Date: Jul 3, 2013 8:03 AM
Author: b92057@yahoo.com
Subject: Re: A Simple Proof of The Four Color Theorem

On Tuesday, July 2, 2013 6:14:12 PM UTC-7, quasi wrote:
> bill wrote:
>

> >quasi wrote:
>
> >>bill wrote:
>
> >>>
>
> >>>Kempe's method was accepted as proof of the FCT until
>
> >>>Heawood created his counter-example.
>
> >>>
>
> >>>Suppose that there was a simple way to 4-color Heawood's graph
>
> >>>>without worrying about the problem of "tangled chains"? Would
>
> >>>that be sufficient for a proof?
>
> >>
>
> >> No.
>
> >>
>
> >> Heawood's graph is a counterexample to Kempe's proposed coloring
>
> >> strategy. According to Kempe's claimed proof, Heawood's graph can
>
> >> be 4-colored by a specific strategy used in the proof. Heawood
>
> >> identifies a specific planar graph which, if one follows Kempe's
>
> >> coloring strategy, then two adjacent vertices will be forced to
>
> >> have the same color. The result is to show that Kempe's proof is
>
> >> invalid as a proof of 4-colorability for planar graphs.
>
> >>
>
> >If Kempe's strategy had been applied to the coloriing with the
>
> >two adjacent vertices with the same color; it would have been
>
> >successful.
>
>
>
> Heawood was just following the coloring strategy which, based
>
> on the supposedly proved claims in Kempe's proof, _had_ to work.
>
> The fact that it didn't work invalidates Kempe's claim, and with
>
> it, the whole proof.
>
>
>
> If Kempe could have fixed the proof by reworking his coloring
>
> strategy, he surely would have done so.
>
>
>
> The fact that Heawood's graph _can_ be 4-colored doesn't resolve
>
> the dilemma. Heawood's graph is just one counterexample to
>
> Kempe's proposed coloring -- one out of infinitely many. If you
>
> could somehow show (in a simple way) that _all_ possible
>
> counterexamples to Kempe's coloring strategy are 4-colorable,
>
> that would achieve your goal of producing a simple proof of the
>
> 4CT.
>
>
>

> >Why do we expect Kempe's strategy to succeed on the first trial
>
> >when no other method is under the same restrictions?
>
>
>
> Because Kempe's proof claimed that his coloring strategy would
>
> _always_ produce a 4-coloring. Kempe's proof said nothing about
>
> multiple trials (whatever that means). Thus, Kempe's proof was
>


The original coloring is C1, The first application of Kempe resulted in C2, The application of Kempe's strategy to coloring C2 would produce coloring C3.
If C3 is not a proper 4-coloring; then C4 will be next
If not C4, then C5, then C6 ... Coo.

Each successive coloring represents a "trial".

For each basic coloring there are 4 sub-colorings as
follows

A

B 0 B

D C

Vertex 0 can be assigned the color A or C or D
or not be assigned any color.




> flawed.
>
>
>

> >If Kempe is to be allowed only one chance; how about a slight
>
> >change to Heawood's coloring before Kempe takes over?
>
>
>
> Once again, your analysis would have to deal with _all_ possible
>
> counterexamples, not just the one Heawood supplied.
>
>
>
>
>
> quasi