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Apr 17, 2016 at 8:29 comment added Gordon Royle For the $K_6$ example, what happens is that all the "other terms" are identically zero. This is because all the minors involved are obtained by contracting exactly 2 of the $k$ edges (and deleting some of the others) and when the 2 edges lie in a triangle, then this creates a loop and hence an identically zero chromatic polynomial. This can cause a problem only when EVERY pair of the $k$ edges lies in a triangle, which can only occur when the $k$ edges lie in a $k+1$-clique.
Apr 17, 2016 at 8:17 comment added Gordon Royle Oxley works with cocircuits, while Woodall/Thomassen each work with sets of edges incident with a vertex. So sometimes Oxley can find a cocircuit strictly smaller than any vertex-star and thereby get a smaller $k$ in the first term. This can make a difference, but only for certain contrived examples.
Apr 16, 2016 at 22:27 comment added Fedor Petrov Also I do not understand why this approach does not prove that $K_6$ is 5-colorable: it has a vertex of degree 5 and all proper minors are 5-colorable.
Apr 16, 2016 at 7:02 comment added Fedor Petrov If cocircuit is defined as inclusion-minimal codependent set, then in general set of edges incident with a vertex is not a cocircuit (well, it is always a disjoint union of cocircuits).
Apr 15, 2016 at 23:11 comment added Gordon Royle @Fedor - removing (deleting and contracting) the edges incident with a vertex in a graph corresponds to removing the elements of a cocircuit in a matroid. In a graph, the edges incident with a vertex are a special type of cocircuit, so the matroid formulation latter is slightly more general even when the matroid is graphic.
Apr 15, 2016 at 18:08 comment added Fedor Petrov I do not understand the formulation of Oxley's result for matroids. What does it mean 'remove a vertex'?
May 12, 2015 at 5:28 vote accept user19906
May 11, 2015 at 14:05 history edited Gordon Royle CC BY-SA 3.0
Added more info regarding BL paper
May 11, 2015 at 12:13 history answered Gordon Royle CC BY-SA 3.0