Given a MPG (Maximal Planar Graph, all faces are triangles, also the infinite face) on v vertices. We can give a +1 or -1 number to the triangles. If we add MOD3 the triangle numbers of the triangles adjacent to a vertex we get a vertex number of 0, 1 or 2 for that vertex. It's always possible to give triangle numbers so that all vertices have a vertex number of 0. It's Heawood's equivalent formulation of the 4 color theorem ( Heawood's vertex character on the dual of a cubic map).
But does there exist a proof or disproof for the following statement: Given all the 2^(2v-4) combinations of triangle numbers, then any set of v-2 vertex numbers has all the 3^(v-2) different combinations of vertex numbers if the two missing vertices are adjacent?
you're implying that "heawood's mistaken proof" was actually correct, iff you've proven the four-color theorem?
> It's always possible to give triangle numbers so that all vertices have a vertex number of 0. It's Heawood's equivalent formulation of the 4 color theorem ( Heawood's vertex character on the dual of a cubic map).
> But does there exist a proof or disproof for the following statement: > Given all the 2^(2v-4) combinations of triangle numbers, then any set of v-2 vertex numbers has all the 3^(v-2) different combinations of vertex numbers if the two missing vertices are adjacent?
> you're implying that "heawood's mistaken proof" was > actually correct, > iff you've proven the four-color theorem?
Could it be that you are confusing Kempe with Heawood? Heawood showed the flaw in Kempe's "proof" but used his Kempe-chaining to prove that five colors are sufficient. If not, where can I find "heawood's mistaken proof" you are talking about?
> > But does there exist a proof or disproof for the > following statement: > > Given all the 2^(2v-4) combinations of triangle > numbers, then any set of v-2 vertex numbers has all > the 3^(v-2) different combinations of vertex numbers > if the two missing vertices are adjacent?
By the way, the statement above is a special case (a degenerated polygon on 2 vertices) of the following more general conjecture: Given a triangulated polygon with vi vertices in it and all the combinations of triangle numbers. Then the inner vertices have all the 3^vi combinations of vertex numbers.
Patrick
(For the definition of triangle numbers and vertex numbers see the first post).
If one of the two conjectures above is proved then a simple classic proof of the 4CT is a corollary of it. So don't expect them easy to prove. But perhaps a disprove is easier (by brute force or anything else) as if one of them is disproved it's not a disproof of the 4CT(*). patrick
(*) They are in fact more general theorems for triangulated graphs than a four coloring of the vertices.
yeah, monsieur Kempe!... anyway, the neccesity of four colors for a map is shown by "behold, the tetrahedron" -- but Bucky didn't know about sufficiency, I guess (perhaps, niether did Spencer-Brown .-)
thus: that is to say, "chromatic abberation" -- so, There!
thus: I am lying about numbertheory, and the number, 1.0000...; who gives a floating fart?
thus: original sources (and "sourcebooks") are really good, such as the below-linked Ouvre de Fermat for number- theory, and Bernoulli/L'Hopital's calculus textbook. (Euclid, not so much, as an encyclopedia, although he did supply new stuff, they say -- and Langlands says that Book 7 needs a lot of work; I do have a nice latter-day textbook on synthetic trigon geometry, but it's in French, so it's hard work.)
thus: of course, and the electrons can't go faster than light *even if* they might already be orbitting the nucleus at such a velocity.
thus: I could see that he got rid of the gamma function, but it'll be a while before that is clear to me; so, I asked about a problem he wrote about, before.
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