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Topic: A New Graph Coloring Conjecture.
Replies: 42   Last Post: Jul 2, 2013 4:08 AM

 Messages: [ Previous | Next ]
 Tucsondrew@me.com Posts: 1,161 Registered: 5/24/13
Re: A New Graph Coloring Conjecture.
Posted: Jun 18, 2013 10:28 PM

On Tuesday, June 18, 2013 5:57:35 PM UTC-7, quasi wrote:
> 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
>
>
>
>
> 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.
>

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, sub-graph 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

Date Subject Author
6/18/13 b92057@yahoo.com
6/18/13 Tucsondrew@me.com
6/18/13 Tucsondrew@me.com
6/18/13 quasi
6/18/13 Tucsondrew@me.com
6/18/13 quasi
6/19/13 trj
6/20/13 b92057@yahoo.com
6/20/13 quasi
6/20/13 b92057@yahoo.com
6/22/13 b92057@yahoo.com
6/23/13 quasi
6/25/13 b92057@yahoo.com
6/25/13 quasi
6/25/13 b92057@yahoo.com
6/26/13 quasi
6/26/13 Brian Q. Hutchings
6/27/13 b92057@yahoo.com
6/27/13 quasi
6/27/13 b92057@yahoo.com
6/27/13 quasi
6/27/13 quasi
6/28/13 b92057@yahoo.com
6/28/13 quasi
6/28/13 b92057@yahoo.com
6/29/13 quasi
7/1/13 b92057@yahoo.com
7/2/13 quasi
6/19/13 b92057@yahoo.com
6/19/13 quasi
6/20/13 b92057@yahoo.com
6/20/13 Tucsondrew@me.com
6/21/13 quasi
6/21/13 quasi
6/22/13 b92057@yahoo.com
6/23/13 quasi
6/19/13 b92057@yahoo.com
6/20/13 quasi
6/20/13 quasi
6/20/13 Butch Malahide
6/20/13 Butch Malahide
6/19/13 b92057@yahoo.com
6/20/13 b92057@yahoo.com