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$k$-chromatic graph is called $k$-critical if removal of any vertex from graph makes it $k - 1$ vertex colorable.

Now i have to prove that if $G$ is a $k$ critical graph then it cannot have $k+1$ vertices.

I can see that the property is true as a triangle is 3 critical. Also we have a 5-cycle as 3 critical but no $4$ vertex graph is $3$ critical.

Can any one help me in what direction i should prove the theorem

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up vote 4 down vote accepted

Suppose you have a k critical graph on k+1 vertices. If there exists a vertex of degree less than k-1, remove it. By assumption you can k-1 color the remaining graph. However if you add back in the vertex removed it can only be adjacent to k-2 colors, so we can complete it to give a k-1 coloring of the whole graph. So we can assume every vertex has degree k-1 or k.

Now choose a pair of degree k-1 not adjacent to one another, they must both be adjacent to all other vertices. (Not all vertices can have degree k, as $K_{k+1}$ is not k chromatic) Remove one, k-1 color the resulting graph then add back in the vertex giving it the same color as the other vertex in the pair. This is a valid k-1 coloring.

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Thanks.. great proof.I have just one doubt.How can we prove that a pair of non adjacent vertex exist with k-1 one degree. Can't we have a graph with just one vertex with k-1 or degree or just one pair of adjacent vertex having k-1 degree – Neel Choudhury Nov 29 '13 at 5:20

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