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| bleuprint... |
 Posted: Mon Nov 02, 2009 6:16 am |
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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? |
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| spudnik... |
| Posted: Mon Nov 02, 2009 4:08 pm |
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you're implying that "heawood's mistaken proof" was actually correct,
iff you've proven the four-color theorem?
[quote]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?
[/quote]
--McSudan Crusades for carbon credits!?! |
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| bleuprint... |
| Posted: Wed Nov 04, 2009 12:23 am |
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[quote]you're implying that "heawood's mistaken proof" was
actually correct,
iff you've proven the four-color theorem?
[/quote]
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?
[quote]
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:[/quote]
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). |
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| bleuprint... |
| Posted: Sat Nov 07, 2009 6:55 am |
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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. |
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| spudnik... |
| Posted: Wed Nov 18, 2009 11:58 am |
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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.
m'brain:
L'Ouvre: http://wlym.com/~animations/fermat/index.html
--HW's Cap'n Trade & Warren "choo-choo" Buffet, together again?...
Dubya wouldn't sign the radical free-trade Kyoto Protocol?...
Rep. Waxman's God-am bill, doesn't institute a tarrif, instead!?! |
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