Princeton University COS 423 Theory of Algorithms Spring 2002 Kevin Wayne COS 423: Theory of...

Princeton University • COS 423 • Theory of Algorithms • Spring 2002 • Kevin Wayne

COS 423: Theory of Algorithms

Princeton University

Spring, 2001

Kevin Wayne

2

Algorithm. (webster.com) A procedure for solving a mathematical problem (as of finding the

greatest common divisor) in a finite number of steps that frequently involves repetition of an operation.

Broadly: a step-by-step procedure for solving a problem or accomplishing some end especially by a computer.

"Great algorithms are the poetry of computation."

Etymology. "algos" = Greek word for pain. "algor" = Latin word for to be cold. Abu Ja'far al-Khwarizmi's = 9th century Arab scholar.

– his book "Al-Jabr wa-al-Muqabilah" evolved into today's high school algebra text

Theory of Algorithms

3

Imagine: A World With No Algorithms

Fast arithmetic. Cryptography.

Quicksort. Databases.

FFT. Signal processing.

Huffman codes. Data compression.

Network flow. Routing Internet packets.

Linear programming. Planning, decision-making.

4

What is COS 423?

Introduction to design and analysis of computer algorithms. Algorithmic paradigms. Analyze running time of programs. Data structures. Understand fundamental algorithmic problems. Intrinsic computational limitations. Models of computation. Critical thinking.

Prerequisites. COS 226 (array, linked list, search tree, graph, heap, quicksort). COS 341 (proof, induction, recurrence, probability).

5

Administrative Stuff

Lectures: (Kevin Wayne) Monday, Wednesday 10:00 - 10:50, COS 104.

TA's: (Edith Elkind, Sumeet Sobti)

Textbook: Introduction to Algorithms (CLR).

Grading: Weekly problem sets. Collaboration, no-collaboration. Class participation, staff discretion. Undergrad / grad.

Course web site: courseinfo.princeton.edu/courses/COS423_S2001/ Fill out questionnaire.

http://courseinfo.princeton.edu/courses/COS423_S2001/

http://courseinfo.princeton.edu/courses/COS423_S2001/

6

Approximate Lecture Outline

Algorithmic paradigms. Divide-and-conquer. Greed. Dynamic programming. Reductions.

Analysis of algorithms. Amortized analysis.

Data structures. Union find. Search trees and extensions.

Graph algorithms. Shortest path, MST. Max flow, matching.

7

Approximate Lecture Outline

NP completeness. More reductions. Approximation algorithms.

Other models of computation. Parallel algorithms. Randomized algorithms.

Miscellaneous. Numerical algorithms. Linear programming.


College Admissions

Sample problem.

Algorithm.

Analysis.

References:

The Stable Marriage Problem by Dan Gusfield and Robert Irving, MIT Press, 1989.

Introduction to Algorithms by Jon Kleinberg and Éva Tardos.

9

College Admissions

Goal: Design a self-reinforcing college admissions process.

Given a set of preferences among colleges and applicants, can we assign applicants to colleges so that for every applicant X, and every college C that X is not attending, either:

C prefers every one of its admitted students to X; X prefers her current situation to the situation in which she is

attending college C.

If this holds, the outcome is STABLE. Individual self-interest prevents any applicant / college to

undermine assignment by joint action.


Love, Marriage, and Lying

Standard disclaimer.

11

Stable Matching Problem

Problem: Given N men and N women, find a "suitable" matching between men and women.

Participants rate members of opposite sex. Each man lists women in order of preference from best to worst.

Zeus Bertha AmyDiane Erika Clare

Yancey Amy ClareDiane Bertha Erika

Xavier Bertha ClareErika Diane Amy

Wyatt Diane AmyBertha Clare Erika

Victor Bertha DianeAmy Erika Clare

Man 1st 2nd 3rd 4th 5th

Men’s Preference List

worstbest

12



Participants rate members of opposite sex. Each man lists women in order of preference from best to worst. Each woman lists men in order of preference.

Erika Yancey ZeusWyatt Xavier Victor

Diane Victor YanceyZeus Xavier Wyatt

Clare Wyatt YanceyXavier Zeus Victor

Bertha Xavier YanceyWyatt Victor Zeus

Amy Zeus WyattVictor Yancey Xavier

Woman 1st 2nd 3rd 4th 5th

Women’s Preference List

worstbest

13



PERFECT MATCHING: everyone is matched monogamously. – each man gets exactly one woman– each woman gets exactly one man

STABILITY: no incentive for some pair of participants to undermine assignment by joint action.

– in matching M, an unmatched pair (m,w) is UNSTABLE if man m and woman w prefer each other to current partners

– unstable pair could each improve by dumping spouses and eloping

STABLE MATCHING = perfect matching with no unstable pairs.(Gale and Shapley, 1962)

14

Lavender assignment is a perfect matching.Are there any unstable pairs?

Men’s Preference List Women’s Preference List

Zeus

Yancey

Xavier

Man

A

B

A

1st

B

A

B

2nd

C

C

C

3rd

Clare

Bertha

Amy

Woman

X

X

Y

1st

Y

Y

X

2nd

Z

Z

Z

3rd

Yes. Bertha and Xavier form an unstable pair. They would prefer each other to current partners.

B

X

Example

15

Example

Green assignment is a stable matching.

A

B

A

C X

X

Y

Y

Y

X

Z

Z

Z


Zeus

Yancey

Xavier

Man 1st

B

A

B

2nd

C

C

3rd

Clare

Bertha

Amy

Woman 1st 2nd 3rd

16

Example

Orange assignment is also a stable matching.

A X

X

Y

Y

Z

Z


Zeus

Yancey

Xavier

Man 1st

A

B

2nd

C

C

3rd

Clare

Bertha

Amy

Woman 1st 2nd 3rd

B

A

B

C

X

Y

Z

17

B

Preference List

Bob

Chris

Adam C

A

B

D

D

Doofus A B C

D

C

A

Stable Roommate Problem

Not obvious that any stable matching exists.

Consider related "stable roommate problem." 2N people. Each person ranks others from 1 to 2N-1. Assign roommate pairs so that no unstable pairs.

C

A

B

D

No stable matching.

All 3 perfect matchings have unstable pair.

E.g., A-C forms unstable pair in lavender matching.

C

A

1st 2nd 3rd

18

Propose-And-Reject Algorithm

Intuitive method that guarantees to find a stable matching.

Initialize each person to be free.

while (some man m is free and hasn't proposed to every woman)

w = first woman on m's list to whom m has not yet proposed

if (w is free)

assign m and w to be engaged

else if (w prefers m to her fiancé m')

assign m and w to be engaged, and m' to be free

else

w rejects m

Gale-Shapley Algorithm (men propose)

19

Implementation and Running Time Analysis

Engagements. Maintain two arrays wife[m], and husband[w]; set equal to 0 if

participant is free. Store list of free men on a stack (queue).

Preference lists. For each man, create a linked list of women, ordered from favorite to

worst.– men propose to women at top of list, if rejected goto next

For each woman, create a "ranking array" such that mth entry in array is woman's ranking of man m.

– allows for queries of the form: does woman w prefer m to m' ?

Resource consumption. Time = (N2). Space = (N2). Optimal.

wmwm to proposed has :),(

20

Men's Preference List

Wyatt

Victor

Man 1st

A

B

2nd

C

D

3rd

C

B

AZeus

Yancey

Xavier C

D

A

B

B

A

D

C

4th

E

E

5th

A

D

E

E

D

C

B

E

Women's Preference List

Bertha

Amy

Man 1st

W

X

2nd

Y

Z

3rd

Y

X

VErika

Diane

Clare Y

Z

V

W

W

V

Z

X

4th

V

W

5th

V

Z

X

Y

Y

X

W

Z

A Worst Case Instance

Number of proposals n(n-1) + 1. Algorithm terminates when last woman gets first proposal.

Number of proposals = n(n-1) + 1 for following family of instances.

21

Proof of Correctness

Observation 1. Men propose to their favorite women first.

Observation 2. Once a woman is matched, she never becomes unmatched. She only "trades up."

Fact 1. All men and women get matched (perfect). Suppose upon termination Zeus is not matched. Then some woman, say Amy, is not matched upon termination. By Observation 2, Amy was never proposed to. But, Zeus proposes to everyone, since he ends up unmatched.

(contradiction)

22

Proof of Correctness

Observation 1. Men propose to their favorite women first.

Observation 2. Once a woman is matched, she never becomes unmatched. She only "trades up."

Fact 2. No unstable pairs. Suppose (Amy, Zeus) is an unstable pair: each prefers

each other to partner in Gale-Shapley matching S*.

Case 1. Zeus never proposed to Amy. Zeus must prefer Bertha to Amy (Observation 1) (Amy, Zeus) is stable. (contradiction)

Case 2. Zeus proposed to Amy. Amy rejected Zeus (right away or later) Amy prefers Yancey to Zeus (women only trade up) (Amy, Zeus) is stable (contradiction)

Bertha-Zeus

Amy-Yancey

S*

23

Understanding the Solution

For a given problem instance, there may be several stable matchings. Do all executions of Gale-Shapley yield the same stable matching?

If so, which one?

Fact 3. Yes. Gale-Shapley finds MAN-OPTIMAL stable matching! Man m is a valid partner of woman w if there exists some stable

matching in which they are married. Man-optimal assignment: every man receives best valid partner.

– simultaneously best for each and every man– there is no stable matching in which any single man individually

does better

24

Proof of Fact 3

Proof. Suppose, for sake of contradiction, some man is

paired with someone other than best partner.– since men propose in decreasing order of

preference, some man is rejected by valid partner Let Yancey be first such man, and let Amy be first valid partner

that rejects him. When Yancey is rejected, Amy forms (or reaffirms) engagement

with man, say Zeus, whom she prefers to Yancey. Let Bertha be Zeus' partner in S. Zeus not rejected by any valid partner at the point when Yancey is

rejected by Amy (since Yancey is first to be rejected by valid partner). Thus, Zeus prefers Amy to Bertha.

But Amy prefers Zeus to Yancey. Thus (Amy, Zeus) is unstable pair in S.

Bertha-Zeus

Amy-Yancey

S

25


Fact 4. Gale-Shapley finds WOMAN-PESSIMAL matching. Each woman married to worst valid partner.

– simultaneously worst for each and every woman.– there is no stable matching in which any single woman

individually does worse

Proof. Suppose (Amy, Zeus) matched in S*, but Zeus is not worst valid

partner for Amy. There exists stable matching S in which Amy is paired with man,

say Yancey, whom she likes less than Zeus. Let Bertha be Zeus' partner in S. Zeus prefers Amy to Bertha (man optimality). (Amy, Zeus) form unstable pair in S. Bertha-Zeus

Amy-Yancey

S

26


Fact 5. The man-optimal stable matching is weakly Pareto optimal. There is no other perfect matching (stable or unstable), where

every man does strictly better.

Proof. Let Amy be last woman in some execution of Gale-Shapley (men

propose) algorithm to receive a proposal. No man is rejected by Amy since algorithm terminates when last

woman receives first proposal. No man matched to Amy will be strictly better off than in man-

optimal stable matching.

27

Extensions: Unacceptable Partners

Yeah, but in real-world every woman is not willing to marry every man, and vice versa?

Some participants declare others as "unacceptable."(prefer to be alone than with given partner)

Algorithm extends to handle partial preference lists.

Matching S unstable if there exists man m and woman w such that: m is either unmatched in S, or strictly prefers w to his partner in S w is either unmatched in S, or strictly prefers m to her partner in S.

Fact 6. Men and women are each partitioned into two sets: those that have partners in all stable matchings; those that have partners in none.

28

Extensions: Sets of Unequal Size

Also, there may be an unequal number of men and women. E.g., |M| = 100 men, |W| = 90 women. Algorithm extends. WLOG, assume |W| < |M|.

Matching S unstable if there exists man m and woman w such that: m is either unmatched in S, or strictly prefers w to his partner in S; w is either unmatched in S, or strictly prefers m to her partner in S.

Fact 7. All women are matched in every stable matching. Men are partitioned into two subsets:

men who are matched in every stable matching; men who are matched in none.

29

Extensions: Limited Polygamy

What about limited polygamy? E.g., Bill wants 3 women. Algorithm extends.

Matching S unstable if there exists man m and woman w such that: either w is unmatched, or w strictly prefers m to her partner; either m does not have all its "places" filled in the matching, or m

strictly prefers w to at least one of its assigned residents.

30

Application: Matching Residents to Hospitals

Sets of unequal size, unacceptable partners, limited polygamy.

Matching S unstable if there exists hospital h and resident r such that: h and r are acceptable to each other; either r is unmatched, or r prefers h to her assigned hospital; either h does not have all its places filled in the matching, or h

prefers r to at least one of its assigned residents.

31

Application: Matching Residents to Hospitals

Matching medical school residents to hospitals. (NRMP) Hospitals ~ Men (limited polygamy allowed). Residents ~ Women. Original use just after WWII (predates computer usage). Ides of March, 13,000+ residents.

Rural hospital dilemma. Certain hospitals (mainly in rural areas) were unpopular and

declared unacceptable by many residents. Rural hospitals were under-subscribed in NRMP matching. How can we find stable matching that benefits "rural hospitals"?

Rural Hospital Theorem: Rural hospitals get exactly same residents in every stable

matching!

32

Deceit: Machiavelli Meets Gale-Shapley

Is there any incentive for a participant to misrepresent his/her preferences?

Assume you know men’s propose-and-reject algorithm will be run. Assume that you know the preference lists of all other participants.

Fact 8. No, for any man yes, for some women!

A X

X

Y

Y

Z

Z

Men’s Preference List Women’s True Preference List

Zeus

Yancey

Xavier

Man 1st

A

B

2nd

C

C

3rd

Clare

Bertha

Amy

Woman 1st 2nd 3rd

B

A

B

C

X

Y

Z

X

Z

Y

Y

Z

X

Amy Lies

Clare

Bertha

Amy

Woman 1st 2nd 3rd

X

Y

Z

X

Y

Z

X

Y

Z

33

Lessons Learned

Powerful ideas learned in COS 423. Isolate underlying structure of problem. Create useful and efficient algorithms.

Sometimes deep social ramifications. Historically, men propose to women. Why not vice versa? Men: propose early and often. Men: be more honest. Women: ask out the guys. Theory can be socially enriching and fun! CS majors get the best partners!!!


Love, Marriage, and Lying: Extra Slides

35

Example With a Unique Stable Matching

Red matching is unique stable matching.

C Y

Z

X

X

Y

X


Zeus

Yancey

Xavier

Man 1st

B

B

2nd

C

A

3rd

Clare

Bertha

Amy

Woman 1st 2nd 3rd

B

C

A

A

Z

Y

Z

36

How to Represent Men and Women

Represent men and women as integers between 0 and N-1. 0 through N-1 since C array indices start at 0. Could use struct if we want to carry around more information,

e.g., name, age, astrological sign.

37

How to Represent Marriages

Use array to keep track of marriages.

int wife[N];

int husb[N];

for (m = 0; m < N; m++)

wife[m] = -1;

for (w = 0; w < N; w++)

husb[w] = -1;

unmatched man if1

womanto matched man if][wife

m

wmwm

unmatched w womanif1

womanto matched man if][husb

wmmw

38

Filling in Some of the Code

while (marriages < N)

for (m = 0; wife[m] != -1; m++) ;

while (-1 == wife[m]) let w be man m’s favorite women to whom he has not yet proposed

if (-1 == husb[w]) husb[m] = w; wife[w] = m; marriages++;

else if (w prefers m to current fiancé) m' = husb[w]; wife[m'] = -1; husb[m] = w; wife[w] = m;

if (w unmatched) m and w get engaged

while (m unmatched)

m' = current fiancé of wm' now unmatchedm and w get engaged

find unmatched man

39


Man 0th 1st 2nd 3rd 4th

0 1 0 3 4 21 3 1 0 2 42 1 4 2 3 03 0 3 2 1 44 1 3 0 4 2

Representing the Preference Lists

Use 2D-array to represent preference lists. 2D-array is array of arrays. mp[m][i] = w if man m’s ith favorite woman is w. wp[w][i] = m if woman w’s ith favorite man is m.

int mp[N][N];

int wp[N][N];

mp[1][0] = 3man 1 likes woman 3 the best

40

Initializing the Preference Lists

Could read from stdin.

We’ll assign random lists for each man and woman. Use randomPermutation function from Lecture P2. Need N random permutations for men, and N for women.

int mp[N][N];

int wp[N][N];

for (m = 0; m < N; m++) randomPermutation(mp[m], N);

for (w = 0; w < N; w++) randomPermutation(wp[w], N);

mp[m] is man m’s preference list array.

41

Dumping

Does woman w=2 prefer man m1=3 to man m2=0?

Yes, m1 appears on woman w’spreference list before m2.

for (i = 0; i < N; i++) {

if (wp[w][i] == m1) YES

if (wp[w][i] == m2) NO

}

search preference list sequentially until m1 or m2 found


Woman 0th 1st 2nd 3rd 4th

0 4 0 1 3 21 2 1 3 0 42 1 2 3 4 03 0 4 3 2 14 3 1 4 2 0

42

Keeping Track of Men’s Proposals

Unmatched man proposes to most favorable woman to whom he hasn’t already proposed.

How do we keep track of which woman a man has proposed to? Men propose in decreasing order of preference. Suffices to keep track of number of proposals in array.

propose[m] = i if man m has proposed to i woman already.

initialize arrayint props[N];

for (i = 0; i < N; i++) props[i] = 0;

43

Keeping Track of Men’s Proposals

Unmatched man proposes to most favorable woman to whom he hasn’t already proposed.

How do we keep track of which woman a man has proposed to? Men propose in decreasing order of preference. Suffices to keep track of number of proposals in array.

propose[m] = i if man m has proposed to i woman already.

while (-1 == wife[n]) {

w = mp[m][props[m]];

props[m]++; . . .}

props[m] is ranking of next woman on preference list

make next proposal to woman mp[m][props[m]]

find next woman to propose to

44

Try Out The Code

#include <stdio.h>#include <stdlib.h>#include <assert.h>#define N 500

int main(void) { int mp[N][N]; /* mp[m][i] = w if man m's ith favorite woman is w */ int wp[N][N]; /* wp[w][i] = m if woman w's ith favorite man is m */ int wife[N]; /* wife[m] = w if m married to w */ int husb[N]; /* husb[w] = m if m married to w */ int props[N]; /* props[m] = i if man m has proposed to i women */

int marriages = 0; /* number of couples matched so far */ int m, w;

/* initialize men */ for (m = 0; m < N; m++) { props[m] = 0; wife[m] = -1; randomPermutation(mp[m], N); }

/* initialize women */ for (w = 0; w < N; w++) { husb[w] = -1; randomPermutation(wp[w], N); }

marriage.c

45

Try Out The Code

while (marriages < N) { /* find first unmatched man */ for (m = 0; wife[m] != -1; m++) ;

printf("man %d proposing:\n", m); /* propose to next women on list until successful */ while (-1 == wife[m]) { w = mp[m][props[m]]; printf(" to woman %d", w); props[m]++; /* woman w unmatched */ if (-1 == husb[w]) { printf(" accepted\t(woman %d previously unmatched)\n", w); husb[w] = m; wife[m] = w; marriages++; }

/* woman w prefers m to current mate */ else if (wr[w][m] < wr[w][husb[w]]) { printf(" accepted\t(woman %d dumps man %d)\n", w, husb[w]); wife[husb[w]] = -1; husb[w] = m; wife[m] = w; } /* otherwise m rejected by w */ else printf(" rejected\t(woman %d prefers %d)\n", w, husb[w]); } }

marriage.c

46

Try Out The Code

Observation: code isREALLY slow for large N.

printf("Stable matching\n"); for (m = 0; m < N; m++) printf("%5d %5d\n", m, wife[m]);

return 0;}

marriage.c% gcc marriage.c% a.out

man 0 proposing: to woman 4 accepted (woman 4 previously unmatched)man 1 proposing: to woman 0 accepted (woman 0 previously unmatched)man 2 proposing: to woman 2 accepted (woman 2 previously unmatched)man 3 proposing: to woman 2 rejected (woman 2 prefers 2) to woman 3 accepted (woman 3 previously unmatched)man 4 proposing: to woman 2 accepted (woman 2 dumps man 2)man 2 proposing: to woman 3 accepted (woman 3 dumps man 3)man 3 proposing: to woman 0 rejected (woman 0 prefers 1) to woman 4 rejected (woman 4 prefers 0) to woman 1 accepted (woman 1 previously unmatched)

Stable matching 0 4 1 0 2 3 3 1 4 2

Unix

47

Why So Slow?

Does woman w=2 prefer man m1=3 to man m2=0?

Need to search through row 2 to find answer.This is repeated many times.If N is large, this can be very expensive.


Woman 0th 1st 2nd 3rd 4th

0 4 0 1 3 21 2 1 3 0 42 1 2 3 4 03 0 4 3 2 14 3 1 4 2 0

48


Man 0th 1st 2nd 3rd 4th

0 1 0 3 4 21 3 1 0 2 42 1 4 2 3 03 0 3 2 1 44 1 3 0 4 2

mp[1][0] = 3man 1 likes woman 3 best

Men’s Rankings

Man 0 1 2 3 4

0 1st 0th 4th 2nd 3rd

1 2nd 1st 3rd 0th 4th

2 4th 0th 2nd 3rd 1st

3 0th 3rd 2nd 1st 4th

4 2nd 0th 4th 1st 3rd

mr[1][3] = 0man 1 likes woman 3 best

An Auxiliary Data Structure

Create a 2D array that stores men’s ranking of women. mr[m][w] = i if man m’s ranking of woman w is i. wr[w][m] = i if woman w’s ranking of man m is i.

49

Men’s Rankings

Man 0 1 2 3 4

0 1st 0th 4th 2nd 3rd

1 2nd 1st 3rd 0th 4th

2 4th 0th 2nd 3rd 1st

3 0th 3rd 2nd 1st 4th

4 2nd 0th 4th 1st 3rd

mr[m][w1] = 2mr[m][w2] = 4

if (mr[m][w1] < mr[m][w2])

YES

else

NO

An Auxiliary Data Structure

Create a 2D array that stores men’s ranking of women. mr[m][w] = i if man m’s ranking of woman w is i. wr[w][m] = i if woman w’s ranking of man m is i.

Does man m = 3 prefer woman w1 = 2 to woman w2 = 4?

50

Check if Marriage is Stable

Check if husb[N] and wife[N] correspond to a stable marriage. Good warmup and useful for debugging. Check every man-woman pair to see if they’re unstable. Use ranking arrays.

int isStable(int husb[], int wife[],

int mr[N][N], int wr[N][N]) {

int m, w;

for (m = 0; m < N; m++)

for (w = 0; w < N; w++)

if (mr[m][w] < mr[m][wife[m]]) &&

(wr[w][m] < wr[w][husb[w]])

return 0;

return 1;

}

isStable

m prefers w to current wife

w prefers m to current husband

51

Check if Marriage is Stable

Check if husb[N] and wife[N] correspond to a stable marriage. Good warmup and useful for debugging. Check every man-woman pair to see if they’re unstable. Use ranking arrays.

Time/space tradeoff for using auxiliary ranking arrays. Disadvantage: requires twice as much memory (storage). Advantage: dramatic speedup in running time

(using 400 MHz Pentium II with N = 10,000). 1 second using ranking arrays. 2 hours by searching preference list sequentially!

Princeton University COS 423 Theory of Algorithms Spring 2002 Kevin Wayne COS 423: Theory of...

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