On Tuesday, July 2, 2013 3:23:46 AM UTC-7, 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 dame color; it would have been successful.
Why do we expect Kempe's strategy to succeed on the first trial when no other method is under the same restrictions?
If Kempe is to be allowed only one chance; how about a slight change to Heawood's coloring before Kempe takes over?
> > However, Heawood's graph _is_ a planar graph, hence it _can_ be > > 4-colored (and probably easily so). So if you show a 4-coloring > > of Heawood's graph, that reveals nothing we don't already know. > > > > quasi