Ma/CS 6b Class 3: Stable Matchings

Transcription

1 Ma/CS 6b Class 3: Stable Matchings α p 5 p 12 p 15 q 1 q 7 q 12 By Adam Sheffer Reminder: Alternating Paths Let G = V 1 V 2, E be a bipartite graph, and let M be a matching of G. A path is alternating for M if it starts with an unmatched vertex of V 1 and every other edge of it is in M. p 0 q 1 p 5 q 7 * Notice that this is definition is different than the one from 6a. p 12 p 15 q 12 q 17 1

2 Reminder: Augmenting Paths Let G = V 1 V 2, E be a bipartite graph, and let M be a matching of G. A path is augmenting for M if it is an alternating path of M, and it ends in an unmatched vertex. p 0 p 5 p 12 p 15 q 1 q 7 q 12 q 17 Neighbor Sets Let G = V 1 V 2, E be a bipartite graph. For any vertex a V 1, we define the neighbor set of a as N a = u V 2 a, u E. N a N b = *u, v+ = *v, w+ a b V 1 V 2 u v w x 2

3 More Neighbor Sets Let G = V 1 V 2, E be a bipartite graph. For any subset A V 1, we define N A = y V 2 x, y E for some x A. N b, c, d = *u, v, w+ N a, e = *u, w, x+ a b c d e V 1 V 2 u v w x Hall's Marriage Theorem Theorem. Let G = V 1 V 2, E be a bipartite graph. There exists a matching of size V 1 in G if and only if for every A V 1, we have A N A. We say that G satisfies Hall s condition if for every A V 1, we have A N A. Philip Hall 3

4 Neighbor Sets and Perfect Matchings Explain why there s no perfect matching in this graph: a b c d u v w x N b, c, d = v, w, so we cannot find a match for all three vertices b, c, d. Proving Hall s Theorem Easy direction. If there exists a subset A V 1 such that A > N A, then there is no matching of size V 1 in G. It remains to prove. If every subset A V 1 satisfies A N A, then there is a matching of size V 1 in G. We now present two different proofs. 4

5 Proof by Contradiction M a maximum matching of G. Assume, for contradiction, that there is a vertex α V 1 that is not matched in M. A the set of vertices a V 1 α such that there is an alternating path from α to a. B V 2 the set of penultimate vertices of an alternating path from α to a vertex of V 1. By definition, every last edge in such a path is in M. Thus, the edges of M contain a matching between A and B, and A = B. An Illustration A V 1 last vertices in alternating path from α. B V 2 the penultimate vertices in these paths. The two last vertices of every path are connected by an edge of M. Thus, A = B. α q 1 p 5 q 7 A = p 5, p 12, p 15 B = q 1, q 7, q 12 p 12 p 15 q 12 q 17 5

6 Proof by Contradiction (cont.) By Hall s condition N A α > A. Thus, there is a vertex B that is adjacent to a vertex of α A α. The vertex is either matched or not. We now show that both cases lead to a contradiction. α q 1 p 5 q 7 A = p 5, p 12, p 15 B = q 1, q 7, q 12 p 12 p 15 q 12 The Case Where is Unmatched If is unmatched, then is not on the alternating path P from α to α. The path P + α, is an augmenting path, contradicting the maximality of M. α q 1 p 5 q 7 p 12 α q 12 6

7 The Case Where is Matched Assume that is matched. Then is the penultimate vertex on an alternating path from α to α contradicting B. α q 1 p 5 α p 15 q 12 q 7 Proof by Induction We prove by induction on V 1 that If every subset A V 1 satisfies A N A, then there is a matching of size V 1 in G. Induction basis. When V 1 obviously holds. = 1 the claim Induction step. We consider two cases: Every nonempty A V 1 satisfies A < N A. There exists A V 1 such that A = N A. 7

8 The First Case Assume that every nonempty A V 1 satisfies A < N(A). Consider an edge a, b E. Remove a and b from G to obtain a graph G = V 1 V 2, E. Every nonempty A V 1 satisfies A N(A). By the induction hypothesis, G contains a matching of size V 1 = V 1 1. Adding the edge a, b to the matching yields a matching of size V 1 in G. The Second Case Assume that there exists a non-empty A V 1 such that A = N A. By the hypothesis, the induced subgraph on A N A contains a matching of size A. Consider the induced subgraph G on V 1 A and V 2 N A. For any subset A V 1 A, since in G A A N(A) N A, then in G A N A. By the hypothesis, G contains a matching of size V 1 A. Thus, G contains a matching of size V 1. 8

9 Illustration: The Induction Step The red vertices in V 1 form a subset A such that A = N a. We find a matching between the red vertices and another between the blue vertices. V 1 V 2 V 1 V 2 Reminder: National Resident Matching Program Every medical student who is about to graduate ranks hospitals in which she wants to do her residency. Every hospital ranks students that it is interested in. Every year, over 20,000 applicants apply to about 1,800 programs. How can we handle this? 9

10 The Corresponding Graph Bipartite graph a vertex in V 1 for each student. A vertex in V 2 for each hospital. An edge between a student and a hospital that are interested in each other. A matching corresponds to assigning students to hospitals. The Problem Problem. We did not consider the rankings of the students and hospitals. We might have chosen the red matching. A B α However, perhaps student A prefers hospital, student B prefers hospital α, and similarly for the hospitals. 10

11 Unstable Matchings We have a bipartite graph G = V 1 V 2, E such that each vertex has a ranking of the vertices that it would like to be connected to. We say that a matching M of G is unstable if there exists an edge a, b E that is not in M and that both a and b prefer to be matched to each other than to their current match (or are unmatched). Etymology Why is this called unstable? The problem was originally formulated for matching men and women to be married. Man A and woman A are married. Man B and woman B are married. If man A prefers woman B and woman B prefers man A, these are unstable marriages! 11

12 Example: An Unstable Matching Rankings: A: 1. α γ. B: 1. α γ. C: α 3. γ α: 1. A 2. B 3. C : 1. A 2. B 3. C γ: 1. A 2. B 3. C A B C α γ The red edges form an unstable matching since both A and α prefer the edge A, α. Stable Marriage Theorem Theorem. For any bipartite graph G = (V 1 V 2, E) with sets of preferences for each vertex, there exists a stable matching. Proven in 1962 by Gale and Shapley. In 2012, the Nobel Prize in Economics was awarded to Shapley for the theory of stable allocations and the practice of market design. 12

13 A Better Matching We say that a matching M in G is better than a matching M, if every vertex of V 2 prefers its matching vertex in M to its matching vertex in M (or has the same matching vertex in both). The matching on the left is better iff α prefer A to B. A α A α B B C γ C γ Proof Idea We prove the stable marriage theorem by showing that for every unstable matching M i, there exists a better matching M i+1. We start with an empty matching M 0 and repeatedly find a better matching. Starting from an empty matching, a better matching can be found at most V 1 V 2 times, so we eventually obtain a stable matching. 13

14 Acceptable Vertices Given a matching M i, we say that a V 1 is acceptable to b V 2 if a, b E M i, and b prefers a to its current match in M. In the figure, B is acceptable to if prefers B to C. A A student S is acceptable to a hospital if it likes S better than its current student. B C α γ Happy Vertices We say that a vertex a V 1 is happy with a matching M i if either a is unmatched in M i, or M i contains the edge a, b, and a prefers b to all of the vertices of V 2 that finds a acceptable. Intuitively, a student is happy if she is assigned to her favorite hospital out of the ones that like her better than their current choice (or is unmatched). 14

15 Building Happy Matchings We only consider matchings where every vertex of V 1 is happy. This is the case for the empty M 0. Given a matching M i : Consider an unmatched vertex a V 1 and let b denote a s favorite among the vertices of V 2 for which a is acceptable. We set M i+1 as M i a, b (possibly also removing an edge a, b M i ). Notice that M i+1 is better than M i and that every vertex of V 1 is happy. Final Details Given a matching M i : What if every unmatched vertex a V 1 has no vertex in V 2 for which a is acceptable. Then M i is a stable matching! What if there are no unmatched vertices in M i? Since every vertex of V 1 is happy, this again means that M i is stable. 15

16 Finding a Stable Matching The proof of the stable marriage theorem presents us with an algorithm for finding a stable matching: Build a sequence of matchings M i such that each is better than the previous one, until we obtain a stable matching. At each step the algorithm picks an unmatched element of V 1 and finds the best match for it (not paying much attention to the preferences of the vertices of V 2 ). The End 16