quasi
Posts:
10,450
Registered:
7/15/05


Re: A New Graph Coloring Conjecture.
Posted:
Jun 18, 2013 9:18 PM


bill wrote:
>Conjecture: A graph is fourcolorable, 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 4colorable.
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 4colorable.
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

