```Date: Jun 18, 2013 9:18 PM
Author: quasi
Subject: Re: A New Graph Coloring Conjecture.

bill wrote:>Conjecture: A graph is four-colorable, if it does not contain>a complete K5 sub graph.I think the following works as a counterexample ...Thus, let G = (V,E) where   V = {a1,b1,c1,d1} U {a2,b2,c2,d2} U {a3,b3,c3,d3}   E = {all edges of each between pairs of vertices in the sets          {a_i,b_i,c_i,d_i} for i = 1,2,3}       U       {the edges of the form a1v2 where v is not equal to a}       U       {the edges of the form a2v3 where v is not equal to a}       U       {the edge a1a3}Suppose G is 4-colorable.Then a1 must have the same color as a2, and similarly, a2 musthave the same color as a3, contradiction, since a1 and a3 areadjacent.Hence G is not 4-colorable.To see that G does not contain a subgraph isomorphic to K5, note that the only vertices of G with degree at least 4 area1,a2,b2,c2,d2,a3,b3,c3,d3. Suppose G has a subgraph H isomorphic to K5.The only vertices of degree at least 4 which are adjacent tob2 are a1,a2,c2,d2, but since a1 is not adjacent to a2, thevertex set of H can't be {a1,a2,b2,c2,d2}. It follows that b2is not a vertex of H. Similarly, c2 and d2 are not vertices ofH.The only vertices of degree at least 4 which are adjacent tob3 are a2,a3,c3,d3, but since a2 is not adjacent to a3, thevertex set of H can't be {a2,a3,b3,c3,d3}. It follows that b3is not a vertex of H. Similarly, c3 and d3 are not vertices ofH.The only remaining vertices having degree at least 4 are a1,a2,a3 but that's not enough vertices for a K5, and besides,there is only one edge between them.Thus, G does not contain a subgraph isomorphic to K5.Therefore, if I haven't made an error, G serves as a counterexample to the stated claim.quasi
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