Tutte theorem

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In the mathematical discipline of graph theory the Tutte theorem, named after William Thomas Tutte, is a characterization of graphs with perfect matchings. It is a generalization of the marriage theorem and is a special case of the Tutte-Berge formula.

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[edit] Tutte's theorem

A given graph G= ( V, E ) has a perfect matching, if and only if, for every subset U of V the number of connected components with an odd number of vertices in the subgraph induced by V \ U is less than or equal to the cardinality of U.[1]

[edit] Proof

To prove that (*)\forall U \subseteq V,\; o(G-U)\le|U| is a necessary condition:

Consider a graph G, with a perfect matching. Let U be an arbitrary subset of V. Delete U. Consider an arbitrary odd component in G-U,\;C . Since G had a perfect matching, at least one vertex in C must be matched to a vertex in U. Hence, each odd component has at least one vertex matched with a vertex in U. Since each vertex in U can be in this relation with at most one connected component (because of it being matched at most once in a perfect matching), o(G-U)\le |U|.

To prove that ( * ) is sufficient:

Let G be an arbitrary graph satisfying ( * ). Consider the expansion of G to G * , a maximally imperfect graph, in the sense that G is a spanning subgraph of G * , but adding an edge to G * will result in a perfect matching. We observe that G * also satisfies condition ( * ) since G * is G with additional edges. Let U be the set of vertices of degree ν − 1 where ν = | V | . By deleting U, we obtain a disjoint union of complete graphs (each component is a complete graph). A perfect matching may now be defined by matching each even component independently, and matching one vertex of an odd component C to a vertex in U and the remaining vertices in C amongst themselves (since an even number of them remain this is possible). The remaining vertices in U may be matched similarly, as U is complete.

This proof is not complete. Deleting U may obtain a disjoint union of complete graphs, but the case where it does not is the more interesting and difficult part of the proof.

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