Saturday, 20 December 2008

soft question - Theorems for nothing (and the proofs for free)

I'd say the Tutte-Berge formula, which is a wonderful result that tells you (almost) everything you want to know about matchings in graphs. Although there are many proofs of this theorem, there is a beautiful proof for free using matroids. Strictly speaking, there is a proof for free of Gallai's Lemma (from which Tutte-Berge follows easily).



Gallai's Lemma.



Let G be a connected graph such that nu(Gx)=nu(G), for all xinV(G). Then |V(G)| is odd and def(G)=1.



Remark: nu(G) is the size of a maximum matching of G, and def(G) denotes the number of vertices of G not covered by a maximum matching.



Proof for free.
In any matroid M define the relation xsimy to mean r(x)=r(y)=1 and r(x,y)=1 or if x=y. (Here, r is the rank function of M). We say that xsimy if and only if xsimy in the dual of M. It is trivial to check that sim (and hence also sim) defines an equivalence relation on the ground set of M.



Now let G satisfy the hypothesis of Gallai's Lemma and let M(G) be the matching matroid of G. By hypothesis, M(G) does not contain any co-loops. Therefore, if x and y are adjacent vertices we clearly have xsimy. But since G is connected, this implies that V(G) consists of a single sim equivalence class. In particular, V(G) has co-rank 1, and so def(G)=1, as required.



Edit. For completeness, I decided to include the derivation of Tutte-Berge from Gallai's lemma. Choose XsubsetV(G) maximal such that def(GX)|X|= def(G). By maximality, every component of GX satisfies the hypothesis in Gallai's lemma. Applying Gallai's lemma to each component, we see that X gives us equality in the Tutte-Berge formula.

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