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Bipartite Matching, Extremal Problems, Matrix Tree Theorem.

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Today’s Plan Proof of Hall’s theorem Algorithms for bipartite matching Extremal problems Matrix tree theorem

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Perfect Matching Does a perfect matching always exist? Of course not. S N(S) If there are more vertices on one side, then of course it is impossible. Let N(S) be the neighbours of vertices in S. If |N(S)| < |S|, then it is impossible to have a perfect matching.

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A Necessary and Sufficient Condition Hall’s Theorem: A bipartite graph G=(V,W;E) has a perfect matching if and only if |N(S)| >= |S| for every subset S of V and W. This is a deep theorem. It tells you exactly when a bipartite graph does not have a perfect matching. (Now you can convince the king.) Is it the only situation when a bipartite graph does not have a perfect matching?

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Proof of Hall’s Theorem (easy direction) One direction is easy, if there is a perfect matching, then |N(S)| >= |S| for every subset S of V. S N(S) Just consider the neighbours of S in the perfect matching. Hall’s Theorem: A bipartite graph G=(V,W;E) has a perfect matching if and only if |N(S)| >= |S| for every subset S of V.

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Proof of Hall’s Theorem (difficult direction) Another direction is more interesting, we need to show whenever |N(S)| >= |S|, then there is a perfect matching! How to prove such kind of statement? Proof by strong induction on the number of edges. Hall’s Theorem: A bipartite graph G=(V,W;E) has a perfect matching if and only if |N(S)| >= |S| for every subset S of V.

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Hall’s Theorem: If |N(S)| >= |S| for every subset S of V, then there is a perfect matching. Proof of Hall’s Theorem Case 1: Every subset S has |N(S)| > |S|. (Easy case) 1.Just delete an edge. 2.By deleting an edge, |N(S)| can decrease by at most 1. 3.Since |N(S)| > |S| before, 4.we still have |N(S)| >= |S| after deleting an edge. 5.Since the graph is smaller (one fewer edge), by induction, 6.there is a perfect matching in this smaller graph, 7.hence there is a perfect matching in the original graph.

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Hall’s Theorem: If |N(S)| >= |S| for every subset S of V, then there is a perfect matching. Proof of Hall’s Theorem Case 2: Suppose there is a subset S with |N(S)| = |S|. S N(S) Divide the graph into two smaller graphs G1 and G2 (so we can apply induction) Find a perfect matching in G2 by induction. Find a perfect matching in G1 by induction. G1 G2 Then we are done.

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Proof of Hall’s Theorem S N(S) Why there is a perfect matching in G2? To apply Hall’s, we want to show for any subset T of S, |N(T) G2| >= |T|. T|N(T) G2| G2 Hall’s Theorem: If |N(S)| >= |S| for every subset S of V, then there is a perfect matching.

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Proof of Hall’s Theorem S N(S) Why there is a perfect matching in G2? For any subset T S, N(T) G2 = N(T). By assumption, |N(T) G2| = |N(T)| >= |T|. Therefore, by induction, there is a perfect matching in G2. T G2 Find a perfect matching in G2 by induction. |N(T) G2|

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Why there is a perfect matching in G1? Proof of Hall’s Theorem S N(S) T N(T) G1 For any subset T, we want to show |N(T) G1| >= |T| to apply induction. 1.Consider T, by assumption, |N(T)| >= |T| 2.Can we conclude that |N(T) G1| >= |T|? 3.No, because N(T) may intersect N(S)! Now what?

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Why there is a perfect matching in G1? Proof of Hall’s Theorem S N(S) T N(T) G1 |S|=|N(S)| For any subset T, we want to show |N(T) G1| >= |T| to apply induction. 1.Consider S T, by assumption, |N(S T)| >= |S T| (the green areas). 2.Since |S|=|N(S)|, |N(S T) – N(S)| >= |S T - S| (the red areas). 3.So |N(T) G1| = |N(S T) – N(S)| > = |S T – S| = |T|, as required.

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Hall’s Theorem: If |N(S)| >= |S| for every subset S of V, then there is a perfect matching. Proof of Hall’s Theorem Case 2: Suppose there is a subset S with |N(S)| = |S|. S N(S) Divide the graph into two smaller graphs G1 and G2 (so we can apply induction) Find a perfect matching in G2 by induction. Find a perfect matching in G1 by induction. G1 G2 Now we are done.

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Today’s Plan Proof of Hall’s theorem Algorithms for bipartite matching Extremal problems Matrix tree theorem

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Greedy method? (add an edge with both endpoints unmatched) First Try

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Key Questions How to tell if a graph does not have a (perfect) matching? How to determine the size of a maximum matching? How to find a maximum matching efficiently?

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Hall’s Theorem [1935]: A bipartite graph G=(A,B;E) has a matching that “saturates” A if and only if |N(S)| >= |S| for every subset S of A. S N(S) Existence of Perfect Matching

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König [1931]: In a bipartite graph, the size of a maximum matching is equal to the size of a minimum vertex cover. What is a good upper bound on the size of a maximum matching? Min-max theoremNP and co-NP Implies Hall’s theorem. Bound for Maximum Matching König [1931]: In a bipartite graph, the size of a maximum matching is equal to the size of a minimum vertex cover.

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Any idea to find a larger matching? Algorithmic Idea?

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Given a matching M, an M-alternating path is a path that alternates between edges in M and edges not in M. An M-alternating path whose endpoints are unmatched by M is an M-augmenting path. Augmenting Path

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What if there is no more M-augmenting path? Prove the contrapositive: A bigger matching an M-augmenting path 1.Consider 2.Every vertex in has degree at most 2 3.A component in is an even cycle or a path 4.Since, an M-augmenting path! If there is no M-augmenting path, then M is maximum! Optimality Condition

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Algorithm Key: M is maximum no M-augmenting path How to find efficiently?

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Finding M-augmenting paths Orient the edges (edges in M go up, others go down) An M-augmenting path a directed path between two unmatched vertices

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Complexity At most n iterations An augmenting path in time by a DFS or a BFS Total running time

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Hall’s Theorem [1935]: A bipartite graph G=(A,B;E) has a matching that “saturates” A if and only if |N(S)| >= |S| for every subset S of A. König [1931]: In a bipartite graph, the size of a maximum matching is equal to the size of a minimum vertex cover. Idea: consider why the algorithm got stuck… Minimum Vertex Cover

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Today’s Plan Proof of Hall’s theorem Algorithms for bipartite matching Extremal problems Matrix tree theorem

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