quasi
Posts:
11,066
Registered:
7/15/05


Re: A Simple Proof of The Four Color Theorem
Posted:
Jul 3, 2013 3:01 PM


bill wrote: >quasi wrote: >>bill wrote: >>> >>>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 4coloring. Kempe's proof said nothing about >>multiple trials (whatever that means). Thus, Kempe's proof was >> flawed. > >The original coloring is C1,
Why does there have to be an original coloring? Does Kempe's presuppose an initial coloring? If not, why isn't Kempe's claimed coloring the first and final one?
>The first application of Kempe resulted in C2. > >The application of Kempe's strategy to coloring C2 >would produce coloring C3.
How does Kempe's strategy depend on the prior coloring?
>If C3 is not a proper 4coloring; then C4 will be next >If not C4, then C5, then C6 ... Coo. > >Each successive coloring represents a "trial".
I don't really understand how the colorings
C_1,C_2,C_3, ...
evolve. But regardless, what guarantees that some C_k is a 4coloring? >For each basic coloring there are 4 subcolorings 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.
I have no idea what the above represents.
But in any case, my previous objection still stands:
You would have to _prove_ that for _all_ possible counterexamples to Kempe's coloring strategy (not just the counterexample Heawood supplied), your revised strategy manages (after finitely many trials) to achieve a 4coloring.
quasi

