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:

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> >quasi wrote:

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> >>bill wrote:

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> >>>

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> >>>Kempe's method was accepted as proof of the FCT until

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> >>>Heawood created his counter-example.

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> >>>

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> >>>Suppose that there was a simple way to 4-color Heawood's graph

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> >>>>without worrying about the problem of "tangled chains"? Would

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> >>>that be sufficient for a proof?

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> >>

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> >> No.

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> >>

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> >> Heawood's graph is a counterexample to Kempe's proposed coloring

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> >> strategy. According to Kempe's claimed proof, Heawood's graph can

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> >> be 4-colored by a specific strategy used in the proof. Heawood

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> >> identifies a specific planar graph which, if one follows Kempe's

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> >> coloring strategy, then two adjacent vertices will be forced to

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> >> have the same color. The result is to show that Kempe's proof is

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> >> invalid as a proof of 4-colorability for planar graphs.

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> >>

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> >If Kempe's strategy had been applied to the coloriing with the

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> >two adjacent vertices with the same color; it would have been

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> >successful.

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> Heawood was just following the coloring strategy which, based

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> on the supposedly proved claims in Kempe's proof, _had_ to work.

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> The fact that it didn't work invalidates Kempe's claim, and with

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> it, the whole proof.

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>

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> If Kempe could have fixed the proof by reworking his coloring

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> strategy, he surely would have done so.

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>

>

> The fact that Heawood's graph _can_ be 4-colored doesn't resolve

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> the dilemma. Heawood's graph is just one counterexample to

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> Kempe's proposed coloring -- one out of infinitely many. If you

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> could somehow show (in a simple way) that _all_ possible

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> counterexamples to Kempe's coloring strategy are 4-colorable,

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> that would achieve your goal of producing a simple proof of the

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> 4CT.

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>

>

> >Why do we expect Kempe's strategy to succeed on the first trial

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> >when no other method is under the same restrictions?

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>

>

> Because Kempe's proof claimed that his coloring strategy would

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> _always_ produce a 4-coloring. Kempe's proof said nothing about

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> 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.

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>

>

> >If Kempe is to be allowed only one chance; how about a slight

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> >change to Heawood's coloring before Kempe takes over?

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>

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> Once again, your analysis would have to deal with _all_ possible

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> counterexamples, not just the one Heawood supplied.

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> quasi