Bipartite Matching

Unweighted Bipartite Matching

Definitions Matching Free Vertex

Definitions Maximum Matching: matching with the largest number of edges

Definition Note that maximum matching is not unique.

Intuition Let the top set of vertices be men Let the bottom set of vertices be women Suppose each edge represents a pair of man and woman who like each other Maximum matching tries to maximize the number of couples!

Applications Matching has many applications. This lecture lets you know how to find maximum matching.

Alternating Path Alternating between matching and non-matching edges. a b cde f g hij d-h-e: alternating path a-f-b-h-d-i: alternating path starts and ends with free vertices f-b-h-e: not alternating path e-j: alternating path starts and ends with free vertices

Idea “Flip” augmenting path to get better matching Note: After flipping, the number of matched edges will increase by 1! 

Idea Theorem (Berge 1975): A matching M in G is maximum iff There is no augmenting path Proof: (  ) If there is an augmenting path, clearly not maximum. (Flip matching and non-matching edges in that path to get a “better” matching!)

Proof for the other direction (  ) Suppose M is not maximum. Let M’ be a maximum matching such that |M’|>|M|. Consider H = M  M’ = (M  M’)-(M  M’) i.e. a set of edges in M or M’ but not both H has two properties:  Within H, number of edges belong to M’ > number of edges belong to M.  H can be decomposed into a set of paths. All paths should be alternating between edges in M and M’. There should exist a path with more edges from M’. Also, it is alternating.

Idea of Algorithm Start with an arbitrary matching While we still can find an augmenting path  Find the augmenting path P  Flip the edges in P

Breadth-First Search Algorithm for Augmented Path Use Breadth-First Search: LEVEL(0) = some unmatched vertex r for odd L > 0, LEVEL(L) = {u|{v,u}  E – M when v  LEVEL(L -1) and when u in no lower level} For even L > 0, LEVEL(L) = {u|{v,u}  M when v  LEVEL(L -1) and u in no lower level} Assume G is bipartite graph with matching M.

Proof of Breadth-First Search Algorithm for Augmented Path Cases (1) If for some odd L >0, LEVEL(L) contains an unmatched vertex u then the Breadth First Search tree T has an augmenting path from r to u (2) Otherwise no augmenting path exists, so M is maximal.

Labelling Algorithm Start with arbitrary matching

Labelling Algorithm Pick a free vertex in the bottom

Labelling Algorithm Run BFS

Labelling Algorithm Alternate unmatched/matched edges

Labelling Algorithm Until a augmenting path is found

Augmenting Tree

Flip!

Repeat Pick another free vertex in the bottom

Repeat Run BFS

Repeat Flip

Answer Since we cannot find any augmenting path, stop!

Overall algorithm Start with an arbitrary matching (e.g., empty matching) Repeat forever  For all free vertices in the bottom, do bfs to find augmenting paths  If found, then flip the edges  If fail to find, stop and report the maximum matching.

Time analysis We can find at most |V| augmenting paths (why?) To find an augmenting path, we use bfs! Time required = O( |V| + |E| ) Total time: O(|V| 2 + |V| |E|)

Improvement We can try to find augmenting paths in parallel for all free nodes in every iteration. Using such approach, the time complexity is improved to O(|V| 0.5 |E|)

Weighted Bipartite Graph

Weighted Matching Score: 6+3+1=10

Maximum Weighted Matching Score: =13

Augmenting Path (change of definition) Any alternating path such that total score of unmatched edges > that of matched edges The score of the augmenting path is  Score of unmatched edges – that of matched edges Note: augmenting path need not start and end at free vertices!

Idea for finding maximum weight matching Theorem: Let M be a matching of maximum weight among matchings of size |M|.  If P is an augmenting path for M of maximum weight,  Then, the matching formed by augmenting M by P is a matching of maximum weight among matchings of size |M|+1.

Overall Algorithm Start with an empty matching Repeat forever  Find an augmenting path P with maximum score  If the score > 0, then flip the edges  Otherwise, stop and report the maximum weight matching.

Time analysis The same! Time required = O(|V| 2 + |V| |E|)

Stable Marriage Problem

Given N men and N women, each person list in order of preference all the people of the opposite sex who would like to marry. Problem:  Engage all the women to all the men in such a way as to respect all their preferences as much as possible.

Stable? A set of marriages is unstable if  two people who are not married both prefer each other than their spouses E.g. Suppose we have A1 B3 C2 D4 E5. This is unstable since  A prefer 2 more than 1  2 prefer A more than C ABCDE E A D B C 2 D E B A C 3 A D B C E 4 C B D A E 5 D B C E A

Naïve solution Starting from a feasible solution. Check if it is stable.  If yes, done!  If not, remove an unstable couple. Is this work?

Naïve solution (2) Does not work! E.g.  A1 B3 C2 D4 E5  A2 B3 C1 D4 E5  A3 B2 C1 D4 E5  A3 B1 C2 D4 E5 ABCDE E A D B C 2 D E B A C 3 A D B C E 4 C B D A E 5 D B C E A

Solution 1. Let X be the first man. 2. X proposes to the best woman in the remaining on his list. (Initially, the first woman on his list!) 3. If α is not engaged  Pair up (X, α). Then, set X=next man and goto If α prefers X more than her fiancee Y,  Pair up (X, α). Then, set X=Y and goto Goto 1

Example ABCDE E A D B C 2 D E B A C 3 A D B C E 4 C B D A E 5 D B C E A ABCDE

Time analysis If there are N men and N women,  O(N 2 ) time