Re: four color theorem

On Jan 22, 12:00 pm, patrick labarque
<patrick.labar...@xxxxxxxxxxxxxxx> wrote:
A simple proof onhttp://home.versateladsl.be/vt649464//Blueprint1a.PDF
based on following issues:
0) Given: any triangulated graph without separating triangles.
1) A Hamilton circuit in it.
2) Association of every vertex, except two, with two adjacent triangles.
3) Orientation of the triangles with +1 or -1.
4) We prove then that the sum of the triangle orientations for each vertex can be a multiple of 3.
5)T he triangulated graph is then 4-colorable (Heawood 1898)
The combination of this issues is NEW.

From "Blueprint"; "... we only have to meet one and the same
condition for
all the vertices (the sum of the V3# for all of the vertices must be
zero)"

Since each triangle contributes to three different vertices, EVERY
E3c
coloring will meet this criteria.

Bill J
.

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