121
Spring CIS 2010
Data Structures and Introduction to Algorithms using JAVA
Lecture Notes 1: Introduction, GCD programs
1999-2010 S.Kannan, V.Tannen, S.Guha & K.Daniilidis 2006 M.Goodrich & R.Tamassia (with permission)
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Website: Textbook:
(RECOMMENDED)
www.seas.upenn.edu/~cis121 Data Structures and Problem Solving using Java THIRD EDITION M. A. Weiss Addison Wesley 2006 ISBN 0-321-32213-4
Grade
Hwk Labs Mid 1 Mid 2 Final 30% 10% 15% 15% 30%
Midterm1: Wed Feb 24 (rooms TBA) Midterm2: Wed Mar 31 (rooms TBA) Final: Mon May 10 (rooms TBA) Homework: (1-2 weeks long)
- programming exercises - problems sets (math-style) - programming project (~3 weeks) No collaboration (see policy on website)
Labs (recitations): attendance and participation are mandatory. They count towards 10% of the grade.
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What is this course about? Problem solving with Java programs (as in cis120)
Data structures (mostly!)
Intro to analysis of algorithms (more algorithms in cis320) Some object-oriented design patterns Some software development good practices .. build better software!
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What makes a program better?
Criteria for comparison? correctness: Programs are designed with a goal in mind. Is this goal (specification) met? efficiency: How long does the program run? What is the response time? (experimental test and/or theoretical analysis) Is the program easy to use? (users cant be underestimated) Will the program be easy to change? (requirements/specifications inevitably change)
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good engineering:
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An introductory example
Problem: compute the greatest common divisor (gcd) of two integers, x and y.
Recall some facts about divisibility of integers Then think about a solution!
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Computing the GCD Idea 1: Enumerate all the numbers smaller than x and y and keep only those that divide both x and y. Then we find the greatest of these and we are done. This seems inefficient since we are not asked for all common divisors, just the greatest. Idea 2: We realize that we can be clever and go through the numbers down instead of up ! The following program implements this idea:
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Computing the GCD
public static int Dgcd(int x, int y) { int i = 0; if (x > y) i = y; else i = x; if (i == 0) return 0; while ((x%i != 0) || (y%i != 0)) // descend while i i--; // does not divide // both x and y return i; }
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// i is set to smallest // of x and y
Hermiones discovery
Idea 3: Hermione Granger points out an algorithm that she discovered in some old medieval manuscript:
public static int Hgcd (int x, int y) { if (x == y) return y; else if (x > y) return Hgcd(x-y, y); // this looks else return Hgcd(x, y-x); // smarter! }
Note that Hgcd is a recursive method (it calls itself).
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Euclids own
Idea 0: Actually, about 2300 years ago Euclid recorded another algorithm for computing the GCD:
public static int Egcd (int x, int y) { if (y == 0) return x; else return Egcd(y, x%y); // the remainder } // !!!
(We thought we were clever with our idea of iterating down through the numbers, but this one does look like magic!)
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Have we solved the problem? Are these programs correct, efficient, well-engineered? Which one is better? These are the kind of questions that this course should help us answer. The GCD problem seemed simple but we shall see that there is a lot more to it than meets the eye. Mathematical correctness: prove that Euclids algorithm nds indeed gcd(a,b) (you may have done this in cse260).
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Correctness experiments (tests)
We run the programs on a few inputs. Here are the results ( means loops or diverges):
Dgcd
Didnt think about negative inputs, eh?
Hgcd
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Egcd
(The last are two explained by Javas integer arithmetic. Division truncates toward 0 i.e. (-8)/3 is -2 not -3. Consequently, to maintain (x/y)*y + x%y == x true, in Java (-8)%3 is -2 and not, as mathematical texts define it, 1.)
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Need for specifications
We made correctness claims about the outputs of the experiments but we never said --- correct with respect to what?
The first moral of the GCD story: After you hear a problem statement and before you begin to code STOP! and make sure you really have a good problem specification !
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Since we have serious problems with inputs that are negative or zero After-the-fact, cop-out specification: If x,y are (strictly) positive integers then gcd(x,y) returns the biggest integer that divides both x and y. Otherwise,the call to gcd(x,y) may result in any behavior (including divergence!) All three programs are correct with respect to this badly engineered specification (and they are badly engineered themselves).
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The good GCD specification
We need the specification for the GCD problem. It should be formulated without thinking about the programs we already wrote. But is the gcd problem well-defined? Does it always have an answer and is the answer unique? After a bit of math (see supplemental slides at end), the resulting (and well-engineered) specification is: If x=y=0 then gcd(x,y) returns (by convention) 0. Otherwise, gcd(x,y) returns the biggest (positive) integer that divides both x and y.
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Efficiency experiments (tests)
What if some of the programs run faster on certain inputs for some reason? We will compare the three programs experimentally on randomly chosen inputs!
import java.util.Random; public class Test { public static int Dgcd(int x, int y) { // copy definition from before } // similarly copy Hgcd and Egcd // continued on next page
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public static void main(String argv[]) { long seed = 15498243; // A little Random rand = new Random(seed); // magic... long startWatch = System.currentTimeMillis(); for (int i = 1; i < 101; i++) { int a = Math.abs(rand.nextInt() % 723431); int b = Math.abs(rand.nextInt() % 723431); int dummy = Dgcd(a,b);} // Then same for Hgcd and Egcd long stopWatch = System.currentTimeMillis(); System.out.println(stopWatch - startWatch); } }
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Results from timing measurements of the three GCD Programs Hgcd
Upper bound on a,b ~500 ~5,000 ~75,000 ~750,000 Time(ms) /100 pairs(a,b) Hgcd 84 96 102 117 Egcd 7 8 9 10 Dgcd 30 194 4,682 49,729
Dgcd Egcd
Looks like Egcd is very good, Hgcd is OK, and D(umb!)gcd is really bad for large numbers. But this is not the whole story (see the graph)!
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running time
input size (#bits)
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The second moral of the GCD story After you have a problem specification and before you begin to code STOP!
and consider the relative efficiency of different algorithms that solve your problem. How to do this is, like, totally, non-obvious. This course will introduce you to the necessary ideas, math and techniques.
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Supplement: some facts about divisibility
Recall that: any integer divides 0 1 divides any integer the absolute value of an integer is
public static int abs(int x) { if (x < 0) return -x; else return x; }
m divides n if and only if abs(m) divides abs(n)
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Supplement: the GCD specification is well-defined Consider the set { all integers that divide both x and y }
this set contains at least 1 and -1 all its elements are in the interval [-k,k] where k = min(abs(x),abs(y)) therefore this set has a largest element, except when x = y = 0 ( then the set consists of all integers!) The convention gcd(0,0)=0 is made in order to extend uniformly properties such as gcd(x,0)=x.
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