```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 nextIf not C4, then C5, then C6 ... Coo.  Each successive coloring represents a "trial". For each basic coloring there are 4 sub-colorings asfollows                         A                 B   0    B                       D    CVertex 0 can be assigned the color A or C or Dor 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
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