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REPRESENTATIVE PROBLEMS

REPRESENTATIVE PROBLEMS
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Published Date:21-07-2017
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1. REPRESENTATIVE PROBLEMS stable matching ‣ five representative problems ‣ Lecture slides by Kevin Wayne Copyright © 2005 Pearson-Addison Wesley Copyright © 2013 Kevin Wayne http://www.cs.princeton.edu/wayne/kleinberg-tardos Last updated on Mar 14, 2014, 5:36 PM1. REPRESENTATIVE PROBLEMS stable matching ‣ five representative problems ‣ SECTION 1.1Matching med-school students to hospitals Goal. Given a set of preferences among hospitals and med-school students, design a self-reinforcing admissions process. Unstable pair: student x and hospital y are unstable if: x prefers y to its assigned hospital. y prefers x to one of its admitted students. Stable assignment. Assignment with no unstable pairs. Natural and desirable condition. Individual self-interest prevents any hospital–student side deal. 3Stable matching problem Goal. Given a set of n men and a set of n women, find a "suitable" matching. Participants rank members of opposite sex. Each man lists women in order of preference from best to worst. Each woman lists men in order of preference from best to worst. favorite least favorite favorite least favorite st nd rd st nd rd 1 2 3 1 2 3 Xavier Amy Bertha Clare Amy Yancey Xavier Zeus Yancey Bertha Amy Clare Bertha Xavier Yancey Zeus Zeus Amy Bertha Clare Clare Xavier Yancey Zeus men's preference list women's preference list 4Perfect matching Def. A matching S is a set of ordered pairs m–w with m ∈ M and w ∈ W s.t. Each man m ∈ M appears in at most one pair of S. Each woman w ∈ W appears in at most one pair of S. Def. A matching S is perfect if S = M = W = n. st nd rd st nd rd 1 2 3 1 2 3 Xavier Amy Bertha Clare Amy Yancey Xavier Zeus Yancey Bertha Amy Clare Bertha Xavier Yancey Zeus Zeus Amy Bertha Clare Clare Xavier Yancey Zeus a perfect matching S = X–C, Y–B, Z–A 5Unstable pair Def. Given a perfect matching S, man m and woman w are unstable if: m prefers w to his current partner. w prefers m to her current partner. Key point. An unstable pair m–w could each improve partner by joint action. st nd rd st nd rd 1 2 3 1 2 3 Xavier Amy Bertha Clare Amy Yancey Xavier Zeus Yancey Bertha Amy Clare Bertha Xavier Yancey Zeus Zeus Amy Bertha Clare Clare Xavier Yancey Zeus Bertha and Xavier are an unstable pair 6Stable matching problem Def. A stable matching is a perfect matching with no unstable pairs. Stable matching problem. Given the preference lists of n men and n women, find a stable matching (if one exists). Natural, desirable, and self-reinforcing condition. Individual self-interest prevents any man–woman pair from eloping. st nd rd st nd rd 1 2 3 1 2 3 Xavier Amy Bertha Clare Amy Yancey Xavier Zeus Yancey Bertha Amy Clare Bertha Xavier Yancey Zeus Zeus Amy Bertha Clare Clare Xavier Yancey Zeus a perfect matching S = X–A, Y–B, Z–C 7Stable roommate problem Q. Do stable matchings always exist? A. Not obvious a priori. Stable roommate problem. 2 n people; each person ranks others from 1 to 2 n – 1. Assign roommate pairs so that no unstable pairs. st nd rd 1 2 3 no perfect matching is stable Adam B C D A–B, C–D ⇒ B–C unstable Bob C A D A–C, B–D ⇒ A–B unstable Chris A B D A–D, B–C ⇒ A–C unstable Doofus A B C Observation. Stable matchings need not exist for stable roommate problem. 8Gale-Shapley deferred acceptance algorithm An intuitive method that guarantees to find a stable matching. GALE–SHAPLEY (preference lists for men and women) ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ INITIALIZE S to empty matching. WHILE (some man m is unmatched and hasn't proposed to every woman) w ← first woman on m's list to whom m has not yet proposed. IF (w is unmatched) Add pair m–w to matching S. ELSE IF (w prefers m to her current partner m') Remove pair m'–w from matching S. Add pair m–w to matching S. ELSE w rejects m. RETURN stable matching S. ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 9Proof of correctness: termination Observation 1. Men propose to women in decreasing order of preference. Observation 2. Once a woman is matched, she never becomes unmatched; she only "trades up." 2 Claim. Algorithm terminates after at most n iterations of while loop. Pf. Each time through the while loop a man proposes to a new woman. 2 There are only n possible proposals. ▪ st nd rd th th st nd rd th th 1 2 3 4 5 1 2 3 4 5 Victor A B C D E Amy W X Y Z V Wyatt B C D A E Bertha X Y Z V W Xavier C D A B E Clare Y Z V W X Yancey D A B C E Diane Z V W X Y Zeus A B C D E Erika V W X Y Z n(n-1) + 1 proposals required 10Proof of correctness: perfection Claim. In Gale-Shapley matching, all men and women get matched. Pf. by contradiction Suppose, for sake of contradiction, that Zeus is not matched upon termination of GS algorithm. 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. ▪ 11Proof of correctness: stability Claim. In Gale-Shapley matching, there are no unstable pairs. Pf. Suppose the GS matching S does not contain the pair A–Z. Case 1: Z never proposed to A. men propose in decreasing order ⇒ Z prefers his GS partner B to A. of preference ⇒ A–Z is stable. A – Y Case 2: Z proposed to A. B – Z ⇒ A rejected Z (right away or later) women only trade up ⋮ ⇒ A prefers her GS partner Y to Z. ⇒ A–Z is stable. In either case, the pair A–Z is stable. ▪ Gale-Shapley matching S 12Summary Stable matching problem. Given n men and n women, and their preferences, find a stable matching if one exists. Theorem. Gale-Shapley 1962 The Gale-Shapley algorithm guarantees to find a stable matching for any problem instance. Q. How to implement GS algorithm efficiently? Q. If there are multiple stable matchings, which one does GS find? 13Efficient implementation 2 Efficient implementation. We describe an O(n ) time implementation. Representing men and women. Assume men are named 1, …, n. Assume women are named 1', …, n'. Representing the matching. Maintain a list of free men (in a stack or queue). Maintain two arrays wifem and husbandw. if m matched to w, then wifem = w and husbandw = m set entry to 0 if unmatched Men proposing. For each man, maintain a list of women, ordered by preference. For each man, maintain a pointer to woman in list for next proposal. 14Efficient implementation (continued) Women rejecting/accepting. Does woman w prefer man m to man m' ? For each woman, create inverse of preference list of men. Constant time access for each query after O(n) preprocessing. st nd rd th th th th th 1 2 3 4 5 6 7 8 pref 8 3 7 1 4 5 6 2 woman prefers man 3 to 6 since inverse3 inverse6 1 2 3 4 5 6 7 8 inverse th th nd th th th rd st 4 8 2 5 6 7 3 1 for i = 1 to n inverseprefi = i 15Understanding the solution For a given problem instance, there may be several stable matchings. Do all executions of GS algorithm yield the same stable matching? If so, which one? st nd rd st nd rd 1 2 3 1 2 3 Xavier Amy Bertha Clare Amy Yancey Xavier Zeus Yancey Bertha Amy Clare Bertha Xavier Yancey Zeus Zeus Amy Bertha Clare Clare Xavier Yancey Zeus an instance with two stable matching: M = A-X, B-Y, C-Z and M' = A-Y, B-X, C-Z 16Understanding the solution Def. Woman w is a valid partner of man m if there exists some stable matching in which m and w are matched. Ex. Both Amy and Bertha are valid partners for Xavier. Both Amy and Bertha are valid partners for Yancey. Clare is the only valid partner for Zeus. st nd rd st nd rd 1 2 3 1 2 3 Xavier Amy Bertha Clare Amy Yancey Xavier Zeus Yancey Bertha Amy Clare Bertha Xavier Yancey Zeus Zeus Amy Bertha Clare Clare Xavier Yancey Zeus an instance with two stable matching: M = A-X, B-Y, C-Z and M' = A-Y, B-X, C-Z 17Understanding the solution Def. Woman w is a valid partner of man m if there exists some stable matching in which m and w are matched. Man-optimal assignment. Each man receives best valid partner. Is it perfect? Is it stable? Claim. All executions of GS yield man-optimal assignment. Corollary. Man-optimal assignment is a stable matching 18Man optimality Claim. GS matching S is man-optimal. Pf. by contradiction Suppose a man is matched with someone other than best valid partner. Men propose in decreasing order of preference ⇒ some man is rejected by valid partner during GS. Let Y be first such man, and let A be the first A – Y valid woman that rejects him. B – Z Let S be a stable matching where A and Y are matched. ⋮ When Y is rejected by A in GS, A forms (or reaffirms) engagement with a man, say Z. stable matching S ⇒ A prefers Z to Y. Let B be partner of Z in S. Z has not been rejected by any valid partner because this is the first (including B) at the point when Y is rejected by A. rejection by a valid partner Thus, Z has not yet proposed to B when he proposes to A. ⇒ Z prefers A to B. Thus A–Z is unstable in S, a contradiction. ▪ 19Woman pessimality Q. Does man-optimality come at the expense of the women? A. Yes. Woman-pessimal assignment. Each woman receives worst valid partner. Claim. GS finds woman-pessimal stable matching S. Pf. by contradiction Suppose A–Z matched in S but Z is not worst valid partner for A. There exists stable matching S in which A is paired with a man, say Y, whom she likes less than Z. ⇒ A prefers Z to Y. A – Y Let B be the partner of Z in S. By man-optimality, B – Z A is the best valid partner for Z. ⋮ ⇒ Z prefers A to B. Thus, A–Z is an unstable pair in S, a contradiction. ▪ stable matching S 20