Re: Four Color Theorem


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.

Click on 'Options" The line of options appears; ie
"Reply | Reply to Author | Forward | Print | Individual Message |
Show original |"
click on "Print", "Cancel" and the mesaage appears with the figure
accurately
displayed

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?

Technically, the graph (X) above is not in G but only in H. The other
vertices have
no effect on the coloring or colorability of graph X.

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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