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Topic: Note to Quasi from Bill
Replies: 1   Last Post: Jun 21, 2013 5:39 AM

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Posts: 12,067
Registered: 7/15/05
Re: Note to Quasi from Bill
Posted: Jun 21, 2013 5:39 AM
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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,
then ...

>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

Which means:

(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).


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