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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
• Spring 2010: David Morrison, Kyle Fox, and Rachit Agarwal
• Fall 2010: Alina Ene c Copyright 1999–2011 Jeff Erickson. Last update July 8, 2011. 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.cs.illinois.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, Spring 2009, Spring 2010,
and Fall 2010. 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.
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.cs.illinois.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, David Morrison, Johnathon Fischer, Ekta Manaktala,
Erin Wolf Chambers, Igor Gammer, Gio Kao, Kevin Milans, Kevin Small, Kyle Fox, Lan
Chen, Michael Bond, Mitch Harris, Nick Hurlburt, Nitish Korula, Rachit Agarwal, 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 students 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.) 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 1400
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—Edgar Ramos,
Herbert Edelsbrunner, Jason Zych, Lenny Pitt, Mahesh Viswanathan, Margaret Fleck, Shang-Hua Teng,
Steve LaValle, and especially Chandra Chekuri, 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.
Finally, “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
ı”
back of the cover page is mine. Additional References
I strongly encourage my students (and other readers) not to restrict themselves to a single textual
reference. Authors and readers bring their own perspectives to the material; no instructor ‘clicks’ with
every student, or even every very strong student. Finding the author that most effectively gets their
intuition into your head take some effort, but that effort pays off handsomely in the long run.
The following references have been particularly valuable sources of inspiration, intuition, examples,
and problems. 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.)
• Thomas Cormen, Charles Leiserson, Ron Rivest, and Cliff Stein. Introduction to Algorithms, third
edition. MIT Press/McGraw-Hill, 2009. (The second edition 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 my undergraduate classes.)
• Jeff Edmonds. How to Think about Algorithms. Cambridge University Press, 2008.
• 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.
• Jon Kleinberg and Éva Tardos. Algorithm Design. Addison-Wesley, 2005. (This is the current
recommended textbook for my graduate algorithms classes.)
• 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 didn’t actually read them until 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.
• Lecture notes, slides, homeworks, exams, and video lectures posted by innumerable colleagues
around the world.
• The Source of All Knowledge (Google) and The Source of All Lies (Wikipedia). Prerequisites
For the most part, these notes assume the reader has mastered the material covered in the ﬁrst
two years of a strong undergraduate computer science curriculum, and has the intellectual maturity to
recognize and repair any remaining gaps in their mastery. (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: High-school algebra, naive set theory, Boolean algebra, ﬁrst-order predicate
logic, sets, functions, relations, modular arithmetic, recursive deﬁnitions, trees (as abstract objects,
not data structures), graphs.
• Elementary discrete probability: uniform vs non-uniform probability distributions, 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, linked lists, binary search trees, at least one balanced search
tree (such as AVL trees, red-black trees, B-trees, skip lists, splay trees, or treaps), binary heaps.
• Fundamental abstract data types: dictionaries, stacks, queues, priority queues; the difference
between this list and the previous list.
• Fundamental algorithms: elementary arithmetic, linear search, binary search, sorting (selection,
insertion, merge-, heap-, quick-, radix, anything but bubble-), pre-/post-/inorder tree traversal. • 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 prerequisite material is covered brieﬂy in these notes, but more as a reminder than a
good introduction. 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 2011 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. Lecture 0: Introduction [F10] Algorithms 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 ( λγος), 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
.
2
¯
¯
known as the writer of the treatise Al-Kitab al-mukhtasar f¯ ¯ ab al-˘abr wa’l-muqabala , from which the
ıhıs¯
g
.
¯
modern word algebra derives. In another treatise, al-Khwarizm¯ popularized the modern decimal system
ı
for writing and manipulating numbers—in particular, the use of a small circle or . ifr to represent a missing
s
quantity—which had originated in India several centuries earlier. This system later became known
in Europe as algorism. 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 cyphers became truly ubiquitous in Western
Europe only after the French revolution 600 years later. The more modern word algorithm is a false
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 © Copyright 2010 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. 1 Lecture 0: Introduction [F10] Algorithms cognate with the Greek word arithmos ( ριθ ός), meaning ‘number’ (and perhaps the aforementioned
άλγος). Thus, until very recently, the word algorithm referred exclusively to pencil-and-paper 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-Khw¯rizm¯ and Fibonacci popularized the cypher.
a
ı
Here is an algorithm, popularized (but almost 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.
4
Euclid and his students almost certainly drew their constructions on an βαξ, a table covered in dust or sand (or perhaps
very small rocks). Over the next several centuries, the Greek abax evolved into the medieval European abacus. 2 Lecture 0: Introduction [F10] Algorithms
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 taught in elementary schools in Eastern Europe in the late 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 ) if x is even x /2 · ( y + y ) + y if x is odd Congressional Apportionment
Here is another good example of an algorithm that comes from outside the world of computing. Article I,
Section 2 of the United States 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
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. 3 Lecture 0: Introduction [F10] Algorithms 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 1992.7 The input array Pop[1 .. n] stores
the populations of the n states, and R is the total number of representatives. Currently, n = 50 and
R = 435. The output array Rep[1 .. n] stores the number of representatives assigned to each state.
APPORTIONCONGRESS(Pop[1 .. n], R):
PQ ← NEWPRIORITYQUEUE
for i ← 1 to n
Rep[i ] ← 1
INSERT P Q, i , Pop[i ]/ 2
R←R−1
while R > 0
s ← EXTRACTMAX( PQ)
Rep[s] ← Rep[s] + 1
INSERT P Q, s, Pop[s] Rep[s] (Rep[s] + 1) R←R−1
return Rep[1 .. n] This pseudocode description assumes that you know how to implement a priority queue that supports
the operations NEWPRIORITYQUEUE, INSERT, and EXTRACTMAX. (The actual law doesn’t assume that, of
course.) The output of the algorithm, and therefore its correctness, does not depend at all on how
the priority queue is implemented. The Census Bureau uses 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.
A bad example
Consider “Martin’s algorithm”:8
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 might 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 actual 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.
7 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.
8
S. Martin, “You Can Be A Millionaire”, Saturday Night Live, January 21, 1978. Appears on Comedy Is Not Pretty, Warner
Bros. Records, 1979. 4 Lecture 0: Introduction [F10] Algorithms 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.
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). 0.3 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. The idiosyncratic
syntactic details of C, C++, C#, Java, Python, Ruby, Erlang, Haskell, OcaML, Scheme, Visual Basic,
Smalltalk, Javascript, Processing, Squeak, Forth, TEX, Fortran, COBOL, Intercal, or Brainfuck are of little
or no importance in algorithm design, and focusing on them will only distract you from what’s really
going on.9 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, there is
natural tendency to describe repeated operations informally: “Do this ﬁrst, then do this second, and
so on.” But 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 says very little about what happens
later iterations. To make the description unambiguous, we must explicitly specify the behavior of every
iteration.
The best way to write down an algorithm is using pseudocode. Pseudocode uses the structure
of formal programming languages and mathematics to break algorithms into primitive steps; but
the primitive steps themselves may be written using mathematics, pure English, or an appropriate
mixture of the two. Well-written pseudocode reveals the internal structure of the algorithm but hides
irrelevant implementation details, making the algorithm much easier to understand, analyze, debug,
and implement.
The precise syntax of pseudocode is a personal choice, but the overriding goal should be clarity and
precision. Ideally, pseudocode should allow any competent programmer to implement the underlying
algorithm, quickly and correctly, in their favorite programming language, without understanding why the
algorithm works. Here are the guidelines I follow and strongly recommend: 9 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. 5 Lecture 0: Introduction [F10] Algorithms
• Be consistent! • 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’.
• Indent everything carefully and consistently; the block structure should be visible from across the
room. This rule is especially important for nested loops and conditionals. Don’t add unnecessary
syntactic sugar like braces or begin/end tags; careful indentation is almost always enough.
• Use mnemonic algorithm and variable names. Short variable names are good, but readability is
more important than concision; except for idioms like loop indices, short but complete words are
better than single letters. Absolutely never use pronouns!
• Use standard mathematical notation for standard mathematical things. For example, write x · y
instead of x ∗ y for multiplication; write x mod y instead of x % y for remainder; write x instead
of sqrt( x ) for square roots; write a b instead of power(a, b) for exponentiation; and write φ instead
of phi for the golden ratio.
• Avoid mathematical notation if English is clearer. For example, ‘Insert a into X ’ may be preferable
to INSERT(X , a) or X ← X ∪ {a}.
• 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.)
• Don’t use a ﬁxed-width typeface to typeset pseudocode; it’s much harder to read than normal
typeset text. Similarly, don’t typeset keywords like ‘for’ or ‘while’ in a different style; the syntactic
sugar is not what you want the reader to look at. On the other hand, I use italics for variables,
SMALL CAPS for algorithms and constants, and a different typeface for literal strings and comments. 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 our algorithms are correct; trusting our instincts, or trying a few test
cases, isn’t good enough. 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 extra work to prove correct. Correctness proofs
almost always involve induction. We like induction. Induction is our friend.10
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.
10 If induction is not your friend, you will have a hard time in this course. 6 Lecture 0: Introduction [F10] Algorithms 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 for accomplishing such a task. The
same problem may have hundreds of different algorithms; the same algorithm may solve hundreds of
different problems.
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”, “Hole in the Bottom of the Sea, “Green Grow the Rushes O”, “Eh, Cumpari!”,
“Alouette”, “Echad Mi Yode’a”, “Chad Gadya”, “Minkurinn í hænsnakofanum”, and “Ist das nicht ein
Schnitzelbank?” For further details, consult your nearest preschooler.
7 Lecture 0: Introduction [F10] Algorithms 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.”
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 priority queue operations, but we
can certainly bound the running time as O(N + RI + (R − n) E ), where N denotes the running time
of NEWPRIORITYQUEUE, I denotes the running time of INSERT, and E denotes the running time of
EXTRACTMAX. Under the reasonable assumption that R > 2n (on average, each state gets at least two
representatives), we can simplify the bound to O(N + R( I + E )). The Census Bureau implements the
priority queue using an unsorted array of size n; this implementation gives us N = I = Θ(1) and
E = Θ(n), so the overall running time is O(Rn). This is good enough for government work, but we
can do better. Implementing the priority queue using a binary heap (or a heap-ordered array) gives us
N = Θ(1) and I = R = O(log n), which implies an overall running time of O(R log n).
Sometimes we are also interested in other computational resources: space, randomness, page faults,
inter-process messages, and so forth. We can use the same techniques to analyze those resources as we
use to analyze running time. 0.5 A Longer Example: Stable Matching 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 internships, and demand immediate responses.
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.11
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
11
The academic job market involves similar gambles, at least in computer science. Some departments start making offers in
February with two-week decision deadlines; other departments don’t even start interviewing until late March; MIT notoriously
waits until May, when all its interviews are over, before making any faculty offers. 8 Lecture 0: Introduction [F10] Algorithms 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.12 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!
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; Gale and Shapley used the metaphor of college
admissions.13 Similar algorithms have since been adopted for other matching markets, including faculty
recruiting in France, university admission in Germany, public school admission in New York and Boston,
and billet assignments for US Navy sailors. The stable matching problem is
The Boston Pool algorithm proceeds in rounds until every position has been ﬁlled. Each round has
two stages:
1. An arbitrary unassigned hospital A offers its position to the best doctor α (according to the
hospital’s preference list) who has not already rejected it.
2. Each doctor plans to accepts the best hospital (according to her preference list) that makes her an
offer. Thus, if α is currently unassigned, she (tentatively) accepts the offer from A. If α already has
an assignment but prefers A, she rejects her existing assignment and (tentatively) accepts the new
offer from A. Otherwise, α rejects the new offer.
For example, suppose four doctors (Dr. Quincy, Dr. Rotwang, Dr. Shephard, and Dr. Tam, represented
by lower-case letters) and four hospitals (Arkham Asylum, Bethlem Royal Hospital, County General
Hospital, and The Dharma Initiative, represented by upper-case letters) rank each other as follows:
q
A
B
C
D r
A
D
C
B s
B
A
C
D t
D
B
C
A AB
tr
st
rq
qs C
t
q
r
s D
s
r
q
t Given these preferences as input, the Boston Pool algorithm might proceed as follows:
12
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.
13
The “Gale-Shapley 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. However, similar statements were previously made by 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.”). We will see many other examples of Stigler’s law in this class. 9 Lecture 0: Introduction [F10] Algorithms
1. Arkham makes an offer to Dr. Tam.
2. Bedlam makes an offer to Dr. Rotwang. 3. County makes an offer to Dr. Tam, who rejects her earlier offer from Arkham.
4. Dharma makes an offer to Dr. Shephard. (From this point on, because there is only one unmatched
hospital, the algorithm has no more choices.)
5. Arkham makes an offer to Dr. Shephard, who rejects her earlier offer from Dharma.
6. Dharma makes an offer to Dr. Rotwang, who rejects her earlier offer from Bedlam.
7. Bedlam makes an offer to Dr. Tam, who rejects her earlier offer from County.
8. County makes an offer to Dr. Quincy.
At this point, all pending offers are accepted, and the algorithm terminates with a matching: (A, s), (B , t ),
(C , q), ( D, r ). You can (and should) verify by brute force that this matching is stable, even though no
doctor was hired by her favorite hospital, and no hospital hired its favorite doctor. In fact, this is the
only stable matching for this list of preferences.
Analyzing the algorithm’s running time is relatively straightforward. Each hospital makes an offer to
each doctor at most once, so 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 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 (n 2 ) time; we leave the details as an easy exercise.
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. Thus, the O(n2 ) upper bound on the
worst-case running time cannot be improved; in this case, we say our analysis is tight.
Correctness
But why is the algorithm correct? Gale and Shapley proved that the Boston Pool algorithm always
computes a stable matching as follows. The algorithm continues as long as there is at least one unﬁlled
position; conversely, when the algorithm terminates (after at most n2 rounds), every position is ﬁlled.
No doctor can accept more than one position, and no hospital can hire more than one doctor. Thus, the
algorithm always computes a matching; it remains only to prove that the matching is stable.
Suppose doctor α is assigned to hospital A in the ﬁnal matching, but prefers B . Because 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.
Lemma 1. During the Boston Pool algorithm, each hospital A is rejected only by doctors that are
infeasible for A.
Proof: We prove the lemma by induction. Consider an arbitrary round of the Boston Pool algorithm, in
which doctor α rejects one hospital A for another hospital B . The rejection implies that α prefers B to A.
10 Lecture 0: Introduction [F10] Algorithms Every doctor that appears higher than α in B ’s preference list has already rejected B and therefore, by
the inductive hypothesis, is infeasible for B .
Now consider an arbitrary matching that assigns α to A. We already established that α prefers B to A.
If B prefers α to its partner, the matching is unstable. On the other hand, if B prefers its partner to α,
then (by our earlier argument) its partner is infeasible, and again the matching is unstable. We conclude
that there is no stable matching that assigns α to A.
Now let best(A) denote the highest-ranked feasible doctor on A’s preference list. Lemma 1 implies
that every doctor that A prefers to its ﬁnal assignment is infeasible for A. On the other hand, the ﬁnal
matching is stable, so the doctor assigned to A is feasible for A. The following result is now immediate:
Corollary 1.1. The Boston Pool algorithm assigns best(A) to A, for every hospital A.
Thus, 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) denote the lowest-ranked feasible hospital on doctor α’s preference list.
Corollary 1.2. 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; 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 .
The previous corollary implies that A prefers α = best(A) to β . Because the matching is stable, α must
therefore prefer her assigned hopital to A. This argument works for any stable assignment, so α prefers
every other feasible match to A; in other words, A = worst(α).
A subtle consequence of these two corollaries, discovered by Dubins and Freeman in 1981, is that a
doctor can potentially improve her assignment by lying about her preferences, but a hospital cannot.
(However, a set of hospitals can collude so that some of their assignments improve.) Partly for this
reason, the National Residency Matching Program reversed its matching algorithm in 1998, so that
potential residents offer to work for hospitals in preference order, 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. In practice, however, this modiﬁcation affected less than 1%
of the resident’s assignments. As far as I know, the precise effect of this change on the patients is an
open problem. 0.6 Why are we here, anyway? This class is ultimately about learning two skills that are crucial for all computer scientists.
1. Intuition: How to think about abstract computation.
2. Language: 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. 11 Lecture 0: Introduction [F10] Algorithms 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 you will enjoy playing with them as much as I do. 12 Lecture 0: Introduction [F10] Algorithms Boethius the algorist versus Pythagoras the abacist.
from Margarita Philosophica by Gregor Reisch (1503) 13 Lecture 0: Introduction [F10] Algorithms 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 pseudolyrics,
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 increases14 . 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.
14 “We’ll drink it out of the hemisemidemiyottapint, boys!” 14 Lecture 0: Introduction [F10] Algorithms 3. Prove that it is possible for the Boston Pool algorithm to execute Ω(n2 ) rounds. (You need to
describe both a suitable input and a sequence of Ω(n2 ) valid proposals.)
4. Describe and analyze an efﬁcient algorithm to determine whether a given set of hospital and
doctor preferences has to a unique stable matching.
5. 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 her assigned match.
• An unmatched doctor and an unmatched hospital appear in each other’s preference lists.
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 Dr. House as their only acceptable intern, then only
House and Harvard will be matched.
6. 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? © Copyright 2010 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. 15 Lecture 1: Recursion [Fa’10] Algorithms 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’.
c Copyright 2011 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. 1 Lecture 1: Recursion [Fa’10] Algorithms 1.2 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).
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 in no time at all.
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
3 This English translation is from W. W. Rouse Ball and H. S. M. Coxeter’s book Mathematical Recreations and Essays. 2 Lecture 1: Recursion [Fa’10] Algorithms recursion recursion The Tower of Hanoi algorithm; ignore everything but the bottom disk The base case for the Tower of Hanoi algorithm. There is no spoon. 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 − 1, src, tmp, dst)
move disk n from src to dst
HANOI(n − 1, 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 (or guessing and checking by induction)
quickly gives us the 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 Lecture 1: Recursion [Fa’10] Algorithms 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 ( n) = T n/ 2 +T n/ 2 + O(n). If we strip out the ﬂoors and ceilings, we get the simpler recurrence T (n) = 2 T (n/2) + O(n). The
“all levels equal” case of the recursion tree method (or its corollary, the Master Theorem) immediately
implies that T (n) = O(n log n).
Aside: Domain Transformations. Wait. . . what? What do you mean ‘immediately’? How can we
just ignore the ﬂoors and ceilings? We could always check that T (n) = O(n log n) 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
4 “The inductive hypothesis” is just a technical nickname for our friend the Recursion Fairy. 4 Lecture 1: Recursion [Fa’10] Algorithms
two 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 similar 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 S O R T I N G E X A M P L
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 Lecture 1: Recursion [Fa’10] Algorithms 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, as we will see later, 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-case behavior 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 Lecture 1: Recursion [Fa’10] Algorithms 1.6 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 Lecture 1: Recursion [Fa’10] Algorithms 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).
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 at most 6 comparisons,
so we need at most 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 11n
5 i 9
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 Lecture 1: Recursion [Fa’10] Algorithms 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. The Egyptian/Russian peasant multiplication algorithm described in the ﬁrst lecture also runs in
O(n2 ) 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 us 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 Lecture 1: Recursion [Fa’10] Algorithms 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 the next 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 Lecture 1: Recursion [Fa’10] Algorithms
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.9 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 original 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 ??).
9 On the other hand, nobody knows whether ﬁnding the minimum number of multiplications is NP-hard, either. 11 Lecture 1: Recursion [Fa’10] Algorithms
(a) Follow these four rules:
•
•
•
• Never move the same disk twice in a row.
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 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 ??, 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 Lecture 1: Recursion [Fa’10] Algorithms 5. A less familiar chapter in the Tower of Hanoi’s history is its brief relocation of the temple from
Benares to Pisa in the early 13th century. The relocation was organized by the wealthy merchantmathematician Leonardo Fibonacci, at the request of the Holy Roman Emperor Frederick II, who
had heard reports of the temple from soldiers returning from the Crusades. The Towers of Pisa
and their attendant monks became famous, helping to establish Pisa as a dominant trading center
on the Italian peninsula.
Unfortunately, almost as soon as the temple was moved, one of the diamond needles began
to lean to one side. To avoid the possibility of the leaning tower falling over from too much
use, Fibonacci convinced the priests to adopt a more relaxed rule: Any number of disks on the
leaning needle can be moved together to another needle in a single move. It was still forbidden
to place a larger disk on top of a smaller disk, and disks had to be moved one at a time onto the
leaning needle or between the two vertical needles.
Thanks to Fibonacci’s new rule, the priests could bring about the end of the universe somewhat
faster from Pisa then they could than could from Benares. Fortunately, the temple was moved
from Pisa back to Benares after the newly crowned Pope Gregory IX excommunicated Frederick II,
making the local priests less sympathetic to hosting foreign heretics with strange mathematical
habits. Soon afterward, a bell tower was erected on the spot where the temple once stood; it too
began to lean almost immediately. The Towers of Pisa. In the ﬁfth move, two disks are taken off the leaning needle. Describe an algorithm to transfer a stack of n disks from one vertical needle to the other
vertical needle, using the smallest possible number of moves. Exactly how many moves does your
algorithm perform?
6. 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.
(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?
13 Lecture 1: Recursion [Fa’10] Algorithms 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. (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?
7. 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.]
8. (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. 14 Lecture 1: Recursion [Fa’10] Algorithms (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 . 9. Describe an algorithm to compute the median of an array A[1 .. 5] of distinct numbers using at
most 6 comparisons.
Instead of writing pseudocode, describe your algorithm using a decision tree: A binary tree
where each internal node is a comparison of the form “A[i ] A[ j ]?” and each leaf is an index
into the array.
A[1]:A[2] <
A[1]:A[3] <
A[2]:A[3] <
A[2] >
A[1] > >
A[1]:A[3] <
A[1] >
A[2]:A[3] < A[3] A[3] >
A[2] Finding the median of a 3-element array using at most 3 comparisons 10. 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?
11. Suppose we are given an array A[1 .. n] with the special property that A[1] ≥ A[2] and A[n − 1] ≤
A[n]. We say that an element A[ x ] is a local minimum if it is less than or equal to both its neighbors,
or more formally, if A[ x − 1] ≥ A[ x ] and A[ x ] ≤ A[ x + 1]. For example, there are six local minima
in the following array:
9772137547334869
We can obviously ﬁnd a local minimum in O(n) time by scanning through the array. Describe and
analyze an algorithm that ﬁnds a local minimum in O(log n) time. [Hint: With the given boundary
conditions, the array must have at least one local minimum. Why?]
15 Lecture 1: Recursion [Fa’10] Algorithms 12. 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.
13. (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 ∪ B ∪ C 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.]
14. (a) Describe an algorithm that sorts an input 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. [Hint: See Lecture 19.]
(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.)
15. 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).
16 Lecture 1: Recursion [Fa’10] Algorithms 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. (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. [Hint: There is
more than one function f such that f (16) = 4.]
(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).]
16. 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.)
17. 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. 17 Lecture 1: Recursion [Fa’10] Algorithms The largest complete subtree of this binary tree has depth 2. 18. 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 2011 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 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms 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) 2 Fast Fourier Transforms 2.1 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] 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] c Copyright 2010 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. 1 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms 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. 2.2 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.
2.2.1 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 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
1
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 · · · 2O(log n) ) bit
operations. The algorithm uses an n-element Fast Fourier Transform, which requires several O(log n)-nit integer multiplications,
∗
which are carried out recursively (of course!), leading to the cascade of logs in the running time. The 2O(log n) factor accounts
∗
for the constant factors accumulated at each of the O(log n) levels of recursion. The fastest algorithm known for integer
∗
multiplication, published by Martin Fürer in 2007, uses O(n log n 2O(log n) ) bit operations. 2 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms convert the polynomials to the more familiar coefﬁcient representation ﬁrst—this takes O(n2 ) 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
2.2.2 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, because 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 must start with 2n + 1 sample values for each polynomial, because that’s how many we
need to uniquely represent their product. Both algorithms run in O(n) time.
Unfortunately, evaluating a polynomial in this representation is no longer straightforward. 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
yj p( x ) =
( x − x k ) ( x j − x k ) k= j
k= j
j =0
Hopefully it’s clear that formula actually describes a polynomial function of x , since each term in the
rightmost sum is a scaled product of monomials. It’s also not hard to verify that p( x j ) = y j for every
index j ; most of the terms of the sum vanish. As I mentioned earlier, the Fundamental Theorem of
Algebra implies that p is the only polynomial that interpolates the points {( x j , y j )}. Lagrange’s formula
can be translated mechanically into an O(n2 )-time algorithm.
2.2.3 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 2.3 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
2 This is where numerical analysis comes from. 3 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms different representation that makes the operation easy, perform that operation, and then switch back.
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.
Lagrange’s formula can be used to convert the sample representation back to the more familiar coefﬁcient
form. 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
x n−1
an−1
yn−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.3 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 hard-coded directly into
the algorithm.4
3 In fact, Lagrange’s formula is just a special case of Cramer’s rule for solving linear systems.
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. On the other hand,
my numerical-analysis colleagues have reasonable cause to shoot me for daring to suggest, even in passing, that anyone
actually invert a matrix at all, ever.
4 4 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms So we can convert from coefﬁcients to samples and back in O(n2 ) time. At ﬁrst lance, this result
seems pointless; we can already add, multiply, or evaluate directly in either representation in O(n2 ) time,
so why bother? But 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. If we
choose a set of sample positions with the right recursive structure, we can perform this conversion more
quickly. 2.4 Divide and Conquer Any polynomial of degree n − 1 can be expressed as a combination of two polynomials of degree (n/2) − 1
as follows:
p( x ) = peven ( x 2 ) + x · podd ( x 2 ).
The coefﬁcients of peven are just the even-degree coefﬁcients of p, and the coefﬁcients of podd are just
the odd-degree coefﬁcients of p. Thus, we can evaluate p( x ) by recursively evaluating peven ( x 2 ) and
podd ( x 2 ) and performing O(1) additional arithmetic operations.
Now call a set X of n values collapsing if either of the following conditions holds:
• X has one element.
• The set X 2 = { x 2 | x ∈ X } has exactly n/2 elements and is (recursively) collapsing.
Clearly the size of any collapsing set is a power of 2. Given a polynomial p of degree n − 1, and a
collapsing set X of size n, we can compute the set { p( x ) | x ∈ X } of sample values as follows:
1. Recursively compute { peven ( x 2 ) | x ∈ X } = { peven ( y ) | y ∈ X 2 }.
2. Recursively compute { podd ( x 2 ) | x ∈ X } = { podd ( y ) | y ∈ X 2 }.
3. For each x ∈ X , compute p( x ) = peven ( x 2 ) + x · podd ( x 2 ).
The running time of this algorithm satisﬁes the familiar recurrence T (n) = 2 T (n/2) + Θ(n), which as
we all know solves to T (n) = Θ(n log n).
Great! Now all we need is a sequence of arbitrarily large collapsible sets. The simplest method to
construct such sets is just to invert the recursive deﬁnition: If X is a collapsible set of size n that does not
contain the number 0, then X = {± x | x ∈ X } is a collapsible set of size 2n. This observation gives us
an inﬁnite sequence of collapsible sets, starting as follows:5
X 1 : = { 1}
X 2 := {1, −1}
X 4 := {1, −1, i , −i }
X 8 := 1, −1, i , −i , 2
2 + 2
2 i, − 2
2 5 − 2
2 i, 2
2 − 2
2 i, − 2
2 + 2
2 i 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 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms 2.5 The Discrete Fourier Transform For any n, the elements of X n are called the complex nth roots of unity; these are the roots of the
polynomial x n − 1 = 0. These n complex values are spaced exactly evenly around the unit circle in the
complex plane. Every nth root of unity is a power of the primitive nth root
ωn = e2πi /n = cos 2π
n + i sin 2π
n . A typical nth root of unity has the form
ωk = e(2πi /n)k = cos
n 2π
n k + i sin 2π
n k. These complex numbers have several useful properties for any integers n and k:
• There are exactly 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
These properties imply immediately that if n is a power of 2, then the set of all nth roots of unity is
collapsible!
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 its coefﬁcient vector).
Thus, given an array P [0 .. n − 1] of coefﬁcients, its discrete Fourier transform is the vector P ∗ [0 .. n − 1]
deﬁned as follows:
n−1
j
P ∗ [ j ] := p(ωn ) = jk
P [ k ] · ωn
k =0 As we already observed, the fact that sets of roots of unity are collapsible implies that we can compute the
discrete Fourier transform in O(n log n) time. The resulting algorithm, called the fast Fourier transform,
was popularized by Cooley and Tukey in 1965.6 The algorithm assumes that n is a power of two; if
necessary, we can just pad the coefﬁcient vector with zeros.
6
Actually, Cooley and Tukey’s 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 Cooley-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.
Tukey apparently developed the 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 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms
FFT( 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] U ∗ ← FFT(U [0 .. n/2 − 1])
V ∗ ← FFT(V [0 .. n/2 − 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
return P ∗ [0 .. n − 1] 2.6 Inverting the FFT We also need to recover the coefﬁcients of the product from the new sample values. Recall that the
transformation from coefﬁcients to sample values is linear; 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 for all integers j and k. Thus, 1
1
1
1
1 ω
2
ω3
ωn
n n 2
4
6
1 ωn
ωn
ωn
V =
6
1 ω3
ωn
ω9
n
n .
.
.
.
.
.
.
.
.
.
.
. n−1
2(n−1)
3(n−1)
1 ωn
ωn
ωn ···
···
···
···
..
.
··· 1 ωn−1 n
2(n−1) ωn 3(n−1) ωn .
. . 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 . 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 m jk = n−1 v jl · w l k =
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 l =0 7 1= 1 n −1
n n
n jl
ωn −lk l =0 = 1, 1 n −1 j k l
=
(ω )
n l =0 n Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms
and if j = k, we have a geometric series
j −k n −1
j
(ωn−k )l m jk =
l =0 = (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.
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]
U ← INVERSEFFT(U [0 .. n/2 − 1])
V ← INVERSEFFT(V [0 .. n/2 − 1])
π
π
ωn ← cos( 2n ) − i sin( 2n )
ω←1 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] 2.7 Fast Multiplication Finally, 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 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∗ ) 8 Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms 2.8 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 ∗ .
Bit-reversal permutation
0000 0010 0010 1010 0110 1010 1100 1010 0110 0111 1110 1110 1110 1000 0001 0001 0001 0011 0101 1001 1010 0101 1001 0101 1011 0111 1101 1101 1100 1001 0011 0010 1101 1011 0111 1011 1101 1011 0111 1111 V* 1100 1000 1110 FFT(n/2) 1100 1001 V 0100 0110 0100 P* 1000 1000 0101 P 0100 0100 0011 U* 0010 0010 1111 1111 Butterﬂy network 0000 0110 FFT(n/2) 0000 0001 U 0000 1111 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, because they allow any two processors to communicate in only O(log n) steps. 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 2010 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 Lecture 3: Backtracking [Fa’10] Algorithms ’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 3 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 multiple alternatives to the next component of the solution, it simply
tries all possible options recursively. 3.1 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.
Obviously, in any solution to the n-Queens problem, there is exactly one queen in each row. So we
will represent our possible solutions using an array Q[1 .. n], where Q[i ] indicates which square in row i
contains a queen, or 0 if no queen has yet been placed in row i . To ﬁnd a solution, we 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).
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 ] = j + r − i ) or (Q[i ] = j − r + i )
legal ← FALSE
if legal
Q[ r ] ← j
RECURSIVENQUEENS(Q[1 .. n], r + 1)
c Copyright 2011 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. 1 Lecture 3: Backtracking [Fa’10] Algorithms ♕
♛
♕
♛
♕
♕
♕
♛
♕
♛
♕
♕
One solution to the 8 queens problem, represented by the array [4,7,3,8,2,5,1,6] 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 backtracking algorithm simply performs a traversal of this tree. ♕ ♕ ♕
♛ ♕ ♕ ♕
♛ ♕
♛ ♕ ♕ ♕ ♕
♛ ♕
♛ ♕
♛ ♕
♛
♕ ♕ ♕
♕
♛ ♕
♛ ♕
♛ ♕ ♕ ♕
♛ ♕
♛
♕
♛ ♕ ♕
♛ ♕
♛ ♕ ♕
♕ ♕
♛ ♕ ♕
♛
♕
♛ ♕ The complete recursion tree for our algorithm for the 4 queens problem. 3.2 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 Lecture 3: Backtracking [Fa’10] Algorithms 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 empty set is a subset of every set X , and 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), which we can solve
using either recursion trees or annihilators (or just guessing) to obtain the upper bound T (n) = O(2n ).
In the worst case, the recursion tree for this algorithm is a complete binary tree with depth n.
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
1 There’s no base case like the vacuous base case! 3 Lecture 3: Backtracking [Fa’10] Algorithms 3.3 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 only need to ﬁgure out what to do with the ﬁrst
element of the input sequence; the Recursion Fairy will 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 the object ‘longest increasing
subsequence’ in terms of the slightly different object ‘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.
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. 4 Lecture 3: Backtracking [Fa’10] Algorithms 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), which as usual
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. 3.4 Optimal Binary Search Trees Our next example combines recursive backtracking with the divide-and-conquer strategy.
Recall that the running time for a successful search in a binary search tree is proportional to the
number of ancestors of the target node.2 As a result, the worst-case search time is proportional to the
depth of the tree. Thus, to minimize the worst-case search time, the height of the tree should be as small
as possible; by this metric, the ideal tree is perfectly balanced.
In many applications of binary search trees, however, it is more important to minimize the total cost
of several searches rather than the worst-case cost of a single search. If x is a more ‘popular’ search
target 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!
2 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 Lecture 3: Backtracking [Fa’10] Algorithms 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]. Our task is to build the binary search tree that
minimizes the total search time, assuming that there will be exactly f [i ] searches for each key A[i ].
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 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! 6 Lecture 3: Backtracking [Fa’10] Algorithms 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 Lecture 3: Backtracking [Fa’10] Algorithms (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 2011 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 4: Efﬁcient Exponential-Time Algorithms [Fa’10]
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
— Martin Gardner, Aha! Insight (1978) 4 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. 4.1 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)).
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’!
Did you notice the punctuation I carefully inserted 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 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 And and and’ and
‘and’, and ‘and’ and ‘and and and Chips’?
c Copyright 2010 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. 1 Algorithms Lecture 4: Efﬁcient Exponential-Time Algorithms [Fa’10] 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
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 ) 2 Lecture 4: Efﬁcient Exponential-Time Algorithms [Fa’10] Algorithms 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,
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; since 1998, at least ﬁfteen different 3SAT
algorithms have been published, each improving the running time by a small amount. As of 2010,
the fastest deterministic algorithm for 3SAT runs in O(1.33334n ) time2 , and the fastest randomized
algorithm runs in O(1.32113n ) expected time3 , but there is good reason to believe that these are not the
best possible.
2 Robin A. Moser and Dominik Scheder. A full derandomization of Schöning’s k-SAT algorithm. ArXiv:1008.4067, 2010.
Kazuo Iwama, Kazuhisa Seto, Tadashi Takai, and Suguru Tamaki. Improved randomized algorithms for 3-SAT. To appear in
Proc. STACS, 2010.
3 3 Lecture 4: Efﬁcient Exponential-Time Algorithms [Fa’10] Algorithms 4.2 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.
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 speed up our algorithm by making several careful modiﬁcations to avoid the worst case of
the running-time 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.
T (n) ≤ O(poly(n)) + max T ( n − 1)
T ( n − 1) + T ( n − 2) = O(1.61803398875n ) The upper bound is derived by solving each case separately using the annihilator method and
taking the larger of the two solutions. The ﬁrst case gives us T (n) = O(poly(n)); 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 every maximal independent set. However, given
any independent set that includes w , removing w and adding v creates another independent set of
the same size. It follows that some maximum independent set includes v , so we don’t need to search
the graph G \ { v }, and the G \ N ( v ) has at most n − 2 nodes. On the other hand, if the degree of v
is at least 2, then G \ N ( v ) has at most n − 3 nodes. T ( n − 1) T ( n − 2)
T (n) ≤ O(poly(n)) + max
= O(1.46557123188n ) T ( n − 1) + T ( n − 3)
4 Lecture 4: Efﬁcient Exponential-Time Algorithms [Fa’10] Algorithms The base of the exponent is the largest root of the characteristic polynomial r 3 − r 2 − 1.
• Now the worst-case is a graph where every node has degree at least 2; we split this worst case
into two subcases. 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 G 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. T (n − 1) T (n − 2) T (n) ≤ O(poly(n)) + max
= O(3n/3 poly(n)) = O(1.44224957031n ) T (n − 1) + T (n − 4) 3 T ( n − 3)
The base of the exponent is 3, the largest root of the characteristic polynomial r 3 − 3. The
third case would give us a bound of O(1.3802775691n ), where the base is the largest root of the
characteristic polynomial r 4 − r 3 − 1.
3 • Now the worst case for our algorithm is a graph with an extraordinarily special structure: Every
node has degree 2. In other words, every component of G is a cycle. But it is easy to prove that the
largest independent set in a cycle of length k has size k/2 . So we can handle this case directly in
polynomial time, without no recursion at all! T ( n − 1) T ( n − 2)
T (n) ≤ O(poly(n)) + max
= O(1.3802775691n ) T ( n − 1) + T ( n − 4)
Again, the base of the exponential running time is the largest root of the characteristic polynomial
r 4 − r 3 − 1.
MAXIMUMINDSETSIZE(G ):
if G = ∅
return 0
else if G has a node v with degree 0 or 1
return 1 + MAXIMUMINDSETSIZE(G \ N ( v )) 〈〈≤ n − 1〉〉 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〉〉
total ← 0
for each component of G
k ← number of vertices in the component
total ← total + k/2
return total As with 3SAT, further improvements are possible but increasingly complex. As of 2010, the fastest
published algorithm for computing maximum independent sets runs in O(1.2210n ) time4 . However,
Fedor V Fomin, Fabrizio Grandoni, and Dieter Kratsch. Measure and conquer: A simple O(20.288n ) independent set
.
algorithm. Proc. SODA, 18–25, 2006.
4 5 Algorithms Lecture 4: Efﬁcient Exponential-Time Algorithms [Fa’10] in an unpublished technical report, Robson describes a computer-generated algorithm that runs in
O(2n/4 poly(n)) = O(1.1889n ) time; just the description of this algorithm requires more than 15 pages.5 Exercises
1. (a) Prove that any n-vertex graph has at most 3n/3 maximal independent sets. [Hint: Modify the
MAXIMUMINDSETSIZE algorithm so that it lists all maximal independent sets.]
(b) Describe an n-vertex graph with exactly 3n/3 maximal independent sets, for every integer n
that is a multiple of 3.
2. 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?] 5
Mike Robson. Finding a maximum independent set in time O(2n/4 ). Technical report 1251-01, LaBRI, 2001. 〈http:
//www.labri.fr/perso/robson/mis/techrep.ps〉. c Copyright 2010 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 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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) 5 Dynamic Programming 5.1
5.1.1 Fibonacci Numbers
Recursive Deﬁnitions Are Recursive Algorithms 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” c Copyright 2010 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. 1 Lecture 5: Dynamic Programming [Fa’10] Algorithms
5.1.2 Memo(r)ization: Remember Everything 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 our recursive algorithm considerably just by writing down the results of our
recursive calls and looking them up again if we need them later. This process was dubbed memoization
by Richard Michie in the late 1960s.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] Memoization clearly decreases the running time of the algorithm, but by how much? 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]. This pattern can be veriﬁed by induction: Each
entry F [i ] is ﬁlled only after its predecessor F [i − 1]. If we ignore the time spent in recursive calls, it
requires only constant time to evaluate the recurrence for each Fibonacci number Fi . But by design, the
recurrence for Fi is evaluated only once! We conclude that MEMFIBO performs only O(n) additions, an
exponential improvement over the naïve recursive algorithm!
5.1.3 Dynamic Programming: Fill Deliberately But once we see how the array F [ ] is ﬁlled, we can replace the recursion with a simple loop that
intentionally ﬁlls the array in order, instead of relying on the complicated recursion to do it for us
‘accidentally’.
ITERFIBO(n):
F [0] ← 0
F [1] ← 1
for i ← 2 to n
F [ i ] ← F [ i − 1] + F [ i − 2]
return F [n] Now the time analysis is immediate: ITERFIBO clearly uses O (n ) additions and stores O(n) integers.
This gives us our ﬁrst explicit dynamic programming algorithm. The dynamic programming paradigm
was developed by Richard Bellman in the mid-1950s, while working at the RAND Corporation. Bellman
deliberately chose the name ‘dynamic programming’ to hide the mathematical character of his work
from his military bosses, who were actively hostile toward anything resembling mathematical research.
Here, the word ‘programming’ does not refer to writing code, but rather to the older sense of planning
or scheduling, typically by ﬁlling in a table. For example, sports programs and theater programs are
schedules of important events (with ads); television programming involves ﬁlling each available time
slot with a show (and ads); degree programs are schedules of classes to be taken (with ads). The Air
2 “My name is Elmer J. Fudd, millionaire. I own a mansion and a yacht.” 2 Lecture 5: Dynamic Programming [Fa’10] Algorithms Force funded Bellman and others to develop methods for constructing training and logistics schedules,
or as they called them, ‘programs’. The word ‘dynamic’ is meant to suggest that the table is ﬁlled in over
time, rather than all at once (as in ‘linear programming’, which we will see later in the semester).3
5.1.4 Don’t Remember Everything After All In many dynamic programming algorithms, it is not necessary to retain all intermediate results through
the entire computation. For example, we can signiﬁcantly reduce the space requirements of our algorithm
ITERFIBO by maintaining only the two newest 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.)
5.1.5 Faster! Faster! Even this algorithm can be improved further, using the following wonderful fact:
01
11 x
y = y
x+y 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 . If we use repeated squaring, computing the nth power
11
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?
5.1.6 Whoa! Not so fast! 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!
3 “I thought dynamic programming was a good name. It was something not even a Congressman could object to. So I used
it as an umbrella for my activities.” 3 Lecture 5: Dynamic Programming [Fa’10] Algorithms The way out of this apparent paradox is to observe that we can’t perform arbitrary-precision
arithmetic in constant time. Multiplying two n-digit numbers using fast Fourier transforms (described
in a different lecture note) requires O(n log n log log n) time. Thus, the matrix-based algorithm’s actual
running time obeys the recurrence T (n) = T ( n/2 ) + O(n log n log log n), which solves to T (n) =
O (n log n log log n ) using recursion trees.
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 (n 2 ). (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), for
which the annihilator method still implies the solution T (n) = O(φ n ). 5.2 Longest Increasing Subsequence In a previous lecture, we developed a recursive algorithm to ﬁnd the length of the longest increasing
subsequence of a given sequence of numbers. Given an array A[1 .. n], the length of the longest
increasing subsequences is computed by the function call LISBIGGER(−∞, A[1 .. n]), where LISBIGGER is
the following recursive algorithm:
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 We can simplify our notation slightly with two simple observations. First, the input variable prev is
always either −∞ or an element of the input array. Second, the second argument of LISBIGGER is always
a sufﬁx of the original input array. If we add a new sentinel value A[0] = −∞ to the input array, we can
identify any recursive subproblem with two array indices.
Thus, we can rewrite the recursive algorithm as follows. Add the sentinel value A[0] = −∞. Let
L IS (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 IS (0, 1). For all i < j , we have if j > n
0
L IS (i , j ) = L IS (i , j + 1) max{ L IS (i , j + 1), 1 + L IS ( j , j + 1)} if A[i ] ≥ A[ j ] otherwise Because each recursive subproblem can be identiﬁed by two indices i and j , we can store the
intermediate values in a two-dimensional array LIS[0 .. n , 1 .. n ]. Since there are O(n2 ) entries in the
table, our memoized algorithm uses O (n 2 ) space.4 Each entry in the table can be computed in O(1)
time once we know its predecessors, so our memoized algorithm runs in O (n 2 ) time.
In fact, we only need half of this array, because we always have i < j . But even if we cared about constant factors in this
class (we don’t), this would be the wrong time to worry about them. The ﬁrst order of business is to ﬁnd an algorithm that
actually works; once we have that, then we can think about optimizing it.
4 4 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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 L IS [i , j ] is ﬁlled in after the entries L IS [i , j + 1] and L IS [ j , j + 1]
in the next columns. But just this partial information is enough to give us an explicit evaluation order.
If we ﬁll in our table one column at a time, from right to left, 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 longest increasing subsequence, and a legal evaluation order Finally, putting everything together, we obtain the following dynamic programming algorithm:
LIS(A[1 .. n]):
A[0] ← −∞
for i ← 0 to n
L IS [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 IS [i , j ] ← LIS [i , j + 1]
else
L IS [i , j ] ← max{ LIS [i , j + 1], 1 + LIS [ j , j + 1]}
return L IS [0, 1] As expected, the algorithm clearly uses O (n 2 ) time and space. However, we can reduce the space to
O(n) by only maintaining the two most recent columns of the table, L IS [·, j ] and L IS [·, j + 1].5
This is not the only recursive strategy we could use for computing longest increasing subsequences
efﬁciently. Here is another recurrence that gives us the O(n) space bound for free. Let L IS (i ) denote
the length of the longest increasing subsequence of A[i .. n] that starts with A[i ]. Our goal is to compute
L IS (0) − 1; we subtract 1 to ignore the sentinel value −∞. To deﬁne L IS (i ) recursively, we only need
to specify the second element in subsequence; the Recursion Fairy will do the rest.
L IS (i ) = 1 + max L IS ( 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 only O (n ) space, and the resulting algorithm runs in
O (n 2 ) time. To transform this memoized recurrence into a dynamic programming algorithm, we only
need to guarantee that L IS ( j ) is computed before L IS (i ) whenever i < j .
5 See, I told you not to worry about constant factors yet! 5 Lecture 5: Dynamic Programming [Fa’10] Algorithms
LIS2(A[1 .. n]):
A[0] = −∞ 〈〈Add a sentinel〉〉 for i ← n downto 0
L IS [i ] ← 1
for j ← i + 1 to n
if A[ j ] > A[i ] and 1 + LIS [ j ] > LIS [i ]
LIS [i ] ← 1 + LIS [ j ]
return L IS [0] − 1 5.3 〈〈Don’t count the sentinel〉〉 The Pattern: Smart Recursion In a nutshell, dynamic programming is recursion without repetition. 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. As
long as we memoize the correct recurrence, an explicit table isn’t really necessary, but if the recursion is
incorrect, nothing works. Dynamic programming is not about ﬁlling in tables.
It’s about smart recursion!
Dynamic programming algorithms are almost always developed in 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:
(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) Analyze space and running time. 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.
(c) 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.
(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
6 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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. You don’t
need to do this on homework or exams.
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! 5.4 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 never work!
Use dynamic programming instead!
What, never?
No, never!
What, never?
Well. . . hardly ever.6
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 the semester. 5.5 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:
6 Greedy methods hardly ever work! So 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! 7 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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.
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:7 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] 7
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. 8 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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.
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; the precise running time is not obvious, but it’s clearly exponential in
m and n. Fortunately, we don’t care about the precise running time of the recursive algorithm. The
recursive running time wouldn’t tell us anything about our eventual dynamic programming algorithm,
so we’re just not going to bother computing it.8
Because each recursive subproblem can be identiﬁed by two indices i and j , we can memoize
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.
j i Dependencies in the memoization table for edit distance, and a legal evaluation order
8 In case you’re curious, the running time of the unmemoized recursive algorithm obeys the following recurrence:
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
T (N ) = max T (n, m) =
n+m=N The annihilator method implies that T (N ) = O((1 +
at most T (n + m) = O((1 + 2)n+m ). O(1) if N = 0, 2 T (N − 1) + T (N − 2) + O(1) otherwise.
2)N ). Thus, the running time of our recursive edit-distance algorithm is 9 Lecture 5: Dynamic Programming [Fa’10] Algorithms Each entry Edit[i , j ] depends only on its three neighboring entries Edit[i − 1, j ] + 1, Edit[i , j − 1] + 1,
and Edit[i − 1, j − 1]. 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. 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 G O R I T H M 0 →1→2→3→4→5→6→7→8→9
A
L
T
R
U
I
S
T
I
C ↓ 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 ↓ ↓ ↓ ↓ ↓ ↓ 6 5 4 4 4 4 ↓ ↓ ↓ ↓ ↓ ↓ 3→4→5→6
3→4→5→6 ↓ 7 6 5 5 5 5 4 ↓ ↓ ↓ ↓ ↓ ↓ ↓ 8 7 6 6 6 6 5 ↓ ↓ ↓ ↓ ↓ ↓ ↓ 4 ↓ 8 7 7 7 7 6 5 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 9 8 8 8 8 7 6 4→5→6 9
10 5 6 5 →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: 10 Lecture 5: Dynamic Programming [Fa’10] Algorithms
A
A L
L G
T A
A G A
A 5.6 L
L
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 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 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 ] 11 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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] 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 ]. 12 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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. 5.7 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}. 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: 13 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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
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? 14 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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} Exercises
1. Suppose you are given an array A[1 .. n] of numbers, which may be positive, negative, or zero.
(a) 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 (b) Describe and analyze an algorithm that ﬁnds the largest product of of elements in a contiguous
subarray A[i .. j ]. You may not assume that all the numbers in the input array are integers.
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 . 15 Lecture 5: Dynamic Programming [Fa’10] Algorithms (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.
(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. It’s almost time to show off your ﬂippin’ sweet dancing skills! Tomorrow is the big dance contest
you’ve been training for your entire life, except for that summer you spent with your uncle in
Alaska hunting wolverines. You’ve obtained an advance copy of the the list of n songs that the
judges will play during the contest, in chronological order.
You know all the songs, all the judges, and your own dancing ability extremely well. For each
integer k, you know that if you dance to the kth song on the schedule, you will be awarded exactly
Score[k] points, but then you will be physically unable to dance for the next Wait[k] songs (that
is, you cannot dance to songs k + 1 through k + Wait[k]). The dancer with the highest total score
at the end of the night wins the contest, so you want your total score to be as high as possible.
Describe and analyze an efﬁcient algorithm to compute the maximum total score you can
achieve. The input to your sweet algorithm is the pair of arrays Score[1 .. n] and Wait[1 .. n].
5. A palindrome is any string that is exactly the same as its reversal, like I, or DEED, or RACECAR, or
AMANAPLANACATACANALPANAMA. 16 Lecture 5: Dynamic Programming [Fa’10] Algorithms (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.
6. Suppose you have a subroutine QUALITY that can compute the ‘quality’ of any given string A[1 .. k]
in O(k) time. 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
SAINTCIBOIREDESACRAMENTDECRISSE can be decomposed into the substrings SAINT + CIBOIRE
+ 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.
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. 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.
9. You are driving a bus along a highway, full of rowdy, hyper, thirsty students and a soda fountain
machine. Each minute that a student is on your bus, that student drinks one ounce of soda. Your
goal is to drop the students off quickly, so that the total amount of soda consumed by all students
is as small as possible.
You know how many students will get off of the bus at each exit. Your bus begins somewhere
along the highway (probably not at either end) and move s at a constant speed of 37.4 miles per
17 Lecture 5: Dynamic Programming [Fa’10] Algorithms hour. You must drive the bus along the highway; however, you may drive forward to one exit then
backward to an exit in the opposite 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. Your input consists of 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).
10. 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
of Nadira, called Dream Dollars, was available in the following denominations: $1, $4, $7, $13,
$28, $52, $91, $365.9
(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.)
11. 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.
9 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. 18 Lecture 5: Dynamic Programming [Fa’10] Algorithms 12. 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.)
-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.
13. 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 .
14. Describe and analyze an efﬁcient algorithm to ﬁnd the length of the longest contiguous substring
that appears both forward and backward in an input string T [1 .. n]. The forward and backward
substrings must not overlap. Here are several examples:
• Given the input string ALGORITHM, your algorithm should return 0.
• Given the input string RECURSION, your algorithm should return 1, for the substring R.
• Given the input string REDIVIDE, your algorithm should return 3, for the substring EDI. (The
forward and backward substrings must not overlap!)
• Given the input string DYNAMICPROGRAMMINGMANYTIMES, your algorithm should return 4, for
the substring YNAM. (In particular, it should not return 6, for the subsequence YNAMIR). 19 Lecture 5: Dynamic Programming [Fa’10] Algorithms § § 15. Dance Dance Revolution is a dance video game, ﬁrst introduced in Japan by Konami in 1998.
Players stand on a platform marked with four arrows, pointing forward, back, left, and right,
arranged in a cross pattern. During play, the game plays a song and scrolls a sequence of n arrows
( , , , or §) from the bottom to the top of the screen. At the precise moment each arrow
reaches the top of the screen, the player must step on the corresponding arrow on the dance
platform. (The arrows are timed so that you’ll step with the beat of the song.) § You are playing a variant of this game called “Vogue Vogue Revolution”, where the goal is to
play perfectly but move as little as possible. When an arrow reaches the top of the screen, if one
of your feet is already on the correct arrow, you are awarded one style point for maintaining your
current pose. If neither foot is on the right arrow, you must move one (and only one) of your feet
from its current location to the correct arrow on the platform. If you ever step on the wrong arrow,
or fail to step on the correct arrow, or move more than one foot at a time, or move either foot
when you are already standing on the correct arrow, all your style points are taken away and you
lose the game.
How should you move your feet to maximize your total number of style points? For purposes
of this problem, assume you always start with you left foot on and you right foot on §, and that
you’ve memorized the entire sequence of arrows. For example, if the sequence is
§ §,
you can earn 5 style points by moving you feet as shown below: § §
§ ➜ ➜ ➜ ➜ ➜ ➜ ➜➜ ➜➜
➜ ➜ ➜ ➜
➜ ➜ ➜ ➜ R
➜
➜ Style point! R
➜L R
➜L ➜ ➜ R
➜L ➜ ➜
R
➜L ➜ ➜ L ➜ R
➜ ➜ Style point! L § R
➜ ➜ ➜
➜ ➜
➜ Begin! R
➜ ➜ § §
§ R
➜ ➜ L L ➜ L Style point! Style point! Style point! (a) Prove that for any sequence of n arrows, it is possible to earn at least n/4 − 1 style points.
(b) Describe an efﬁcient algorithm to ﬁnd the maximum number of style points you can earn
during a given VVR routine. The input to your algorithm is an array Arrow[1 .. n] containing
the sequence of arrows.
16. 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.
17. [Stolen from abhi shelat.] You have mined a large slab of marble from your quarry. For simplicity,
suppose the marble slab is a rectangle measuring n inches in height and m inches in width. You
want to cut the slab into smaller rectangles of various sizes—some for kitchen countertops, some
20 Lecture 5: Dynamic Programming [Fa’10] Algorithms for large sculpture projects, others for memorial headstones. You have a marble saw that can make
either horizontal or vertical cuts across any rectangular slab. At any given time, you can query the
spot price p x , y of an x × y marble rectangle, for any positive integers x and y . These prices will
vary with demand, so do not make any assumptions about them; in particular, larger rectangles
may have much smaller spot prices. Given the spot prices, describe an algorithm to compute how
to subdivide an n × m marble slab to maximize your proﬁt.
18. 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]. 19. 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|. 21 Lecture 5: Dynamic Programming [Fa’10] Algorithms (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 ].
20. 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 21. 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. 22. 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.
23. Let T be a rooted binary tree with n vertices, and let k ≤ n be a positive integer. We would like to
mark k vertices in T so that every vertex has a nearby marked ancestor. More formally, we deﬁne
the clustering cost of any subset K of vertices as
cost(K ) = max cost( v, K ),
v 22 Lecture 5: Dynamic Programming [Fa’10] Algorithms where the maximum is taken over all vertices v in the tree, and if v ∈ K
0
cost( v, K ) = ∞
if v is the root of T and v ∈ K 1 + cost(parent( v )) otherwise Describe and analyze a dynamic-programming algorithm to compute the minimum clustering
cost of any subset of k vertices in T . For full credit, your algorithm should run in O(n2 k2 ) time. 1 1
2 1
2 2 1
2 3 2 2 3
2 1 1 3 2
1
2 1
2 2
3 1
2 1 1 1
2 2 2
2 3 3 3 2
3 A subset of 5 vertices with clustering cost 3 24. 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 2 1 1 3 3 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
label from its parent.. The cost of an labeling is the number of nodes that have smaller labels than
23 Lecture 5: Dynamic Programming [Fa’10] Algorithms 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.)
25. 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:
•
•
•
• 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.
CU
G
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G C
CA UUACA CAUGAGUA AAUGU U
A A
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C U
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U A
A A
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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 (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 24 Lecture 5: Dynamic Programming [Fa’10] Algorithms of the gap lengths.10 Describe and analyze an algorithm that computes the minimum possible
score of a secondary structure for a given RNA sequence.
26. 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.]
27. 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.)
28. 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?]
For full credit, your algorithm should run in O(n4 ) time; faster algorithms are worth extra
credit. The fastest algorithm I know for this problem runs in O(n3/2 log n) time, but this requires
several tricks, in both the algorithm and its analysis. 10 This score function has absolutely no connection to reality; I just made it up. Real RNA structure prediction requires much
more complicated scoring functions. 25 Lecture 5: Dynamic Programming [Fa’10] Algorithms — Randall Munroe, xkcd (http://xkcd.com/399/)
Reproduced under a Creative Commons Attribution-NonCommercial 2.5 License c Copyright 2010 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. 26 Algorithms Lecture 6: Advanced Dynamic Programming Tricks [Fa’10]
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) 6 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. 6.1 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.
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 optimal editing sequence for that preﬁx. Speciﬁcally, any 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 ) such that some optimal edit sequence from A[1 .. i ] into B [1 .. j ] contains an optimal edit sequence from A[1 .. m/2] to B [1 .. Half(i , j )].
We can deﬁne this function recursively 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
(Because there there may be more than one optimal edit sequence, this is not the only correct deﬁnition.)
A simple inductive argument implies that Half(m, n) is indeed 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.
c Copyright 2011 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. 1 Lecture 6: Advanced Dynamic Programming Tricks [Fa’10] Algorithms
Edit
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5 Finally, to compute the optimal editing sequence that transforms A into B , we recursively compute
the optimal sequences transforming A[1 .. m/2] into B [1 .. Half(m, n)] and transforming A[m/2 + 1 .. m]
into B [Half(m, n) + 1 .. n]. The recursion bottoms out when one string has only constant length, in which
case we can determine the optimal editing sequence in linear time using our old dynamic programming
algorithm. The running time of the resulting 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. 6.2 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. 2 Lecture 6: Advanced Dynamic Programming Tricks [Fa’10] Algorithms Fix the two input strings A[1 .. n] and B [1 .. m]. For any indices i and j , let LCS(i , j ) denote the
length of the longest common subsequence of the preﬁxes A[1 .. i ] and B [1 .. j ]. This function can be
deﬁned recursively as follows: if i = 0 or j = 0
0
LCS(i , j ) = LCS(i − 1, j − 1) + 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. «
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5 6 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: 3 Lecture 6: Advanced Dynamic Programming Tricks [Fa’10] Algorithms 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 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. 6.3 Saving Time: Monotonicity
The SMAWK matrix-searching algorithm is a better example here; the problem is more general, the
algorithm is simple, and the proof is self-contained. Next time! 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 relatively 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,
4 Algorithms Lecture 6: Advanced Dynamic Programming Tricks [Fa’10] 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
intermediate pivot values are stored at the internal nodes. An algorithm discovered by 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).]
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 Lecture 6: Advanced Dynamic Programming Tricks [Fa’10] Algorithms 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.
(b) Modify your algorithm from part (a) so that it runs in O(n log n) time when T is 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 2011 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 Lecture 7: Greedy Algorithms [Fa’10] 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) 7 Greedy Algorithms 7.1 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 a ﬁle 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.
c Copyright 2011 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. 1 Lecture 7: Greedy Algorithms [Fa’10] 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 k f [π(k)] · Σcost(π) = 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. 7.2 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 offers all of its classes
on the same day every week, called ‘Soberday’ by the students.1 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 interest 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 2 Lecture 7: Greedy Algorithms [Fa’10] Algorithms A maximal conﬂict-free schedule for a set of classes. 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.
This problem has a fairly simple recursive solution, based on the observation that either you take
class 1 or you don’t. Let B4 denote the set of classes that end before class 1 starts, and let L8 denote 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.2
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.)
1 but interestingly, not by the faculty
But you should still work out the details yourself. The dynamic programming algorithm can be used to ﬁnd the “best”
schedule for several different deﬁnitions of “best”, but the greedy algorithm I’m about to describe only works when “best”
means “biggest”. Also, you need the practice.
2 3 Lecture 7: Greedy Algorithms [Fa’10] Algorithms 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] 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 Lecture 7: Greedy Algorithms [Fa’10] Algorithms 7.3 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.
• 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. 7.4 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:3
n f [i ] · depth(i ).
i =1 In 1951, as a PhD student at MIT, David Huffman developed the following greedy algorithm to produce
such an optimal code:4
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:5
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.
3
This looks almost exactly like the cost of a binary search tree, but the optimization problem is very different: code trees are
not required to keep the keys in any particular order.
4
Huffman was a student in an information theory class taught by Robert Fano, who was a close colleague of Claude Shannon,
the father of information theory. Fano and Shannon had previously developed a different greedy algorithm for producing
preﬁx codes—split the frequency array into two subarrays as evenly as possible, and then recursively builds a code for each
subarray—but these Fano-Shannon codes were known not to be optimal. Fano posed the (then open) problem of ﬁnding an
optimal encoding to his class; Huffman solved the problem as a class project, in lieu of taking a ﬁnal exam.
5
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 Lecture 7: Greedy Algorithms [Fa’10] Algorithms To keep things simple, let’s forget about the forty-four spaces, nineteen apostrophes, nineteen commas,
three hyphens, and only one period, and just encode the letters. Here’s the frequency table:
A C D 3 3 2 F G H 26 5 E 3 8 N O R U V W X 13 2 16 I L 9 6 27 22 2 S T 5 8 451 Y Z 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 E F G H 3 3 26 5 3 I 8 L N O R S T U V W X Y D
Z 13 2 16 9 6 27 22 2 5 8 45 3 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. For example, the code for A is 110000, and the code
for S is 00.
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 If we use this code, 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 in the example message, 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 6 Lecture 7: Greedy Algorithms [Fa’10] Algorithms 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.
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 7 Lecture 7: Greedy Algorithms [Fa’10] Algorithms
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.
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 8 Lecture 7: Greedy Algorithms [Fa’10] Algorithms Exercises
1. Consider the following alternative greedy algorithms for the class scheduling problem. For each
algorithm, either prove or disprove that it always constructs an optimal schedule. Assume that all
four algorithms break ties arbitrarily.
(a) Choose the course that ends latest, discard all conﬂicting classes, and recurse.
(b) Choose the course that starts ﬁrst, discard all conﬂicting classes, and recurse.
(c) Choose the course with shortest duration, discard all conﬂicting classes, and recurse.
(d) Choose a course that conﬂicts with the fewest other courses, discard all conﬂicting classes, and
recurse.
2. Now consider a weighted version of the class scheduling problem, where different classes offer
different number of credit hours (totally unrelated to the duration of the class lectures). Your goal
is now to choose a set of non-conﬂicting classes that give you the largest possible number of credit
hours, given an array of start times, end times, and credit hours as input.
(a) Prove that the greedy algorithm described in the notes — Choose the class that ends ﬁrst and
recurse — does not always return an optimal schedule.
(b) Describe an algorithm to compute the optimal schedule in O(n2 ) time.
3. 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. 4. 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
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.
5. 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
9 Lecture 7: Greedy Algorithms [Fa’10] Algorithms A set of intervals stabbed by four points (shown here as vertical segments) 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 4 5 5 4 4
3 5 3 3
1
2 A proper coloring of a set of intervals using ﬁve colors. 6. Suppose you are a simple shopkeeper living in a country with n different types of coins, with
values 1 = c [1] < c [2] < · · · < c [n]. (In the U.S., for example, n = 6 and the values are 1, 5, 10,
25, 50 and 100 cents.) Your beloved and benevolent dictator, El Generalissimo, has decreed that
whenever you give a customer change, you must use the smallest possible number of coins, so as
not to wear out the image of El Generalissimo lovingly engraved on each coin by servants of the
Royal Treasury.
(a) In the United States, there is a simple greedy algorithm that always results in the smallest
number of coins: subtract the largest coin and recursively give change for the remainder. El
Generalissimo does not approve of American capitalist greed. Show that there is a set of coin
values for which the greedy algorithm does not always give the smallest possible of coins.
(b) Now suppose El Generalissimo decides to impose a currency system where the coin denominations are consecutive powers b0 , b1 , b2 , . . . , b k of some integer b ≥ 2. Prove that despite El
Generalissimo’s disapproval, the greedy algorithm described in part (a) does make optimal
change in this currency system.
(c) Describe and analyze an efﬁcient algorithm to determine, given a target amount A and a
sorted array c [1 .. n] of coin denominations, the smallest number of coins needed to make
A cents in change. Assume that c [1] = 1, so that it is possible to make change for any
amount A.
7. Suppose you have just purchased a new type of hybrid car that uses fuel extremely efﬁciently, but
can only travel 100 miles on a single battery. The car’s fuel is stored in a single-use battery, which
must be replaced after at most 100 miles. The actual fuel is virtually free, but the batteries are
expensive and can only be installed by licensed battery-replacement technicians. Thus, even if you
decide to replace your battery early, you must still pay full price for the new battery to be installed.
Moreover, because these batteries are in high demand, no one can afford to own more than one
battery at a time.
10 Lecture 7: Greedy Algorithms [Fa’10] Algorithms Suppose you are trying to get from San Francisco to New York City on the new Inter-Continental
Super-Highway, which runs in a direct line between these two cities. There are several fueling
stations along the way; each station charges a different price for installing a new battery. Before
you start your trip, you carefully print the Wikipedia page listing the locations and prices of every
fueling station on the ICSH. Given this information, how do you decide the best places to stop for
fuel?
More formally, suppose you are given two arrays D[1 .. n] and C [1 .. n], where D[i ] is the
distance from the start of the highway to the i th station, and C [i ] is the cost to replace your
battery at the i th station. Assume that your trip starts and ends at fueling stations (so D[1] = 0
and D[n] is the total length of your trip), and that your car starts with an empty battery (so you
must install a new battery at station 1).
(a) Describe and analyze a greedy algorithm to ﬁnd the minimum number of refueling stops
needed to complete your trip. Don’t forget to prove that your algorithm is correct.
(b) But what you really want to minimize is the total cost of travel. Show that your greedy
algorithm in part (a) does not produce an optimal solution when extended to this setting.
(c) Describe an efﬁcient algorithm to compute the locations of the fuel stations you should stop
at to minimize the total cost of travel.
8. 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.
(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.)
11 Lecture 7: Greedy Algorithms [Fa’10] Algorithms (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. This subroutine is not difﬁcult to write, but it’s not the interesting part of the
problem. c Copyright 2011 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 Lecture 8: Matroids [Fa’10] Algorithms 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 8 Matroids 8.1 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 .
c Copyright 2010 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. 1 Lecture 8: Matroids [Fa’10] Algorithms • 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 Lecture 8: Matroids [Fa’10] Algorithms 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 Lecture 8: Matroids [Fa’10] Algorithms 8.2 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 Lecture 8: Matroids [Fa’10] Algorithms 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 Lecture 8: Matroids [Fa’10] Algorithms 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 2010 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 Lecture 9: Randomized Algorithms [Fa’10] 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) 9 Randomized Algorithms 9.1 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? 9.2 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?” c Copyright 2010 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. 1 Lecture 9: Randomized Algorithms [Fa’10] 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! 9.3 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 Lecture 9: Randomized Algorithms [Fa’10] 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)] = 9.4 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 Lecture 9: Randomized Algorithms [Fa’10] 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. 9.5 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 Lecture 9: Randomized Algorithms [Fa’10] Algorithms 9.6 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. (See Section 9.8
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. 9.7 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.
5 Lecture 9: Randomized Algorithms [Fa’10] Algorithms 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 event that can prevent a comparison
between Bi and N j is choosing some intermediate pivot bolt Bk , with i < k < j , 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 .
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
6 Lecture 9: Randomized Algorithms [Fa’10] Algorithms 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
n n−k+1 = n+4
= n+4
k =2 1 [reorder summations] k i =1 n−k+1
k
n = n+4 [substitute k = j − i + 1] k i = 1 k =2 k =2
n 1 ( n + 1)
k =2 1
k n − 1
k =2 = n + 4((n + 1)(H n − 1) − (n − 1))
= n + 4(nH n − 2n + H n )
= 4nH n − 7n + 4H n
Sure enough, it’s Θ(n log n). 9.8 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) = 1 n n k =1 T ( k − 1) + T ( n − k ) + 2 n − 1 = 2 n−1
n k =0 T (k) + 2n − 1 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 7 Lecture 9: Randomized Algorithms [Fa’10] Algorithms both sides of the recurrence by n to get rid of the fractions.
n−1 T ( k ) + 2 n2 − n n T ( n) = 2
k =0
n−2 T (k) + 2(n − 1)2 − (n − 1) (n − 1) T (n − 1) = 2
k =0
n−2 T ( k ) + 2 n2 − 5 n + 3 =2
k =0 n T (n) − (n − 1) T (n − 1) = 2 T (n − 1) + 4n − 3
n+1
3
T ( n) =
T ( n − 1) + 4 −
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 ( n + 1) = t ( n − 1) +
= t ( n − 1) + T ( n − 1) 4
n+1
7
n+1 n
−
− +4− 3
n 3
n( n + 1)
3
n 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
1
The last step uses the recursive deﬁnition of the harmonic numbers: H n = H n + 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 Lecture 9: Randomized Algorithms [Fa’10] Algorithms Exercises
Probability
Several of these problems refer to decks of playing cards. A standard (Anglo-American) deck of 52 playing
cards contains 13 cards in each of four suits: spades (m), hearts (n), diamonds (o), and clubs (p). Within
each suit, the 13 cards have distinct ranks: 2, 3, 4, 5, 6, 7, 8, 9, 10, jack (J ), queen (Q), king (K ), and
ace (A). For purposes of these problems, the ranks are ordered A < 2 < 3 < · · · < 9 < 10 < J < Q < K ;
thus, for example, the jack of spades has higher rank thank the eight of diamonds.
1. Clock Solitaire is played with a standard deck of playing cards. 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.
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?
2. Professor Jay is about to perform a public demonstration with two decks of cards, one with red
backs (‘the red deck’) and one with blue backs (‘the blue deck’). Both decks lie face-down on a
table in front of Professor Jay, shufﬂed so that every permutation of each deck is equally likely.
To begin the demonstration, Professor Jay turns over the top card from each deck. If one of
these two cards is the three of clubs (3p), the demonstration ends immediately. Otherwise, the
good Professor repeatedly hurls the cards he just turned over into the thick, pachydermatous outer
melon layer of a nearby watermelon, and then turns over the next card from the top of each deck.
The demonstration ends the ﬁrst time a 3p is turned over. Thus, if 3p is the last card in both
decks, the demonstration ends with 102 cards embedded in the watermelon, that most prodigious
of household fruits.
(a) What is the exact expected number of cards that Professor Jay hurls into the watermelon?
(b) For each of the statements below, give the exact probability that the statement is true of the
ﬁrst pair of cards Professor Jay turns over.
i.
ii.
iii.
iv. Both cards are threes.
One card is a three, and the other card is a club.
If (at least) one card is a heart, then (at least) one card is a diamond.
The card from the red deck has higher rank than the card from the blue deck. (c) For each of the statements below, give the exact probability that the statement is true of the
last pair of cards Professor Jay turns over.
i.
ii.
iii.
iv. Both cards are threes.
One card is a three, and the other card is a club.
If (at least) one card is a heart, then (at least) one card is a diamond.
The card from the red deck has higher rank than the card from the blue deck. 9 Lecture 9: Randomized Algorithms [Fa’10] Algorithms 3. Penn and Teller agree to play the following game. Penn shufﬂes a standard deck of playing cards
so that every permutation is equally likely. Then Teller draws cards from the deck, one at a
time without replacement, until he draws the three of clubs (3♣), at which point the remaining
undrawn cards instantly burst into ﬂames.
The ﬁrst time Teller draws a card from the deck, he gives it to Penn. From then on, until the
game ends, whenever Teller draws a card whose value is smaller than the last card he gave to
Penn, he gives the new card to Penn.6 To make the rules unambiguous, they agree beforehand
that A = 1, J = 11, Q = 12, and K = 13.
(a) What is the expected number of cards that Teller draws?
(b) What is the expected maximum value among the cards Teller gives to Penn?
(c) What is the expected minimum value among the cards Teller gives to Penn?
(d) What is the expected number of cards that Teller gives to Penn? [Hint: Let 13 = n.]
4. 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.
5. 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.
6 Speciﬁcally, he hurls it directly into the back of Penn’s right hand. 10 Lecture 9: Randomized Algorithms [Fa’10] Algorithms Randomized Algorithms
6. 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 your only source of randomness.
(d) What is the exact expected number of times that your FAIRCOIN algorithm calls ONEINTHREE?
7. (a) Suppose you have access to a function FAIRCOIN that returns a single random bit, chosen
uniformly and independently from the set {0, 1}, in O(1) time. Describe and analyze an
algorithm RANDOM(n), which returns an integer chosen uniformly and independently at
random from the set {1, 2, . . . , n}.
(b) Suppose you have access to a function FAIRCOINS(k) that returns k random bits (or equivalently, a random integer chosen uniformly and independently from the set {0, 1, . . . , 2k − 1})
in O(1) time, given any non-negative integer k as input. Describe and analyze an algorithm
RANDOM(n), which returns an integer chosen uniformly and independently at random from
the set {1, 2, . . . , n}.
For each of the remaining problems, you may 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). 8. 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 ( )? 11 Lecture 9: Randomized Algorithms [Fa’10] Algorithms 9. 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?]
10. 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 ?
11. 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)] (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.]
12 Lecture 9: Randomized Algorithms [Fa’10] Algorithms (d) Describe and analyze an implementation of RANDOMPERMUTATION that runs in expected
worst-case time O(n).
(e) Describe and analyze an implementation of RANDOMPERMUTATION that runs in expected
worst-case time O(n log n), using fair coin ﬂips (instead of RANDOM) as the only source of
randomness.
(f) 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.
12. 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).
13. 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.]
13 Lecture 9: Randomized Algorithms [Fa’10] Algorithms (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?
14. 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.
15. 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.
16. 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!]
14 Lecture 9: Randomized Algorithms [Fa’10] Algorithms ∨
∧ ∧
∨ ∨
∧ ∧ ∧ ∨
∧ ∨ ∧ ∧ ∧ ∧ (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!]
17. 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 2010 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. 15 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 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) 10 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. 10.1
10.1.1 Treaps
Deﬁnitions 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.
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.
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.
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,
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’.
2
J. Vuillemin, A unifying look at data structures. Commun. ACM 23:229–239, 1980. c Copyright 2010 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. 1 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms
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6 A treap. The top half of each node shows its search key and the bottom half shows its priority.
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A G H I L M O R T A geometric interpretation of the same treap. 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
10.1.2 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.
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 ).
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
3 E. M. McCreight. Priority search trees. SIAM J. Comput. 14(2):257–276, 1985.
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.
4 2 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms
y x right x y
left A right rotation at x and a left rotation at y are inverses.
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8 Left to right: After inserting (S , −1), rotate it up to ﬁx the heap property.
Right to left: Before deleting (S , −1), rotate it down to make it a leaf. 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).
10.1.3 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 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 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. The order of the arguments is unimportant; the subsets X (i , k) and X (k, i ) are identical. The
subset X (1, n) = X (n, 1) contains all n nodes in the treap.
Lemma 1. For all i = k, node 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:
6 See Larry Niven and Jerry Pournelle, The Gripping Hand, Pocket Books, 1994. 4 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 1
k − i + 1 [i = k]
i
=0
Pr[Ak = 1] =
|k − i | + 1 1 if i < k
if i = k
if i > k i−k+1 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).
10.1.4 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 5 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 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. 10.2 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.
10.2.1 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
7 William Pugh. Skip lists: A probabilistic alternative to balanced trees. Commun. ACM 33(6):668–676, 1990. 6 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 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 +∞ 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.
10.2.2 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
7 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 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
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 ≥1 Clearly, if < , then Pr[ L ( x ) ≥ ] > Pr[ L ( x ) ≥
number of levels as follows:
E[ L ( x )] = Pr[ L ≥ ]
. 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 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms
10.2.3 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. (a) The left spine of a binary tree is a path starting at the root and following only left-child
pointers down to a leaf. What is the expected number of nodes in the left spine of an n-node
treap?
(b) What is the 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?]
(c) 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. 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 ? 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 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms 3. 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
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?
4. 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.
10 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms
(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)?] 5. 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 right 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.
(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.
6. 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 11 Lecture 10: Treaps and Skip Lists [Fa’10] Algorithms (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.) c Copyright 2010 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 Lecture 11: Tail Inequalities [Fa’10] Algorithms If you hold a cat by the tail
you learn things you cannot learn any other way.
— Mark Twain 11 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. 11.1 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 Lecture 11: Tail Inequalities [Fa’10] Algorithms 11.2 Independence 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. 11.3 Chernoff Bounds
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 exp( t X i ) =
i E[exp( t X i )]
i We can bound the individual expectations E[exp( t X 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 Lecture 11: Tail Inequalities [Fa’10] Algorithms 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 small 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] < 11.4 e −δ
(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 Lecture 11: Tail Inequalities [Fa’10] Algorithms 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 Lecture 11: Tail Inequalities [Fa’10] Algorithms 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 2010 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 Lecture 12: Hash Tables [Fa’10] Algorithms 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 28,731,569½.
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) 12
12.1 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. 12.2 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
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 .)
© Copyright 2010 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. 1 Lecture 12: Hash Tables [Fa’10] Algorithms 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.
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. 2 Lecture 12: Hash Tables [Fa’10] Algorithms α := n
m 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! Later in this lecture, I’ll describe how to achieve constant expected worst-case
search time using secondary hash tables. 12.3 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 Latin for “Beware of cannibals.” 3 Lecture 12: Hash Tables [Fa’10] Algorithms 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. 12.4 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 Lecture 12: Hash Tables [Fa’10] Algorithms 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
k Pr[X j = k] = n k k!
1 < 1−
k 1
n n ≤ k 1 1− 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 for all j (c + 1) ln n
ln ln n . A somewhat more complicated argument implies that if we throw n balls randomly into n bins, then
with high probability, the most popular bin contains at least Ω(log n/ log log n) balls.
5 Lecture 12: Hash Tables [Fa’10] Algorithms However, if we make the hash table large enough, we can expect every ball to land in its own 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 i =1 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 n−1 = m 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. 12.5 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. The simple uniform hashing assumption implies 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.
Linearity of expectation implies that n E[n2 ]
j n = E n n Xi jXk j =
i =1 k =1 n E[X i j X k j ] =
i =1 k =1 i =1 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
6 Lecture 12: Hash Tables [Fa’10] Algorithms 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 n 1 2 n2 1
= 2− .
n 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). 12.6 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}.
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 7 Lecture 12: Hash Tables [Fa’10] Algorithms 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
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 12.7 Cuckoo Hashing To be written!
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. 8 Lecture 12: Hash Tables [Fa’10] Algorithms 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).
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. © Copyright 2010 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 Lecture 13: Randomized Minimum Cut [Fa’10] Algorithms 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) 13
13.1 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 fairly complex; we will see some of these later in the semester.
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
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.
© Copyright 2010 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. 1 Lecture 13: Randomized Minimum Cut [Fa’10] Algorithms original vertices u and v . Equivalently, collapsing the edge shrinks the edge down to nothing, pulling the
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 }. Any edge in an n-vertex graph can be collapsed in O(n) time, assuming the graph is represented as
an adjacency list; I’ll leave the precise implementation details as an easy exercise.
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. 13.2 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 Because each collapse requires 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). Lecture 13: Randomized Minimum Cut [Fa’10] Algorithms 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
= n i =
i i =3 13.3 i =1
n 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− 2 N n( n − 1) ≤ 1 − e−2N /n(n−1) , by The World’s Most Useful Inequality 1 + x ≤ e x . By making N larger, we can make this probability
n
arbitrarily close to 1, but never equal to 1. In particular, 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.) 13.4 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
3 Lecture 13: Randomized Minimum Cut [Fa’10] Algorithms 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−2 i =n/ 2+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 edges of the minimum cut are CONTRACTed
and (2) the recursive call to BETTERGUESS returns the minimum cut of the CONTRACTed graph. Thus, 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 + 1). The same argument implies that X 2 is the
minimum cut with probability at least 1/2 · P (n/ 2 + 1). Because these two events are independent, we
have the following recurrence, with base case P (n) = 1 for all n ≤ 6.
P ( n) ≥ 1 − 1 − 1
2 P n
2 2 +1 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.
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
3 n
2 +1 = O(n2 log n) ˜
David R. Karger∗ and Cliff Stein. An O(n2 ) algorithm for minimum cuts. Proc. 25th STOC, 757–765, 1993. 4 Lecture 13: Randomized Minimum Cut [Fa’10] Algorithms 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 c ln2 n ln 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! 13.5 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 − 2 p ( k − 1) = p ( k − 1) − 2 p(k − 1)2
4 with base case p(0) = 1. Let ¯ (k) be the function that satisﬁes this recurrence with equality; clearly,
p
p(k) ≥ ¯ (k). Substituting the function z (k) = 4/ ¯ (k) − 1 into this recurrence implies (after a bit of
p
p
algebra) gives a new recurrence
z ( k ) = z ( k − 1) + 2 + 1
z ( k − 1) with base case z (0) = 3. Clearly z (k) > 1 for all k, so we have a conservative upper bound z (k) <
z (k − 1) + 3, which implies (by induction) that z (k) ≤ 3k + 3. Substituting ¯ (k) = 4/(z (k) + 1) into this
p
solution, we conclude that
1
p(k) ≥ ¯ (k) >
p
= Ω(1/k).
3k + 6
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 Lecture 13: Randomized Minimum Cut [Fa’10] Algorithms 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. Consider the following generalization of the BETTERGUESS algorithm, where we pass in a real
parameter 0 < α < 1 in addition to the graph G .
BETTERGUESS(G , α):
n ← number of vertices in G
if n > 8
X 1 ← BETTERGUESS(CONTRACT(G , n/α), α)
X 2 ← BETTERGUESS(CONTRACT(G , n/α), α)
return min{X 1 , X 2 }
else
use brute force Assume for this question that the input graph G has a unique minimum cut.
(a) What is the running time of the modiﬁed algorithm, as a function of n and α? [Hint: Consider
the cases α < 2, α = 2, and α > 2 separately.]
(b) What is the probability that CONTRACT(G , n/α) does not contract any edge in the minimum
cut in G ? Give both an exact expression involving both n and α, and a simple approximation
in terms of just α. [Hint: When α = 2, the probability is approximately 1/2.]
(c) Estimate the probability that BETTERGUESS(G , α) returns the minimum cut in G , by adapting
the solution to the Karger-Stein recurrence. [Hint: Consider the cases α < 2, α = 2, and
α > 2 separately.]
(d) Suppose we iterate BETTERGUESS(G , α) until we are guaranteed to see the minimum cut with
high probability. What is the running time of the resulting algorithm? For which value of α is
this running time minimized?
(e) Suppose we modify BETTERGUESS(G , α) further, to recurse four times instead of only twice.
Now what is the best choice of α? What is the resulting running time?
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). 6 Lecture 14: Amortized Analysis [Sp’10] Algorithms 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) 14
14.1 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. 14.2 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.
c Copyright 2010 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. 1 Lecture 14: Amortized Analysis [Sp’10] Algorithms
14.2.1 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.
14.2.2 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 Lecture 14: Amortized Analysis [Sp’10] Algorithms 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.
14.2.3 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.
14.2.4 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 Lecture 14: Amortized Analysis [Sp’10] Algorithms 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. 14.3 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 Lecture 14: Amortized Analysis [Sp’10] Algorithms 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. 14.4 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 Lecture 14: Amortized Analysis [Sp’10] Algorithms 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? 14.5 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 remember 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 integer i such that 2i divides n.
(b) f (n) is the largest power of 2 that divides n.
(c) f (n) = n if n is a power of 2, and f (n) = 1 otherwise.
(d) 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
6 Lecture 14: Amortized Analysis [Sp’10] Algorithms push them onto stack Si +1 to make room. (Thus, if Si +1 is already full, we ﬁrst recursively move
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:
7 Lecture 14: Amortized Analysis [Sp’10] Algorithms • Push: add a new item to the left end of the list;
• 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
8 Lecture 14: Amortized Analysis [Sp’10] Algorithms 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.
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 2010 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 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) 15
15.1 Scapegoat and Splay Trees
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 c Copyright 2010 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. 1 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms Galperin* and Ron Rivest in 1993 [10]. The second is the splay tree, discovered by Danny Sleator and
Bob Tarjan in 1981 [16, 14]. 15.2 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 ). 15.3 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 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. 15.4 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 15.5 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!) 15.6 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms
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 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 n j e h
g m f l
x 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 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. 15.7 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 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 )). 15.8 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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms [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 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms 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. 10 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms (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
( 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!]
11 Lecture 15: Scapegoat and Splay Trees [Sp’10] Algorithms (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 )?
(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 2010 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 Lecture 16: Disjoint Sets [Sp’10] Algorithms
E pluribus unum (Out of many, one) — Ofﬁcial 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) 16 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. 16.1 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.
c Copyright 2010 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. 1 Lecture 16: Disjoint Sets [Sp’10] Algorithms
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. 16.2 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
2 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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. 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.
3 Lecture 16: Disjoint Sets [Sp’10] Algorithms Theorem 1. A sequence of m MAKEWEIGHTEDSET operations and n WEIGHTEDUNION operations takes
O(m + n log n) time in the worst case.
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 ). 16.3 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 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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. 16.4 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.3
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.
1 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.
3
For an even simpler analysis, see: Seth Pettie. Applications of forbidden 0-1 matrices to search tree and path compressionbased data structures. Proc. 21st Ann. ACM-SIAM Symp. Discrete Algorithms, 1457–1467, 2010.
2 5 Lecture 16: Disjoint Sets [Sp’10] Algorithms
COMPRESS( x , y ):
〈〈 y must be an ancestor of x 〉〉
if x = y
COMPRESS(parent( x ), y )
parent( x ) ← parent( y ) SHATTER( x ):
if parent( x ) = x
SHATTER(parent( x ))
parent( x ) ← x 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+ .
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
6 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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. 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 7 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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,
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 16.5 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 ∗ r ≤ m + + 2 n. 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.
8 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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.
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.4 16.6 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 1928.5 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
4
Robert E. Tarjan. A class of algorithms which require non-linear time to maintain disjoint sets. J. Comput. Syst. Sci.
19:110–127, 1979.
5
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. 9 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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.6 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
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).
c 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. 16.7 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 Pettie7 , 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)).
6 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!
7
Splay trees, Davenport-Schinzel sequences, and the deque conjecture. Proceedings of the 19th Annual ACM-SIAM
Symposium on Discrete Algorithms, 1115–1124, 2008. 10 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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).
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. 11 Lecture 16: Disjoint Sets [Sp’10] Algorithms 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.
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
4, 42
8, 15, 16, 23, 30
9, 10, 11, 12, 13, 14 It’s bigger than 1.
It’s between 4 and 42.
It’s between 8 and 15.
It’s 11; you win! Describe a strategy that allows you to win in O(α(n)) rounds! c Copyright 2010 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 Lecture 17: Basic Graph Properties [Sp’10] Algorithms 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 17
17.1 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; I will usually write u v instead of {u. v } to denote the undirected edge
between u and v . In a directed graph, the edges are ordered pairs of vertices; I will usually write u v
instead of (u. v ) to denote the directed edge from u to v . We will usually be concerned only 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).
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 components, which are its maximal connected subgraphs. Two
vertices are in the same component if and only if there is a path between them.
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. c Copyright 2010 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. 1 Lecture 17: Basic Graph Properties [Sp’10] Algorithms A cycle is a path that starts and ends at the same vertex, and has at least one edge. An undirected
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 .
Directed graphs can contain directed paths and directed cycles. A directed graph is strongly
connected if there is a directed path from any vertex to any other. A directed graph is acyclic if it does
not contain a directed cycle; directed acyclic graphs are often called dags. 17.2 Abstract Representations and Examples The most common way to visually represent graphs is 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. However, embeddings are not the only useful representation of graphs. 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.
Another good example is the dependency graph of a recursive algorithm. Dependency graphs are
directed acyclic graphs. The vertices are all the distinct recursive subproblems that arise when executing
the algorithm on a particular input. There is an edge from one subproblem to another if evaluating 2 Lecture 17: Basic Graph Properties [Sp’10] Algorithms f d e a g b h b d a c h fg i
c e (a) i a bc de (b) 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). the second subproblem requires a recursive evaluation of the ﬁrst subproblem. For example, for the
Fibonacci recurrence if n = 0,
0
Fn = 1 Fn−1 + Fn−2 if n = 1, otherwise, the vertices of the dependency graph are the integers 0, 1, 2, . . . , n, and for each integer i from 2 to n,
and the edges are (i − 1) i and (i − 2) i for every integer i between 2 and n. For the edit distance
recurrence i
if j = 0 j if i = 0 Edit(i , j ) = Edit(i − 1, j ) + 1, min
otherwise Edit(i , j − 1) + 1, Edit(i − 1, j − 1) + A[i ] = B [ j ]
the dependency graph is an m × n grid with diagonals. Dynamic programming works efﬁciently for any
recurrence that has a small dependency graph; a proper evaluation order ensures that each subproblem
is visited after its predecessors.
A slightly more frivolous example of a graph is the conﬁguration graph of a game, puzzle, or
mechanism like tic-tac-toe, checkers, the Rubik’s Cube, the Towers of Hanoi, or a Turing machine. The
vertices of the conﬁguration graph are all the valid conﬁgurations of the puzzle; there is an edge from
one conﬁguration to another if it is possible to transform one conﬁguration into the other with a simple
move. (Obviously, the precise deﬁnition depends on what moves are allowed.) Even for reasonably
simple mechanisms, the conﬁguration graph can be extremely complex, and we typically only have
access to local information about the graph. The conﬁguration graph of the 4-disk Tower of Hanoi Finally, the ﬁnite-state automata used in formal language theory are just labeled directed graphs. A
deterministic ﬁnite-state automaton is usually formally deﬁned as a 5-tuple M = (Q, Σ, δ, q0 , A), where Q
3 Lecture 17: Basic Graph Properties [Sp’10] Algorithms is a ﬁnite set of states, Σ is a ﬁnite set called the alphabet, δ : Q × Σ → Q is a transition function, q0 ∈ Q is
the initial state, and F ⊆ Q is the set of accepting states. But it is often useful to think of M as a directed
graph G M whose vertices are the states Q, and whose edges have the form q δ(q, x ) for every state
q ∈ Q and character x ∈ Σ. Then basic questions about the language accepted by M can be phrased as
questions about the graph G M . For example, the language accepted by M is empty if and only if there is
no path in G M from q0 to an accepting state.
It’s important not to confuse these examples/representations of graphs with the actual deﬁnition: A
graph is a pair of sets (V, E ), where V is an arbitrary ﬁnite set, and E is a set of pairs (either ordered or
unordered) of elements of V . 17.3 Graph Data Structures 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, in which each entry 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
01100
10111
11011
01101
01110
00011
00000
00000
00000 f
0
0
0
1
1
0
0
0
0 g
0
0
0
0
0
0
0
1
1 hi
00
00
00
00
00
00
10
01
10 a
b
c
d
e
f
g
h
i b
a
a
b
b
d
h
g
g c
c
b
c
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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—adjacency lists are usually a better choice. 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
2 See footnote 1. 4 Lecture 17: Basic Graph Properties [Sp’10] Algorithms 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 is only useful
for vertices with very large degree, because the constant overhead in both the space and search time is
larger for hash tables than for simple linked lists.
At this point, you might reasonably 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. One compelling answer is that many
graphs are implicitly represented by adjacency matrices. For example, intersection graphs are usually
represented as an array or list of the underlying geometric 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. Similarly, we can apply any graph algorithm to a conﬁguration graph as though
it were given to us as an adjacency list, provided we can enumerate all possible moves from a given
conﬁguration in constant time each.
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. 17.4 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 the stack is not empty
v ← POP
if v is unmarked
mark v
for each edge vw
PUSH(w ) RECURSIVEDFS( v ):
if v is unmarked
mark v
for each edge vw
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.) A stack
is a particular type of bag, but certainly not the only one. Here is the generic traversal algorithm: 5 Lecture 17: Basic Graph Properties [Sp’10] Algorithms TRAVERSE(s):
put s in bag
while the bag is not empty
take v from the bag
if v is unmarked
mark v
for each edge vw
put w into the bag This traversal algorithm clearly marks each vertex in the graph at most once. In order to show that
it visits every node in the graph at least once, we modify the algorithm slightly; the modiﬁcations are
highlighted in red. Instead of keeping vertices in the bag, the modiﬁed algorithm stores pairs of vertices.
This allows us to remember, whenever we visit a vertex v for the ﬁrst time, which previously-visited
vertex p put v into the bag. This vertex p is called the parent of v .
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 vw
put ( v, w ) into the bag () (†)
() Lemma 1. TRAVERSE(s) marks every vertex in any connected graph exactly once, and the set of pairs
( v, parent( v )) with parent( v ) = ∅ deﬁnes a spanning tree of the graph.
Proof: The algorithm obviously marks s. Let v be any vertex other than s, and let s · · · u v be the
directed 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 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. Thus, by induction on the shortest-path distance from s, the algorithm marks every
vertex in the graph. This implies that parent( v ) is well-deﬁned for every vertex v .
The algorithm clearly marks every vertex at most once, so it must mark every vertex exactly once.
Call any pair ( v, parent( v )) with parent( v ) = ∅ a parent edge. For any node v , the directed 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. Because each
vertex is marked at most once, the for loop (†) is executed at most V times. Each edge uv is put into the
bag exactly twice; once as the pair (u, v ) and once as the pair ( v, u), so line ( ) is executed at most 2 E
times. Finally, we can’t take more things out of the bag than we put in, so line ( ) is executed at most
2 E + 1 times. 17.5 Examples Let’s ﬁrst assume that the graph is represented by an adjacency list, so that the overhead of the for
loop (†) is only constant time per edge.
6 Lecture 17: Basic Graph Properties [Sp’10] Algorithms • If we implement the ‘bag’ using a stack, we recover our original depth-ﬁrst search algorithm. Each
execution of ( ) or ( ) takes constant time, so the overall running time is O (V + E ). If the graph
is connected, we have V ≤ E + 1, and 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 start vertex and on the order that neighbors are visited in the for loop (†), but in
general, depth-ﬁrst spanning trees are long and skinny.
• If we use a queue instead of a stack, we get breadth-ﬁrst search. Again, each execution of ( )
or ( ) takes constant time, so the overall running time for connected graphs 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. We’ll see shortest paths again in a
future lecture. Again, exact shape of a breadth-ﬁrst spanning tree depends on the start vertex and
on the order that neighbors are visited in the for loop (†), but in general, shortest path trees are
short and bushy.
d b 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. • Now 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 ( ), the resulting algorithm is reasonably
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. Surprisingly, as long as all the edge weights are distinct, the resulting tree does
not depend on the start vertex or the order that neighbors are visited; in this case, there is only
one minimum spanning tree. We’ll see minimum spanning trees again in the next lecture.
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 run in
O(V 2 ) time, and ‘shortest-ﬁrst search’ runs in O(V 2 + E log E ) = O(V 2 log V ) time. 17.6 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 ) Some textbooks claim that this wrapper can only be used with depth-ﬁrst search; they’re wrong. 7 Lecture 17: Basic Graph Properties [Sp’10] Algorithms 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).
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.
8 Lecture 17: Basic Graph Properties [Sp’10] Algorithms 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.]
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
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. 9 Lecture 17: Basic Graph Properties [Sp’10] Algorithms 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
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
4 Most other variants of draughts have ‘ﬂying kings’, which behave very differently than what’s described here. 10 Lecture 17: Basic Graph Properties [Sp’10] Algorithms should run in O(n) time, where n is the total number of pieces. [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.] 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. c Copyright 2010 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 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms 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 18
18.1 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 → 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.
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 ) 18.2 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
c Copyright 2011 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. 1 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms 8 5
10
2 18 3 12 30 16 14
4 26 A weighted graph and its minimum spanning tree. 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.1
Lemma 1. The minimum spanning tree contains every safe edge.
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. u e
v
e’ Proving that every safe edge is in the minimum spanning tree. The ‘bad’ component of F is highlighted. Lemma 2. The minimum spanning tree contains no useless edge.
Proof: Adding any useless edge to F would introduce a cycle.
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 we must
describe how to identify new safe and new useless edges, at each iteration of our generic template.
1 Lemmas 1 and 2 are actually special cases of two more general observations. 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 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms 18.3 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 18 12
14 14
4 16 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. We can ﬁnd all the safe edge in the graph in O( E ) time as follows. First, we count the components
of F using whatever-ﬁrst search, using the TRAVERSEALL wrapper function. As we count, we label every
vertex with its component number; that is, every vertex in the ﬁrst component traversed by TRAVERSEALL
gets label 1, every vertex in the second component gets label 2, and so on.
If F has only one component, we’re done. Otherwise, we compute an array S [1 .. V ] of edges, where
S [i ] is the minimum-weight edge with one endpoint in the i th component (or a sentinel value NULL if
there are less than i components). To compute this array, we consider each edge uv in the input graph G .
If the endpoints u and v have the same label, then uv is useless. Otherwise, we compare the weight of
uv to the weights of S [label(u)] and S [label( v )] and update the array entries if necessary. ˙
B OR U VKA(V, E ):
F = (V, ∅)
TRAVERSEALL( F ) 〈〈count and label components〉〉
while F has more than one component
ADDALLSAFEEDGES(V, E )
TRAVERSEALL( F )
return F ADDALLSAFEEDGES(V, E ):
for i ← 1 to V
S [i ] ← NULL 〈〈sentinel: w (NULL) := ∞〉〉
for each edge uv ∈ E
if label(u) = label( v )
if w (uv ) < w (S [label(u)])
S [label(u)] ← uv
if w (uv ) < w (S [label( v )])
S [label( v )] ← uv
for i ← 1 to V
if S [i ] = NULL
add S [i ] to F Each call to TRAVERSEALL requires O(V ) time, because the forest F has at most V − 1 edges. Assuming
the graph is represented by an adjacency list, the rest of each iteration of the main while loop requires
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 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms O( E ) time, because we spend constant time on each edge. Because the graph is connected, we have
V ≤ E + 1, so each iteration of the while loop takes O( E ) time.
Each iteration 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 F initially has V components, the while loop
iterates at most 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
˙
recent algorithms to compute minimum spanning trees are all generalizations of Boruvka’s algorithm,
not the other two classical algorithms described next. 18.4 Jarník’s (‘Prim’s’) Algorithm The next oldest minimum spanning tree algorithm was ﬁrst described by the Czech mathematician Vojtˇch
e
˙
Jarník in a 1929 letter to Boruvka; Jarník published his discovery the following year. The algorithm was
independently rediscovered by Kruskal in 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 ‘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 priority queue. When we
pull the minimum-weight edge out of the priority queue, 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 priority queue. In
other words, Jarník’s algorithm is another instance of the generic graph traversal algorithm we saw last
4 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms time, using a priority queue as the ‘bag’! If we implement the algorithm this way, the algorithm runs in
O( E log E ) = O (E log V ) time. 18.5 Improving Jarník’s Algorithm We can improve Jarník’s algorithm using a more advanced priority queue data structure called a
Fibonacci heap, ﬁrst described by Michael Fredman and Robert Tarjan in 1984. Fibonacci heaps support
the standard priority queue operations INSERT, EXTRACTMIN, and DECREASEKEY. However, unlike standard
binary heaps, which require O(log n) time for every operation, Fibonacci heaps support INSERT and
DECREASEKEY in constant amortized time. The amortized cost of EXTRACTMIN is still O(log n).
To apply this faster data structure, we keep vertices in the priority queue instead of edge, where
the key for each vertex v is either the minimum-weight edge between v and the evolving tree T , or
∞ if there is no such edge. We can INSERT all the vertices into the priority queue at the beginning of
the algorithm; then, whenever we add a new edge to T , we may need to decrease the keys of some
neighboring vertices.
To make the description easier, we break the algorithm into two parts. JARNÍKINIT initializes the
priority queue; JARNÍKLOOP is the main algorithm. The input consists of the vertices and edges of the
graph, plus the start vertex s. For each vertex v , we maintain both its key key( v ) and the incident edge
edge( v ) such that w (edge( v )) = key( v ).
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 neighbor u of v
if u ∈ T and key(u) > w (uv )
/
edge(u) ← uv
DECREASEKEY(u, w (uv )) The operations 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. Thus, if we use a Fibonacci heap,
˙
the improved algorithm runs in O (E + V log V ) time, which is faster than Boruvka’s algorithm unless
E = O(V ).
In practice, however, this improvement is rarely faster than the naive implementation using a binary
heap, unless the graph is extremely large and dense. The Fibonacci heap algorithms are quite complex,
and the hidden constants in both the running time and space are signiﬁcant—not outrageous, but
certainly bigger than the hidden constant 1 in the O(log n) time bound for binary heap operations. 18.6 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 5 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms elaborate’.3 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 18 12 30 26 30 30 16 26 30 26 18 16 14 14 14 26 Kruskal’s algorithm run on the example graph. Thick edges are in F . Dashed edges are useless. 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 increasing weight
F ← (V, ∅)
for each vertex v ∈ V
MAKESET( v )
for i ← 1 to | E |
uv ← i th lightest edge in E
if FIND(u) = FIND( v )
UNION(u, v )
add uv to F
return F
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
4
Live at the Assembly Hall! Only $49.95 on Pay-Per-View! 6 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms 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.]
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. 7 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms 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 ∞.) Describe and analyze an algorithm to 11 1
6 t 4 3 10 2 7 5
s 9
8 12 The bottleneck distance between s and t is 5. 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).
8 Lecture 18: Minimum Spanning Trees [Sp’10] Algorithms ˙
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 2011 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 Lecture 19: Shortest Paths [Sp’10] Algorithms 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) 19
19.1 Shortest Paths
Introduction Suppose we are given a weighted directed graph G = (V, E , w ) with two special vertices, and we want to
ﬁnd the shortest path from a source vertex s to a target vertex t . That is, we want to ﬁnd the directed
path p starting at s and ending at t that minimizes the function
w ( p ) := w (u v ).
u v∈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?’, I 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, because 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, because the presence of a negative cycle might imply 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 we 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 .
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
At one time, there was a speed trap on I-70 just east of the Indiana/Ohio border, but only for eastbound trafﬁc.
c Copyright 2010 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. 1 Lecture 19: Shortest Paths [Sp’10] Algorithms Almost 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 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 some vertices, we can always choose one shortest 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 a d x
If s u y v a b c d v and s a x y d u are shortest paths,
then s a b c d u is also a shortest path. Shortest path trees and minimum spanning trees are very different. For one thing, there is only one
minimum spanning tree (if edge weights are distinct), but each source vertex induces a different shortest
path tree. 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
16 30 18 3 12 14 30 16 14 4 26 4 26 A minimum spanning tree and a shortest path tree (rooted at the topmost vertex) of the same graph. 19.2 Warning! Throughout this lecture, we will explicitly consider only directed graphs. All of the algorithms described
in this lecture also work for undirected graphs with some minor modiﬁcations, but only if negative edges
are prohibited. Dealing with negative edges in undirected graphs is considerably more subtle. We cannot
simply replace every undirected edge with a pair of directed edges, because this would transform any
negative edge into a short negative cycle. Subpaths of an undirected shortest path that contains a
negative edge are not necessarily shortest paths; consequently, the set of all undirected shortest paths
from a single source vertex may not deﬁne a tree, even if shortest paths are unique.
s
1 u s
1 –1 v 1 u s
1 –1 v 1 1 u –1 v An undirected graph where shortest paths from s are unique but do not deﬁne a tree. A complete treatment of undirected graphs with negative edges is beyond the scope of this lecture (if
not the entire course). I will only mention that a single shortest path in an undirected graph with negative
edges can be computed in O(V E + V 2 log V ) time, by a reduction to maximum weighted matching.
2 Lecture 19: Shortest Paths [Sp’10] Algorithms 19.3 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 Lester Ford in 1956,
and independently by George 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 v path, or ∞ if there is no such path. • pred( v ) is the predecessor of v in the tentative shortest s
vertex. v path, or NULL if there is no such In fact, 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.
At the beginning of the 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 four simple claims:
1. For every vertex v , dist( v ) is always greater than or equal to the shortest-path distance from s to v .
This claim can be veriﬁed by induction on the number of relaxations.
2. If no edge is tense, then for every vertex v , dist( v ) is the shortest-path distance from s to v . This
claim can be veriﬁed by induction on the length of the true shortest path, or on the rank of the
true shortest path distance if all edge weights are non-negative.
3. 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 . This claim can be proved by induction on the number of edges in the path. (Why
does pred( v ) have its current value?)
4. 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.
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.
3 Speciﬁcally, Dantzig showed that the shortest path problem can be phrased as a linear programming problem, and then
described an interpretation of his simplex method in terms of the original graph. His description is equivalent to Ford’s
relaxation strategy. 3 Lecture 19: Shortest Paths [Sp’10] Algorithms 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 FIFO queue, and a priority queue. 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 prove to be much more efﬁcient. 19.4 Dijkstra’s Algorithm If we implement the bag using a priority queue, 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 George 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.) On the other hand, in
practice, Dijkstra’s algorithm is usually quite fast even for graphs with negative edges. 19.5 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
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
Most textbooks, and indeed Dijkstra’s original paper, present a version of Dijkstra’s algorithm that gives incorrect results
for graphs with negative edges, because it never visits the same vertex more than once. 4 Lecture 19: Shortest Paths [Sp’10] Algorithms
∞ ∞
3 2
1 ∞
0 ∞ ∞ 0
12 8
∞ 2
1 5
∞ 10 6 ∞ 3 ∞
0 12 8 6 3 0
s 6 3
0
s ∞ 7
3 2
1
0 3 8
14 4 9 0
12 8
3 4 ∞ 3 12
7
3 4
6 0
s 9
5
4 10 7 6 2
1 5
4 10 3 4 0
s 4 12
7 ∞ 3 4 12
5 4 10 7 ∞ 2
1 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. 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, Racetrack, . . . ) 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. 19.6 Shimbel’s Algorithm (‘Bellman-Ford’) If we replace the heap in Dijkstra’s algorithm with a FIFO queue, we obtain an algorithm ﬁrst described
by Shimbel in 1955, then independently rediscovered by Moore in 1957, by Woodbury and Dantzig in
1957, and by Bellman in 1958. Because Bellman explicitly used Ford’s formulation of relaxing edges,
this algorithm is usually called ‘Bellman-Ford’, although some early sources refer to ‘Bellman-Shimbel’.
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 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.)
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 at most the length of the shortest
path s v consisting of at most i edges.
Since a shortest path can only pass through each vertex once, either the algorithm halts before the
5 Lecture 19: Shortest Paths [Sp’10] Algorithms 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 ).
e
∞
−8
d
∞ −8 2
f
∞ 1
0 −1 6 0 3 −3 −3 4
6 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
3
c f
7 b
2 −1 6
a 2
1 5 8
∞
c 4 d
4 f
∞ b
4 −3 −3 −8 2
1 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 end-of-phase token.
Shaded vertices are in the queue at the end of each phase. The bold edges describe the evolving shortest path tree. 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 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 Lecture 19: Shortest Paths [Sp’10] Algorithms 19.7 Shimbel’s Algorithm as Dynamic Programming Shimbel’s algorithm can also be recast as a dynamic programming algorithm. Let disti ( v ) denote the
length of the shortest path s v consisting of at most i edges. It’s not hard to see that this function obeys
the following recurrence: 0
if i = 0 and v = s ∞
if i = 0 and v = s
disti ( v ) = disti −1 ( v ), min
otherwise
min (disti −1 (u) + w (u v ))
u v∈E For the moment, let’s assume the graph has no negative cycles; our goal is to compute distV −1 ( t ). We can
clearly memoize this two-parameter function into a two-dimensional array. A straightforward dynamic
programming evaluation of this recurrence looks like this:
SHIMBELDP(s)
dist[0, s] ← 0
for every vertex v = s
dist[0, v ] ← ∞
for i ← 1 to V − 1
for every vertex v
dist[i , v ] ← dist[i − 1, v ]
for every edge u v
if dist[i , v ] > dist[i − 1, u] + w (u v )
dist[i , v ] ← dist[i − 1, u] + w (u v ) Now let us make two minor changes to this algorithm. First, we remove one level of indentation
from the last three lines. This may change the order in which we examine edges, but the modiﬁed
algorithm still computes disti ( v ) for all i and v . Second, we change the indices in the last two lines from
i − 1 to i . This change may cause the distances dist[i , v ] to approach the true shortest-path distances
more quickly than before, but the algorithm is still correct.
SHIMBELDP2(s)
dist[0, s] ← 0
for every vertex v = s
dist[0, v ] ← ∞
for i ← 1 to V − 1
for every vertex v
dist[i , v ] ← dist[i − 1, v ]
for every edge u v
if dist[i , v ] > dist[i , u] + w (u v )
dist[i , v ] ← dist[i , u] + w (u v ) Now notice that the iteration index i is completely redundant! We really only need to keep a onedimensional array of distances, which means we don’t need to scan the vertices in each iteration of the
main loop.
SHIMBELDP3(s)
dist[s] ← 0
for every vertex v = s
dist[ v ] ← ∞
for i ← 1 to V − 1
for every edge u v
if dist[ v ] > dist[u] + w (u v )
dist[ v ] ← dist[u] + w (u v ) 7 Lecture 19: Shortest Paths [Sp’10] Algorithms The resulting algorithm is (almost) identical to our earlier algorithm SHIMBEL! The ﬁrst three lines
initialize the shortest path distances, and the last two lines check whether an edge is tense, and if so,
relaxes it. The only thing missing is the explicit maintenance of predecessors, but that’s easy to add. 19.8 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 data structure 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 Exercises
1. 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.
2. 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). Speciﬁcally, 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.]
3. 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.]
4. 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.
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 8 Lecture 19: Shortest Paths [Sp’10] Algorithms 5 4 17 8 0 42 1 23 9 14 7 16 A looped tree. (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.
5. (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.
6. 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.
7. 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.
8. Negative edges cause problems in shortest-path algorithms because of the possibility of negative
cycles. But what if the input graph has no cycles?
in the worst case, so nobody’s ever bothered to publish it. 9 Lecture 19: Shortest Paths [Sp’10] Algorithms (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.
9. 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
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.
10. 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.]
11. 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 0.5 Area 51, AZ 10 Langley, VA Lecture 19: Shortest Paths [Sp’10] Algorithms 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.)
12. 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
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/69/)
Reproduced under a Creative Commons Attribution-NonCommercial 2.5 License c Copyright 2010 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 20: 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) 20 All-Pairs Shortest Paths 20.1 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). 20.2 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.
c Copyright 2010 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. 1 Algorithms Lecture 20: 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. 20.3 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! 20.4 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 ), so the new weight of every edge is
w (u v ) = dist(s, u) + w (u v ) − dist(s, v ).
2 Algorithms Lecture 20: All-Pairs Shortest Paths 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
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
dist[u, ·] ← DIJKSTRA(V, E , w , u)
for every vertex 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 ). 20.5 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
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
3 Algorithms Lecture 20: All-Pairs Shortest Paths 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
for all vertices v
if u = v
dist[u, v, 0] ← 0
else
dist[u, v, 0] ← ∞
for k ← 1 to V − 1
for all vertices u
for all vertices v
dist[u, v, k] ← ∞
for all vertices x
if dist[u, v, k] > dist[u, x , k − 1] + w ( x v )
dist[u, v, k] ← dist[u, x , k − 1] + w ( x v ) In fact, this algorithm is almost the same as running Shimbel’s algorithm once from each source
vertex. The main difference is the innermost loop, which in Shimbels 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. Other differences are just as in the dynamic programming development of Shimbel’s
algorithm—we don’t need the inner loop over vertices v , and we only need a two-dimensional table.
After the kth iteration of the main loop in the following algorithm, the value of dist[u, v ] lies between the
true shortest path distance from u to v and the value dist[u, v, k] computed in the previous algorithm.
SHIMBELAPSP(V, E , w ):
for all vertices u
for all vertices v
if u = v
dist[u, v ] ← 0
else
dist[u, v ] ← ∞
for k ← 1 to V − 1
for all vertices u
for all edges x v
if dist[u, v ] > dist[u, x ] + w ( x v )
dist[u, v ] ← dist[u, x ] + w ( x v ) 4 Algorithms 20.6 Lecture 20: All-Pairs Shortest Paths 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 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
for all vertices v
dist[u, v, 0] ← w (u v )
for i ← 1 to lg V
〈〈k = 2i 〉〉
for all vertices u
for all vertices v
dist[u, v, i ] ← ∞
for all vertices x
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] This algorithm is not the same as V invocations of any single-source algorithm; we can’t replace the
innermost loop over vertices with a loop over edges. However, we can remove the last dimension of
the table, using dist[u, v ] everywhere in place of dist[u, v, i ], just as in Shimbel’s algorithm, thereby
reducing the space from O(V 3 ) to O(V 2 ).
FASTSHIMBELAPSP(V, E , w ):
for all vertices u
for all vertices v
dist[u, v ] ← w (u v )
for i ← 1 to lg V
for all vertices u
for all vertices v
for all vertices x
if dist[u, v ] > dist[u, x ] + dist[ x , v ]
dist[u, v ] ← dist[u, x ] + dist[ x , v ] 20.7 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.)
5 Algorithms Lecture 20: All-Pairs Shortest Paths
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.
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 20.8 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 .
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 20: All-Pairs Shortest Paths 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 e
iat
ed
rm ≤ r-1
e
int des
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 ).
Once again, we should immediately see that a dynamic programming algorithm will implement this
recurrence in Θ(V 3 ) time.
FLOYDWARSHALL(V, E , w ):
for all vertices u
for all vertices v
dist[u, v, 0] ← w (u v )
for r ← 1 to V
for all vertices u
for all vertices 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] Just like our earlier algorithms, we can simplify the algorithm by removing the third dimension of
the memoization table. Also, because the vertex numbering was chosen arbitrary, so there’s no reason to
refer to it explicitly in the pseudocode.
FLOYDWARSHALL2(V, E , w ):
for all vertices u
for all vertices v
dist[u, v ] ← w (u v )
for all vertices r
for all vertices u
for all vertices v
if dist[u, v ] > dist[u, r ] + dist[ r, v ]
dist[u, v ] ← dist[u, r ] + dist[ r, v ] Now compare this algorithm with FASTSHIMBELAPSP. Instead of O(log V ) passes through all triples of
vertices, FLOYDWARSHALL2 only requires a single pass, but only because it uses a different nesting order
for the three for-loops!
7 Algorithms Lecture 20: All-Pairs Shortest Paths 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. (The resulting algorithm is not the same as Floyd-Warshall!)
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 (FLOYDWARSHALL2) to return the correct
shortest-path distances, even if the graph has negative cycles.
3. All of the shortest-path algorithms described in this note can also be modiﬁed to return an explicit
description of some negative cycle, instead of simply reporting that a negative cycle exists.
(a) Describe how to modify Johnson’s algorithm to return either the matrix of shortest-path
distances or a negative cycle.
(b) Describe how to modify the Floyd-Warshall algorithm (FLOYDWARSHALL2) to return either the
matrix of shortest-path distances or a negative cycle.
If the graph contains more than one negative cycle, your algorithms may choose one arbitrarily. 8 Algorithms Lecture 20: All-Pairs Shortest Paths 4. 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,
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 ).
5. 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.)
6. 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
3 Automata theory is a prerequisite for the undergraduate 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 9 Algorithms Lecture 20: All-Pairs Shortest Paths (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.
(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 2010 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 Lecture 21: Maximum Flows and Minimum Cuts [Fa’10]
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) Harris and Ross’s map of the Warsaw Pact rail network 21 Figure 2
From Harris and Ross [1955]: Schematic diagram of the railway network of the Western Soviet Union and Eastern European countries, with a maximum ﬂow of value 163,000 tons from
Russia to Eastern Europe, and a cut of capacity 163,000 tons indicated as ‘The bottleneck’. Maximum Flows and Minimum Cuts h m x-ﬂow m and th S. m
In the mid-1950s, Air Force researchersTT.e E.aHarris in-cut F. eoreRoss published a classiﬁed report studying
the rail network that IlinkedNtheepSovietNUnion1954,its dsatelliteson [1954] gave (nin Eastern Europe. The network
n the RA D R ort of 19 ovember to For and Fulker countries ext to deﬁning
the
uﬂ
was modeled as a graph maximmmforoundirroblteem graphu,ggesyiing tthatsitmgeographic iregions,ﬂandtih105 edges, representing
with 44 w p ec d and s s sa t ng he heplmxaximtumdﬂfow v) tuhee imax- al wom ne
vertices, representing e me ho or t al s equ o t cut theore
minimum in ty of rail eparating sou e and t minal. was oof is not c weight,
links between those regionscapacithe a cut snetwork.rcEacheredgeTheir prgiven aonstructive, representing the rate
but for planar graphs, with source and sink on the outer boundary, they give a polynomialat which material could, cbestshipped ofrom rone of 26 May 1955,theacker [1955a] showed that thby trial and error, they
time on ructive meth d. In a eport region to Rob next. Essentially e
max-ﬂow min- amount of e deriv d also rom the be x-disjoint ersion Menger’
determined both the maximumcut theorem can bstuff ethat fcould vertemovedvfromofRussias into Europe, as well as
theorem.
the cheapest way to disrupt ttheirnetworkord anremoving linkserved tin less ax-ﬂow min-cuterms, blowing up train
As for he d ected case, F by d Fulkerson [1955] obs (or hat the m abstract t
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irected gr
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tracks), which they called lhsolofs Dantfzor d1951a] onapnhs.ger solzuitionsd fFulk(includingwthe yﬁgure gat the top of the page)
‘thealbottleneck’. i Theirt results thrsonan1955]tatiooned,robleem etndtn e
the resu t
ig [
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oh
1
were only declassiﬁed in 1999.
This one of the ﬁrst recorded applications of 25 maximum ﬂow and minimum cut problems. For
the
both problems, the input is a directed graph G = (V, E ), along with special vertices s and t called the
source and target. As in the previous lectures, I will use u v to denote the directed edge from vertex u
1
Both the map and the story were taken from Alexander Schrijver’s fascinating survey ‘On the history of combinatorial
optimization (till 1960)’. c Copyright 2011 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. 1 Lecture 21: Maximum Flows and Minimum Cuts [Fa’10] Algorithms to vertex v . Intuitively, the maximum ﬂow problem asks for the largest amount of material that can
be transported from one vertex to another; the minimum cut problem asks for the minimum damage
needed to separate two vertices. 21.1 Flows An ( s , t )-ﬂow (or just a ﬂow if the source and target are clear from context) is a function f : E →
that satisﬁes the following conservation constraint at every vertex v except possibly s and t :
f (u v ) =
u ≥0 f ( v w ).
w In English, the total ﬂow into v is equal to the total ﬂow out of v . To keep the notation simple, we deﬁne
f (u v ) = 0 if there is no edge u v in the graph. The value of the ﬂow f , denoted | f |, is the total net
ﬂow out of the source vertex s:
| f | := f (s w ) −
w f (u s).
u It’s not hard to prove that | f | is also equal to the total net ﬂow into the target vertex t , as follows. To
simplify notation, let ∂ f ( v ) denote the total net ﬂow out of any vertex v :
∂ f ( v ) := f (u v ) −
u f ( v w ).
w The conservation constraint implies that ∂ f ( v ) = 0 or every vertex v except s and t , so
∂ f ( v ) = ∂ f (s) + ∂ f ( t ).
v On the other hand, any ﬂow that leaves one vertex must enter another vertex, so we must have
v ∂ f ( v ) = 0. It follows immediately that | f | = ∂ f (s ) = −∂ f ( t ).
Now suppose we have another function c : E → ≥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 Lecture 21: Maximum Flows and Minimum Cuts [Fa’10] Algorithms 21.2 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 → ≥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 (u v ) by the conservation constraint f (u v ) f (v w) − removing duplicate edges u v ∈S w v ∈S w∈T = 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. 21.3 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:
3 Lecture 21: Maximum Flows and Minimum Cuts [Fa’10] Algorithms 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 → ≥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 →
capacity, as follows: c (u v ) − f (u v ) if u v ∈ E
c f (u v ) = f ( v u) 0 , called the residual 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 10 5/20
10/10 t
5
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.
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 4 Lecture 21: Maximum Flows and Minimum Cuts [Fa’10] Algorithms 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 . push through the augmenting path in G f . We deﬁne a new ﬂow function f : E → f (u v ) + F
f (u v ) = f (u v ) − F f (u v ) as follows: 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 the unique 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 21: Maximum Flows and Minimum Cuts [Fa’10] 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 O( E ) 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 O( E ) path ﬂows and cycle ﬂows.
(e) Prove that every ﬂow with value 0 can be written as the sum of O( E ) 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 2011 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 Lecture 22: Max-Flow Algorithms [Fa’10] 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) 22
22.1 Max-Flow Algorithms
Ford-Fulkerson augmenting path Ford and Fulkerson’s proof of the Maxﬂow Mincut theorem, described in the previous lecture note,
translates immediately to an algorithm to compute maximum ﬂows: 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 the Ford-Fulkerson algorithm runs in O (E | f ∗ |) time in the worst case.
The following example shows that this running time analysis is essentially tight. Consider the 4-node
network illustrated below, where X is some large integer. The maximum ﬂow in this network is clearly
2X . Moreover, the Ford-Fulkerson could alternate between pushing 1 unit of ﬂow along the augmenting
path s u v t and then pushing 1 unit of ﬂow along the augmenting path s v u t , leading to a
running time of Θ(X ) = Ω(| f ∗ |).
v
X X
s t 1
X X u
A bad example for the Ford-Fulkerson algorithm. Ford and Fulkerson’s algorithm works quite well in many practical situations, or in settings where
the maximum ﬂow value | f ∗ | is small, but without further constraints on the augmenting paths, this is
not an efﬁcient algorithm in general. The example network above can be described using only O(log X )
bits; thus, the running time of Ford-Fulkerson is exponential in the input size.
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, although still in exponential time. However, if we
allow irrational capacities, the algorithm can actually 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 six-node network shown on the next page. Six of the nine 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.)
c Copyright 2011 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. 1 Lecture 22: Max-Flow Algorithms [Fa’10] Algorithms s
X X X 1 ϕ 1
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,
and 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 .
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! 22.2 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:
1 ...or perhaps the laws of physics. Yeah, right. Whatever. Like reality actually matters in this class. 2 Lecture 22: Max-Flow Algorithms [Fa’10] Algorithms 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, one 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, which is actually a polynomial function of the input size. 22.3 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.
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. 3 Lecture 22: Max-Flow Algorithms [Fa’10] Algorithms 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.
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. 22.4 Push-Relabel
To be written! 22.5 Further Progress This is nowhere near the end of the story for maximum-ﬂow algorithms. Decades of further research
have led to a number of even faster algorithms, some of which are summarized in the table below.3 All
of the algorithms listed below compute a maximum ﬂow in several iterations. Each algorithm has two
variants: a simpler version that performs each iteration by brute force, and a faster variant that uses
sophisticated data structures to maintain a spanning tree of the ﬂow network, so that each iteration can
be performed (and the spanning tree updated) in logarithmic time. There is no reason to believe that the
best algorithms known so far are optimal; indeed, maximum ﬂows are still a very active area of research.
3 To keep the table short, I have deliberately omitted algorithms whose running time depends on the maximum capacity, the
sum of the capacities, or the maximum ﬂow value. Even with this restriction, the table is incomplete! 4 Lecture 22: Max-Flow Algorithms [Fa’10] Algorithms
Technique
Blocking ﬂow
Network simplex Direct With dynamic trees O(V 3 )
O(V 2 E ) O(V E log V )
O(V E log V ) Push-relabel (generic)
Push-relabel (FIFO)
Push-relabel (highest label)
Pseudoﬂow O(V 2 E )
O(V 3 )
O(V 2 E )
O(V 2 E ) — O(V 2 log(V 2 / E ))
— O(V E log V ) Sources
[Dinits; Sleator and Tarjan]
[Dantzig; Goldfarb and Hao;
Goldberg, Grigoriadis, and Tarjan]
[Goldberg and Tarjan]
[Goldberg and Tarjan]
[Cheriyan and Maheshwari; Tunçel]
[Hochbaum] Several purely combinatorial maximum-ﬂow algorithms and their running times. 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 Lecture 22: Max-Flow Algorithms [Fa’10] 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 2011 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 23: Applications of Maximum Flow [Fa’10]
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) 23
23.1 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. 23.2 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.
c Copyright 2011 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. 1 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms 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
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! 23.3 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 . 2 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms 23.4 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
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, where for each girl x and
boy y , we have c ( x ) = 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 register).
Every binary assignment problem can be reduced 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.
The value of the maximum ﬂow clearly is at most C V , where C is an upper bound on all the vertex
and edge capacities, and n is the number of vertices, so the off-the-shelf Ford-Fulkerson algorithm runs
in O (E | f ∗ |) = O (C V E ) time. If C = O(1), this is more efﬁcient than any specialized ﬂow algorithm. 23.5 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
1 Most authors refer to ﬁnding a maximum-weight matching in a bipartite graph as the assignment problem. 3 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms 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 75–59
71–63
69–66
63–72
49–86 28
28
27
27
27 NYY BOS TOR DET 3
3
8
7
3 BAL 8
2 7
7
0 3
4
0
0 2
7
4 0
0 0 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 .
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 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms 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 ) =
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 3 NYY NYY
BOS BAL 8
s 7 5 NYY
TOR t
7 2
7 1 BOS BAL
BOS
BAL
TOR 13 TOR The ﬂow graph for the 1996 American League East standings. Unlabeled edges have inﬁnite capacity. 23.6 Project Selection In our ﬁnal example, suppose we are given a set of n projects that we could possibly perform; for
simplicity, we identify each project by an integer between 1 and n. Some projects cannot be started until
certain other projects are completed. This set of dependencies is described by a directed acyclic graph,
where an edge i j indicates that project i depends on project j . Finally, each project i has an associated
proﬁt pi which is given to us if the project is completed; however, some projects have negative proﬁts,
which we interpret as positive costs. We can choose to ﬁnish any subset X of the projects that includes
all its dependents; that is, for every project x ∈ X , every project that x depends on is also in X . Our goal
is to ﬁnd a valid subset of the projects whose total proﬁt is as large as possible. In particular, if all of the
jobs have negative proﬁt, the correct answer is to do nothing.
At a high level, our task to partition the projects into two subsets S and T , the jobs we Select and the
jobs we Turn down. So intuitively, we’d like to model our problem as a minimum cut problem in a certain
5 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms
-2 -3 -5 -8 4 6 2 3 A dependency graph for a set of projects. Circles represent proﬁtable projects; squares represent costly projects. graph. But in which graph? How do we enforce prerequisites? We want to maximize proﬁt, but we only
know how to ﬁnd minimum cuts. And how do we convert negative proﬁts into positive capacities?
We deﬁne a new graph G by adding a source vertex s and a target vertex t to the dependency graph,
with an edge s j for every proﬁtable job (with p j > 0), and an edge i t for every costly job (with
pi < 0). Intuitively, we can think of s as a new job (“To the bank!”) with proﬁt/cost 0 that we must
perform last. We assign edge capacities as follows:
• c (s j ) = p j for every proﬁtable job j ;
• c (i t ) = − pi for every costly job i ;
• c (i j ) = ∞ for every dependency edge i j .
All edge-capacities are positive, so this is a legal input to the maximum cut problem.
Now consider an (s, t )-cut (S , T ) in G . If the capacity S , T is ﬁnite, then for every dependency edge
i j , projects i and j are on the same side of the cut, which implies that S is a valid solution. Moreover,
we claim that selecting the jobs in S earns us a total proﬁt of C − S , T , where C is the sum of all the
positive proﬁts. This claim immediately implies that we can maximize our total proﬁt by computing a
minimum cut in G .
t
3 2
∞ -2
∞ 5 -3 -5 ∞ ∞ 4 -8 ∞
∞ 6
4 8 2 6 ∞ ∞ 2 3
3 s The ﬂow network for the example dependency graph, along with its minimum cut.
The cut has capacity 13 and C = 15, so the total proﬁt for the selected jobs is 2. We prove our key claim as follows. For any subset A of projects, we deﬁne three functions:
cost(A) := − pi =
i ∈ A : p i <0 beneﬁt(A) := c (i t )
i ∈A pj =
j ∈ A : p i >0 proﬁt(A) := c (s j )
j ∈A pi = beneﬁt(A) − cost(A).
i ∈A By deﬁnition, C = beneﬁt(S ) + beneﬁt( T ). Because the cut (S , T ) has ﬁnite capacity, only edges of the
form s j and i t can cross the cut. By construction, every edge s j points to a proﬁtable job and each
6 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms edge i t points from a costly job. Thus, S , T = cost(S ) + beneﬁt( T ). We immediately conclude that
C − S , T = beneﬁt(S ) − cost(S ) = proﬁt(S ), as claimed. 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 → ≥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 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
7 Lecture 23: Applications of Maximum Flow [Fa’10] Algorithms 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. The UIUC Faculty Senate has decided to convene a committee to determine whether Chief Illiniwek
should become the ofﬁcial mascot symbol of the University of Illinois Global Campus. Exactly
one faculty member must be chosen from each academic department to serve on this committee.
Some faculty members have appointments in multiple departments, but each committee member
will represent only one department. For example, if Prof. Blagojevich is afﬁliated with both the
Department of Corruption and the Department of Stupidity, and he is chosen as the Stupidity representative, then someone else must represent Corruption. Finally, University policy requires that
any committee on virtual mascots symbols must contain the same number of assistant professors,
associate professors, and full professors. Fortunately, the number of departments is a multiple of 3.
Describe an efﬁcient algorithm to select the membership of the Global Illiniwek Committee.
Your input is a list of all UIUC faculty members, their ranks (assistant, associate, or full), and their
departmental afﬁliation(s). There are n faculty members and 3k departments.
7. You’re organizing the First Annual UIUC Computer Science 72-Hour Dance Exchange, to be held
all day Friday, Saturday, and Sunday. Several 30-minute sets of music will be played during the
event, and a large number of DJs have applied to perform. You need to hire DJs according to the
following constraints.
• Exactly k sets of music must be played each day, and thus 3k sets altogether.
• Each set must be played by a single DJ in a consistent music genre (ambient, bubblegum,
dubstep, horrorcore, hyphy, trip-hop, Nitzhonot, Kwaito, J-pop, Nashville country, . . . ).
• Each genre must be played at most once per day.
• Each candidate DJ has given you a list of genres they are willing to play.
• Each DJ can play at most three sets during the entire event.
Suppose there are n candidate DJs and g different musical genres available. Describe and analyze
an efﬁcient algorithm that either assigns a DJ and a genre to each of the 3k sets, or correctly
reports that no such assignment is possible. 8 Algorithms Lecture 23: Applications of Maximum Flow [Fa’10] 8. The University of Southern North Dakota at Hoople has hired you to write an algorithm to schedule
their ﬁnal exams. Each semester, USNDH offers n different classes. There are r different rooms
on campus and t different time slots in which exams can be offered. You are given two arrays
E [1 .. n] and S [1 .. r ], where E [i ] is the number of students enrolled in the i th class, and S [ j ] is
the number of seats in the j th room. At most one ﬁnal exam can be held in each room during each
time slot. Class i can hold its ﬁnal exam in room j only if E [i ] < S [ j ].
Describe and analyze an efﬁcient algorithm to assign a room and a time slot to each class (or
report correctly that no such assignment is possible).
9. Suppose you are running a web site that is visited by the same set of people every day. Each visitor
claims membership in one or more demographic groups; for example, a visitor might describe
himself as male, 35–45 years old, a father, a resident of Illinois, an academic, a blogger, and
a fan of Joss Whedon.4 Your site is supported by advertisers. Each advertiser has told you
which demographic groups should see its ads and how many of its ads you must show each day.
Altogether, there are n visitors, k demographic groups, and m advertisers.
Describe an efﬁcient algorithm to determine, given all the data described in the previous
paragraph, whether you can show each visitor exactly one ad per day, so that every advertiser
has its desired number of ads displayed, and every ad is seen by someone in an appropriate
demographic group.
10. 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
(a) Describe an efﬁcient algorithm that either rounds A in this fashion, or reports correctly that
no such rounding is possible.
(b) Suppose we are guaranteed that none of the entries in the input matrix A are integers.
Describe an even faster algorithm that either rounds A or reports correctly that no such
rounding is possible. For full credit, your algorithm must run in O(mn) time. [Hint: Don’t
use ﬂows.]
11. Ad-hoc networks are made up of low-powered wireless devices. In principle5 , 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.
4
5 Har har har! Mine is an evil laugh! Now die!
but not really in practice 9 Algorithms Lecture 23: Applications of Maximum Flow [Fa’10] 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.
12. 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.] c Copyright 2011 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 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms “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 → , and capacity function c : E → . 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 → 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 ( t s) = ∞.
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. c Copyright 2010 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. 1 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms 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 →
f :E → : be a feasible (s, t )-ﬂow in the original graph G . Consider the following function
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 → 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 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms 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 }) → 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 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms Similar reductions allow us to several other variants of the maximum ﬂow problem using the same
path-augmentation techniques. For example, we could associate capacities and demands with the
vertices instead of (or in addition to) the edges, as well as a range of excesses with every vertex, instead
of a single excess value. 18.3 Minimum-Cost Flows Now imagine that each edge e in the network has both a capacity c (e) and a cost $(e). The cost function
describes the cost of sending a unit of ﬂow through the edges; thus, the cost any ﬂow f is deﬁned as
follows:
$( f ) =
$(e) · f (e).
e∈ E The minimum-cost maximum-ﬂow problem is to compute a maximum ﬂow of minimum cost. If the
network has only one maximum ﬂow, that’s what we want, but if there is more than one maximum ﬂow,
we want the maximum ﬂow whose cost is as small as possible. Costs can either be positive, negative, or
zero. However, if an edge u v and its reversal v u both appear in the graph, their costs must sum to
zero: $(u v ) = −$( v u). Otherwise, we could make an inﬁnite proﬁt by pushing ﬂow back and forth
along the edge!
Each augmentation step in the standard Ford-Fulkerson algorithm both increases the value of the
ﬂow and changes its cost. If the total cost of the augmenting path is positive, the cost of the ﬂow
decreases; conversely, if the total cost of the augmenting path is negative, the cost of the ﬂow decreases.
We can also change the cost of the ﬂow without changing its value, by augmenting along a directed cycle
in the residual graph. Again, augmenting along a negative-cost cycle decreases the cost of the ﬂow, and
augmenting along a positive-cost cycle increases the cost of the ﬂow.
It follows immediately that a ﬂow f is a minimum-cost maximum ﬂow in G if and only if the residual
graph G f has no directed paths from s to t and no negative-cost cycles.
We can compute a min-cost max-ﬂow using the so-called cycle cancelling algorithm ﬁrst proposed by
Morton Klein in 1967. The algorithm has two phases; in the ﬁrst, we compute an arbitrary maximum
ﬂow f , using any method we like. The second phase repeatedly decreases the cost of f , by augmenting
f along a negative-cost cycle in the residual graph G f , until no such cycle exists. As in Ford-Fulkerson,
the amount of ﬂow we push around each cycle is equal to the minimum residual capacity of any edge on
the cycle.
In each iteration of the second phase, we can use a modiﬁcation of Shimbel’s shortest path algorithm
(often called “Bellman-Ford”) to ﬁnd a negative-cost cycle in O(V E ) time. To bound the number of
iterations in the second phase, we assume that both the capacity and the cost of each edge is an integer,
and we deﬁne
C = max c (e)
and
D = max |$(e)|.
e∈ E e∈ E The cost of any feasible ﬂow is clearly between − EC D and EC D, and each augmentation step decreases
the cost of the ﬂow by a positive integer, and therefore by at least 1. We conclude that the second phase
requires at most 2 EC D iterations, and therefore runs in O (V E 2 C D ) time. As with the raw Ford-Fulkerson
algorithm, this running time is exponential in the complexity of the input, and it may never terminate if
the capacities and/or costs are irrational.
Like Ford-Fulkerson, more careful choices of which cycle to cancel can lead to more efﬁcient
algorithms. Unfortunately, some obvious choices are NP-hard to compute, including the cycle with most
negative cost and the negative cycle with the fewest edges. In the late 1980s, Andrew Goldberg and Bob
Tarjan developed a min-cost ﬂow algorithm that repeatedly cancels the so-called minimum-mean cycle,
which is the cycle whose average cost per edge is smallest. By combining an algorithm of Karp to compute
4 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms minimum-mean cycles in O( EV ) time, efﬁcient dynamic tree data structures, and other sophisticated
techniques that are (unfortunately) beyond the scope of this class, their algorithm achieves a running
time of O (E 2 V log2 V ). The fastest min-cost max-ﬂow algorithm currently known,1 due to James Orlin,
reduces the problem to O( E log V ) iterations of Dijkstra’s shortest-path algorithm; Orlin’s algorithm runs
in O (E 2 log V + E V log2 V ) time. 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 → ,
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 algorithms2 , 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
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
1 at least, among algorithms whose running times do not depend on C and D
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 ).
2 5 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms 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).
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.
6 Lecture 18: Extensions of Maximum Flow [Fa’10] Algorithms 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 → + . 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. (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.
7 Algorithms Lecture 18: Extensions of Maximum Flow [Fa’10] (c) Describe and analyze an efﬁcient algorithm to compute the smallest set of monotone paths
that covers every marked cell.
3. Suppose we are given a set of boxes, each speciﬁed by their height, width, and depth in centimeters.
All three side lengths of every box lie strictly between 10cm and 20cm. As you should expect, one
box can be placed inside another if the smaller box can be rotated so that its height, width, and
depth are respectively smaller than the height, width, and depth of the larger box. Boxes can be
nested recursively. Call a box is visible if it is not inside another box.
Describe and analyze an algorithm to nest the boxes so that the number of visible boxes is as
small as possible.
4. Let G be a directed ﬂow network whose edges have costs, but which contains no negative-cost
cycles. Prove that one can compute a minimum-cost maximum ﬂow in G using a variant of
Ford-Fulkerson that repeatedly augments the (s, t )-path of minimum total cost in the current
residual graph. What is the running time of this algorithm? c Copyright 2010 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 Lecture 25: Linear Programming [Fa ’10] Algorithms 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) 25 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 ∈ d 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 ) ∈ n×d , a column vector b ∈ n , and a row vector c ∈ d .
Each coordinate of the vector x is called a variable. Each of the linear inequalities is called a constraint.
c Copyright 2010 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. 1 Lecture 25: Linear Programming [Fa ’10] Algorithms 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.
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). 25.1 The Geometry of Linear Programming A point x ∈ d 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.
1
2 Confusingly, some authors call this standard form.
Confusingly, some authors call this standard form. 2 Lecture 25: Linear Programming [Fa ’10] Algorithms Any linear equation in d variables deﬁnes a hyperplane in d ; think of a line when d = 2, or a plane
when d = 3. This hyperplane divides d 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
(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 d (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. 3 Lecture 25: Linear Programming [Fa ’10] Algorithms A two-dimensional polyhedron (white) that is unbounded downward but bounded upward. 25.2 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.
4 Lecture 25: Linear Programming [Fa ’10] Algorithms 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! 25.3 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
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! 25.4 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 ( ) ⇐⇒ 3 min y · b
s.t. yA ≥ 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. 5 Lecture 25: Linear Programming [Fa ’10] Algorithms 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 ⇐⇒ max − b · y
s.t. −A y ≤ −c
y ≥0 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. 25.5 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∗ ?
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
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 Lecture 25: Linear Programming [Fa ’10] Algorithms In fact, there is. Let’s multiply each of the constraints in our LP by a new non-negative scalar value yi :
4 x1 + x2 + 3 x3 maximize
subject to 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 .
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. 25.6 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
hyperplanes 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).
7 Lecture 25: Linear Programming [Fa ’10] Algorithms To develop some intuition, let’s ﬁrst consider the very special case where x ∗ = (0, 0, . . . , 0). Let ei
denote the i th standard basis vector, whose i th coordinate is 1 and all other coordinates are 0. Because
x i∗ = 0 for all i , our non-degeneracy assumption implies the strict inequality ai · x ∗ < bi for all i . Thus,
any sufﬁciently small ball around the origin does not intersect any other constraint hyperplane ai · x = bi .
Thus, for all i , and for any sufﬁciently small δ > 0, the vector δ ei is feasible. Because x ∗ is the unique
optimum, we must have δ ci = c · (δ ei ) < c · x ∗ = 0. We conclude that ci < 0 for all i .
Now let y = (0, 0, . . . , 0) as well. We immediately observe that yA ≥ c and y ≥ 0; in other words,
y is a feasible solution for the dual linear program . But y · b = 0 = c · x ∗ , so the weak duality theorem
implies that y is an optimal solution to , and the proof is complete for this very special case.
Now let us consider the more general case. Let x ∗ be the optimal solution for the linear program Π;
our non-degeneracy assumption 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 constraints and
possibly changing coordinates), we can assume that these are the ﬁrst d constraints. Thus, we have
ai · x ∗ = bi
∗ ai · x < bi for all i ≤ d ,
for all i ≥ d + 1, where ai denotes the i th row of A. Let A• denote the d × d matrix containing the ﬁrst d rows of A. Our
non-degeneracy assumption implies that A• has full rank, and thus has a well-deﬁned inverse V = A−1 .
•
Now deﬁne a vector y ∈ n by setting
y j := c · v j for all j ≤ d , y j := 0 for all j ≥ d + 1, where v j denotes the j th column of V = A−1 . Note that ai · v j = 0 if i = j , and ai · v j = 1 if i = j .
•
To simplify notation, let y• = ( y1 , y2 , . . . , yd ) and let b• = ( b1 , b2 , . . . , bd ) = A• x ∗ . Because yi = 0
for all i ≥ d + 1, we immediately have
y · b = y• · b• = cV b• = cA−1 b• = c · x ∗
•
and
yA = y• A• = cVA• = cA−1 A• = c .
•
The point x ∗ lies on exactly d constraint hyperplanes; moreover, any sufﬁciently small ball around x ∗
˜
intersects only those d constraint hyperplanes. Consider the point x = x ∗ − v j , for some index
∗
˜
1 ≤ j ≤ d and some sufﬁciently small > 0. We have ai · x = ai · x − (ai · v j ) = bi for all i = j , and
˜
˜
a j · x = a j · x ∗ − (a j · v j ) = b j − < b j . Thus, x is a feasible point for Π. Because x ∗ is the unique
˜
optimum for Π, we must have c · x = c · x ∗ − (c · v j ) < c · x ∗ . We conclude that y j = c · v j > 0 for all j .
We have shown that yA ≥ c and y ≥ 0, so y is a feasible solution for the dual linear program . We
have also shown that y · b = c · x ∗ , so by the Weak Duality Theorem, y is also an optimal solution for ,
and the proof is complete!
We can also give a useful geometric interpretation to the vector y• ∈ d . Each linear equation
ai · x = bi deﬁnes a hyperplane in d with normal vector ai . The normal vectors a1 , . . . , ad are linearly
independent (by non-degeneracy) and therefore describe a coordinate frame for the vector space d .
d
The deﬁnition of y• implies that c = y• A• = i =1 yi ai . In other words, y• lists the coefﬁcients of the
objective vector c in the coordinate frame a1 , . . . , ad . 8 Lecture 25: Linear Programming [Fa ’10] Algorithms 25.7 Complementary Slackness Complementary Slackness Theorem. Let x ∗ be an optimal solution to a canonical linear program Π,
and let y ∗ be an optimal solution to its dual . Then for every index i , we have yi∗ > 0 if and only if
ai · x ∗ = bi . Symmetrically, for every index j , we have x ∗ > 0 if and only if y ∗ · a j = c j .
j
To be written Exercises
1. (a) Describe how to transform any linear program written in general form into an equivalent
linear program written in slack form.
d maximize
j =1
d subject to
j =1
d
j =1
d
j =1 cj x j
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 max c · x
s.t. Ax = b
x≥0 for each i = p + q + 1 .. n (b) Describe precisely how to dualize a linear program written in slack form.
(c) Describe precisely how to dualize a linear program written in general form:
In all cases, keep the number of variables in the resulting linear program as small as possible.
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 that any self-dual linear program Π is syntactically 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) Give a linear-programming formulation of the maximum-cardinality bipartite matching 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. Give a linear-programming formulation of the minimum-cost feasible circulation problem. Here you
are given a ﬂow network whose edges have both capacities and costs, and your goal is to ﬁnd a
feasible circulation (ﬂow with value 0) whose cost is as small as possible.
9 Lecture 25: Linear Programming [Fa ’10] Algorithms 5. 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.]
6. Helly’s theorem states that for any collection of convex bodies in d , 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.]
7. Given points ( x 1 , y1 ), ( x 2 , y2 ), . . . , ( x n , yn ) in the plane, the linear regression problem asks for real
numbers a and b such that the line y = a x + b ﬁts the points as closely as possible, according to
some criterion. The most common ﬁt criterion is minimizing the L2 error, deﬁned as follows:7
n ( yi − a x i − b)2 . 2 (a, b) =
i =1 But there are several other ﬁt criteria, some of which can be optimized via linear programming.
(a) The L1 error (or total absolute deviation) of the line y = a x + b is deﬁned as follows:
n
1 (a, b) = yi − a x i − b .
i =1 Describe a linear program whose solution (a, b) describes the line with minimum L1 error.
(b) The L∞ error (or maximum absolute deviation) of the line y = a x + b is deﬁned as follows:
∞ (a, b) n = max yi − a x i − b .
i =1 Describe a linear program whose solution (a, b) describes the line with minimum L∞ error. 7
This measure is also known as sum of squared residuals, and the algorithm to compute the best ﬁt is normally called
(ordinary/linear) least squares ﬁtting. c Copyright 2010 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 Lecture 26: Linear Programming Algorithms [Fa’10]
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 26 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 deterministic simplex variant is known to run
in sub-exponential time in the worst case.1 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. 26.1 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 d , 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 d 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.
1 However, there are randomized variants of the simplex algorithm that run in subexponential expected time, most notably
the RANDOMFACET algorithm analyzed by Gil Kalai in 1992, and independently by Jiˇí Matoušek, Micha Sharir, and Emo Welzl in
r
1996. No randomized variant is known to run in polynomial time. In particular, in 2010, Oliver Friedmann, Thomas Dueholm
Hansen, and Uri Zwick proved that the worst-case expected running time of RANDOMFACET is superpolynomial.
c Copyright 2010 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. 1 Algorithms Lecture 26: Linear Programming Algorithms [Fa’10] 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
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.2
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. 26.2 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. Many 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, or even in expectation.3
2 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.
3
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. This long-standing conjecture was
ﬁnally disproved in 2010 by Fransisco Santos, who described an counterexample with 43 variables, 86 facets, and diameter 44. 2 Lecture 26: Linear Programming Algorithms [Fa’10] Algorithms 26.3 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
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. 26.4 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
3 Lecture 26: Linear Programming Algorithms [Fa’10] Algorithms 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
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 Lecture 26: Linear Programming Algorithms [Fa’10] 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 26.5 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 Lecture 26: Linear Programming Algorithms [Fa’10] 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) +
· T (n − 1, b + 1) otherwise
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 Lecture 26: Linear Programming Algorithms [Fa’10] 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 Lecture 26: Linear Programming Algorithms [Fa’10] 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 2010 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 Lecture 27: Lower Bounds [Sp’10] Algorithms It was a Game called Yes and No, where Scrooge’s nephew had to think of something, and the rest
must ﬁnd out what; he only answering to their questions yes or no, as the case was. The brisk ﬁre of
questioning to which he was exposed, elicited from him that he was thinking of an animal, a live animal,
rather a disagreeable animal, a savage animal, an animal that growled and grunted sometimes, and
talked sometimes, and lived in London, and walked about the streets, and wasn’t made a show of, and
wasn’t led by anybody, and didn’t live in a menagerie, and was never killed in a market, and was not
a horse, or an ass, or a cow, or a bull, or a tiger, or a dog, or a pig, or a cat, or a bear. At every fresh
question that was put to him, this nephew burst into a fresh roar of laughter; and was so inexpressibly
tickled, that he was obliged to get up off the sofa and stamp. At last the plump sister, falling into a
similar state, cried out :
“ I have found it out ! I know what it is, Fred ! I know what it is !”
“ What is it ?” cried Fred.
“ It’s your Uncle Scro-o-o-o-oge !”
Which it certainly was. Admiration was the universal sentiment, though some objected that the
reply to “ Is it a bear?” ought to have been “Yes;" inasmuch as an answer in the negative was sufﬁcient
to have diverted their thoughts from Mr Scrooge, supposing they had ever had any tendency that way.
— Charles Dickens, A Christmas Carol (1843) 27
27.1 Lower Bounds
Huh? Whuzzat? So far in this class we’ve been developing algorithms and data structures to solve certain problems as
quickly as possible. Starting with this lecture, we’ll turn the tables, by proving that certain problems
cannot be solved as quickly as we might like them to be.
Let TA(X ) denote the running time of algorithm A given input X . For most of the semester, we’ve
been concerned with the the worst-case running time of A as a function of the input size:
TA(n) := max TA(X ).
|X |=n The worst-case complexity of a problem Π is the worst-case running time of the fastest algorithm for
solving it:
TΠ (n) := min TA(n) = min
max TA(X ).
A solves Π A solves Π |X |=n Any algorithm A that solves Π immediately implies an upper bound on the complexity of Π; the inequality
TΠ (n ) ≤ TA (n ) follows directly from the deﬁnition of TΠ . Just as obviously, faster algorithms give us
better (smaller) upper bounds. In other words, whenever we give a running time for an algorithm, what
we’re really doing—and what most computer scientists devote their entire careers doing1 —is bragging
about how easy some problem is.
Now, instead of bragging about how easy problems are, we will argue that certain problems are
hard, by proving lower bounds on their complexity. This is considerably harder than proving an upper
bound, because it’s no longer enough to examine a single algorithm. To prove an inequality of the form
1 This sometimes leads to long sequences of results that sound like an obscure version of “Name that Tune”:
Lennes: “I can triangulate that polygon in O(n2 ) time.”
Shamos: “I can triangulate that polygon in O(n log n) time.”
Tarjan: “I can triangulate that polygon in O(n log log n) time.”
Seidel: “I can triangulate that polygon in O(n log∗ n) time.” [Audience gasps.]
Chazelle: “I can triangulate that polygon in O(n) time.” [Audience gasps and applauds.]
“Triangulate that polygon!” c Copyright 2011 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. 1 Lecture 27: Lower Bounds [Sp’10] Algorithms TΠ (n) = Ω( f (n)), we must prove that every algorithm that solves Π has a worst-case running time
Ω( f (n)), or equivalently, that no algorithm runs in o( f (n)) time. 27.2 Decision Trees Unfortunately, there is no formal deﬁnition of the phrase ‘all algorithms’!2 So when we derive lower
bounds, we ﬁrst have to specify precisely what kinds of algorithms we will consider and precisely how to
measure their running time. This speciﬁcation is called a model of computation.
One rather powerful model of computation—and the only model we’ll talk about in this lecture—is
the decision tree model. A decision tree is, as the name suggests, a tree. Each internal node in the tree
is labeled by a query, which is just a question about the input. The edges out of a node correspond to the
possible answers to that node’s query. Each leaf of the tree is labeled with an output. To compute with a
decision tree, we start at the root and follow a path down to a leaf. At each internal node, the answer to
the query tells us which node to visit next. When we reach a leaf, we output its label.
For example, the guessing game where one person thinks of an animal and the other person tries to
ﬁgure it out with a series of yes/no questions can be modeled as a decision tree. Each internal node is
labeled with a question and has two edges labeled ‘yes’ and ‘no’. Each leaf is labeled with an animal.
Does it live in the water?
NO S YE Does it have wings? fly spider Does it have wings? gnu NO NO YE
S frog NO S S YE YE YE fish Does it have more than four legs? NO S Does it have scales? eagle A decision tree to choose one of six animals. Here’s another simple and familiar example, called the dictionary problem. Let A be a ﬁxed array
with n numbers. Suppose we want to determine, given a number x , the position of x in the array A, if
any. One solution to the dictionary problem is to sort A (remembering every element’s original position)
and then use binary search. The (implicit) binary search tree can be used almost directly as a decision
tree. Each internal node in the search tree stores a key k; the corresponding node in the decision tree
stores the question ‘Is x < k?’. Each leaf in the search tree stores some value A[i ]; the corresponding
node in the decision tree asks ‘Is x = A[i ]?’ and has two leaf children, one labeled ‘i ’ and the other ‘none’.
We deﬁne the running time of a decision tree algorithm for a given input to be the number of queries
in the path from the root to the leaf. For example, in the ‘Guess the animal’ tree above, T (frog) = 2. Thus,
the worst-case running time of the algorithm is just the depth of the tree. This deﬁnition ignores other
kinds of operations that the algorithm might perform that have nothing to do with the queries. (Even
the most efﬁcient binary search problem requires more than one machine instruction per comparison!)
2 Complexity-theory snobs purists sometimes argue that ‘all algorithms’ is just a synonym for ‘all Turing machines’. This is
utter nonsense; Turing machines are just another model of computation. Turing machines might be a reasonable abstraction of
physically realizable computation—that’s the Church-Turing thesis—but it has a few problems. First, computation is an abstract
mathematical process, not a physical process. Algorithms that use physically unrealistic components (like exact real numbers,
or unbounded memory) are still mathematically well-deﬁned and still provide useful intuition about real-world computation.
Moreover, Turing machines don’t accurately reﬂect the complexity of physically realizable algorithms, because (for example)
they can’t do arithmetic or access arbitrary memory locations in constant time. At best, they estimate algorithmic complexity
up to polynomial factors (although even that is unknown). 2 Lecture 27: Lower Bounds [Sp’10] Algorithms x<11? 7 11 13 17 19 NO YES - S
YE
- 5 - YES YES YES
4 YES S
YE YE YES YES 3 6 - 7 - x=19? 8 NO - NO 2 x=17? NO - x=13?
NO 1 x<19? NO 19 x=11?
NO 17 x=7? NO 13 x=5? x<13? NO 11 x=3? NO 7 x=2? NO 5 19 x<7? NO 3 13 NO 2 7 NO S
YE x<3? 17
S 5 3 x<17?
NO S YE YES x<5?
11 S 5 YE 3 YES 2 - Left: A binary search tree for the ﬁrst eight primes.
Right: The corresponding binary decision tree for the dictionary problem (- = ‘none’). But the number of decisions is certainly a lower bound on the actual running time, which is good enough
to prove a lower bound on the complexity of a problem.
Both of the examples describe binary decision trees, where every query has only two answers. We
may sometimes want to consider decision trees with higher degree. For example, we might use queries
like ‘Is x greater than, equal to, or less than y ?’ or ‘Are these three points in clockwise order, colinear, or
in counterclockwise order?’ A k-ary decision tree is one where every query has (at most) k different
answers. From now on, I will only consider k -ary decision trees where k is a constant. 27.3 Information Theory Most lower bounds for decision trees are based on the following simple observation: The answers to
the queries must give you enough information to specify any possible output. If a problem has N
different outputs, then obviously any decision tree must have at least N leaves. (It’s possible for several
leaves to specify the same output.) Thus, if every query has at most k possible answers, then the depth
of the decision tree must be at least logk N = Ω(log N ).
Let’s apply this to the dictionary problem for a set S of n numbers. Since there are n + 1 possible
outputs, any decision tree must have at least n + 1 leaves, and thus any decision tree must have depth at
least logk (n + 1) = Ω(log n). So the complexity of the dictionary problem, in the decision-tree model of
computation, is Ω(log n). This matches the upper bound O(log n) that comes from a perfectly-balanced
binary search tree. That means that the standard binary search algorithm, which runs in O(log n) time,
is optimal—there is no faster algorithm in this model of computation. 27.4 But wait a second. . . We can solve the membership problem in O(1) expected time using hashing. Isn’t this inconsistent with
the Ω(log n) lower bound?
No, it isn’t. The reason is that hashing involves a query with more than a constant number of
outcomes, speciﬁcally ‘What is the hash value of x ?’ In fact, if we don’t restrict the degree of the decision
tree, we can get constant running time even without hashing, by using the obviously unreasonable query
‘For which index i (if any) is A[i ] = x ?’. No, I am not cheating — remember that the decision tree model
allows us to ask any question about the input!
This example illustrates a common theme in proving lower bounds: choosing the right model of
computation is absolutely crucial. If you choose a model that is too powerful, the problem you’re studying
may have a completely trivial algorithm. On the other hand, if you consider more restrictive models, the
problem may not be solvable at all, in which case any lower bound will be meaningless! (In this class,
we’ll just tell you the right model of computation to use.) 3 Lecture 27: Lower Bounds [Sp’10] Algorithms 27.5 Sorting Now let’s consider the classical sorting problem — Given an array of n numbers, arrange them in
increasing order. Unfortunately, decision trees don’t have any way of describing moving data around, so
we have to rephrase the question slightly:
Given a sequence 〈 x 1 , x 2 , . . . , x n 〉 of n distinct numbers, ﬁnd the permutation π such that
x π(1) < x π(2) < · · · < x π(n) .
Now a k-ary decision-tree lower bound is immediate. Since there are n! possible permutations π, any
decision tree for sorting must have at least n! leaves, and so must have depth Ω(log(n!)). To simplify the
lower bound, we apply Stirling’s approximation
n! = n n 2πn 1 + Θ e 1
n > n
e n . This gives us the lower bound
logk (n!) > logk n
e n = n logk n − n logk e = Ω(n log n). This matches the O(n log n) upper bound that we get from mergesort, heapsort, or quicksort, so those
algorithms are optimal. The decision-tree complexity of sorting is Θ(n log n).
Well. . . we’re not quite done. In order to say that those algorithms are optimal, we have to demonstrate that they ﬁt into our model of computation. A few minutes of thought will convince you that
they can be described as a special type of decision tree called a comparison tree, where every query is of
the form ‘Is x i bigger or smaller than x j ?’ These algorithms treat any two input sequences exactly the
same way as long as the same comparisons produce exactly the same results. This is a feature of any
comparison tree. In other words, the actual input values don’t matter, only their order. Comparison trees
describe almost all well-known sorting algorithms: bubble sort, selection sort, insertion sort, shell sort,
quicksort, heapsort, mergesort, and so forth—but not radix sort or bucket sort. 27.6 Finding the Maximum and Adversaries Finally let’s consider the maximum problem: Given an array X of n numbers, ﬁnd its largest entry.
Unfortunately, there’s no hope of proving a lower bound in this formulation, since there are an inﬁnite
number of possible answers, so let’s rephrase it slightly.
Given a sequence 〈 x 1 , x 2 , . . . , x n 〉 of n distinct numbers, ﬁnd the index m such that x m is the
largest element in the sequence.
We can get an upper bound of n − 1 comparisons in several different ways. The easiest is probably to
start at one end of the sequence and do a linear scan, maintaining a current maximum. Intuitively, this
seems like the best we can do, but the information-theoretic bound is only log2 n . And in fact, this
bound is tight! We can locate the maximum element by asking only log2 n ‘unreasonable’ questions
like “Is the index of the maximum element odd?” No, this is not cheating—the decision tree model
allows arbitrary questions.
To prove a non-trivial lower bound for this problem, we must do two things. First, we need to
consider a more reasonable model of computation, by restricting the kinds of questions the algorithm is
allowed to ask. We will consider the comparison tree model, where every query must have the form “Is 4 Lecture 27: Lower Bounds [Sp’10] Algorithms x i > x j ?”. Since most algorithms3 for ﬁnding the maximum rely on comparisons to make control-ﬂow
decisions, this does not seem like an unreasonable restriction.
Second, we will use something called an adversary argument. The idea is that an all-powerful
malicious adversary pretends to choose an input for the algorithm. When the algorithm asks a question
about the input, the adversary answers in whatever way will make the algorithm do the most work.
If the algorithm does not ask enough queries before terminating, then there will be several different
inputs, each consistent with the adversary’s answers, that should result in different outputs. In this case,
whatever the algorithm outputs, the adversary can ‘reveal’ an input that is consistent with its answers,
but contradicts the algorithm’s output, and then claim that that was the input that he was using all
along.
For the maximum problem, the adversary originally pretends that x i = i for all i , and answers all
comparison queries accordingly. Whenever the adversary reveals that x i < x j , he marks x i as an item
that the algorithm knows (or should know) is not the maximum element. At most one element x i is
marked after each comparison. Note that x n is never marked. If the algorithm does less than n − 1
comparisons before it terminates, the adversary must have at least one other unmarked element x k = x n .
In this case, the adversary can change the value of x k from k to n + 1, making x k the largest element,
without being inconsistent with any of the comparisons that the algorithm has performed. In other
words, the algorithm cannot tell that the adversary has cheated. However, x n is the maximum element in
the original input, and x k is the largest element in the modiﬁed input, so the algorithm cannot possibly
give the correct answer for both cases. Thus, in order to be correct, any algorithm must perform at least
n − 1 comparisons.
The adversary argument we described has two very important properties. First, no algorithm can
distinguish between a malicious adversary and an honest user who actually chooses an input in advance
and answers all queries truthfully. But much more importantly, the adversary makes absolutely no
assumptions about the order in which the algorithm performs comparisons. The adversary forces
any comparison-based algorithm4 to either perform n − 1 comparisons, or to give the wrong answer for
at least one input sequence. Exercises
0. Simon bar Kokhba thinks of an integer between 1 and 1,000,000 (or so he claims). You are
trying to determine his number by asking as few yes/no questions as possible. How many yes/no
questions are required to determine Simon’s number in the worst case? Give both an upper bound
(supported by an algorithm) and a lower bound. 1. Consider the following multi-dictionary problem. Let A[1 .. n] be a ﬁxed array of distinct integers.
Given an array X [1 .. k], we want to ﬁnd the position (if any) of each integer X [i ] in the array A.
In other words, we want to compute an array I [1 .. k] where for each i , either I [i ] = 0 (so zero
means ‘none’) or A[ I [i ]] = X [i ]. Determine the exact complexity of this problem, as a function of
n and k, in the binary decision tree model.
3 but not all—see Exercise ??
In fact, the n − 1 lower bound for ﬁnding the maximum holds in a more powerful model called algebraic decision trees,
which are binary trees where every query is a comparison between two polynomial functions of the input values, such as ‘Is
2
17
52
42
x 1 − 3 x 2 x 3 + x 4 bigger or smaller than 5 + x 1 x 3 x 5 − 2 x 7 ?’
4 5 Lecture 27: Lower Bounds [Sp’10] Algorithms 2. We say that an array A[1 .. n] is k-sorted if it can be divided into k blocks, each of size n/k, such
that the elements in each block are larger than the elements in earlier blocks, and smaller than
elements in later blocks. The elements within each block need not be sorted.
For example, the following array is 4-sorted:
1243 7685 10 11 9 12 15 13 16 14 (a) Describe an algorithm that k-sorts an arbitrary array in O(n log k) time.
(b) Prove that any comparison-based k-sorting algorithm requires Ω(n log k) comparisons in the
worst case.
(c) Describe an algorithm that completely sorts an already k-sorted array in O(n log(n/k)) time.
(d) Prove that any comparison-based algorithm to completely sort a k-sorted array requires
Ω(n log(n/k)) comparisons in the worst case.
In all cases, you can assume that n/k is an integer. 3. Recall the nuts-and-bolts problem from the lecture on randomized algorithms. We are given n
bolts and n nuts of different sizes, where each bolt exactly matches one nut. Our goal is to ﬁnd
the matching nut for each bolt. The nuts and bolts are too similar to compare directly; however,
we can test whether any nut is too big, too small, or the same size as any bolt.
(a) Prove that in the worst case, Ω(n log n) nut-bolt tests are required to correctly match up the
nuts and bolts.
(b) Now suppose we would be happy to ﬁnd most of the matching pairs. Prove that in the worst
case, Ω(n log n) nut-bolt tests are required even to ﬁnd n/2 arbitrary matching nut-bolt pairs.
(c) Prove that in the worst case, Ω(n + k log n) nut-bolt tests are required to ﬁnd k arbitrary
matching pairs. [Hint: Use an adversary argument for the Ω(n) term.]
(d) Describe a randomized algorithm that ﬁnds k matching nut-bolt pairs in O(n + k log n)
expected time. 4. Suppose you want to determine the largest number in an n-element set X = { x 1 , x 2 , . . . , x n },
where each element x i is an integer between 1 and 2m − 1. Describe an algorithm that solves this
problem in O(n + m) steps, where at each step, your algorithm compares one of the elements x i
with a constant. In particular, your algorithm must never actually compare two elements of X !
[Hint: Construct and maintain a nested set of ‘pinning intervals’ for the numbers that you have
not yet removed from consideration, where each interval but the largest is either the upper half or
lower half of the next larger block.] c Copyright 2011 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 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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) 28
28.1 Adversary Arguments
Three-Card Monte Until Times Square was sanitized into TimesSquareLandTM , 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, the two of spades, 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 never wins, because the dealer cheats. The dealer actually holds at
least 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 (which might be the
two of clubs, but most tourists won’t notice that wasn’t one o the original cards). 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!1 28.2 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
1
Even if the dealer is a sloppy magician, he’ll cheat anyway. The dealer is almost always surrounded by shills; these are the
‘tourists’ who look like they’re actually winning, who turn over cards when the dealer ‘isn’t looking’, who casually mention how
easy the game is to win, and so on. The shills protect the dealer from any angry tourists who notice the dealer cheating, and
shake down any tourists who refuse to pay after making a bet. You cannot win this game, ever. c Copyright 2011 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. 1 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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
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. 28.3 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!) 2 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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.
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. 28.4 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. 28.5 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. 3 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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.
• 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! 28.6 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 topology2 —but it is
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.
2 No, this will not be on the ﬁnal exam. 4 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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. 28.7 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 .
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
3n
n
+
−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. 28.8 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: 5 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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.
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. 6 Lecture 28: Adversary Arguments [Sp’10] Algorithms 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.
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?]
7 Lecture 28: Adversary Arguments [Sp’10] Algorithms (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 2011 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 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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 29
29.1 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
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. © Copyright 2011 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. 1 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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! 29.2 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.
Perhaps the single most important open questions in theoretical computer science, if not all of
mathematics, is whether the complexity classes P and NP are actually different. Intuitively, it seems
obvious to most people 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. The prominent physicist Richard Feynman, one of the early pioneers of quantum computing,
apparently had trouble accepting that P = NP was actually an open question. But nobody knows how
to prove it. The Clay Mathematics Institute lists P versus NP as the ﬁrst of its seven Millennium Prize
Problems, offering a $1,000,000 reward for its solution.
A more subtle but still open question is whether the complexity classes NP and co-NP are different.
Even if we can verify every YES answer quickly, there’s no reason to believe 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. 29.3 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 Lecture 29: NP-Hard Problems [Fa’10] Algorithms coNP NP
P What we think the world looks like. 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.
Calling 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 (and therefore all) of them seems incredibly unlikely.
NP-hard
coNP 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 1973.4 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. 29.4 Formal Deﬁnition (HC SVNT DRACONES) More formally, a problem Π is deﬁned to be NP-hard if and only if, for every problem Π in NP, there is a
polynomial-time Turing reduction from Π to Π—a Turing reduction just means a reduction that can
4 Levin ﬁrst reported his results at seminars in Moscow in 1971; news of Cook’s result did not reach the Soviet Union until at
least 1973, after Levin’s announcement of his results had been published. In accordance with Stigler’s Law, this result is often
called ‘Cook’s Theorem’. Cook won the Turing award for his proof. Levin was denied his PhD at Moscow University for political
reasons; he emigrated to the US in 1978 and earned a PhD at MIT a year later. 3 Lecture 29: NP-Hard Problems [Fa’10] Algorithms be executed on a Turing machine. Polynomial-time Turing reductions are also called oracle reductions
or Cook reductions. It is elementary, but extremely tedious, to prove that any algorithm that can be
executed on a random-access machine5 in time T (n) can be simulated on a single-tape Turing machine
in time O( T (n)2 ), so polynomial-time Turing reductions don’t actually need to be described using Turing
machines. We’ll simply take that step on faith.
Complexity-theory researchers prefer to deﬁne NP-hardness in terms of polynomial-time many-one
reductions, which are also called Karp reductions. Karp reductions are deﬁned over languages: sets of
strings over a ﬁxed alphabet Σ. (Without loss of generality, we can assume that Σ = {0, 1}.) A many-one
reduction from one language Π ⊆ Σ∗ to another language Π ⊆ Σ∗ is an function f : Σ∗ → Σ∗ such that
x ∈ Π if and only if f ( x ) ∈ Π. Then we could deﬁne a language Π to be NP-hard if and only if, for any
language Π ∈ NP, there is a many-one reduction from Π to Π that can be computed in polynomial time.
Every Karp reduction is a Cook reduction, but not vice versa; speciﬁcally, any Karp reduction from
Π to Π is equivalent to transforming the input to Π into the input for Π , invoking an oracle (that is,
a subroutine) for Π , and then returning the answer verbatim. However, as far as we know, not every
Cook reduction can be simulated by a Karp reduction.
Complexity theorists prefer Karp reductions primarily because NP is closed under Karp reductions,
but is not closed under Cook reductions (unless NP=co-NP, which is considered unlikely). 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. As a trivial example, consider the problem UNSAT: Given a boolean
formula, is it always false? On the other hand, many-one reductions apply only to decision problems (or
more formally, to languages); formally, no optimization or construction problem is Karp-NP-hard.
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.
Fortunately, none of these subtleties raise their ugly heads in practice, so you can wake up now. 29.5 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, right next to your Mom’s birthday
and the actual rules of Monopoly.6
To prove that problem A is NP-hard, reduce a known NP-hard problem to A.
5 Random-access machines are a model of computation that more faithfully models physical computers. A random-access
machine has unbounded random-access memory, modeled as an array M [0 .. ∞] where each address M [i ] holds a single w -bit
integer, for some ﬁxed integer w , and can read to or write from any memory addresses in constant time. RAM algorithms are
formally written in assembly-like language, using instructions like ADD i , j , k (meaning “ M [i ] ← M [ j ] + M [k]”), INDIR i , j
(meaning “ M [i ] ← M [ M [ j ]]”), and IFZGOTO i , (meaning “if M [i ] = 0, go to line ”). But in practice, RAM algorithms can
be faithfully described in higher-level pseudocode, as long as we’re careful about arithmetic precision.
6
If a player lands on an available property and declines (or is unable) to buy it, that property is immediately auctioned off
to the highest bidder; the player who originally declined the property may bid, and bids may be arbitrarily higher or lower
than the list price. Players still collect rent while in Jail, but not if the property is mortgaged. A player landing on Free Parking
does not win anything. A player landing on Go gets $200, no more. Finally, Jeff always gets the car. 4 Lecture 29: NP-Hard Problems [Fa10] Algorithms 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 ANDs. For example, we can transform the example circuit
into a formula as follows:
y1 x1 y4 x2
y5 x3
x4 z y3 y2 y6 x5 ( 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 ) ∧ (z = y4 ∧ y7 ∧ y6 ) ∧ z
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 internal gate variables yi and the output variable z .
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:
O(n) boolean circuit −
−−→ boolean formula
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.
5 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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. 29.6 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 )
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 6 Lecture 29: NP-Hard Problems [Fa’10] Algorithms
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) −−→ 3CNF formula
boolean circuit −
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. 29.7 Maximum Independent Set (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¯
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
7 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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. 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)) 29.8 =⇒ 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. 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.
8 Lecture 29: NP-Hard Problems [Fa’10] Algorithms
O(n) −−→ complement graph G
graph G −
MAXCLIQUE
trivial largest independent set ←−− largest clique
− 29.9 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
− 29.10 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 Φ,
we produce a graph GΦ 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.
X T X F a a The truth gadget and a variable gadget for a. • 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. A straightforward case analysis implies that
if all three literal nodes in the clause gadget are colored FALSE, then some edge in the gadget must
be monochromatic. Since the variable gadgets force each literal node to be colored either TRUE or
9 Lecture 29: NP-Hard Problems [Fa’10] Algorithms FALSE, in any valid 3-coloring, at least one of the three literal nodes is colored TRUE. On the other
hand, for any coloring of the literal nodes where at least one literal node is colored TRUE, there is
a valid 3-coloring of the clause gadget.
a
b T c A clause gadget for (a ∨ b ∨ ¯).
c I need to emphasize here that the ﬁnal graph GΦ contains exactly one node T , exactly one node F ,
¯
and exactly two nodes a and a for each variable.
Now the proof of correctness is just brute force case analysis. 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) −−→ graph
3CNF formula −
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.
10 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 29.11 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. Determining whether a
graph contains a Hamiltonian cycle, on the other hand, is NP-hard.
To prove this, we describe 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 uv 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 only
three ways. Eventually, these options will correspond to one or both 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 vertex 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 (v,x,•) (v,y,•) (v,z,•) (x,v,•) (y,v,•) (z,v,•) … (v,w,•) (w,v,•) 3
k The vertex chain for v : all edge gadgets involving v are strung together and joined to the k cover vertices. An example of the output from our reduction is shown on the next page.
It is now straightforward but tedious 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.
11 Lecture 29: NP-Hard Problems [Fa’10] Algorithms (u,v,•) 1
(v,u,•) u v (u,w,•) (v,w,•) (v,x,•) w x (w,u,•) (w,v,•) (x,v,•) (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; we conclude that 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. 29.12 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 positive 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 another 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; and 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 12 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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:
au := 1110004 = 1344
a v := 1101104 = 1300
aw := 1011014 = 1105
a x := 1000114 = 1029 buv
buw
b vw
bv x
bw x := 0100004 = 256
:= 0010004 = 64
:= 0001004 = 16
:= 0000104 = 4
:= 0000014 = 1 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? idn’t we just prove that P=NP? Hey, where’s our million dollars? Alas,
life is not so simple. 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 a value of t that is exponentially larger than
the size of our original input graph, which would force our dynamic programming algorithm to run in
exponential time.
Algorithms like this are said to run in pseudo-polynomial time, and any NP-hard problem with
such an algorithm is called weakly NP-hard. Equivalently, a weakly NP-hard problem is one that can
be solved in polynomial time when all input numbers are represented in unary (as a sum of 1s), but
becomes NP-hard when all input numbers are represented in binary. If a problem is NP-hard even when
all the input numbers are represented in unary, we say that the problem is strongly NP-hard. 29.13 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
13 Lecture 29: NP-Hard Problems [Fa’10] Algorithms proofs for these problems in detail, but you can ﬁnd most of them in Garey and Johnson’s classic Scary
Black Book of NP-Completeness.7
• 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 problem can be proved
NP-hard by reduction from the general circuit satisﬁability problem, by replacing each crossing
with a small series of gates. (The details of the reduction are an easy8 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 problem 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 problem can be proved
NP-hard by reduction from PLANARCIRCUITSAT.9
• 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 problem
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 problem 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 three-element subsets,
such that every subset has exactly the same sum? Despite the similar names, this problem is very
different from PARTITION; sorry, I didn’t make up the names. This problem 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.
• 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 problem 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 problem 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 problem is a generalization of the HAMILTONIANPATH problem. Of course, the corresponding
shortest path problem is in P.
7
Michael Garey and David Johnson. Computers and Intractability: A Guide to the Theory of NP-Completeness. W. H. Freeman
and Co., 1979.
8
or at least nondeterministically easy
9
Surprisingly, PLANARNOTALLEQUAL3SAT is solvable in polynomial time! 14 Lecture 29: NP-Hard Problems [Fa’10] Algorithms • 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 problem can be proved
NP-hard by reduction to HAMILTONIANPATH.
In addition to these dry but useful problems, 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)10 , Tetris (by reduction from 3PARTITION)11 , and Sudoku (by reduction from 3SAT)12 . 29.14 On Beyond Zebra P and NP are only the ﬁrst two steps in an enormous hierarchy of complexity classes. To close these
notes, let me describe a few more classes of interest.
Polynomial Space. 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. A
problem is PSPACE-complete if it is both PSPACE-hard and in PSPACE. If any PSPACE-hard problem is
in NP, then PSPACE=NP; similarly, if any PSPACE-hard problem is in P, then PSPACE=P.
The canonical PSPACE-complete problem is the quantiﬁed boolean formula problem, or QBF : Given
a boolean formula Φ that may include any number of universal or existential quantiﬁers, but no free
variables, is Φ equivalent to TRUE? For example, the following expression is a valid input to QBF:
¯
¯
∃a : ∀ b : ∃c : (∀d : a ∨ b ∨ c ∨ d ) ⇔ (( b ∧ ¯) ∨ (∃e : (a ⇒ e) ∨ (c = a ∧ e))).
c
SAT is provably equivalent the special case of QBF where the input formula contains only existential
quantiﬁers. QBF remains PSPACE-hard even when the input formula must have all its quantiﬁers at
the beginning, the quantiﬁers strictly alternate between ∃ and ∀, and the quantiﬁed proposition is in
conjunctive normal form, with exactly three literals in each clause, for example:
¯
∃a : ∀ b : ∃c : ∀d : (a ∨ b ∨ c ) ∧ ( b ∨ ¯ ∨ d ) ∧ (a ∨ c ∨ d ) ∧ (a ∨ ¯ ∨ d )
c¯
b¯
This restricted version of QBF can also be phrased as a two-player strategy question. Suppose
two players, Alice and Bob, are given a 3CNF predicate with free variables x 1 , x 2 , . . . , x n . The players
alternately assign values to the variables in order by index—Alice assigns a value to x 1 , Bob assigns a
value to x 2 , and so on. Alice eventually assigns values to every variable with an odd index, and Bob
eventually assigns values to every variable with an even index. Alice wants to make the expression
TRUE, and Bob wants to make it FALSE. Assuming Alice and Bob play perfectly, who wins this game?
10
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
11
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.
12
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. 15 Lecture 29: NP-Hard Problems [Fa’10] Algorithms Not surprisingly, most two-player games13 like tic-tac-toe, reversi, checkers, go, chess, and mancala—or
more accurately, appropriate generalizations of these constant-size games to arbitrary board sizes—are
PSPACE-hard.
Another canonical PSPACE-hard problem is NFA totality: Given a non-deterministic ﬁnite-state
automaton M over some alphabet Σ, does M accept every string in Σ∗ ? The closely related problems
NFA equivalence (Do two given NFAs accept the same language?) and NFA minimization (Find the smallest
NFA that accepts the same language as a given NFA) are also PSPACE-hard, as are the corresponding
questions about regular expressions. (The corresponding questions about deterministic ﬁnite-state
automata are all solvable in polynomial time.)
Exponential time. The next signiﬁcantly larger complexity class, EXP (also called EXPTIME), is the
c
set of decision problems that can be solved in exponential time, that is, using at most 2n steps for some
constant 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. 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. A problem is EXP-complete if it is both EXP-hard and in EXP If any
.
EXP-hard problem is in PSPACE, then EXP=PSPACE; similarly, if any EXP-hard probelm is in NP, then
EXP=NP. We do know that P = EXP; in particular, no EXP-hard problem is in P.
Natural generalizations of many interesting 2-player games—like checkers, chess, mancala, and
go—are actually EXP-hard. The boundary between PSPACE-complete games and EXP-hard games is
rather subtle. For example, there are three ways to draw in chess (the standard 8 × 8 game): stalemate
(the player to move is not in check but has no legal moves), repeating the same board position three
times, or moving ﬁfty times without capturing a piece. The n × n generalization of chess is either in
PSPACE or EXP-hard depending on how we generalize these rules. If we declare a draw after (say) n3
capture-free moves, then every game must end after a polynomial number of moves, so we can simulate
all possible games from any given position using only polynomial space. On the other hand, if we ignore
the capture-free move rule entirely, the resulting game can last an exponential number of moves, so
there no obvious way to detect a repeating position using only polynomial space; indeed, this version of
n × n chess is EXP-hard.
Excelsior! Naturally, even exponential time is not the end of the story. NEXP is the class of decision
problems that can be solve in nondeterministic exponential time; equivalently, a decision problem is in
NEXP if and only if, for every YES instance, there is a proof of this fact that can be checked in exponential
time. EXPSPACE is the set of decision problems that can be solved using exponential space. Even these
larger complexity classes have hard and complete problems; for example, if we add the intersection
operator ∩ to the syntax of regular expressions, deciding whether two such expressions describe the
same language is EXPSPACE-hard. Beyond EXPSPACE are complexity classes with doubly-exponential
resource bounds (EEXP, NEEXP, and EEXPSPACE), then triply exponential resource bounds (EEEXP,
NEEEXP, and EEEXPSPACE), and so on ad inﬁnitum.
All these complexity classes can be ordered by inclusion as follows:
P ⊆ NP ⊆ PSPACE ⊆ EXP ⊆ NEXP ⊆ EXPSPACE ⊆ EEXP ⊆ NEEXP ⊆ EEXPSPACE ⊆ EEEXP ⊆ · · · ,
Most complexity theorists strongly believe that every inclusion in this sequence is strict; that is, no two
of these complexity classes are equal. However, the strongest result that has been proved is that every
13 For a good (but now slightly dated) overview of known results on the computational complexity of games and puzzles, see
Erik D. Demaine and Robert Hearn’s survey “Playing Games with Algorithms: Algorithmic Combinatorial Game Theory” at
http://arxiv.org/abs/cs.CC/0106019. 16 Lecture 29: NP-Hard Problems [Fa’10] Algorithms class in this sequence is strictly contained in the class three steps later in the sequence. For example, we
have proofs that P=EXP and PSPACE=EXPSPACE, but not whether P=PSPACE or NP=EXP
.
The limit of this series of increasingly exponential complexity classes is the class ELEMENTARY of
decision problems that can be solved using time or space bounded by any so-called elementary function
of n. A function of n is elementary if it is equal to 2 ↑k p(n) for some integer k and some polynomial
p(n), where
x
if k = 0,
2 ↑k x :=
2↑k−1 x
2
otherwise.
You might be tempted to conjecture that every natural decidable problem can be solved in elementary
time, but then you would be wrong. Consider the extended regular expressions deﬁned by recursively
combining (possibly empty) strings over some ﬁnite alphabet by concatenation, union (+), Kleene
closure (∗), and negation. For example, the extended regular expression (0 + 1)∗ 00(0 + 1)∗ represents
the set of strings in {0, 1}∗ that do not contain two 0s in a row. It is possible to determine algorithmically
whether two extended regular expressions describe identical languages, by recursively converting each
expression into an equivalent NFA, converting each NFA into a DFA, and then minimizing the DFA.
Unfortunately, however, this equivalence problem cannot be decided in only elementary time!! 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.
(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
( x ∧ y ∧ z) ∨ ( y ∧ z) ∨ ( x ∧ y ∧ z)
is in disjunctive normal form. DNF-SAT asks, given a boolean formula in disjunctive normal form,
whether that formula is satisﬁable.
(a) Describe a polynomial-time algorithm to solve DNF-SAT.
(b) What is the error in the following argument that P=NP? 17 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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,
( x ∨ y ∨ z ) ∧ ( x ∨ y ) ⇐⇒ ( x ∧ y ) ∨ ( y ∧ x ) ∨ (z ∧ x ) ∨ (z ∧ y )
Now we can use the algorithm 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) 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 boolean formula Φ in
conjunctive normal form where each clause contains at most 3 literals and each variable
appears in at most 3 clauses, does Φ have a satisfying assignment?
6. There’s something special about the number 3.
(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.
7. There’s nothing special about the number 3.
(a) The problem 12PARTITION is deﬁned as follows: Given a set S of 12n positive integers,
determine whether the elements of S can be split into n subsets of 12 elements each whose
sums are all equal. Prove that 12PARTITION is NP-hard. [Hint: Reduce from 3PARTITION. It
may be easier to consider multisets ﬁrst.]
(b) The problem 12COLOR is deﬁned as follows: Given an undirected graph G , determine whether
we can color each vertex with one of twelve colors, so that every edge touches two different
colors. Prove that 12COLOR is NP-hard. [Hint: Reduce from 3COLOR.]
(c) The problem 12SAT is deﬁned as follows: Given a boolean formula Φ in conjunctive normal
form, with exactly twelve literals per clause, determine whether Φ has a satisfying assignment.
Prove that 12SAT is NP-hard. [Hint: Reduce from 3SAT.]
8. (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.]
18 Lecture 29: NP-Hard Problems [Fa’10] Algorithms (a) Gadget for planar 3-colorability. (b) Gadget for degree-4 planar 3-colorability. 9. 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.]
3
1
3 4
0 0 2 2
4 A careful 5-coloring. 10. 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?
11. The RECTANGLETILING problem is deﬁned as follows: Given one large rectangle and several smaller
rectangles, determine whether the smaller rectangles can be placed inside the large rectangle with
no gaps or overlaps. Prove that RECTANGLETILING is NP-complete.
2
6
7
1 2 3 4 5 7 6 3 4
5 1 A positive instance of the RECTANGLETILING problem. 12. Let G = (V, E ) be a graph. A dominating set in G is a subset S of the vertices such that every vertex
in G is either in S or adjacent to a vertex in S . The DOMINATINGSET problem asks, given a graph G
and an integer k as input, whether G contains a dominating set of size k. Either prove that this
problem is NP-hard or describe a polynomial-time algorithm to solve it.
13. 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
19 Lecture 29: NP-Hard Problems [Fa’10] Algorithms A dominating set of size 3 in the Peterson graph. 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.
14. (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).]
15. For each problem below, either describe a polynomial-time algorithm or prove that the problem is
NP-complete.
(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?
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. Consider the following solitaire game. The puzzle consists of an n × m grid of squares, where each
square may be empty, occupied by a red stone, or occupied by a blue stone. The goal of the puzzle
is to remove some of the given stones so that the remaining stones satisfy two conditions: (1)
every row contains at least one stone, and (2) no column contains stones of both colors. For some
initial conﬁgurations of stones, reaching this goal is impossible.
20 Lecture 29: NP-Hard Problems [Fa’10] Algorithms A solvable puzzle and one of its many solutions. An unsolvable puzzle. Prove that it is NP-hard to determine, given an initial conﬁguration of red and blue stones,
whether the puzzle can be solved.
18. 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.
19. 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
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.
20. You’re in charge of choreographing a musical for your local community theater, and it’s time to
ﬁgure out the ﬁnal pose of the big show-stopping number at the end. (“Streetcar!”) You’ve decided
that each of the n cast members in the show will be positioned in a big line when the song ﬁnishes,
all with their arms extended and showing off their best spirit ﬁngers.
The director has declared that during the ﬁnal ﬂourish, each cast member must either point
both their arms up or point both their arms down; it’s your job to ﬁgure out who points up and
who points down. Moreover, in a ﬁt of unchecked power, the director has also given you a list of
arrangements that will upset his delicate artistic temperament. Each forbidden arrangement is a
subset of the cast members paired with arm positions; for example: “Marge may not point her
arms up while Ned, Apu, and Smithers point their arms down.”
Prove that ﬁnding an acceptable arrangement of arm positions is NP-hard.
21. The next time you are at a party, one of the guests will suggest everyone play a round of Three-Way
Mumbledypeg, a game of skill and dexterity that requires three teams and a knife. The ofﬁcial
Rules of Three-Way Mumbledypeg (ﬁxed during the Holy Roman Three-Way Mumbledypeg 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 21 Lecture 29: NP-Hard Problems [Fa’10] Algorithms 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 Mumbledypeg teams, or prove that the problem is NP-hard.
22. The party you are attending is going great, but now it’s time to line up for The Algorithm
March (
)! This dance was originally developed by the Japanese comedy
duo Itsumo Kokokara (
) for the children’s television show PythagoraSwitch
(
). The Algorithm March is performed by a line of people; each person in line
starts a speciﬁc sequence of movements one measure later than the person directly in front of
them. Thus, the march is the dance equivalent of a musical round or canon, like “Row Row Row
Your Boat”.
Proper etiquette dictates that each marcher must know the person directly in front of them in
line, lest a minor mistake during lead to horrible embarrassment between strangers. Suppose you
are given a complete list of which people at your party know each other. Prove that it is NP-hard
to determine the largest number of party-goers that can participate in the Algorithm March. You
may assume there are no ninjas at your party.
23. (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.] 22 Lecture 29: NP-Hard Problems [Fa’10] Algorithms [General solutions give you a 50% tip.]
— Randall Munroe, xkcd (http://xkcd.com/287/)
Reproduced under a Creative Commons Attribution-NonCommercial 2.5 License © Copyright 2011 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 Lecture 30: Approximation Algorithms [Fa’10] 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. 30
30.1 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 i with the earliest ﬁnishing time Total[i ].
GREEDYLOADBALANCE( T [1 .. n], m):
for i ← 1 to m
Total[i ] ← 0
for j ← 1 to n
mini ← arg mini Total[i ]
A[ j ] ← mini
Total[mini] ← Total[mini] + T [ j ]
return A[1 .. m]
© Copyright 2011 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. 1 Lecture 30: Approximation Algorithms [Fa’10] 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 OPT 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.
OPT ≥ max T [ j ] and j OPT ≥ n 1
m 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 machine i . Our ﬁrst trivial observation implies
that T [ j ] ≤ OPT. To ﬁnish the proof, we must show that Total[i ] − T [ j ] ≤ OPT. 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 ] ≤ OPT
j =1 by our second trivial observation. We conclude that the makespan Total[i ] is at most 2 · OPT. 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 Lecture 30: Approximation Algorithms [Fa’10] 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 machine, we must have j ≥ m + 1. As in the previous proof, we know that Total[i ] − T [ j ] ≤ OPT.
In any schedule, at least two of the ﬁrst m + 1 jobs, say jobs k and , must be assigned to the same
machine. Thus, T [k] + T [ ] ≤ OPT. Since max{k, } ≤ m + 1 ≤ j , and the jobs are sorted in decreasing
order by duration, we have
T [ j ] ≤ T [m + 1] ≤ T [max{k, }] = min { T [k], T [ ]} ≤ OPT/2.
We conclude that the makespan Total[i ] is at most 3 · OPT/2, as claimed. 30.2 Generalities Consider an arbitrary optimization problem. Let OPT(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
OPT(X )
A(X ) ≤ α(n) and A(X )
OPT(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 OPT(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. 30.3 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 Obviously this algorithm doesn’t compute the optimal vertex cover—that would imply P=NP!—but
it does compute a reasonably close approximation.
1 A heuristic is an algorithm that doesn’t work. 3 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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 OPT 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 |/OPT. It follows that Gi −1
has at least one node with degree at least |Gi −1 |/OPT. Since di is the maximum degree of any node
in Gi −1 , we have
| G i −1 |
di ≥
OPT
Moreover, for any j ≥ i − 1, the subgraph G j has no more edges than Gi −1 , so di ≥ |G j |/OPT. This
observation implies that
OPT OPT di ≥
i =1 i =1 | G i −1 |
OPT OPT ≥
i =1 |GOPT |
OPT OPT = |GOPT | = |G | − di .
i =1 In other words, the ﬁrst OPT iterations of GREEDYVERTEXCOVER remove at least half the edges of G . Thus,
after at most OPT lg|G | ≤ 2 OPT 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(OPT 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(OPT · log n). Details
are left as an exercise for the reader. 30.4 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
2 A matroid (see the lecture note on greedy algorithms) is a special type of set system. 4 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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 ∗ ). 30.5 Vertex Cover, Again The greedy approach doesn’t always lead to the best approximation algorithms. Consider the following
alternate heuristic for vertex cover:
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 5 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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. 30.6 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. 30.7 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
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 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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. Theorem 5. A depth-ﬁrst ordering of the minimum spanning tree gives a 2-approximation of the shortest
traveling salesman tour.
Proof: Let OPT denote the cost of the optimal TSP tour, let MST 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 · MST. 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 · MST.
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, MST ≥ OPT. We conclude that A ≤ 2 · OPT; 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
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.
Theorem 6. Given a weighted graph that obeys the triangle inequality, the Christoﬁdes heuristic
computes a (3/2)-approximation of the shortest traveling salesman tour.
7 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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. Proof: Let A denote the length of the tour computed by the Christoﬁdes heuristic; let OPT denote the
length of the optimal tour; let MST denote the total length of the minimum spanning tree; let MOM
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 MST + MOM. By the
triangle inequality, taking a shortcut past a previously visited vertex can only shorten the tour. Thus,
A ≤ MST + MOM.
By the triangle inequality, the optimal tour of the odd-degree vertices of T cannot be longer than
OPT. 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
OPT/2. On the other hand, both matchings have length at least MOM. Thus, MOM ≤ OPT/2.
Finally, recall our earlier observation that MST ≤ OPT.
Putting these three inequalities together, we conclude that A ≤ 3 · OPT/2, as claimed. 30.8 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
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.
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
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 Lecture 30: Approximation Algorithms [Fa’10] Algorithms The ﬁrst ﬁve iterations of Gonzalez’s k-center clustering algorithm. been chosen. The algorithm includes an extra iteration to compute the ﬁnal clustering radius rk (and
the next center ck+1 ).
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 7. GONZALEZKCENTER computes a 2-approximation to the optimal k-center clustering.
Proof: Let OPT denote the optimal k-center clustering radius for P . For any index i , let ci and ri denote
the i th center point and i th clustering radius computed by GONZALEZKCENTER.
By construction, each center point c j has distance at least r j −1 from any center point ci with i < j .
Moreover, for any i < j , we have ri ≥ r j . Thus, |ci c j | ≥ rk for all indices i and j .
On the other hand, at least one cluster in the optimal clustering contains at least two of the points
c1 , c2 , . . . , ck+1 . Thus, by the triangle inequality, we must have |ci c j | ≤ 2 · OPT for some indices i and j .
We conclude that rk ≤ 2 · OPT, as claimed. 30.9 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
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:
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 d 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 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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. 30.10 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
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 .
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.
8
Do, or do not. There is no ‘try’. (Are old one thousand when years you, alphabetical also in order talk will you.) 10 Lecture 30: Approximation Algorithms [Fa’10] Algorithms 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 8. 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 .
(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.
11 Lecture 30: Approximation Algorithms [Fa’10] Algorithms (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 9. 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 Lecture 30: Approximation Algorithms [Fa’10] 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 OPT · 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 · OPT.
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 · OPT.
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 Lecture 30: Approximation Algorithms [Fa’10] 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 Lecture 30: Approximation Algorithms [Fa’10] 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 · OPT + 1)/3 bins, where OPT 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 OPT bins.
• FIRSTFIT places at most OPT − 1 items outside the ﬁrst OPT 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 Lecture 30: Approximation Algorithms [Fa’10] 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. © Copyright 2011 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 Appendix I: Proof by Induction [Fa’10] Algorithms Jeder Genießende meint, dem Baume habe es an der Frucht gelegen;
aber ihm lag am Samen.
[Everyone who enjoys thinks that the fundamental thing about trees is the fruit,
but in fact it is the seed.]
— Friedrich Wilhelm Nietzsche,
Vermischte Meinungen und Sprüche [Mixed Opinions and Maxims] (1879)
In view of all the deadly computer viruses that have been spreading lately,
Weekend Update would like to remind you:
When you link up to another computer,
you’re linking up to every computer that that computer has ever linked up to.
— Dennis Miller, “Saturday Night Live”, (c. 1985)
Anything that, in happening, causes itself to happen again, happens again.
— Douglas Adams (2005) Proof by Induction
Induction is a method for proving universally quantiﬁed propositions—statements about all elements
of a (usually inﬁnite) set. Induction is also the single most useful tool for reasoning about, developing,
and analyzing algorithms. These notes give several examples of inductive proofs, along with a standard
boilerplate and some motivation to justify (and help you remember) why induction works. 1 Prime Divisors: Proof by Smallest Counterexample A divisor of a positive integer n is a positive integer p such that the ratio n/ p is an integer. The integer 1
is a divisor of every positive integer (because n/1 = n), and every integer is a divisor of itself (because
n/n = 1). A proper divisor of n is any divisor of n other than n itself. A positive integer is prime if it
has exactly two divisors, which must be 1 and itself; equivalently; a number is prime if and only if 1 is
its only proper divisor. A positive integer is composite if it has more than two divisors (or equivalently,
more than one proper divisor). The integer 1 is neither prime nor composite, because it has exactly one
divisor, namely itself.
Let’s prove our ﬁrst theorem:
Theorem 1. Every integer greater than 1 has a prime divisor.
The very ﬁrst thing that you should notice, after reading just one word of the theorem, is that this
theorem is universally quantiﬁed—it’s a statement about all the elements of a set, namely, the set of
positive integers larger than 1. If we were forced at gunpoint to write this sentence using fancy logic
notation, the ﬁrst character would be the universal quantiﬁer ∀, pronounced ‘for all’. Fortunately, that
won’t be necessary.
There are only two ways to prove a universally quantiﬁed statement: directly or by contradiction.
Let’s say that again, louder: There are only two ways to prove a universally quantiﬁed statement:
directly or by contradiction. Here are the standard templates for these two methods, applied to
Theorem 1:
Direct proof: Let n be an arbitrary integer greater than 1. . . . blah blah blah . . .
Thus, n has at least one prime divisor.
c Copyright 2011 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. 1 Appendix I: Proof by Induction [Fa’10] Algorithms Proof by contradiction: For the sake of argument, assume there is an integer greater
than 1 with no prime divisor.
Let n be an arbitrary integer greater than 1 with no prime divisor. . . . blah blah blah . . .
But that’s just silly. Our assumption must be incorrect. The shaded boxes . . . blah blah blah . . . indicate missing proof details (which you have to ﬁll in).
Most people ﬁnd proofs by contradiction easier to discover than direct proofs, so let’s try that ﬁrst.
Proof by contradiction: For the sake of argument, assume there is an integer greater
than 1 with no prime divisor.
Let n be an arbitrary integer greater than 1 with no prime divisor.
Since n is a divisor of n, and n has no prime divisors, n cannot be prime.
Thus, n must have at least one divisor d such that 1 < d < n.
Let d be an arbitrary divisor of n such that 1 < d < n.
Since n has no prime divisors, d cannot be prime.
Thus, d has at least one divisor d such that 1 < d < d .
Let d be an arbitrary divisor of d such that 1 < d < d .
Because d /d is an integer, n/d = (n/d ) · (d /d ) is also an integer.
Thus, d is also a divisor of n.
Since n has no prime divisors, d cannot be prime.
Thus, d has at least one divisor d such that 1 < d < d .
Let d be an arbitrary divisor of d such that 1 < d < d .
Because d /d is an integer, n/d = (n/d ) · (d /d ) is also an integer.
Thus, d is also a divisor of n.
Since n has no prime divisors, d cannot be prime. . . . blah HELP! blah I’M STUCK IN AN INFINITE LOOP! blah . . .
But that’s just silly. Our assumption must be incorrect. We seem to be stuck in an inﬁnite loop, looking at smaller and smaller divisors d > d > d > · · · ,
none of which are prime. But this loop can’t really be inﬁnite. There are only n − 1 positive integers
smaller than n, so the proof must end after at most n − 1 iterations. But how do we turn this observation
into a formal proof? We need a single, self-contained proof for all integers n; we’re not allowed to write
longer proofs for bigger integers. The trick is to jump directly to the smallest counterexample.
Proof by smallest counterexample: For the sake of argument, assume that there is
an integer greater than 1 with no prime divisor.
Let n be the smallest integer greater than 1 with no prime divisor.
Since n is a divisor of n, and n has no prime divisors, n cannot be prime.
Thus, n has a divisor d such that 1 < d < n.
Let d be a divisor of n such that 1 < d < n.
Because n is the smallest counterexample, d has a prime divisor.
Let p be a prime divisor of d .
Because d / p is an integer, n/ p = (n/d ) · (d / p) is also an integer.
Thus, p is also a divisor of n.
But this contradicts our assumption that n has no prime divisors!
So our assumption must be incorrect. Hooray, our ﬁrst proof! We’re done!
Um. . . well. . . no, we’re deﬁnitely not done. That’s a ﬁrst draft up there, not a ﬁnal polished proof.
We don’t write proofs just to convince ourselves; proofs are primarily a tool to convince other people.
(In particular, ‘other people’ includes the people grading your homeworks and exams.) And while
2 Appendix I: Proof by Induction [Fa’10] Algorithms proofs by contradiction are usually easier to write, direct proofs are almost always easier to read. So
as a service to our audience (and our grade), let’s transform our minimal-counterexample proof into a
direct proof.
Let’s ﬁrst rewrite the indirect proof slightly, to make the structure more apparent. First, we break
the assumption that n is the smallest counterexample into three simpler assumptions: (1) n is an integer
greater than 1; (2) n has no prime divisors; and (3) there are no smaller counterexamples. Second,
instead of dismissing the possibility than n is prime out of hand, we include an explicit case analysis.
Proof by smallest counterexample: Let n be an arbitrary integer greater than 1.
For the sake of argument, suppose n has no prime divisor.
Assume that every integer k such that 1 < k < n has a prime divisor.
There are two cases to consider: Either n is prime, or n is composite.
• Suppose n is prime.
Then n is a prime divisor of n.
• Suppose n is composite.
Then n has a divisor d such that 1 < d < n.
Let d be a divisor of n such that 1 < d < n.
Because no counterexample is smaller than n , d has a prime divisor.
Let p be a prime divisor of d .
Because d / p is an integer, n/ p = (n/d ) · (d / p) is also an integer.
Thus, p is a prime divisor of n.
In each case, we conclude that n has a prime divisor.
But this contradicts our assumption that n has no prime divisors!
So our assumption must be incorrect. Now let’s look carefully at the structure of this proof. First, we assumed that the statement we want
to prove is false. Second, we proved that the statement we want to prove is true. Finally, we concluded
from the contradiction that our assumption that the statement we want to prove is false is incorrect, so
the statement we want to prove must be true.
But that’s just silly. Why do we need the ﬁrst and third steps? After all, the second step is a proof
all by itself! Unfortunately, this redundant style of proof by contradiction is extremely common, even in
professional papers. Fortunately, it’s also very easy to avoid; just remove the ﬁrst and third steps!
Proof by induction: Let n be an arbitrary integer greater than 1.
Assume that every integer k such that 1 < k < n has a prime divisor.
There are two cases to consider: Either n is prime or n is composite.
• First, suppose n is prime.
Then n is a prime divisor of n.
• Now suppose n is composite.
Then n has a divisor d such that 1 < d < n.
Let d be a divisor of n such that 1 < d < n.
Because no counterexample is smaller than n , d has a prime divisor.
Let p be a prime divisor of d .
Because d / p is an integer, n/ p = (n/d ) · (d / p) is also an integer.
Thus, p is a prime divisor of n.
In both cases, we conclude that n has a prime divisor. This style of proof is called induction.1 The assumption that there are no counterexamples smaller
than n is called the induction hypothesis. The two cases of the proof have different names. The
1 Many authors use the high-falutin’ name the principle of mathematical induction, to distinguish it from inductive reasoning,
the informal process by which we conclude that pigs can’t whistle, horses can’t ﬂy, and NP-hard problems cannot be solved
in polynomial time. We already know that every proof is mathematical (and arguably, all mathematics is proof), so as a
description of a proof technique, the adjective ‘mathematical’ is simply redundant. 3 Appendix I: Proof by Induction [Fa’10] Algorithms ﬁrst case, which we argue directly, is called the base case. The second case, which actually uses the
induction hypothesis, is called the inductive case. You may ﬁnd it helpful to actually label the induction
hypothesis, the base case(s), and the inductive case(s) in your proof.
The following point cannot be emphasized enough: The only difference between a proof by induction and a proof by smallest counterexample is the way we write down the argument. The essential
structure of the proofs are exactly the same. The core of our original indirect argument is a proof of the
following implication for all n:
n has no prime divisor =⇒ some number smaller than n has no prime divisor.
The core of our direct proof is the following logically equivalent implication:
every number smaller than n has a prime divisor =⇒ n has a prime divisor
The left side of this implication is just the induction hypothesis.
The proofs we’ve been playing with have been very careful and explicit; until you’re comfortable
writing your own proofs, you should be equally careful. A more mature proof-writer might express the
same proof more succinctly as follows:
Proof by induction: Let n be an arbitrary integer greater than 1. Assume that every
integer k such that 1 < k < n has a prime divisor. If n is prime, then n is a prime divisor
of n. On the other hand, if n is composite, then n has a proper divisor d . The induction
hypothesis implies that d has a prime divisor p. The integer p is also a divisor of n. A proof in this more succinct form is still worth full credit, provided the induction hypothesis is written
explicitly and the case analysis is obviously exhaustive.
A professional mathematician would write the proof even more tersely:
Proof: Induction. And you can write that tersely, too, when you’re a professional mathematician. 2 The Axiom of Induction Why does this work? Well, let’s step back to the original proof by smallest counterexample. How do we
know that a smallest counterexample exists? This seems rather obvious, but in fact, it’s impossible to
prove without using the following fact:
Every non-empty set of positive integers has a smallest element.
Every set X of positive integers is the set of counterexamples to some proposition P (n) (speciﬁcally, the
proposition n ∈ X ). Thus, the Axiom of Induction can be rewritten as follows:
If the proposition P (n) is false for some positive integer n,
then
the proposition ( P (1) ∧ P (2) ∧ · · · ∧ P (n − 1) ∧ ¬ P (n)) is true for some positive integer n.
Equivalently, in English:
If some statement about positive integers has a counterexample,
then
that statement has a smallest counterexample.
4 Appendix I: Proof by Induction [Fa’10] Algorithms We can write this implication in contrapositive form as follows:
If the proposition ( P (1) ∧ P (2) ∧ · · · ∧ P (n − 1) ∧ ¬ P (n)) is false for every positive integer n,
then
the proposition P (n) is true for every positive integer n.
Finally, let’s rewrite the ﬁrst half of this statement in a logically equivalent form, by replacing ¬( p ∧ ¬q)
with p → q.
If the implication ( P (1) ∧ P (2) ∧ · · · ∧ P (n − 1)) → P (n) is true for every positive integer n,
then
the proposition P (n) is true for every positive integer n.
This formulation is usually called the Axiom of Induction. In a proof by induction that P (n) holds for
all n, the conjunction ( P (1) ∧ P (2) ∧ · · · ∧ P (n − 1)) is the inductive hypothesis.
A proof by induction for the proposition “ P (n) for every positive integer n” is nothing but a direct proof of the more complex proposition “( P (1) ∧ P (2) ∧ · · · ∧ P (n − 1)) → P (n) for every positive integer n”. Because it’s a direct proof, it must start by considering an arbitrary positive integer,
which we might as well call n. Then, to prove the implication, we explicitly assume the hypothesis
( P (1) ∧ P (2) ∧ · · · ∧ P (n − 1)) and then prove the conclusion P (n) for that particular value of n. The
proof almost always breaks down into two or more cases, each of which may or may not actually use
the inductive hypothesis.
Here is the boilerplate for every induction proof. Read it. Learn it. Use it.
Theorem: P (n) for every positive integer n.
Proof by induction: Let n be an arbitrary positive integer.
Assume inductively that P (k) is true for every positive integer k < n.
There are several cases to consider:
• Suppose n is . . . blah blah blah . . .
Then P (n) is true. • Suppose n is . . . blah blah blah . . .
The inductive hypothesis implies that . . . blah blah blah . . .
Thus, P (n) is true.
In each case, we conclude that P (n) is true. Some textbooks distinguish between several different types of induction: ‘regular’ induction versus
‘strong’ induction versus ‘complete’ induction versus ‘structural’ induction versus ‘transﬁnite’ induction
versus ‘Noetherian’ induction. Distinguishing between these different types of induction is pointless
hairsplitting; I won’t even deﬁne them. Every ‘different type’ of induction proof is provably equivalent
to a proof by smallest counterexample. (Later we will consider inductive proofs of statements about
partially ordered sets other than the positive integers, for which ‘smallest’ has a different meaning, but
this difference will prove to be inconsequential.) 3 Stamps and Recursion Let’s move on to a completely different example. 5 Appendix I: Proof by Induction [Fa’10] Algorithms Theorem 2. Given an unlimited supply of 5-cent stamps and 7-cent stamps, we can make any amount
of postage larger than 23 cents.
We could prove this by contradiction, using a smallest-counterexample argument, but let’s aim for
a direct proof by induction this time. We start by writing down the induction boilerplate, using the
standard induction hypothesis: There is no counterexample smaller than n.
Proof by induction: Let n be an arbitrary integer greater than 23.
Assume that for any integer k such that 23 < k < n, we can make k cents in postage. . . . blah blah blah . . .
Thus, we can make n cents in postage. How do we ﬁll in the details? One approach is to think about what you would actually do if you really
had to make n cents in postage. For example, you might start with a 5-cent stamp, and then try to make
n − 5 cents in postage. The inductive hypothesis says you can make any amount of postage bigger than
23 cents and less than n cents. So if n − 5 > 23, then you already know that you can make n − 5 cents
in postage! (You don’t know how to make n − 5 cents in postage, but so what?)
Let’s write this observation into our proof as two separate cases: either n > 28 (where our approach
works) or n ≤ 28 (where we don’t know what to do yet).
Proof by induction: Let n be an arbitrary integer greater than 23.
Assume that for any integer k such that 23 < k < n, we can make k cents in postage.
There are two cases to consider: Either n > 28 or n ≤ 28.
• Suppose n > 28.
Then 23 < n − 5 < n.
Thus, the induction hypothesis implies that we can make n − 5 cents in postage.
Adding one more 5-cent stamp gives us n cents in postage.
• Now suppose n ≤ 28. . . . blah blah blah . . .
In both cases, we can make n cents in postage. What do we do in the second case? Fortunately, this case considers only ﬁve integers: 24, 25, 26,
27, and 28. There might be a clever way to solve all ﬁve cases at once, but why bother? They’re
small enough that we can ﬁnd a solution by brute force in less than a minute. To make the proof more
readable, I’ll unfold the nested cases and list them in increasing order.
Proof by induction: Let n be an arbitrary integer greater than 23.
Assume that for any integer k such that 23 < k < n, we can make k cents in postage.
There are six cases to consider: n = 24, n = 25, n = 26, n = 27, n = 28, and n > 28. •
•
•
•
•
• 24 = 7 + 7 + 5 + 5
25 = 5 + 5 + 5 + 5 + 5
26 = 7 + 7 + 7 + 5
27 = 7 + 5 + 5 + 5 + 5
28 = 7 + 7 + 7 + 7
Suppose n > 28.
Then 23 < n − 5 < n. Thus, the induction hypothesis implies that we can make n − 5 cents in postage.
Adding one more 5-cent stamp gives us n cents in postage.
In all cases, we can make n cents in postage. Voilà! An induction proof! More importantly, we now have a recipe for discovering induction proofs.
6 Appendix I: Proof by Induction [Fa’10] Algorithms 1. Write down the boilerplate. Write down the universal invocation (‘Let n be an arbitrary. . . ’),
the induction hypothesis, and the conclusion, with enough blank space for the remaining details.
Don’t be clever. Don’t even think. Just write. This is the easy part. To emphasize the common
structure, the boilerplate will be indicated in green for the rest of this handout.
2. Think big. Don’t think how to solve the problem all the way down to the ground; you’ll only
make yourself dizzy. Don’t think about piddly little numbers like 1 or 5 or 10100 . Instead, think
about how to reduce the proof about some absfoluckingutely ginormous value of n to a proof about
some other number(s) smaller than n. This is the hard part.
3. Look for holes. Look for cases where your inductive argument breaks down. Solve those cases
directly. Don’t be clever here; be stupid but thorough.
4. Rewrite everything. Your ﬁrst proof is a rough draft. Rewrite the proof so that your argument is
easier for your (unknown?) reader to follow.
The cases in an inductive proof always fall into two categories. Any case that uses the inductive
hypothesis is called an inductive case. Any case that does not use the inductive hypothesis is called a
base case. Typically, but not always, base cases consider a few small values of n, and the inductive cases
consider everything else. Induction proofs are usually clearer if we present the base cases ﬁrst, but I
ﬁnd it much easier to discover the inductive cases ﬁrst. In other words, I recommend writing induction
proofs backwards.
Well-written induction proofs very closely resemble well-written recursive programs. We computer
scientists use induction primarily to reason about recursion, so maintaining this resemblance is extremely useful—we only have to keep one mental pattern, called ‘induction’ when we’re writing proofs
and ‘recursion’ when we’re writing code. Consider the following C and Scheme programs for making n
cents in postage:
void postage(int n)
{
assert(n>23);
switch ($n$)
{
case 24: printf("7+7+5+5");
case 25: printf("5+5+5+5+5");
case 26: printf("7+7+7+5");
case 27: printf("7+5+5+5+5");
case 28: printf("7+7+7+7");
default:
postage(n-5);
printf("+5");
}
} break;
break;
break;
break;
break; (define (postage n)
(cond ((= n 24) (5 5 7 7))
((= n 25) (5 5 5 5 5))
((= n 26) (5 7 7 7))
((= n 27) (5 5 5 5 7))
((= n 28) (7 7 7 7))
((> n 28) (cons 5 (postage (- n 5)))))) 7 Appendix I: Proof by Induction [Fa’10] Algorithms The C program begins by declaring the input parameter (“Let n be an arbitrary integer. . . ") and
asserting its range (“. . . greater than 23."). (Scheme programs don’t have type declarations.) In both
languages, the code branches into six cases: ﬁve that are solved directly, plus one that is handled
by invoking the inductive hypothesis recursively. 4 More on Prime Divisors Before we move on to different examples, let’s prove another fact about prime numbers:
Theorem 3. Every positive integer is a product of prime numbers.
First, let’s write down the boilerplate. Hey! I saw that! You were thinking, weren’t you? Stop that
this instant! Don’t make me turn the car around. First we write down the boilerplate.
Proof by induction: Let n be an arbitrary positive integer.
Assume that any positive integer k < n is a product of prime numbers.
There are some cases to consider: . . . blah blah blah . . .
Thus, n is a product of prime numbers. Now let’s think about how you would actually factor a positive integer n into primes. There are
a couple of different options here. One possibility is to ﬁnd a prime divisor p of n, as guaranteed by
Theorem 1, and recursively factor the integer n/ p. This argument works as long as n ≥ 2, but what
about n = 1? The answer is simple: 1 is the product of the empty set of primes. What else could it be?
Proof by induction: Let n be an arbitrary positive integer.
Assume that any positive integer k < n is a product of prime numbers.
There are two cases to consider: either n = 1 or n ≥ 2. • If n = 1, then n is the product of the elements of the empty set, each of which is
prime, green, sparkly, vanilla, and hemophagic. • Suppose n > 1. Let p be a prime divisor of n, as guaranteed by Theorem 2. The
inductive hypothesis implies that the positive integer n/ p is a product of primes,
and clearly n = (n/ p) · p.
In both cases, n is a product of prime numbers. But an even simpler method is to factor n into any two proper divisors, and recursively handle them
both. This method works as long as n is composite, since otherwise there is no way to factor n into
smaller integers. Thus, we need to consider prime numbers separately, as well as the special case 1.
Proof by induction: Let n be an arbitrary positive integer.
Assume that any positive integer k < n is a product of prime numbers.
There are three cases to consider: either n = 1, n is prime, or n is composite. • If n = 1, then n is the product of the elements of the empty set, each of which is
prime, red, broody, chocolate, and lycanthropic. • If n is prime, then n is the product of one prime number, namely n.
• Suppose n is composite. Let d be any proper divisor of n (guaranteed by the deﬁnition of ‘composite’), and let m = n/d . Since both d and m are positive integers
smaller than n, the inductive hypothesis implies that d and m are both products of
prime numbers. We clearly have n = d · m.
In both cases, n is a product of prime numbers. 8 Appendix I: Proof by Induction [Fa’10] Algorithms 5 Summations Here’s an easy one.
n 3i = Theorem 4.
i =0 3n+1 − 1
2 for every non-negative integer n. First let’s write down the induction boilerplate, which empty space for the details we’ll ﬁll in later.
Proof by induction: Let n be an arbitrary non-negative integer.
k 3i = Assume inductively that 3 k +1 − 1
2 i =0 for every non-negative integer k < n. There are some number of cases to consider: . . . blah blah blah . . .
n 3i = We conclude that 3n+1 − 1
2 i =0 . Now imagine you are part of an inﬁnitely long assembly line of mathematical provers, each assigned
to a particular non-negative integer. Your task is to prove this theorem for the integer 8675310. The
regulations of the Mathematical Provers Union require you not to think about any other integer but
your own. The assembly line starts with the Senior Master Prover, who proves the theorem for the
case n = 0. Next is the Assistant Senior Master Prover, who proves the theorem for n = 1. After him
is the Assistant Assistant Senior Master Prover, who proves the theorem for n = 2. Then the Assistant
Assistant Assistant Senior Master Prover proves the theorem for n = 3. As the work proceeds, you start
to get more and more bored. You attempt strike up a conversation with Jenny, the prover to your left,
but she ignores you, preferring to focus on the proof. Eventually, you fall into a deep, dreamless sleep.
An undetermined time later, Jenny wakes you up by shouting, “Hey, doofus! It’s your turn!” As you
look around, bleary-eyed, you realize that Jenny and everyone to your left has ﬁnished their proofs,
and that everyone is waiting for you to ﬁnish yours. What do you do?
What you do, after wiping the drool off your chin, is stop and think for a moment about what you’re
trying to prove. What does that
notation actually mean? Intuitively, we can expand the notation as
follows:
8675310 3i = 30 + 31 + · · · + 38675309 + 38675310 .
i =0 Notice that this expression also contains the summation that Jenny just ﬁnished proving something
about:
8675309 3i = 30 + 31 + · · · + 38675308 + 38675309 .
i =0 Putting these two expressions together gives us the following identity:
8675310 8675309 3i + 38675310 3i =
i =0 i =0 In fact, this recursive identity is the deﬁnition of . Jenny just proved that the summation on the right
is equal to (38675310 − 1)/2, so we can plug that into the right side of our equation:
8675310 8675309 3i + 38675310 = 3i =
i =0 i =0 9 38675310 − 1
2 + 38675310 . Appendix I: Proof by Induction [Fa’10] Algorithms And it’s all downhill from here. After a little bit of algebra, you simplify the right side of this equation
to (38675311 − 1)/2, wake up the prover to your right, and start planning your well-earned vacation.
Let’s insert this argument into our boilerplate, only using a generic ‘big’ integer n instead of the
speciﬁc integer 8675310:
Proof by induction: Let n be an arbitrary non-negative integer.
k 3i = Assume inductively that 3 k +1 − 1
2 i =0 for every non-negative integer k < n. There are two cases to consider: Either n is big or n is small . • If n is big , then
n−1 n 3i =
i =0 3i + 3n [deﬁnition of i =0
n =
= 3 −1
2 + 3n [induction hypothesis, with k = n − 1] 3 n+1 − 1 [algebra] 2 • On the other hand, if n is small , then . . . blah blah blah . . .
n 3i = In both cases, we conclude that 3 n+1 − 1
2 i =0 . Now, how big is ‘big’, and what do we do when n is ‘small’? To answer the ﬁrst question, let’s look at
where our existing inductive argument breaks down. In order to apply the induction hypothesis when
k = n − 1, the integer n − 1 must be non-negative; equivalently, n must be positive. But that’s the only
assumption we need: The only case we missed is n = 0. Fortunately, this case is easy to handle directly.
Proof by induction: Let n be an arbitrary non-negative integer.
k 3i = Assume inductively that 3 k +1 − 1
2 i =0 for every non-negative integer k < n. There are two cases to consider: Either n = 0 or n ≥ 1.
n 3i = 30 = 1, and • If n = 0, then
i =0 3n+1 − 1 = 2 31 − 1
2 = 1. • On the other hand, if n ≥ 1, then
n−1 n 3i =
i =0 3i + 3n [deﬁnition of i =0
n =
= 3 −1
2 + 3n 3 n+1 − 1 [algebra] 2
n 3i = In both cases, we conclude that
i =0 [induction hypothesis, with k = n − 1] 3 n+1 − 1
2 10 . Appendix I: Proof by Induction [Fa’10] Algorithms IH Here is the same proof, written more tersely; the non-standard symbol == indicates the use of the
induction hypothesis.
Proof by induction: Let n be an arbitrary non-negative integer, and assume inductively
k that i =0 3i = (3k+1 − 1)/2 for every non-negative integer k < n. The base case n = 0 is
trivial, and for any n ≥ 1, we have
n−1 n 3i =
i =0 IH 3i + 3n == 3n − 1
2 i =0 + 3n = 3 n+1 − 1
2 . This is not the only way to prove this theorem by induction; here is another:
Proof by induction: Let n be an arbitrary non-negative integer, and assume inductively
k that i =0 3i = (3k+1 − 1)/2 for every non-negative integer k < n. The base case n = 0 is
trivial, and for any n ≥ 1, we have
n
i =0 n−1 n 3 i = 30 + 3 i = 30 + 3 ·
i =1 IH 3i == 30 + 3 ·
i =0 3n − 1
2 = 3 n+1 − 1
2 . In the remainder of these notes, I’ll give several more examples of induction proofs. In some cases, I
give multiple proofs for the same theorem. Unlike the earlier examples, I will not describe the thought
process that lead to the proof; in each case, I followed the basic outline on page 7. 6 Tiling with Triominos The next theorem is about tiling a square checkerboard with triominos. A triomino is a shape composed
of three squares meeting in an L-shape. Our goal is to cover as much of a 2n × 2n grid with triominos
as possible, without any two triominos overlapping, and with all triominos inside the square. We can’t
cover every square in the grid—the number of squares is 4n , which is not a multiple of 3—but we can
cover all but one square. In fact, as the next theorem shows, we can choose any square to be the one
we don’t want to cover. Almost tiling a 16 × 16 checkerboard with triominos. Theorem 5. For any non-negative integer n, the 2n × 2n checkerboard with any square removed can
be tiled using L-shaped triominos.
Here are two different inductive proofs for this theorem, one ‘top down’, the other ‘bottom up’. 11 Appendix I: Proof by Induction [Fa’10] Algorithms Proof by top-down induction: Let n be an arbitrary non-negative integer. Assume that
for any non-negative integer k < n, the 2k × 2k grid with any square removed can be tiled
using triominos. There are two cases to consider: Either n = 0 or n ≥ 1. • The 20 × 20 grid has a single square, so removing one square leaves nothing, which
we can tile with zero triominos. • Suppose n ≥ 1. In this case, the 2n × 2n grid can be divided into four smaller 2n−1 ×
2n−1 grids. Without loss of generality, suppose the deleted square is in the upper
right quarter. With a single L-shaped triomino at the center of the board, we can
cover one square in each of the other three quadrants. The induction hypothesis
implies that we can tile each of the quadrants, minus one square.
In both cases, we conclude that the 2n × 2n grid with any square removed can be tiled
with triominos. Top-down inductive proof of Theorem 4. Proof by bottom-up induction: Let n be an arbitrary non-negative integer. Assume
that for any non-negative integer k < n, the 2k × 2k grid with any square removed can be
tiled using triominos. There are two cases to consider: Either n = 0 or n ≥ 1. • The 20 × 20 grid has a single square, so removing one square leaves nothing, which
we can tile with zero triominos. • Suppose n ≥ 1. Then by clustering the squares into 2 × 2 blocks, we can transform
any 2n × 2n grid into a 2n−1 × 2n−1 grid. Suppose square (i , j ) has been removed
from the 2n × 2n grid. The induction hypothesis implies that the 2n−1 × 2n−1 grid with
block ( i /2 , j /2 ) removed can be tiled with double-size triominos. Each doublesize triomono can be tiled with four smaller triominos, and block ( i /2 , j /2 ) with
square (i , j ) removed is another triomino.
In both cases, we conclude that the 2n × 2n grid with any square removed can be tiled
with triominos. Second proof of Theorem 4. 7 Binary Numbers Exist Theorem 6. Every non-negative integer can be written as the sum of distinct powers of 2. 12 Appendix I: Proof by Induction [Fa’10] Algorithms Intuitively, this theorem states that every number can be represented in binary. (That’s not a proof,
by the way; it’s just a restatement of the theorem.) I’ll present four distinct inductive proofs for this
theorem. The ﬁrst two are standard, by-the-book induction proofs.
Proof by top-down induction: Let n be an arbitrary non-negative integer. Assume that
any non-negative integer less than n can be written as the sum of distinct powers of 2.
There are two cases to consider: Either n = 0 or n ≥ 1. • The base case n = 0 is trivial—the elements of the empty set are distinct and sum
to zero. • Suppose n ≥ 1. Let k be the largest integer such that 2k ≤ n, and let m = n − 2k .
Observe that m < 2k+1 − 2k = 2k . Because 0 ≤ m < n, the inductive hypothesis
implies that m can be written as the sum of distinct powers of 2. Moreover, in the
summation for m, each power of 2 is at most m, and therefore less than 2k . Thus,
m + 2k is the sum of distinct powers of 2.
In either case, we conclude that n can be written as the sum of distinct powers of 2.
Proof by bottom-up induction: Let n be an arbitrary non-negative integer. Assume
that any non-negative integer less than n can be written as the sum of distinct powers
of 2. There are two cases to consider: Either n = 0 or n ≥ 1. • The base case n = 0 is trivial—the elements of the empty set are distinct and sum
to zero. • Suppose n ≥ 1, and let m = n/2 . Because 0 ≤ m < n, the inductive hypothesis
implies that m can be written as the sum of distinct powers of 2. Thus, 2m can also
be written as the sum of distinct powers of 2, each of which is greater than 20 . If
n is even, then n = 2m and we are done; otherwise, n = 2m + 20 is the the sum of
distinct powers of 2.
In either case, we conclude that n can be written as the sum of distinct powers of 2. The third proof deviates slightly from the induction boilerplate. At the top level, this proof doesn’t
actually use induction at all! However, a key step requires its own (straightforward) inductive proof.
Proof by algorithm: Let n be an arbitrary non-negative integer. Let S be a multiset
containing n copies of 20 . Modify S by running the following algorithm:
while S has more than one copy of any element 2i
Remove two copies of 2i from S
Insert one copy of 2i +1 into S Each iteration of this algorithm reduces the cardinality of S by 1, so the algorithm must
eventually halt. When the algorithm halts, the elements of S are distinct. We claim that
just after each iteration of the while loop, the elements of S sum to n.
Proof by induction: Consider an arbitrary iteration of the loop. Assume inductively that just after each previous iteration, the elements of S sum to n.
Before any iterations of the loop, the elements of S sum to n by deﬁnition. The
induction hypothesis implies that just before the current iteration begins, the
elements of S sum to n. The loop replaces two copies of some number 2i with
their sum 2i +1 , leaving the total sum of S unchanged. Thus, when the iteration
ends, the elements of S sum to n. Thus, when the algorithm halts, the elements of S are distinct powers of 2 that sum to n.
We conclude that n can be written as the sum of distinct powers of 2. 13 Appendix I: Proof by Induction [Fa’10] Algorithms The fourth proof uses so-called ‘weak’ induction, where the inductive hypothesis can only be applied
at n−1. Not surprisingly, tying all but one hand behind our backs makes the resulting proof longer, more
complicated, and harder to read. It doesn’t help that the algorithm used in the proof is overly speciﬁc.
Nevertheless, this is the ﬁrst approach that occurs to most students who have not truly accepted the
Recursion Fairy into their hearts.
Proof by baby-step induction: Let n be an arbitrary non-negative integer. Assume
that any non-negative integer less than n can be written as the sum of distinct powers
of 2. There are two cases to consider: Either n = 0 or n ≥ 1. • The base case n = 0 is trivial—the elements of the empty set are distinct and sum
to zero. • Suppose n ≥ 1. The inductive hypothesis implies that n − 1 can be written as the
sum of distinct powers of 2. Thus, n can be written as the sum of powers of 2, which
are distinct except possibly for two copies of 20 . Let S be this multiset of powers
of 2.
Now consider the following algorithm: i←0
while S has more than one copy of 2i
Remove two copies of 2i from S
Insert one copy of 2i +1 into S i ← i+1
Each iteration of this algorithm reduces the cardinality of S by 1, so the algorithm
must eventually halt. We claim that for every non-negative integer i , the following invariants are satisﬁed after the i th iteration of the while loop (or before the
algorithm starts if i = 0): – The elements of S sum to n.
Proof by induction: Let i be an arbitrary non-negative integer. Assume that
for any non-negative integer j ≤ i , after the j th iteration of the while loop, the
elements of S sum to n. If i = 0, the elements of S sum to n by deﬁnition of S .
Otherwise, the induction hypothesis implies that just before the i th iteration,
the elements of S sum to n; the i th iteration replaces two copies of 2i with 2i +1 ,
leaving the sum unchanged. – The elements in S are distinct, except possibly for two copies of 2i .
Proof by induction: Let i be an arbitrary non-negative integer. Assume that
for any non-negative integer j ≤ i , after the j th iteration of the while loop, the
elements of S are distinct except possibly for two copies of 2 j . If i = 0, the
invariant holds by deﬁnition of S . So suppose i > 0. The induction hypothesis
implies that just before the i th iteration, the elements of S are distinct except
possibly for two copies of 2i . If there are two copies of 2i , the algorithm replaces
them both with 2i +1 , and the invariant is established; otherwise, the algorithm
halts, and the invariant is again established. The second invariant implies that when the algorithm halts, the elements of S are
distinct.
In either case, we conclude that n can be written as the sum of distinct powers of 2. Repeat after me: “Doctor! Doctor! It hurts when I do this!” 14 Appendix I: Proof by Induction [Fa’10] Algorithms 8 Irrational Numbers Exist Theorem 7. 2 is irrational. Proof: I will prove that p2 = 2q2 (and thus p/q = 2) for all positive integers p and q.
Let p and q be arbitrary positive integers. Assume that for any positive integers i < p
and j < q, we have i 2 = 2 j 2 . Let i = p/2 and j = q/2 . There are three cases to
consider: • Suppose p is odd. Then p2 = (2i + 1)2 = 4i 2 + 4i + 1 is odd, but 2q2 is even.
• Suppose p is even and q is odd. Then p2 = 4i 2 is divisible by 4, but 2q2 =
2(2 j + 1)2 = 4(2 j 2 + 2 j ) + 2 is not divisible by 4.
• Finally, suppose p and q are both even. The induction hypothesis implies that
i 2 = 2 j 2 . Thus, p2 = 4i 2 = 8 j 2 = 2q2 .
In every case, we conclude that p2 = 2q2 . For some reason, this proof is almost always presented as a proof by inﬁnite descent. Notice that
the induction hypothesis assumed that both p and q were as small as possible. Notice also that the ‘base
cases’ included every pair of integers p and q where at least one of the integers is odd. 9 Fibonacci Parity The Fibonacci numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . . are recursively deﬁned as follows: if n = 0
0
Fn = 1 Fn−1 + Fn−2 if n = 1 if n ≥ 2 Theorem 8. For all non-negative integers n, Fn is even if and only if n is divisible by 3.
Proof: Let n be an arbitrary non-negative integer. Assume that for all non-negative
integers k < n, Fk is even if and only if n is divisible by 3. There are three cases to
consider: n = 0, n = 1, and n ≥ 2. • If n = 0, then n is divisible by 3, and Fn = 0 is even.
• If n = 1, then n is not divisible by 3, and Fn = 1 is odd.
• If n ≥ 2, there are two subcases to consider: Either n is divisible by 3, or it isn’t.
– Suppose n is divisible by 3. Then neither n − 1 nor n − 2 is divisible by 3.
Thus, the inductive hypothesis implies that both Fn−1 and Fn−2 are odd. So
Fn is the sum of two odd numbers, and is therefore even.
– Suppose n is not divisible by 3. Then exactly one of the numbers n − 1 and
n − 2 is divisible by 3. Thus, the inductive hypothesis implies that exactly one
of the numbers Fn−1 and Fn−2 is even, and the other is odd. So Fn is the sum
of an even number and an odd number, and is therefore odd.
In all cases, Fn is even if and only if n is divisible by 3. 15 Appendix I: Proof by Induction [Fa’10] Algorithms 10 Recursive Functions Theorem 9. Suppose the function F : → is deﬁned recursively by setting F (0) = 0 and F (n) =
1 + F ( n/2 ) for every positive integer n. Then for every positive integer n, we have F (n) = 1 + log2 n .
Proof: Let n be an arbitrary positive integer. Assume that F (k) = 1 + log2 k for every
positive integer k < n. There are two cases to consider: Either n = 1 or n ≥ 2. • Suppose n = 1. Then F (n) = F (1) = 1 + F ( 1/2 ) = 1 + F (0) = 1 and 1 + log2 n =
1 + log2 1 = 1 + 0 = 1.
• Suppose n ≥ 2. Because 1 ≤ n/2 < n, the induction hypothesis implies that
F ( n/2 ) = 1 + log2 n/2 . The deﬁnition of F (n) now implies that F (n) = 1 +
F ( n/2 ) = 2 + log2 n/2 .
Now there are two subcases to consider: n is either even or odd. – If n is even, then n/2 = n/2, which implies
F (n) = 2 + log2 n/2
= 2 + log2 (n/2)
= 2 + (log2 n) − 1
= 2 + log2 n − 1
= 1 + log2 n .
– If n is odd, then n/2 = (n − 1)/2, which implies
F (n) = 2 + log2 n/2
= 2 + log2 ((n − 1)/2)
= 1 + log2 (n − 1)
= 1 + log2 n
by the algebra in the even case. Because n > 1 and n is odd, n cannot be a
power of 2; thus, log2 n = log2 (n − 1) .
In all cases, we conclude that F (n) = 1 + log2 n . 11 Trees Recall that a tree is a connected undirected graph with no cycles. A subtree of a tree T is a connected
subgraph of T ; a proper subtree is any tree except T itself.
Theorem 10. In every tree, the number of vertices is one more than the number of edges.
This one is actually pretty easy to prove directly from the deﬁnition of ‘tree’: a connected acyclic graph.
Proof: Let T be an arbitrary tree. Choose an arbitrary vertex v of T to be the root, and
direct every edge of T outward from v . Because T is connected, every node except v
has at least one edge directed into it. Because T is acyclic, every node has at most one
edge directed into it, and no edge is directed into v . Thus, for every node x = v , there is
exactly one edge directed into x . We conclude that the number of edges is one less than
the number of nodes. 16 Appendix I: Proof by Induction [Fa’10] Algorithms But we can prove this theorem by induction as well, in several different ways. Each inductive proof
is structured around a different recursive deﬁnition of ‘tree’. First, a tree is either a single node, or two
trees joined by an edge.
Proof: Let T be an arbitrary tree. Assume that in any proper subtree of T , the number of
vertices is one more than the number of edges. There are two cases to consider: Either
T has one vertex, or T has more than one vertex. • If T has one vertex, then it has no edges.
• Suppose T has more than one vertex. Because T is connected, every pair of vertices is joined by a path. Thus, T must contain at least one edge. Let e be an
arbitrary edge of T , and consider the graph T \ e obtained by deleting e from T .
Because T is acyclic, there is no path in T \ e between the endpoints of e. Thus,
T has at least two connected components. On the other hand, because T is connected, T \ e has at most two connected components. Thus, T \ e has exactly two
connected components; call them A and B .
Because T is acyclic, subgraphs A and B are also acyclic. Thus, A and B are subtrees of T , and therefore the induction hypothesis implies that | E (A)| = |V (A)| − 1
and | E (B )| = |V (B )| − 1.
Because A and B do not share any vertices or edges, we have |V ( T )| = |V (A)| +
|V (B )| and | E ( T )| = | E (A)| + | E (B )| + 1.
Simple algebra now implies that | E ( T )| = |V ( T )| − 1.
In both cases, we conclude that the number of vertices in T is one more than the number
of edges in T . Second, a tree is a single node connected by edges to a ﬁnite set of trees.
Proof: Let T be an arbitrary tree. Assume that in any proper subtree of T , the number of
vertices is one more than the number of edges. There are two cases to consider: Either
T has one vertex, or T has more than one vertex. • If T has one vertex, then it has no edges.
• Suppose T has more than one vertex. Let v be an arbitrary vertex of T , and let
d be the degree of v . Delete v and all its incident edges from T to obtain a new
graph G . This graph has exactly d connected components; call them G1 , G2 , . . . , Gd .
Because T is acyclic, every subgraph of T is acyclic. Thus, every subgraph Gi is a
proper subtree of G . So the induction hypothesis implies that | E (Gi )| = |V (Gi )| − 1
for each i . We conclude that
d | E ( T )| = d + d | E (Gi )| = d +
i =1 d (|V (Gi )| − 1) =
i =1 |V (Gi )| = |V ( T )| − 1.
i =1 In both cases, we conclude that the number of vertices in T is one more than the number
of edges in T . But you should never attempt to argue like this:
Not a Proof: The theorem is clearly true for the 1-node tree. So let T be an arbitrary
tree with at least two nodes. Assume inductively that the number of vertices in T is one
more than the number of edges in T . Suppose we add one more leaf to T to get a new
tree T . This new tree has one more vertex than T and one more edge than T . Thus, the
number of vertices in T is one more than the number of edges in T . 17 Appendix I: Proof by Induction [Fa’10] Algorithms This is not a proof. Every sentence is true, and the connecting logic is correct, but it does not imply
the theorem, because it doesn’t explicitly consider all possible trees. Why should the reader believe that
their favorite tree can be recursively constructed by adding leaves to a 1-node tree? It’s true, of course,
but that argument doesn’t prove it. Remember: There are only two ways to prove any universally
quantiﬁed statement: Directly (“Let T be an arbitrary tree. . . ") or by contradiction (“Suppose some
tree T doesn’t. . . ").
Here is a correct inductive proof using the same underlying idea. In this proof, I don’t have to prove
that the proof considers arbitrary trees; it says so right there on the ﬁrst line! As usual, the proof very
strongly resembles a recursive algorithm, including a subroutine to ﬁnd a leaf.
Proof: Let T be an arbitrary tree. Assume that in any proper subtree of T , the number of
vertices is one more than the number of edges. There are two cases to consider: Either T
has one vertex, or T has more than one vertex. • If T has one vertex, then it has no edges.
• Otherwise, T must have at least one vertex of degree 1, otherwise known as a leaf.
Proof: Consider a walk through the graph T that starts at an arbitrary
vertex and continues as long as possible without repeating any edge.
The walk can never visit the same vertex more than once, because T
is acyclic. Whenever the walk visits a vertex of degree at least 2, it can
continue further, because that vertex has at least one unvisited edge.
But the walk must eventually end, because T is ﬁnite. Thus, the walk
must eventually reach a vertex of degree 1.
Let be an arbitrary leaf of T , and let T be the tree obtained by deleting
Then we have the identity from T . | E ( T )| = | E ( T )| − 1 = |V ( T )| − 2 = |V ( T )| − 1,
where the ﬁrst and third equalities follow from the deﬁnition of T , and the second
equality follows from the inductive hypothesis.
In both cases, we conclude that the number of vertices in T is one more than the number
of edges in T . 12 Strings Recall that a string is any ﬁnite sequence of symbols. More formally, a string is either empty or a single
symbol followed by a string. Let x · y denote the concatenation of strings x and y , and let reverse( x )
denote the reversal of string x . For example, now · here = nowhere and reverse(stop) = pots.
Theorem 11. For all strings x and y , we have reverse( x · y ) = reverse( y ) · reverse( x ). 18 Appendix I: Proof by Induction [Fa’10] Algorithms Proof: Let x and y be arbitrary strings. Assume for any proper substring w of x that
reverse(w · y ) = reverse( y ) · reverse(w ). There are two cases to consider: Either x is the
empty string, or not. • If x is the empty string, then reverse( x ) is also the empty string, so reverse( x · y ) =
reverse( y ) = reverse( y ) · reverse( x ).
• Suppose x is not the empty string. Then x = a · w , for some character a and some
string w , and therefore
reverse( x · y ) = reverse((a · w ) · y ) = reverse(a · (w · y ))
= reverse(w · y ) · a
= reverse( y ) · reverse(w ) · a [induction hypothesis] = reverse( y ) · reverse( x ).
In both cases, we conclude that reverse( x · y ) = reverse( y ) · reverse( x ). 13 Regular Languages Theorem 12. Every regular language is accepted by a non-deterministic ﬁnite automaton.
Proof: In fact, we will show something stronger: Every regular language is accepted by
an NFA with exactly one accepting state.
Let R be an arbitrary regular expression over the ﬁnite alphabet Σ. Assume that for
any sub-expression S of R, the corresponding regular language is accepted by an NFA with
S
one accepting state, denoted
. There are six cases to consider—three base cases
and three recursive cases—mirroring the recursive deﬁnition of a regular expression. • If R = ∅, then L (R) = ∅ is accepted by an NFA with no transitions:
ε • If R = , then L (R) = { } is accepted by the NFA . . • If R = a for some character a ∈ Σ, then L (R) = {a} is accepted by the NFA a . • Suppose R = S T for some regular expressions S and T . The inductive hypothesis
S
T
implies that S and T are accepted by NFAs
and
, respectively.
S
T
Then L ( T ) = {uv | u ∈ L (S ), v ∈ L ( T )} is accepted by the NFA
.
• Suppose R = S + T for some regular expressions S and T . The inductive hypothesis
S
T
implies that S and T are accepted by NFAs
and
, respectively.
Then L (R) = L (S ) ∪ L ( T ) is accepted by the NFA ε S ε ε T ε . • Finally, suppose R = S ∗ for some regular expression S . The inductive hypothesis
S
implies that S is accepted by an NFA
. Then the language L (R) = L (S )∗ is
ε accepted by the NFA S . ε In every case, L ( x ) is accepted by an NFA with one accepting state. c Copyright 2011 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. 19 Appendix II: Solving Recurrences [Fa’10] Algorithms Change is certain. Peace is followed by disturbances; departure of evil men by their
return. Such recurrences should not constitute occasions for sadness but realities for
awareness, so that one may be happy in the interim.
— I Ching [The Book of Changes] (c. 1100 BC)
“. . . 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]
Wil Wheaton: Embrace the dark side!
Sheldon: That’s not even from your franchise!
— “The Wheaton Recurrence”, Bing Bang Theory, April 12, 2010 Solving Recurrences
1 Introduction A recurrence is a recursive description of a function, or in other words, a description of 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?
c Copyright 2011 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. 1 Appendix II: Solving Recurrences [Fa’10] Algorithms 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).
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!
1 . . . except of course during exams, where you aren’t supposed to use any outside sources 2 ..., Appendix II: Solving Recurrences [Fa’10] Algorithms 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 − 3) + 7
= ···
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 −1 [inductive hypothesis] − 1) [initial recurrence for T (n − (k − 1))] = 2k T (n − k) + (2k − 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
0
F1
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 3 Appendix II: Solving Recurrences [Fa’10] Algorithms 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. 2.3 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. 4 Appendix II: Solving Recurrences [Fa’10] Algorithms
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.
T (n) = 2 T (n/2) + n = 21 T (n/21 ) + 1 · n
T ( n) = 2 k − 1 T ( n/ 2 k −1 ) + ( k − 1) n [k = 1, given recurrence]
[inductive hypothesis] = 2 k − 1 2 T ( n/ 2 k ) + n/ 2 k − 1 + ( k − 1) n
k [substitution] k = 2 T (n/2 ) + 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
5 Appendix II: Solving Recurrences [Fa’10] Algorithms 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
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 lg n + n ≤ an lg n? [induction hypothesis] 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 lg n + n [induction hypothesis]
[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 = a n lg lg n − a n + n
≤ a n lg lg n [induction hypothesis]
[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 = b n lg lg n − b n + n
≥ b n lg lg n [induction hypothesis]
[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 ). 6 Appendix II: Solving Recurrences [Fa’10] Algorithms 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).
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 (??) 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) f(n) a
f(n/b) f(n/b) a +
f(n/b) a f(n/b²)
f(n/b²)
f(n/b²)
f(n/b²) f(n/b) a f(n/b²)
f(n/b²)
f(n/b²)
f(n/b²) f(n/b²)
f(n/b²)
f(n/b²)
f(n/b²) a ⋅ f(n/b) a + f(n/b²)
f(n/b²)
f(n/b²)
f(n/b²) a² ⋅ f(n/b²)
+ +
L L L L L L L 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 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 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 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 L a ⋅f(n/b ) 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 (??)
is the same:
L T (n) = n.
i =0 The recursion tree has L = log2 n levels, so T (n ) = Θ(n log n ).
7 Appendix II: Solving Recurrences [Fa’10] Algorithms
• 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 (??) 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?
• 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 (??) 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 not 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 8 Appendix II: Solving Recurrences [Fa’10] Algorithms
∞ 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
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 ).
1 3 • 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
9 Appendix II: Solving Recurrences [Fa’10] Algorithms 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.
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 The Nuclear Bomb Finally, let me describe without proof a powerful generalization of the recursion tree method, ﬁrst
published by Lebanese researchers Mohamad Akra and Louay Bazzi in 1998. Consider a general
divide-and-conquer recurrence of the form
k T ( n) = ai T (n/ bi ) + f (n),
i =1 where k is a constant, ai > 0 and bi > 1 are constants for all i , and f (n) = Ω(nc ) and f (n) = O(nd ) for
some constants 0 < c ≤ d . (As usual, we assume the standard base case T (Θ(1)) = Θ(1)).) Akra and
Bazzi prove that this recurrence has the closed-form asymptotic solution
n T ( n) = Θ nρ 1+
1 f (u)
u ρ +1 du , where ρ is the unique real solution to the equation
k
i =1 ρ ai / bi = 1. In particular, the Akra-Bazzi theorem immediately implies the following form of the Master Theorem: log b a
)
if c < log b a −
Θ(n
T (n) = aT (n/ b) + nc =⇒ T (n) = 10 Θ(nc log n) if c = log b a Θ(nc )
if c > log b a + Appendix II: Solving Recurrences [Fa’10] Algorithms The Akra-Bazzi theorem does not require that the parameters ai and bi are integers, or even rationals;
ρ
on the other hand, even when all parameters are integers, the characteristic equation i ai / bi = 1 may
have no analytical solution.
Here are a few examples of recurrences that are difﬁcult (or impossible) for recursion trees, but have
easy solutions using the Akra-Bazzi theorem.
• Randomized quicksort: T (n ) = T (3n /4) + T (n /4) + n
The equation (3/4)ρ + (1/4)ρ = 1 has the unique solution ρ = 1, and therefore
n T ( n) = Θ n 1 +
1 1
u = O(n log n). du • Deterministic selection: T (n ) = T (n /5) + T (7n /10) + n
The equation (1/5)ρ + (7/10)ρ = 1 has no analytical solution. However, we easily observe that
(1/5) x + (7/10) x is a decreasing function of x , and therefore 0 < ρ < 1. Thus, we have
n
1 f (u)
uρ+1 u1−ρ n u−ρ du = du =
1 and therefore n 1−ρ =
u=1 n1−ρ − 1
1−ρ = Θ(n1−ρ ), T (n) = Θ(nρ · (1 + Θ(n1−ρ )) = Θ(n).
1 3 • Randomized search trees: T (n ) = 4 T (n /4) + 4 T (3n /4) + 1
The equation 1 ( 1 )ρ + 3 ( 3 )ρ = 1 has the unique solution ρ = 0, and therefore
44
44
n T ( n) = Θ 1 +
1 1
u du = Θ(log n). • Ham-sandwich trees: T (n ) = T (n /2) + T (n /4) + 1. Recall that we could only prove the very
weak bounds n
T ( n)
n using recursion trees. The equation (1/2)ρ + (1/4)ρ = 1 has the
unique solution ρ = log2 ((1 + 5)/2) ≈ 0.69424, which can be obtained by setting x = 2ρ and
solving for x . Thus, we have
n
1 and therefore 1
uρ+1 du = u−ρ
−ρ n =
u=1 1 − n−ρ
ρ = Θ(1) T (n) = Θ (nρ (1 + Θ(1))) = Θ(nlg φ ). The obvious advantage of the Akra-Bazzi method is that it can solve almost any divide-and-conquer recurrence with just a few lines of calculation. (There are a few nasty exceptions like T (n) = n T ( n) + n
where we have to fall back on recursion trees.) On the other hand, the steps appear to be magic, which
makes the method hard to remember, and for most divide-and-conquer recurrences, the much simpler
recursion tree method is sufﬁcient. 11 Appendix II: Solving Recurrences [Fa’10] Algorithms 5 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)
12 40 24
384
−+
n · 2n −
−
+
2n + (2 − 8 + 4) · 4
=
248
2
4
8
= ( n − 3) · 2 n + 4
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! 5.1 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)
12 Appendix II: Solving Recurrences [Fa’10] Algorithms 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 t