
Re: A New Graph Coloring Conjecture.
Posted:
Jun 18, 2013 10:28 PM


On Tuesday, June 18, 2013 5:57:35 PM UTC7, quasi wrote: > 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. >
You are correct.
My original analysis of the question made an incorrect assumption. Which was that the absence of a K5, and That of a K3,3, subgraph implies planarity.
I then looked up Kuratowski's theorem, and realized my mistake. My correction edit yield no good either.
So, I am wondering, What it is that the OP, bill, is investigating?
> > quasi
ZG

