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Tutorial 6 of CSCI2110 Bipartite Matching. Tutor: Zhou Hong ( 周宏 ) hzhou@cse.cuhk.edu.hk. Outline. Maximum Bipartite Matching B-Matching Tic-Tac-Toe (Optional) Midterm Review. Maximum Matching. The bipartite matching problem: Find a matching with the maximum number of edges.

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## Tutorial 6 of CSCI2110 Bipartite Matching

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**Tutorial 6 of CSCI2110Bipartite Matching**Tutor: Zhou Hong (周宏) hzhou@cse.cuhk.edu.hk**Outline**• Maximum Bipartite Matching • B-Matching • Tic-Tac-Toe (Optional) • Midterm Review**Maximum Matching**The bipartite matching problem: Find a matching with the maximum number of edges. G=(AB,E) B A In the following, we assume |A|=|B| for simplicity. E A perfect matchingis 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.**Oracle of Perfect Matching**output input Perfect Perfect Oracle Oracle Matching Matching No Perfect Matching Our Goal: find a maximum matching by querying perfect matching oracle as few as possible**Bipartite Graph Contains Perfect Matching**If G already has a perfect matching M, then M must be a maximum matching. We done with only one query! B A However, what shall we do if G does not contain a perfect matching?**Matching of Size n-1**To get an ideal of the reduction step, first, we assume G has a maximum matching of size n-1. B A**First Attempt**• 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. B A Drawback of This Approach? We need to try different pair of vertices, in the worst case, O(n2) queries are required.**A Better Solution**• 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. B A A matching of size n-1 in original graph guarantees a perfect matching in new graph**A Better Solution**• 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. B A On the other hand, a perfect matching in new graph will guarantee a matching of size n-1 in original graph**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. At most n How many queries do we need? Using binary search, we can improve to log(n) queries!**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**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**B-Matching**Suppose we have a bipartite graph G=(AB,E), for each vertex v, there is a degree bound bv. B A A b-matchingis a subset of edges so that each vertex vAB has degree at most bv. Matching is a special case of b-matching, with bv=1 for all v.**Reduction to Maximum Matching**• For each vertex v with degree bound bv, make bv copies of vertex v. • Connect the duplicate vertices according to the original graph.**Reduction to Maximum Matching**• For each vertex v with degree bound bv, make bv 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.**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**Winning Set of Tic-Tac-Toe**Therefore, there are totally 8 winning sets**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**Generalized Tic-Tac-Toe**• Generalize to nxn grid • There are n2 squares • Winning sets are still the whole horizontal, vertical or diagonal rows, totally 2n+2**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**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 … … Note that, player 1 can never mark both squares in a pair. … …**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 alwayshave such a trivial strategy to avoid losing, consider the standard (3x3) Tic-Tac-Toe!**Thank You!**Q & A ?

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