Date: Jun 21, 2013 5:39 AM
Subject: Re: Note to Quasi from Bill
Some context restored ...
>What are MA vertices?
>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,
>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.
>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.
>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.
>(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
(1) It satisfies your stated hypothesis, that is, it has no
subgraph isomorphic to K5
(2) It fails your stated conclusion, that is, it's not
4-colorable (and hence is also non-planar).