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 must

have the same color as a3, contradiction, since a1 and a3 are

adjacent.

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 are

a1,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 to

b2 are a1,a2,c2,d2, but since a1 is not adjacent to a2, the

vertex set of H can't be {a1,a2,b2,c2,d2}. It follows that b2

is not a vertex of H. Similarly, c2 and d2 are not vertices of

H.

The only vertices of degree at least 4 which are adjacent to

b3 are a2,a3,c3,d3, but since a2 is not adjacent to a3, the

vertex set of H can't be {a2,a3,b3,c3,d3}. It follows that b3

is not a vertex of H. Similarly, c3 and d3 are not vertices of

H.

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