Date: Jun 21, 2013 5:39 AM
Author: quasi
Subject: Re: Note to Quasi from Bill
Some context restored ...

>quasi:

>What are MA vertices?

>

>bill:

>mutually adjacent vertices.

>So take 4 MA vertices.

>Then connect vertex 5 to each of the first four.

That would make a K5 which fails to satisfy your stated

hypothesis. And if you keep 5 vertices but drop an edge,

then ...

>quasi:

>Any non-complete graph with 5 vertices is 4-colorable.

Thus, it would also satisfy your conclusion. So it wouldn't be

a counterexample. A counterexample must satisfy the hypothesis

but fail the conclusion.

>bill:

>Not if the graph is non-planar.

Your conjecture never specified anything about planarity.

Perhaps you want to revise your conjecture? If so, please

state it fully and precisely.

Every non-complete graph with exactly 5 vertices is planar.

>bill:

>If there are no forceable sets of four vertices the graph

>must be 4 colorable.

What are you claiming. Can you state it precisely?

What are forcible sets of 4 vertices? Do you mean subgraphs

isomorphic to K4?

Are you restricting to planar graphs?

If so, then of course your graph will be 4 colorable (by the

4 Color Theorem).

If you are not restricting to planar graphs, then your claim

is almost certainly false.

But in any case, your claim needs to be made more precise.

>quasi:

>(as relates to your original claim):

>Except that I've already shown you a counterexample.

>Is your example planar?

Of course not.

The graph I posted yields a _counterexample_ to your original

conjecture.

Which means:

(1) It satisfies your stated hypothesis, that is, it has no

subgraph isomorphic to K5

but

(2) It fails your stated conclusion, that is, it's not

4-colorable (and hence is also non-planar).

quasi