Re: I have a paper on graph coloring

Proginoskes wrote:
bill wrote:
Proginoskes wrote:

bill wrote:
That is, I almost have a paper on graph coloring, because while I have
everything worked out as far as I can tell, I have no idea how to put
it together in order to submit it for publication.

What I need now is probably someone who has actually published

something on graph theory and is willing to coauthor this paper with
me.

So yeah, if anyone out there is interested, or knows someone who might
be, I would really appreciate the help if I can get it. Reply to this
or email me directly at the address I'm posting from if you'd like to
help.

It's not the Four-Color Theorem "proof" you came up with it, is it? If
so, it won't fly, and I've repeatedly explained why.

My proof will not "fly" IF
1) Each of the 240 possible 4-colorings of P extends to a proper
4-coloring of (G-v)

I don't see how this helps. There's no contradiction if there is a
4-coloring of P which doesn't extend to G-v; such a thing could exist.

I am wrong!
The ONLY significant difference in the 4-colorings of P is which pairs
of vertices; ie {13}, {24}, {35}, {41} or {52}, can have the same color
in
a given configuration. For instance - Say that v_1 & v_3 can be the
same color given configuration. Then the proof fails if pair {24} or
pair
{25} can have the same color in that configuration: and succeeds when
neither pair {24} nor pair {25} can have the same color! [For
clarity:
If v_3 and v_5 are the same color, then pair {35} has the same color.]

..
AND
2) no 3-coloring of P extends to a proper 4-coloring of (G-v).

Well, it really boils down to whether (2) is true or false. Anyone who
attempts to use this proof method has the burdon of showing that (2)
doesn't happen.

The proof succeeds if there IS a 3-coloring of P extends to a proper
4-coloring of (G-v)!

For the sake of discussion, let's agree that condition 1) is true.
Then
pairs {13} and{24} are true! But 2) is true only if pairs {13} & {24}
are mutually exclusive as are pairs {13} and {25}! That is:
ABACD & ABCBD are possible, but neither ABABD nor CBCBD are
possible!.

If the question was to be decided soley on its merits, I could
probably
prove that pairs {13} and {24} are not mutually exclusive?

---Bill J


--- Christopher Heckman

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