Re: Four Color Theorem
- From: "bill" <b92057@xxxxxxxxx>
- Date: 13 Jul 2006 17:55:46 -0700
Proginoskes wrote:
bill wrote:So what! This has nothing to do with my need to make the cycle not
Proginoskes wrote:
bill wrote:
Proginoskes wrote:
bill wrote:
a.khanm@xxxxxxxxx wrote:
bill wrote:
Hypothesis: A simple loopless maximal planar graph with a 5-degree
vertex cannot be a
minimal counter example to the four color theorem!.
It follows from the four colour theorem. However, proving it directly
is another matter. A possible way is induction on the minimum vertex
degree of a maximal planar graph.
Any triangulation with minimum degree 1,2,3 or 4 is cannot be a minimal
counter example by induction on the number of vertices. Assume that
none of the triangulations with minimum vertex degree n-1 is a minimal
counter example and prove the result for n.
This would prove the general version of your statement and the four
colour theorem. But this will take some doing as inductive arguments to
prove four colour theorem and its reformulations have always failed.
My proof is based on the assumption that the following graph is not 4-chroma.
[The drawing below is correct in the "print" mode}
If you're going to post plain text which is supposed to line up, use a
proportional font, like Courier.
Actually, that should be FIXED font.
But it lines up perfectly on my screen! Did you try the print mode?
I'm using Google Groups to post, and there is no "print mode"
available.
Did you find the "print" mode"? It's under "Options" as is "Remove".
Yes. But relying on "print" mode is still bad practice for Usenet.
In a fixed font, the following would appear as a straight (dashed)
line, with 'ABC' under it:
\
\
\
\
\
\
\
ABC \
If the line appears broken, then you're not using the Usenet standard.
|..............| Internal chord
|--------------| External {B-C-B} Kempe chain
D <== A-----B-----C-----B------D ==> A
|--------------|--------------| External {A-C) & {C-D}
Kempe chains
That one came out okay.
The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D} chains
at common 'C' vertices.
So? What about the other vertices in the graph G?
If the graph above is 4-Chroma, then the other vertices in G are irrelevant.
If its not, the the other vertices in G cannot make it 4-chroma. At least that is
my contention.
But a coloring is a global object, not a local one. Consider the
analogy of a graph which is a cycle of length 2*10^100 + 3 and choose a
vertex v. If you look at all the vertices within a distance of 10^100
from v, you will get a graph that only needs 2 colors. But the overall
graph needs 3 colors.
3-colorable!
regards
---Bill J
It looks like you made Kempe's mistake again.
Maybe not ... (See the note below.)
What is Kempe's mistake? When did I make it before?
BTW; you suggested the B-C-B chain idea!
I'm keeping half an eye out on a few dozen threads in four math
newsgroups. You'll have to be more specific than that.
--- Christopher Heckman
.
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