Re: Snark Theorem Equivalent to the 4CT.

On Jul 7, 8:00 pm, Proginoskes <CCHeck...@xxxxxxxxx> wrote:
On Jul 7, 4:15 pm, bill <b92...@xxxxxxxxx> wrote:

The "snark theorem" (paraphrased) says the statement that
"all snarks are non-planar" is equivalent to the 4CT.

The exsitence of a planar snark would prove the 4CT to be false; ie

1, A snark in not 3 edge colorable
2. Therefore the snark is cannot be 4 face colorable,

and the 4CT is false.

But is the absence of a planar snark sufficient to prove the 4CT true?

If some snark S happens to be planar, then that graph cannot be 3-edge
colored. This means that the regions of the dual of S cannot be
colored with 4 colors. That would mean the 4CT is false.

If the 4CT is false, there is a counterexample G. Let H be the
triangulated version of G (in case G isn't); then H isn't 4-colorable,
either. The dual of H is a 3-regular graph which cannot be 3-edge
colored (because a 3-edge coloring would produce a 4-coloring of H).
The dual of H is 2-connected, since a bridge of the dual of H
corresponds to a loop in H. Thus the dual of H would be a snark, which
is also planar.

Thus, proving "there is no planar snark" is equivalent to the 4CT, so
it is both necessary and sufficient for the 4CT to be true.

--- Christopher Heckman

I am confused! Let C be the 2-connected cubic dual of H.
If C is not 3 edge colorable, then C is a snark.
If C is a snark and C is planar, then the 4CT is false
If C is a snark and C is non-planar, then the 4CT is false.


Bill J

.

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  • Re: Snark Theorem Equivalent to the 4CT.
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