis holds, this subset has logarithmic size, and Miller’s algorithm runs in polynomial time. The Riemann Hypothesis is
a century-old open problem about the distribution of prime numbers. A solution would be at least as signiﬁcant as proving
Fermat’s Last Theorem or P=NP
. 8 Algorithms Non-Lecture M: Number-Theoretic Algorithms Because n is not a Carmichael number, F is a proper subset of Z ∗ . Given any two elements a, b ∈ F ,
Zn
∗
their product a · b mod n in Z n is also an element of F :
Z
(a · b)n−1 ≡ a n−1 b n−1 ≡ 1 · 1 ≡ 1 (mod n)
We also easily observe that 1 is an element of F , and the multiplicative inverse (mod n) of any element
of F is also in F . Thus, F is a proper subgroup of Z ∗ , that is, a proper subset that is also a group under
Zn
the same binary operation. A standard result in group theory states that if F is a subgroup of a ﬁnite
group G , the number of elements of F divides the number of elements of G . (We used a special case of
this result in our proof of Euler’s Theorem.) In our setting, this means that | F | divides φ (n). Since we
already know that | F | < φ (n), we must have | F | ≤ φ (n)/2. Thus, at most half the elements of Z ∗ pass
Zn
the Fermat test.
The case of Carmichael numbers is more complicated, but the main idea is the same: at most half
the possible values of a pass the 1 test. See CLRS for further details. 9 Algorithms N Non-Lecture N: Convex Hulls Convex Hulls N.1 Deﬁnitions We are given a set P of n points in the plane. We want to compute something called the convex hull
of P . Intuitively, the convex hull is what you get by driving a nail into the plane at each point and
then wrapping a piece of string around the nails. More formally, the convex hull is the smallest convex
polygon containing the points:
• polygon: A region of the plane bounded by a cycle of line segments, called edges, joined end-to-end
in a cycle. Points where two successive edges meet are called vertices.
• convex: For any two points p, q inside the polygon, the line segment pq is completely inside the
polygon.
• smallest: Any convex proper subset of the convex hull excludes at least one point in P . This
implies that every vertex of the convex hull is a point in P .
We can also deﬁne the convex hull as the largest convex polygon whose vertices are all points in P , or
the unique convex polygon that contains P and whose vertices are all points in P . Notice that P might
have interior points that are not vertices of the convex hull. A set of points and its convex hull.
Convex hull vertices are black; interior points are white. Just to make things concrete, we will represent the points in P by their Cartesian coordinates,
in two arrays X [1 .. n] and Y [1 .. n]. We will represent the convex hull as a circular linked list of
vertices in counterclockwise order. if the i th point is a vertex of the convex hull, ne x t [i ] is index of
the next vertex counterclockwise and pr ed [i ] is the index of the next vertex clockwise; otherwise,
ne x t [i ] = pr ed [i ] = 0. It doesn’t matter which vertex we choose as the ‘head’ of the list. The decision
to list vertices counterclockwise instead of clockwise is arbitrary.
To simplify the presentation of the convex hull algorithms, I will assume that the points are in general
position, meaning (in this context) that no three points lie on a common line. This is just like assuming
that no two elements are equal when we talk about sorting algorithms. If we wanted to really implement
these algorithms, we would have to handle colinear triples correctly, or at least consistently. This is fairly
easy, but deﬁnitely not trivial. N.2 Simple Cases Computing the convex hull of a single point is trivial; we just return that point. Computing the convex
hull of two points is also trivial.
For three points, we have two different possibilities — either the points are listed in the array in
clockwise order or counterclockwise order. Suppose our three points are (a, b), (c , d ), and (e, f ), given
in that order, and for the moment, let’s also suppose that the ﬁrst point is furthest to the left, so a < c 1 Algorithms Non-Lecture N: Convex Hulls ←−−−−−→
and a < f . Then the three points are in counterclockwise order if and only if the line (a, b)(c , d ) is less
←−−−−−→
than the slope of the line (a, b)(e, f ):
counterclockwise ⇐⇒ d−b
c−a < f −b
e−a Since both denominators are positive, we can rewrite this inequality as follows:
counterclockwise ⇐⇒ ( f − b)(c − a) > (d − b)(e − a)
This ﬁnal inequality is correct even if (a, b) is not the leftmost point. If the inequality is reversed, then
the points are in clockwise order. If the three points are colinear (remember, we’re assuming that never
happens), then the two expressions are equal.
(e,f)
(a,b) (c,d)
Three points in counterclockwise order. Another way of thinking about this counterclockwise test is that we’re computing the cross-product
of the two vectors (c , d ) − (a, b) and (e, f ) − (a, b), which is deﬁned as a 2 × 2 determinant:
counterclockwise ⇐⇒ c−a
e−a d−b
>0
f −b We can also write it as a 3 × 3 determinant:
1a
counterclockwise ⇐⇒ 1 c
1e b
d >0
f All three boxed expressions are algebraically identical.
This counterclockwise test plays exactly the same role in convex hull algorithms as comparisons play
in sorting algorithms. Computing the convex hull of of three points is analogous to sorting two numbers:
either they’re in the correct order or in the opposite order. N.3 Jarvis’s Algorithm (Wrapping) Perhaps the simplest algorithm for computing convex hulls simply simulates the process of wrapping a
piece of string around the points. This algorithm is usually called Jarvis’s march, but it is also referred to
as the gift-wrapping algorithm.
Jarvis’s march starts by computing the leftmost point (that is, the point whose x -coordinate is
smallest), because we know that the left most point must be a convex hull vertex. Finding clearly takes
linear time. 2 Algorithms Non-Lecture N: Convex Hulls l p=l l
p p
p p
l l l
p The execution of Jarvis’s March. Then the algorithm does a series of pivoting steps to ﬁnd each successive convex hull vertex, starting
with and continuing until we reach again. The vertex immediately following a point p is the point
that appears to be furthest to the right to someone standing at p and looking at the other points. In
other words, if q is the vertex following p, and r is any other input point, then the triple p, q, r is in
counter-clockwise order. We can ﬁnd each successive vertex in linear time by performing a series of O(n)
counter-clockwise tests.
JARVISMARCH(X [1 .. n], Y [1 .. n]):
←1
for i ← 2 to n
if X [i ] < X [ ]
←i
p←
repeat
q ← p+1
〈〈Make sure p = q〉〉
for i ← 2 to n
if CCW( p, i , q)
q←i
nex t [ p] ← q; pr ev [q] ← p
p←q
until p = Since the algorithm spends O(n) time for each convex hull vertex, the worst-case running time is
O(n2 ). However, this naïve analysis hides the fact that if the convex hull has very few vertices, Jarvis’s
march is extremely fast. A better way to write the running time is O(nh), where h is the number of
convex hull vertices. In the worst case, h = n, and we get our old O(n2 ) time bound, but in the best case
h = 3, and the algorithm only needs O(n) time. Computational geometers call this an output-sensitive
algorithm; the smaller the output, the faster the algorithm. N.4 Divide and Conquer (Splitting) The behavior of Jarvis’s marsh is very much like selection sort: repeatedly ﬁnd the item that goes in the
next slot. In fact, most convex hull algorithms resemble some sorting algorithm.
For example, the following convex hull algorithm resembles quicksort. We start by choosing a pivot
point p. Partitions the input points into two sets L and R, containing the points to the left of p, including 3 Algorithms Non-Lecture N: Convex Hulls p itself, and the points to the right of p, by comparing x -coordinates. Recursively compute the convex
hulls of L and R. Finally, merge the two convex hulls into the ﬁnal output.
The merge step requires a little explanation. We start by connecting the two hulls with a line segment
between the rightmost point of the hull of L with the leftmost point of the hull of R. Call these points
p and q, respectively. (Yes, it’s the same p.) Actually, let’s add two copies of the segment pq and call
them bridges. Since p and q can ‘see’ each other, this creates a sort of dumbbell-shaped polygon, which
is convex except possibly at the endpoints off the bridges. p p q p q p p q q p q q Merging the left and right subhulls. We now expand this dumbbell into the correct convex hull as follows. As long as there is a clockwise
turn at either endpoint of either bridge, we remove that point from the circular sequence of vertices and
connect its two neighbors. As soon as the turns at both endpoints of both bridges are counter-clockwise,
we can stop. At that point, the bridges lie on the upper and lower common tangent lines of the two
subhulls. These are the two lines that touch both subhulls, such that both subhulls lie below the upper
common tangent line and above the lower common tangent line.
Merging the two subhulls takes O(n) time in the worst case. Thus, the running time is given by
the recurrence T (n) = O(n) + T (k) + T (n − k), just like quicksort, where k the number of points in R.
Just like quicksort, if we use a naïve deterministic algorithm to choose the pivot point p, the worst-case
running time of this algorithm is O(n2 ). If we choose the pivot point randomly, the expected running
time is O(n log n).
There are inputs where this algorithm is clearly wasteful (at least, clearly to us). If we’re really
unlucky, we’ll spend a long time putting together the subhulls, only to throw almost everything away
during the merge step. Thus, this divide-and-conquer algorithm is not output sensitive. A set of points that shouldn’t be divided and conquered. N.5 Graham’s Algorithm (Scanning) Our third convex hull algorithm, called Graham’s scan, ﬁrst explicitly sorts the points in O(n log n) and
then applies a linear-time scanning algorithm to ﬁnish building the hull.
4 Algorithms Non-Lecture N: Convex Hulls We start Graham’s scan by ﬁnding the leftmost point , just as in Jarvis’s march. Then we sort the
points in counterclockwise order around . We can do this in O(n log n) time with any comparison-based
sorting algorithm (quicksort, mergesort, heapsort, etc.). To compare two points p and q, we check
whether the triple , p, q is oriented clockwise or counterclockwise. Once the points are sorted, we
connected them in counterclockwise order, starting and ending at . The result is a simple polygon with
n vertices. l A simple polygon formed in the sorting phase of Graham’s scan. To change this polygon into the convex hull, we apply the following ‘three-penny algorithm’. We
have three pennies, which will sit on three consecutive vertices p, q, r of the polygon; initially, these are
and the two vertices after . We now apply the following two rules over and over until a penny is
moved forward onto :
• If p, q, r are in counterclockwise order, move the back penny forward to the successor of r .
• If p, q, r are in clockwise order, remove q from the polygon, add the edge pr , and move the middle
penny backward to the predecessor of p. The ‘three-penny’ scanning phase of Graham’s scan. Whenever a penny moves forward, it moves onto a vertex that hasn’t seen a penny before (except
the last time), so the ﬁrst rule is applied n − 2 times. Whenever a penny moves backwards, a vertex is
removed from the polygon, so the second rule is applied exactly n − h times, where h is as usual the
number of convex hull vertices. Since each counterclockwise test takes constant time, the scanning
phase takes O(n) time altogether. 5 Algorithms N.6 Non-Lecture N: Convex Hulls Chan’s Algorithm (Shattering) The last algorithm I’ll describe is an output-sensitive algorithm that is never slower than either Jarvis’s
march or Graham’s scan. The running time of this algorithm, which was discovered by Timothy Chan in
1993, is O(n log h). Chan’s algorithm is a combination of divide-and-conquer and gift-wrapping.
First suppose a ‘little birdie’ tells us the value of h; we’ll worry about how to implement the little
birdie in a moment. Chan’s algorithm starts by shattering the input points into n/h arbitrary1 subsets,
each of size h, and computing the convex hull of each subset using (say) Graham’s scan. This much of
the algorithm requires O((n/h) · h log h) = O(n log h) time. Shattering the points and computing subhulls in O(n log h) time. Once we have the n/h subhulls, we follow the general outline of Jarvis’s march, ‘wrapping a string
around’ the n/h subhulls. Starting with p = , where is the leftmost input point, we successively ﬁnd
the convex hull vertex the follows p and counterclockwise order until we return back to again.
The vertex that follows p is the point that appears to be furthest to the right to someone standing
at p. This means that the successor of p must lie on a right tangent line between p and one of the
subhulls—a line from p through a vertex of the subhull, such that the subhull lies completely on the
right side of the line from p’s point of view. We can ﬁnd the right tangent line between p and any subhull
in O(log h) time using a variant of binary search. (This is a practice problem in the homework!) Since
there are n/h subhulls, ﬁnding the successor of p takes O((n/h) log h) time altogether.
Since there are h convex hull edges, and we ﬁnd each edge in O((n/h) log h) time, the overall running
time of the algorithm is O(n log h). Wrapping the subhulls in O(n log h) time. Unfortunately, this algorithm only takes O(n log h) time if a little birdie has told us the value of h in
advance. So how do we implement the ‘little birdie’? Chan’s trick is to guess the correct value of h; let’s
denote the guess by h∗ . Then we shatter the points into n/h∗ subsets of size h∗ , compute their subhulls,
and then ﬁnd the ﬁrst h∗ edges of the global hull. If h < h∗ , this algorithm computes the complete convex
hull in O(n log h∗ ) time. Otherwise, the hull doesn’t wrap all the way back around to , so we know our
guess h∗ is too small.
Chan’s algorithm starts with the optimistic guess h∗ = 3. If we ﬁnish an iteration of the algorithm
i
and ﬁnd that h∗ is too small, we square h∗ and try again. Thus, in the i th iteration, we have h∗ = 32 . In
1
In the ﬁgures, in order to keep things as clear as possible, I’ve chosen these subsets so that their convex hulls are disjoint.
This is not true in general! 6 Algorithms Non-Lecture N: Convex Hulls the ﬁnal iteration, h∗ < h2 , so the last iteration takes O(n log h∗ ) = O(n log h2 ) = O(n log h) time. The
total running time of Chan’s algorithm is given by the sum
k k
i O(n log 32 ) = O(n) ·
i =1 2i
i =1 for some integer k. Since any geometric series adds up to a constant times its largest term, the total
running time is a constant times the time taken by the last iteration, which is O(n log h). So Chan’s
algorithm runs in O(n log h) time overall, even without the little birdie. Exercises
1. The convex layers of a point set X are deﬁned by repeatedly computing the convex hull of X and
removing its vertices from X , until X is empty.
(a) Describe an algorithm to compute the convex layers of a given set of n points in the plane in
O(n2 ) time.
(b) Describe an algorithm to compute the convex layers of a given set of n points in the plane in
O(n log n) time. The convex layers of a set of points. 2. Let X be a set of points in the plane. A point p in X is Pareto-optimal if no other point in X is both
above and to the right of p. The Pareto-optimal points can be connected by horizontal and vertical
lines into the staircase of X , with a Pareto-optimal point at the top right corner of every step. See
the ﬁgure on the next page.
(a) QUICKSTEP: Describe a divide-and-conquer algorithm to compute the staircase of a given set
of n points in the plane in O(n log n) time.
(b) SCANSTEP: Describe an algorithm to compute the staircase of a given set of n points in the
plane, sorted in left to right order, in O(n) time.
(c) NEXTSTEP: Describe an algorithm to compute the staircase of a given set of n points in the
plane in O(nh) time, where h is the number of Pareto-optimal points.
(d) SHATTERSTEP: Describe an algorithm to compute the staircase of a given set of n points in the
plane in O(n log h) time, where h is the number of Pareto-optimal points.
In all these problems, you may assume that no two points have the same x - or y -coordinates.
7 Algorithms Non-Lecture N: Convex Hulls The staircase (thick line) and staircase layers (all lines) of a set of points. 3. The staircase layers of a point set are deﬁned by repeatedly computing the staircase and removing
the Pareto-optimal points from the set, until the set becomes empty.
(a) Describe and analyze an algorithm to compute the staircase layers of a given set of n points
in O(n log n) time.
(b) An increasing subsequence of a point set X is a sequence of points in X such that each point is
above and to the right of its predecessor in the sequence. Describe and analyze an algorithm
to compute the longest increasing subsequence of a given set of n points in the plane in
O(n log n) time. [Hint: There is a one-line solution that uses part (a). But why is is correct?] c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 8 Algorithms Non-Lecture O: Line Segment Intersection
Spengler: There’s something very important I forgot to tell you.
Venkman: What?
Spengler: Don’t cross the streams.
Venkman: Why?
Spengler: It would be bad.
Venkman: I’m fuzzy on the whole good/bad thing. What do you mean, “bad"?
Spengler: Try to imagine all life as you know it stopping instantaneously and every
molecule in your body exploding at the speed of light.
Stantz: Total protonic reversal.
Venkman: Right. That’s bad. Okay. All right. Important safety tip. Thanks, Egon.
— Ghostbusters (1984) O Line Segment Intersection O.1 Introduction In this lecture, I’ll talk about detecting line segment intersections. A line segment is the convex hull of
two points, called the endpoints (or vertices) of the segment. We are given a set of n line segments, each
speciﬁed by the x - and y -coordinates of its endpoints, for a total of 4n real numbers, and we want to
know whether any two segments intersect.
To keep things simple, just as in the previous lecture, I’ll assume the segments are in general position.
• No three endpoints lie on a common line.
• No two endpoints have the same x -coordinate. In particular, no segment is vertical, no segment is
just a point, and no two segments share an endpoint.
This general position assumption lets us avoid several annoying degenerate cases. Of course, in any real
implementation of the algorithm I’m about to describe, you’d have to handle these cases. Real-world
data is full of degeneracies! Degenerate cases of intersecting segments that we’ll pretend never happen:
Overlapping colinear segments, endpoints inside segments, and shared endpoints. O.2 Two segments The ﬁrst case we have to consider is n = 2. (The problem is obviously trivial when n ≤ 1!) How do we
tell whether two line segments intersect? One possibility, suggested by a student in class, is to construct
the convex hull of the segments. Two segments intersect if and only if the convex hull is a quadrilateral
whose vertices alternate between the two segments. In the ﬁgure below, the ﬁrst pair of segments has a
triangular convex hull. The last pair’s convex hull is a quadrilateral, but its vertices don’t alternate. Some pairs of segments. 1 Algorithms Non-Lecture O: Line Segment Intersection Fortunately, we don’t need (or want!) to use a full-ﬂedged convex hull algorithm just to test two
segments; there’s a much simpler test.
Two segments ab and cd intersect if and only if
←
→
• the endpoints a and b are on opposite sides of the line cd , and
←
→
• the endpoints c and d are on opposite sides of the line ab.
To test whether two points are on opposite sides of a line through two other points, we use the same
counterclockwise test that we used for building convex hulls. Speciﬁcally, a and b are on opposite sides
←
→
of line cd if and only if exactly one of the two triples a, c , d and b, c , d is in counterclockwise order. So
we have the following simple algorithm.
INTERSECT(a, b, c , d ):
if CCW(a, c , d ) = CCW( b, c , d )
return FALSE
else if CCW(a, b, c ) = CCW(a, b, d )
return FALSE
else
return TRUE Or even simpler:
INTERSECT(a, b, c , d ):
return CCW(a, c , d ) = CCW( b, c , d ) ∧ CCW(a, b, c ) = CCW(a, b, d ) O.3 A Sweep Line Algorithm To detect whether there’s an intersection in a set of more than just two segments, we use something
called a sweep line algorithm. First let’s give each segment a unique label. I’ll use letters, but in a real
implementation, you’d probably use pointers/references to records storing the endpoint coordinates.
Imagine sweeping a vertical line across the segments from left to right. At each position of the sweep
line, look at the sequence of (labels of) segments that the line hits, sorted from top to bottom. The only
times this sorted sequence can change is when the sweep line passes an endpoint or when the sweep line
passes an intersection point. In the second case, the order changes because two adjacent labels swap
places.1 Our algorithm will simulate this sweep, looking for potential swaps between adjacent segments.
The sweep line algorithm begins by sorting the 2n segment endpoints from left to right by comparing
their x -coordinates, in O(n log n) time. The algorithm then moves the sweep line from left to right,
stopping at each endpoint.
We store the vertical label sequence in some sort of balanced binary tree that supports the following
operations in O(log n) time. Note that the tree does not store any explicit search keys, only segment
labels.
• Insert a segment label.
• Delete a segment label.
• Find the neighbors of a segment label in the sorted sequence.
O(log n) amortized time is good enough, so we could use a scapegoat tree or a splay tree. If we’re willing
to settle for an expected time bound, we could use a treap or a skip list instead.
1 Actually, if more than two segments intersect at the same point, there could be a larger reversal, but this won’t have any
effect on our algorithm. 2 Algorithms Non-Lecture O: Line Segment Intersection B F EE
BB
CD
D EE
FF
BD
D C A A E D
B
A BB
AC
C B
C
D ! The sweep line algorithm in action. The boxes show the label sequence stored in the binary tree.
The intersection between F and D is detected in the last step. Whenever the sweep line hits a left endpoint, we insert the corresponding label into the tree in
O(log n) time. In order to do this, we have to answer questions of the form ‘Does the new label X go
above or below the old label Y?’ To answer this question, we test whether the new left endpoint of X is
above segment Y, or equivalently, if the triple of endpoints left(Y), right(Y), left(X) is in counterclockwise
order.
Once the new label is inserted, we test whether the new segment intersects either of its two neighbors
in the label sequence. For example, in the ﬁgure above, when the sweep line hits the left endpoint of F,
we test whether F intersects either B or E. These tests require O(1) time.
Whenever the sweep line hits a right endpoint, we delete the corresponding label from the tree in
O(log n) time, and then check whether its two neighbors intersect in O(1) time. For example, in the
ﬁgure, when the sweep line hits the right endpoint of C, we test whether B and D intersect.
If at any time we discover a pair of segments that intersects, we stop the algorithm and report the
intersection. For example, in the ﬁgure, when the sweep line reaches the right endpoint of B, we discover
that F and D intersect, and we halt. Note that we may not discover the intersection until long after the
two segments are inserted, and the intersection we discover may not be the one that the sweep line
would hit ﬁrst. It’s not hard to show by induction (hint, hint) that the algorithm is correct. Speciﬁcally, if
the algorithm reaches the nth right endpoint without detecting an intersection, none of the segments
intersect.
For each segment endpoint, we spend O(log n) time updating the binary tree, plus O(1) time
performing pairwise intersection tests—at most two at each left endpoint and at most one at each
right endpoint. Thus, the entire sweep requires O(n log n) time in the worst case. Since we also spent
O(n log n) time sorting the endpoints, the overall running time is O(n log n).
Here’s a slightly more formal description of the algorithm. The input S [1 .. n] is an array of line
segments. The sorting phase in the ﬁrst line produces two auxiliary arrays:
• label[i ] is the label of the i th leftmost endpoint. I’ll use indices into the input array S as the labels,
so the i th vertex is an endpoint of S [label[i ]].
• isleft[i ] is TRUE if the i th leftmost endpoint is a left endpoint and FALSE if it’s a right endpoint.
The functions INSERT, DELETE, PREDECESSOR, and SUCCESSOR modify or search through the sorted
label sequence. Finally, INTERSECT tests whether two line segments intersect. 3 Algorithms Non-Lecture O: Line Segment Intersection
ANYINTERSECTIONS(S [1 .. n]):
sort the endpoints of S from left to right
create an empty label sequence
for i ← 1 to 2n
← label[i ]
if isleft[i ]
INSERT( )
if INTERSECT(S [ ], S [SUCCESSOR( )])
return TRUE
if INTERSECT(S [ ], S [PREDECESSOR( )])
return TRUE
else
if INTERSECT(S [SUCCESSOR( )], S [PREDECESSOR( )])
return TRUE
DELETE(label[i ])
return FALSE Note that the algorithm doesn’t try to avoid redundant pairwise tests. In the ﬁgure below, the top
and bottom segments would be checked n − 1 times, once at the top left endpoint, and once at the right
endpoint of every short segment. But since we’ve already spent O(n log n) time just sorting the inputs,
O(n) redundant segment intersection tests make no difference in the overall running time. The same pair of segments might be tested n − 1 times. Exercises
1. Let X be a set of n rectangles in the plane, each speciﬁed by a left and right x -coordinate and a
top and bottom y -coordinate. Thus, the input consists of four arrays L [1 .. n], R[1 .. n], T [1 .. n],
and B [1 .. n], where L [i ] < R[i ] and T [i ] > B [i ] for all i .
(a) Describe and analyze and algorithm to determine whether any two rectangles in X intersect,
in O(n log n) time.
(b) Describe and analyze an algorithm to ﬁnd a point that lies inside the largest number of
rectangles in X , in O(n log n) time.
(c) Describe and analyze an algorithm to compute the area of the union of the rectangles in X ,
in O(n log n) time.
2. Describe and analyze a sweepline algorithm to determine, given n circles in the plane, whether
any two intersect, in O(n log n) time. Each circle is speciﬁed by its center and its radius, so the
input consists of three arrays X [1 .. n], Y [1 .. n], and R[1 .. n]. Be careful to correctly implement
the low-level primitives.
3. Describe an algorithm to determine, given n line segments in the plane, a list of all intersecting
pairs of segments. Your algorithm should run in O(n log n + k log n) time, where n is the number
of segments, and k is the number of intersecting pairs.
4 Algorithms Non-Lecture O: Line Segment Intersection 4. This problem asks you to compute skylines of various cities. In each city, all the buildings have a
signature geometric shape. Describe an algorithm to compute a description of the union of n such
shapes in O(n log n) time.
(a) Manhattan: Each building is a rectangle whose bottom edge lies on the x -axis, speciﬁed by
the left and right x -coordinates and the top y -coordinate. The Manhattan skyline (b) Giza: Each building is a right isosceles triangle whose base lies on the x -axis, speciﬁed by the
( x , y )-coordinates of its apex. The Egyptian skyline (c) Nome: Each building is a semi-circular disk whose center lies on the x -axis, speciﬁed by its
center x -coordinate and radius.
5. [Adapted from CLRS, problem 33-3.] A group of n Ghostbusters are teaming up to ﬁght n ghosts.
Each Ghostbuster is armed with a proton pack that shoots a stream along a straight line until it
encounters (and neutralizes) a ghost. The Ghostbusters decide that they will ﬁre their proton
packs simultaneously, with each Ghostbuster aiming at a different ghost. Crossing the streams
would be bad—total protonic reversal, yadda yadda—so it is vital that each Ghostbuster choose
his target carefully. Assume that each Ghostbuster and each ghost is a single point in the plane,
and that no three of these 2n points are collinear.
(a) Prove that there is a set of n disjoint line segments, each joining one Ghostbuster to one
ghost. [Hint: Consider the matching of minimum total length.]
(b) Prove that the non-collinearity assumption is necessary in part (a).
(c) A partitioning line is a line in the plane such that the number of ghosts on one side of the line
is equal to the number of Ghostbusters on the same side of that line. Describe an algorithm
to compute a partitioning line in O(n log n) time.
(d) Prove that there is a partitioning line with exactly n/2 ghosts and exactly n/2 Ghostbusters on either side. This is a special case of the so-called ham sandwich theorem. Describe
an algorithm to compute such a line as quickly as possible.
(e) Describe a randomized algorithm to ﬁnd an approximate ham-sandwich line—that is, a
partitioning line with at least n/4 ghosts on each side—in O(n log n) time.
(f) Describe an algorithm to compute a set of n disjoint line segments, each joining one Ghostbuster to one ghost, as quickly as possible.
c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 5 Algorithms Non-Lecture P: Polygon Triangulation If triangles had a god, they would give him three sides.
— Charles Louis de Secondat Montesquie (1721)
Down with Euclid! Death to triangles!
— Jean Dieudonné (1959) P Polygon Triangulation P
.1 Introduction Recall from last time that a polygon is a region of the plane bounded by a cycle of straight edges joined
end to end. Given a polygon, we want to decompose it into triangles by adding diagonals: new line
segments between the vertices that don’t cross the boundary of the polygon. Because we want to keep
the number of triangles small, we don’t allow the diagonals to cross. We call this decomposition a
triangulation of the polygon. Most polygons can have more than one triangulation; we don’t care which
one we compute. Two triangulations of the same polygon. Before we go any further, I encourage you to play around with some examples. Draw a few polygons
(making sure that the edges are straight and don’t cross) and try to break them up into triangles. P
.2 Existence and Complexity If you play around with a few examples, you quickly discover that every triangulation of an n-sided
has n − 2 triangles. You might even try to prove this observation by induction. The base case n = 3 is
trivial: there is only one triangulation of a triangle, and it obviously has only one triangle! To prove
the general case, let P be a polygon with n edges. Draw a diagonal between two vertices. This splits P
into two smaller polygons. One of these polygons has k edges of P plus the diagonal, for some integer k
between 2 and n − 2, for a total of k + 1 edges. So by the induction hypothesis, this polygon can be
broken into k − 1 triangles. The other polygon has n − k + 1 edges, and so by the induction hypothesis,
it can be broken into n − k − 1 tirangles. Putting the two pieces back together, we have a total of
(k − 1) + (n − k − 1) = n − 2 triangles. Breaking a polygon into two smaller polygons with a diagonal. 1 Algorithms Non-Lecture P: Polygon Triangulation This is a ﬁne induction proof, which any of you could have discovered on your own (right?), except
for one small problem. How do we know that every polygon has a diagonal? This seems patently
obvious, but it’s surprisingly hard to prove, and in fact many incorrect proofs were actually published as
late as 1975. The following proof is due to Meisters in 1975.
Lemma 1. Every polygon with more than three vertices has a diagonal.
Proof: Let P be a polygon with more than three vertices. Every vertex of a P is either convex or concave,
depending on whether it points into or out of P , respectively. Let q be a convex vertex, and let p and r
be the vertices on either side of q. For example, let q be the leftmost vertex. (If there is more than one
leftmost vertex, let q be the the lowest one.) If pr is a diagonal, we’re done; in this case, we say that the
triangle pqr is an ear.
If pr is not a diagonal, then pqr must contain another vertex of the polygon. Out of all the vertices
←
→
inside pqr , let s be the vertex furthest away from the line pr . In other words, if we take a line parallel
←
→
←
→
to pr through q, and translate it towards pr , then then s is the ﬁrst vertex that the line hits. Then the
line segment qs is a diagonal.
r’ s’ p q’ r
q p’ The leftmost vertex q is the tip of an ear, so pr is a diagonal.
The rightmost vertex q is not, since p q r contains three other vertices. In this case, q s is a diagonal. P
.3 Existence and Complexity Meister’s existence proof immediately gives us an algorithm to compute a diagonal in linear time. The
input to our algorithm is just an array of vertices in counterclockwise order around the polygon. First,
we can ﬁnd the (lowest) leftmost vertex q in O(n) time by comparing the x -coordinates of the vertices
(using y -coordinates to break ties). Next, we can determine in O(n) time whether the triangle pqr
contains any of the other n − 3 vertices. Speciﬁcally, we can check whether one point lies inside a
triangle by performing three counterclockwise tests. Finally, if the triangle is not empty, we can ﬁnd the
vertex s in O(n) time by comparing the areas of every triangle pqs; we can compute this area using
the counterclockwise determinant.
Here’s the algorithm in excruciating detail. We need three support subroutines to compute the area
of a polygon, to determine if three poitns are in counterclockwise order, and to determine if a point is
inside a triangle.
〈〈Return twice the signed area of P [i ] P [ j ] P [k]〉〉
AREA(i , j , k):
return ( P [k]. y − P [i ]. y )( P [ j ]. x − P [i ]. x ) − ( P [k]. x − P [i ]. x )( P [ j ]. y − P [i ]. y )
〈〈Are P [i ], P [ j ], P [k] in counterclockwise order?〉〉
CCW(i , j , k):
return AREA(i , j , k) > 0
〈〈Is P [i ] inside P [ p] P [q] P [ r ]?〉〉
INSIDE(i , p, q, r ):
return CCW(i , p, q) and CCW(i , q, r ) and CCW(i , r, p) 2 Algorithms Non-Lecture P: Polygon Triangulation
FINDDIAGONAL( P [1 .. n]):
q←1
for i ← 2 to n
if P [i ]. x < P [q]. x
q←i
p ← q − 1 mod n
r ← q + 1 mod n
ear ← TRUE
s←p
for i ← 1 to n
if i ≤ p and i = q and i = r and INSIDE(i , p, q, r )
ear ← FALSE
if AREA(i , r, p) > AREA(s, r, p)
s←i
if ear = TRUE
return ( p, r )
else
return (q, s) Once we have a diagonal, we can recursively triangulate the two pieces. The worst-case running
time of this algorithm satisﬁes almost the same recurrence as quicksort:
T (n) ≤ max T (k + 1) + T (n − k + 1) + O(n).
2≤k≤n−2 So we can now triangulate any polygon in O(n2 ) time. P
.4 Faster Special Cases For certain special cases of polygons, we can do much better than O(n2 ) time. For example, we can
easily triangulate any convex polygon by connecting any vertex to every other vertex. Since we’re given
the counterclockwise order of the vertices as input, this takes only O(n) time. Triangulating a convex polygon is easy. Another easy special case is monotone mountains. A polygon is monotone if any vertical line intersects
the boundary in at most two points. A monotone polygon is a mountain if it contains an edge from the
rightmost vertex to the leftmost vertex. Every monotone polygon consists of two chains of edges going
left to right between the two extreme vertices; for mountains, one of these chains is a single edge. A monotone mountain and a monotone non-mountain. 3 Algorithms Non-Lecture P: Polygon Triangulation Triangulating a monotone mounting is extremely easy, since every convex vertex is the tip of an ear,
except possibly for the vertices on the far left and far right. Thus, all we have to do is scan through the
intermediate vertices, and when we ﬁnd a convex vertex, cut off the ear. The simplest method for doing
this is probably the three-penny algorithm used in the “Graham’s scan” convex hull algorithm—instead
of ﬁlling in the outside of a polygon with triangles, we’re ﬁlling in the inside, both otherwise it’s the
same process. This takes O(n) time. Triangulating a monotone mountain. (Some of the triangles are very thin.) We can also triangulate general monotone polygons in linear time, but the process is more complicated. A good way to visuialize the algorithm is to think of the polygon as a complicated room. Two
people named Tom and Bob are walking along the top and bottom walls, both starting at the left end
and going to the right. At all times, they have a rubber band stretched between them that can never
leave the room.
T B A rubber band stretched between a vertex on the top and a vertex on the bottom of a monotone polygon. Now we loop through all the vertices of the polygon in order from left to right. Whenever we see
a new bottom vertex, Bob moves onto it, and whenever we see a new bottom vertex Tom moves onto
it. After either person moves, we cut the polygon along the rubber band. (In fact, this will only cut
the polygon along a single diagonal at any step.) When we’re done, the polygon is decomposed into
triangles and boomerangs—nonconvex polygons consisting of two straight edges and a concave chain. A
boomerang can only be triangulated in one way, by joining the apex to every vertex in the concave chain. 4 Algorithms Non-Lecture P: Polygon Triangulation Triangulating a monotone polygon by walking a rubber band from left to right. I don’t want to go into too many implementation details, but a few observations shoudl convince
you that this algorithm can be implemented to run in O(n) time. Notice that at all times, the rubber
band forms a concave chain. The leftmost edge in the rubber band joins a top vertex to a bottom vertex.
If the rubber band has any other vertices, either they are all on top or all on bottom. If all the other
vertices are on top, there are just three ways the rubber band can change:
1. The bottom vertex changes, the rubber band straighens out, and we get a new boomerang.
2. The top vertex changes and the rubber band gets a new concave vertex.
3. The top vertex changes, the rubber band loses some vertices, and we get a new boomerang.
Deciding between the ﬁrst case and the other two requires a simple comparison between x -coordinates.
Deciding between the last two requires a counterclockwise test. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 5 Algorithms Non-Lecture Q: Reductions
Reduce big troubles to small ones, and small ones to nothing.
— Chinese proverb
I have yet to see any problem, however complicated, which, when you looked at
it in the right way, did not become still more complicated.
— Poul Anderson, New Scientist (September 25, 1969) Q Reductions Q.1 Introduction An extremely common technique for deriving algorithms is reduction—instead of solving a problem
directly, we use an algorithm for some other related problem as a subroutine or black box.
For example, when we talked about nuts and bolts, we argued that once the nuts and bolts are sorted,
we can match each nut to its bolt in linear time. Thus, since we can sort nuts and bolts in O(n log n)
expected time, then we can also match them in O(n log n) expected time:
Tmatch (n) ≤ Tsort (n) + O(n) = O(n log n) + O(n) = O(n log n).
Let’s consider (as we did in the previous lecture) a decision tree model of computation, where every
query is a comparison between a nut and a bolt—too big, too small, or just right? The output to the
matching problem is a permutation π, where for all i , the i th nut matches the π(i )th bolt. Since there
are n! permutations of n items, any nut/bolt comparison tree that matches n nuts and bolts has at least
n! leaves, and thus has depth at least log3 (n!) = Ω(n log n).
Now the same reduction from matching to sorting can be used to prove a lower bound for sorting
nuts and bolts, just by reversing the inequality:
Tsort (n) ≥ Tmatch (n) − O(n) = Ω(n log n) − O(n) = Ω(n log n).
Thus, any nut-bolt comparison tree that sorts n nuts and bolts has depth Ω(n log n), and our randomized
quicksort algorithm is optimal.1
The rest of this lecture assumes some familiarity with computational geometry. Q.2 Sorting to Convex Hulls Here’s a slightly less trivial example. Suppose we want to prove a lower bound for the problem of
computing the convex hull of a set of n points in the plane. To do this, we demonstrate a reduction from
sorting to convex hulls.
To sort a list of n numbers {a, b, c , . . .}, we ﬁrst transform it into a set of n points {(a, a2 ), ( b, b2 ),
(c , c 2 ), . . .}. You can think of the original numbers as a set of points on a horizontal real number line,
and the transformation as lifting those point up to the parabola y = x 2 . Then we compute the convex
hull of the parabola points. Finally, to get the ﬁnal sorted list of numbers, we output the ﬁrst coordinate
of every convex vertex, starting from the leftmost vertex and going in counterclockwise order.
1 We could have proved this lower bound directly. The output to the sorting problem is two permutations, so there are n!2
possible outputs, and we get a lower bound of log3 (n!2 ) = Ω(n log n). 1 Algorithms Non-Lecture Q: Reductions Reducing sorting to computing a convex hull. Transforming the numbers into points takes O(n) time. Since the convex hull is output as a circular
doubly-linked list of vertices, reading off the ﬁnal sorted list of numbers also takes O(n) time. Thus,
given a black-box convex hull algorithm, we can sort in linear extra time. In this case, we say that there
is a linear time reduction from sorting to convex hulls. We can visualize the reduction as follows:
O(n) set of n numbers −
−−→ set of n points
CONVEXHULL
O(n) −
sorted list of numbers ←− convex polygon
(I strongly encourage you to draw a picture like this whenever you use a reduction argument, at least
until you get used to them.) The reduction gives us the following inequality relating the complexities of
the two problems:
Tsort (n) ≤ Tconvex hull (n) + O(n)
Since we can compute convex hulls in O(n log n) time, our reduction implies that we can also sort in
O(n log n) time. More importantly, by reversing the inequality, we get a lower bound on the complexity
of computing convex hulls.
Tconvex hull (n) ≥ Tsort (n) − O(n)
Since any binary decision tree requires Ω(n log n) time to sort n numbers, it follows that any binary
decision tree requires Ω(n log n) time to compute the convex hull of n points. Q.3 Watch the Model! This result about the complexity of computing convex hulls is often misquoted as follows:
Since we need Ω(n log n) comparisons to sort, we also need Ω(n log n) comparisons (between
x -coordinates) to compute convex hulls.
Although this statement is true, it’s completely trivial, since it’s impossible to compute convex hulls
using any number of comparisons! In order to compute hulls, we must perform counterclockwise tests
on triples of points.
The convex hull algorithms we’ve seen — Graham’s scan, Jarvis’s march, divide-and-conquer, Chan’s
shatter — can all be modeled as binary2 decision trees, where every query is a counterclockwise test on
three points. So our binary decision tree lower bound is meaningful, and several of those algorithms are
optimal.
This is a subtle but important point about deriving lower bounds using reduction arguments. In
order for any lower bound to be meaningful, it must hold in a model in which the problem can be solved!
2 or ternary, if we allow colinear triples of points 2 Algorithms Non-Lecture Q: Reductions Often the problem we are reducing from is much simpler than the problem we are reducing to, and thus
can be solved in a more restrictive model of computation. Q.4 Element Uniqueness (A Bad Example) The element uniqueness problem asks, given a list of n numbers x 1 , x 2 , . . . , x n , whether any two of them
are equal. There is an obvious and simple algorithm to solve this problem: sort the numbers, and then
scan for adjacent duplicates. Since we can sort in O(n log n) time, we can solve the element uniqueness
problem in O(n log n) time.
We also have an Ω(n log n) lower bound for sorting, but our reduction does not give us a lower
bound for element uniqueness. The reduction goes the wrong way! Inscribe the following on the back of
your hand3 :
To prove that problem A is harder than problem B, reduce B to A.
There isn’t (as far as I know) a reduction from sorting to the element uniqueness problem. However,
using other techniques (which I won’t talk about), it is possible to prove an Ω(n log n) lower bound
for the element uniqueness problem. The lower bound applies to so-called algebraic decision trees. An
algebraic decision tree is a ternary decision tree, where each query asks for the sign of a constant-degree
polynomial in the variables x 1 , x 2 , . . . , x n . A comparison tree is an example of an algebraic decision tree,
using polynomials of the form x i − x j . The reduction from sorting to element uniqueness implies that
any algebraic decision tree requires Ω(n log n) time to sort n numbers. But since algebraic decision trees
are ternary decision trees, we already knew that. Q.5 Closest Pair The simplest version of the closest pair problem asks, given a list of n numbers x 1 , x 2 , . . . , x n , to ﬁnd the
closest pair of elements, that is, the elements x i and x j that minimize | x i − x j |.
There is an obvious reduction from element uniqueness to closest pair, based on the observation that
the elements of the input list are distinct if and only if the distance between the closest pair is bigger
than zero. This reduction implies that the closest pair problem requires Ω(n log n) time in the algebraic
decision tree model.
trivial set of n numbers −
−−→ set of n numbers
CLOSESTPAIR
O(1) TRUE or FALSE ←−− closest pair x i , x j
−
There are also higher-dimensional closest pair problems; for example, given a set of n points in the
plane, ﬁnd the two points that closest together. Since the one-dimensional problem is a special case
of the 2d problem — just put all n point son the x -axis — the Ω(n log n) lower bound applies to the
higher-dimensional problems as well. Q.6 3SUM to Colinearity. . . Unfortunately, lower bounds are relatively few and far between. There are thousands of computational
problems for which we cannot prove any good lower bounds. We can still learn something useful about
the complexity of such a problem by from reductions, namely, that it is harder than some other problem.
3 right under all those rules about logarithms, geometric series, and recurrences 3 Algorithms Non-Lecture Q: Reductions Here’s an example. The problem 3SUM asks, given a sequence of n numbers x 1 , . . . , x n , whether any
three of them sum to zero. There is a fairly simple algorithm to solve this problem in O(n2 ) time (hint,
hint). This is widely believed to be the fastest algorithm possible. There is an Ω(n2 ) lower bound for
3SUM, but only in a fairly weak model of computation.4
Now consider a second problem: given a set of n points in the plane, do any three of them lie on a
common non-horizontal line? Again, there is an O(n2 )-time algorithm, and again, this is believed to be
the best possible. The following reduction from 3SUM offers some support for this belief. Suppose we
are given an array A of n numbers as input to 3SUM. Replace each element a ∈ A with three points (a, 0),
(−a/2, 1), and (a, 2). Thus, we replace the n numbers with 3n points on three horizontal lines y = 0,
y = 1, and y = 2.
If any three points in this set lie on a common non-horizontal line, they consist of one point on each
of those three lines, say (a, 0), (− b/2, 1), and (c , 2). The slope of the common line is equal to both
− b/2 − a and c + b/2; since these two expressions are equal, we must have a + b + c = 0. Similarly,
is any three elements a, b, c ∈ A sum to zero, then the resulting points (a, 0), (− b/2, 1), and (c , 2) are
colinear.
So we have a valid reduction from 3SUM to the colinear-points problem:
O(n) −−→ set of 3n points
set of n numbers −
COLINEAR?
trivial −
TRUE or FALSE ←−− TRUE or FALSE
T3SUM (n) ≤ Tcolinear (3n) + O(n) =⇒ Tcolinear (n) ≥ T3SUM (n/3) − O(n). Thus, if we could detect colinear points in o(n2 ) time, we could also solve 3SUM in o(n2 ) time, which
seems unlikely. Conversely, if we could prove an Ω(n2 ) lower bound for 3SUM in a sufﬁciently powerful
model of computation, it would imply an Ω(n2 ) lower bound for the colinear points problem as well.
The existing Ω(n2 ) lower bound for 3SUM does not imply a lower bound for ﬁnding colinear points,
because the model of computation is too weak. It is possible to prove an Ω(n2 ) lower bound directly
using an adversary argument, but only in a fairly weak decision-tree model of computation.
Note that in order to prove that the reduction is correct, we have to show that both yes answers
and no answers are correct: the numbers sum to zero if and only if three points lie on a line. Even
though the reduction itself only goes one way, from the ‘easier’ problem to the ‘harder’ problem,
the proof of correctness must go both ways.
Anka Gajentaan and Mark Overmars5 deﬁned a whole class of computational geometry problems
that are harder than 3SUM; they called these problems 3SUM-hard. A sub-quadratic algorithm for any
3SUM-hard problem would imply a subquadratic algorithm for 3SUM. I’ll ﬁnish the lecture with two more
examples of 3SUM-hard problems. Q.7 . . . to Segment Splitting . . . Consider the following segment splitting problem: Given a collection of line segments in the plane, is
there a line that does not hit any segment and splits the segments into two non-empty subsets?
4
The Ω(n2 ) lower bound holds in a decision tree model where every query asks for the sign of a linear combination of three
of the input numbers. For example, ‘Is 5 x 1 + x 42 − 17 x 5 positive, negative, or zero?’ See my paper ‘Lower bounds for linear
satisﬁability problems’ (http://www.uiuc.edu/~jeffe/pubs/linsat.html) for the gory(!) details.
5
A. Gajentaan and M. Overmars, On a class of O(n2 ) problems in computational geometry, Comput. Geom. Theory Appl.
5:165–185, 1995. ftp://ftp.cs.ruu.nl/pub/RUU/CS/techreps/CS-1993/1993-15.ps.gz 4 Algorithms Non-Lecture Q: Reductions To show that this problem is 3SUM-hard, we start with the collection of points produced by our last
reduction. Replace each point by a ‘hole’ between two horizontal line segments. To make sure that the
only way to split the segments is by passing through three colinear holes, we build two ‘gadgets’, each
consisting of ﬁve segments, to cap off the left and right ends as shown in the ﬁgure below. Top: 3n points, three on a non-horizontal line.
Bottom: 3n + 13 segments separated by a line through three colinear holes. This reduction could be performed in linear time if we could make the holes inﬁnitely small, but
computers can’t really deal with inﬁnitesimal numbers. On the other hand, if we make the holes too big,
we might be able to thread a line through three holes that don’t quite line up. I won’t go into details, but
it is possible to compute a working hole size in O(n log n) time by ﬁrst computing the distance between
the closest pair of points.
Thus, we have a valid reduction from 3SUM to segment splitting (by way of colinearity):
O ( n) O(n log n) trivial trivial −−→ set of 3n + 13 segments
set of n numbers −
−−→ set of 3n points −−−
SPLITTABLE? TRUE or FALSE ←−− TRUE or FALSE ←−−−− TRUE or FALSE
−
−
T3SUM (n) ≤ Tsplit (3n + 13) + O(n log n) Q.8 =⇒ Tsplit (n) ≥ T3SUM n − 13
3 − O(n log n). . . . to Motion Planning Finally, suppose we want to know whether a robot can move from one position and location to another.
To make things simple, we’ll assume that the robot is just a line segment, and the environment in
which the robot moves is also made up of non-intersecting line segments. Given an initial position and
orientation and a ﬁnal position and orientation, is there a sequence of translations and rotations that
moves the robot from start to ﬁnish?
To show that this motion planning problem is 3SUM-hard, we do one more reduction, starting from
the set of segments output by the previous reduction algorithm. Speciﬁcally, we use our earlier set of line
segments as a ‘screen’ between two large rooms. The rooms are constructed so that the robot can enter
or leave each room only by passing through the screen. We make the robot long enough that the robot
can pass from one room to the other if and only if it can pass through three colinear holes in the screen.
(If the robot isn’t long enough, it could get between the ‘layers’ of the screen.) See the ﬁgure below: 5 Algorithms Non-Lecture Q: Reductions start end The robot can move from one room to the other if and only if the screen between the rooms has three colinear holes. Once we have the screen segments, we only need linear time to compute how big the rooms should
be, and then O(1) time to set up the 20 segments that make up the walls. So we have a fast reduction
from 3SUM to motion planning (by way of colinearity and segment splitting):
O ( n) O(n log n) O(n) −−→ set of 3n points −−−
−−→ set of 3n + 13 segments −
−−→ set of 3n + 33 segments
set of n numbers −
MOVABLE?
trivial trivial TRUE or FALSE ←−− TRUE or FALSE ←−−−−
−
− T3SUM (n) ≤ Tmotion (3n + 33) + O(n log n) TRUE or FALSE =⇒ trivial ←−− TRUE or FALSE
− Tmtion (n) ≥ T3SUM n − 33
3 − O(n log n). Thus, a sufﬁciently powerful Ω(n2 ) lower bound for 3SUM would imply an Ω(n2 ) lower bound for
motion planning as well. The existing Ω(n2 ) lower bound for 3SUM does not imply a lower bound for
this problem the model of computation in which the lower bound holds is too weak to even solve
the motion planning problem. In fact, the best lower bound anyone can prove for this motion planning
problem is Ω(n log n), using a (different) reduction from element uniqueness. But the reduction does
give us evidence that motion planning ‘should’ require quadratic time. c Copyright 2009 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms/ for the most recent revision. 6 CS 373 Homework 0 (due 1/26/99) Spring 1999 CS 373: Combinatorial Algorithms, Spring 1999
http://www-courses.cs.uiuc.edu/ cs373
Homework 0 (due January 26, 1999 by the beginning of class) Name:
Net ID: Alias: Neatly print your name (ﬁrst name ﬁrst, with no comma), your network ID, and a short alias into
the boxes above. Do not sign your name. Do not write your Social Security number. Staple this
sheet of paper to the top of your homework. Grades will be listed on the course web site by alias,
so your alias should not resemble your name (or your Net ID). If you do not give yourself an alias,
you will be stuck with one we give you, no matter how much you hate it.
Everyone must do the problems marked ◮. Problems marked £ are for 1-unit grad students
and others who want extra credit. (There’s no such thing as “partial extra credit”!) Unmarked
problems are extr