Berges theorem theorem 1 a matching m in a graph g is

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Berge’s Theorem Theorem 1. A matching M in a graph G is maximum if and only if there is no augmenting path in G with respect to M . Proof. ( ) We prove the contrapositive. If P is an augmenting path with respect to M , then augmentation increases the size of the matching. Thus M is not maximum. ( ) We prove the contrapositive. Let M be a matching in G that is not maximum. (Our goal is to find an augmenting path for M .) Let M 0 be a maximum matching. Then M 0 contains more edges than M . Proof (cont’d). Let H be the spanning subgraph of G consisting of edges in M or M 0 , but not both. Since each vertex of G is matched at most once in each of M and M 0 , the maximum degree of H is at most 2. So each component of H is a path or a cycle, and if some vertex has degree 2 then one of the incident edges is in M and the other is in M 0 . Let H 1 be a component of H . The edges of H 1 alternate along the path or cycle between those in M and those in M 0 . So if H 1 is a cycle then it has even length. Hence X H 1 a cycle | M E ( H 1 ) | = X H 1 a cycle | M 0 E ( H 1 ) | . Proof (cont’d). We have | M | = | M M 0 | + X H 1 a cycle | M E ( H 1 ) | + X H 1 a path | M E ( H 1 ) | | M 0 | = | M M 0 | + X H 1 a cycle | M 0 E ( H 1 ) | + X H 1 a path | M 0 E ( H 1 ) | .
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Since | M 0 | > | M | , X H 1 a path | M 0 E ( H 1 ) | > X H 1 a path | M E ( H 1 ) | . Therefore there exists a path H * 1 of H which has more edge in M 0 than in M . Since the edges of H * 1 alternate between M and M 0 , it is an augmenting path with respect to M . 2 Matchings in bipartite graphs onig-Hall Theorem For a graph G and X V ( G ), let N ( X ) be the set of vertices adjacent to at least one vertex of X . That is, N ( X ) = v X N ( v ) . Theorem 2. Let G be a bipartite graph, with bipartition { V 1 , V 2 } . Then there is a matching in G with all vertices in V 1 matched if and only if | N ( S ) | ≥ | S | for every set S V 1 . Note that if | V 1 | = | V 2 | , then this theorem characterises when G has a perfect matching. Proof. ( ) Suppose that there is a matching in G that matches all vertices in V 1 . Then for all S V 1 , we have | N ( S ) | ≥ | S | since different vertices of S are matched to different vertices of V 2 . ( ) Let G be a bipartite graph such that | N ( S ) | ≥ | S | for all S V 1 . Suppose to the contrary that there is no matching in G such that all vertices in V 1 are matched. Let M be a maximum matching. Then at least one vertex of V 1 , say, v , is unmatched with respect to M . Let X be the set of vertices of G that are connected to v by an alternating path with respect to M . Then v X . Since M is a maximum matching, v is the only vertex of X that is unmatched under M . Proof (cont’d). Let X 1 := V 1 X and X 2 := V 2 X . Since every vertex of X \{ v } is matched under M , by the definition of X , it follows that X 1 \{ v } is matched to X 2 . Hence | X 2 | = | X 1 | - 1 and X 2 N ( X 1 ). On the other hand, every vertex w N ( X 1 ) is connected to v by an alternating path (which is obtained by the alternating path from v to a neighbor x of w followed by the edge xw ). Hence N ( X 1 ) X which implies N ( X 1 ) X 2 since G is bipartite. Thus N ( X 1 ) = X 2 , and | N ( X 1 ) | = | X 2 | = | X 1 | - 1 < | X 1 | , which is a contradiction.
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