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Tutorial 6 of CSCI2110 Bipartite Matching Tutor: Zhou Hong ( 周宏 ) hzhou@cse.cuhk.edu.hk

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Outline Maximum Bipartite Matching B-Matching Tic-Tac-Toe (Optional) Midterm Review

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The bipartite matching problem: Find a matching with the maximum number of edges. A perfect matching is a matching in which every vertex is matched (i.e. of degree 1). Reduce to Perfect Matching: Once you know how to solve perfect matching, you can also do maximum matching. Maximum Matching A B E In the following, we assume |A|=|B| for simplicity.

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Perfect Matching Oracle Oracle of Perfect Matching Perfect Matching Oracle input output No Perfect Matching Our Goal: find a maximum matching by querying perfect matching oracle as few as possible

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Bipartite Graph Contains Perfect Matching A B If G already has a perfect matching M, then M must be a maximum matching. We done with only one query! However, what shall we do if G does not contain a perfect matching?

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Matching of Size n-1 A B To get an ideal of the reduction step, first, we assume G has a maximum matching of size n-1.

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First Attempt A B By adding an appropriate edge, G will contain a perfect matching. We can find it by one query. Then, remove new added edge from the perfect matching, we done. Drawback of This Approach? We need to try different pair of vertices, in the worst case, O(n 2 ) queries are required.

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A Better Solution A B Add two dummy vertices connecting to all vertices on the opposite. In the new graph, we find a perfect matching by one query. Remove the new added edges, we have a maximum matching of original graph. A matching of size n-1 in original graph guarantees a perfect matching in new graph

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A Better Solution A B Add two dummy vertices connecting to all vertices on the opposite. In the new graph, we find a perfect matching by one query. Remove the new added edges, we have a maximum matching of original graph. On the other hand, a perfect matching in new graph will guarantee a matching of size n-1 in original graph

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Complete Reduction No perfect matching in the new graph implies no matching of size n-1 in the original graph. What shall we do? Repeat the previous procedure until we find a perfect matching In each iteration –We add two dummy vertices connecting with all vertices on the opposite. –Make a query with the updated graph. How many queries do we need? At most n Using binary search, we can improve to log(n) queries!

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Residence Assignment Assignment Requirements: Each student can be assigned to at most one room Each room can accommodate a specific number of students Our Goal: Maximize the number of residents

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Basic Idea For a shared room with b beds, treat it as b single rooms In the new setting –Applying for a shared room becomes applying for all b duplicate single rooms –Then, we only need to find a maximum matching

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A B B-Matching Matching is a special case of b-matching, with b v =1 for all v.

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Reduction to Maximum Matching For each vertex v with degree bound b v, make b v copies of vertex v. Connect the duplicate vertices according to the original graph.

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Reduction to Maximum Matching For each vertex v with degree bound b v, make b v copies of vertex v. Connect the duplicate vertices according to the original graph. Find a maximum matching in the new graph. Map the resulting matching back to the original graph.

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Tic-Tac-Toe A paper-and-pencil game. – Two players (X and O) – They take turns marking spaces in a 3x3 grid Player who succeeds in placing three respective marks in a horizontal, vertical, or diagonal row wins the game Best play from both parties leads to a draw A game won by the first player X

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Winning Set of Tic-Tac-Toe Therefore, there are totally 8 winning sets

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Bipartite Graph Now, let’s construct a bipartite graph – On one side, the vertices corresponding to the squares in the grid (totally 9) – On the other side, the vertices corresponding to the winning sets (totally 8) – An edge between a square and a winning set indicate that the square is in the winning set

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123 456 789 1 5 2 3 4 6 7 8 9

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Generalized Tic-Tac-Toe Generalize to nxn grid – There are n 2 squares – Winning sets are still the whole horizontal, vertical or diagonal rows, totally 2n+2

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B-Matching Consider a b-matching in the bipartite graph between squares and winning sets – Degree bound for square is 1 – Degree bound for winning set is 2 We claim that – if there exists a b-matching such that each winning set incidents with exactly 2 edges – then player 2 has a trivial strategy to avoid losing

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Example of 5x5 Grid In case of 5x5 grid Tic-Tac-Toe, there exists a b-matching as required – (not shown) Player 2 has a trivial strategy to force a draw – Pair squares in the grid according to the b-matching – If player 1 marks a square in a pair, then marks the other square in the pair, otherwise marks randomly 12345 678910 1112131415 1617181920 2122232425 12345 678910 1112131415 1617181920 2122232425 12345 678910 1112131415 1617181920 2122232425 3 4 … … … … Note that, player 1 can never mark both squares in a pair.

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Existence of The Trivial Strategy V.S. Hall’s Theorem For each winning set, make a copy of it in the bipartite graph Then, the desired b-matching is corresponding to a matching saturating both original and duplicate winning sets Generalized Hall’s Theorem: A bipartite graph G=(V,W;E) has a matching saturating W if and only if |N(S)| >= |S| for every subset S of W. Player 2 does not always have such a trivial strategy to avoid losing, consider the standard (3x3) Tic-Tac-Toe!

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Thank You! Q & A ?

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