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| Author |
Message |
a.khanm@gmail.com
science forum beginner
Joined: 02 Jul 2006
Posts: 14
|
 Posted: Mon Jul 03, 2006 7:49 am Post subject:
Re: Four Color Theorem
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|
Quite true. But when I said "inductive argument" I had something
different in mind. For instance if there were a feasible way (a finite
set of rules) of generating any arbitrary triangulation on n vertices
say Tn inductively from a triangulation on n-1 vertices say Tn-1 then
four colour theorem could be proved inductively by extending the four
colouring of
Tn-1 to Tn.
In the past people have used two simple rules:
1. adding a vertex to an existing face of Tn-1 (i.e., adding a vertex
of degree 3)
2. adding a vertex to an existing edge Tn-1 (i.e., adding a vertex of
degree 4)
Clearly Tn obtained by any of these rules is reducible. However,
someone by the name of R.E. Kenyon Jr. has used the operation of
expanding an existing vertex of Tn-1 to generate all triangulations
inductively. See
http://www.xenodochy.org/article/fourcolor.html
But this is quite trivial and takes us right back where we started.
Proginoskes wrote:
| Quote: | a.khanm@gmail.com 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.
But a variation of induction is the backbone to reduction, a technique
that both Appel and Haken, and Robertson Sanders Seymour Thomas used to
find a proof.
--- Christopher Heckman |
|
|
| Back to top |
|
 |
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Wed Jul 05, 2006 11:30 pm Post subject:
Re: Four Color Theorem
|
|
|
a.khanm@gmail.com wrote:
| Quote: | 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}
|..............| Internal chord
|--------------| External {B-C-B} Kempe chain
D <== A-----B-----C-----B------D ==> A
|--------------|--------------| External {A-C) & {C-D}
Kempe chains
The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D}
chains
at common 'C' vertices.
----Bill J |
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| Back to top |
|
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Thu Jul 06, 2006 5:35 am Post subject:
Re: Four Color Theorem
|
|
|
bill wrote:
| Quote: | a.khanm@gmail.com 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.
| Quote: | |..............| Internal chord
|--------------| External {B-C-B} Kempe chain
D <== A-----B-----C-----B------D ==> A
|--------------|--------------| External {A-C) & {C-D}
Kempe chains
The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D} chains
at common 'C' vertices.
|
It looks like you made Kempe's mistake again.
--- Christopher Heckman |
|
| Back to top |
|
|
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Thu Jul 06, 2006 6:45 am Post subject:
Re: Four Color Theorem
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|
Proginoskes wrote:
| Quote: | bill wrote:
a.khanm@gmail.com 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.
|
But it lines up perfectly on my screen! Did you try the print mode?
| Quote: |
|..............| Internal chord
|--------------| External {B-C-B} Kempe chain
D <== A-----B-----C-----B------D ==> A
|--------------|--------------| External {A-C) & {C-D}
Kempe chains
The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D} chains
at common 'C' vertices.
It looks like you made Kempe's mistake again.
|
What is Kempe's mistake? When did I make it before?
BTW; you suggested the B-C-B chain idea!
| Quote: |
--- Christopher Heckman |
|
|
| Back to top |
|
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Thu Jul 06, 2006 7:15 am Post subject:
Re: Four Color Theorem
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|
|
bill wrote:
| Quote: | Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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.
| Quote: | 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.
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.
| Quote: | |..............| 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.
| Quote: | 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?
| Quote: | It looks like you made Kempe's mistake again.
|
Maybe not ... (See the note below.)
| Quote: | 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 |
|
| Back to top |
|
|
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Thu Jul 06, 2006 6:41 pm Post subject:
Re: Four Color Theorem
|
|
|
Proginoskes wrote:
| Quote: | bill wrote:
Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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
| Quote: | 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.
| Quote: | 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 |
|
|
| Back to top |
|
|
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Sat Jul 08, 2006 7:30 pm Post subject:
Re: Four Color Theorem
|
|
|
Proginoskes wrote:
| Quote: | bill wrote:
Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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.
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?
It looks like you made Kempe's mistake again.
Maybe not ... (See the note below.)
|
What note?
| Quote: |
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 |
|
|
| Back to top |
|
|
Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Sun Jul 09, 2006 4:51 am Post subject:
Re: Four Color Theorem
|
|
|
bill wrote:
| Quote: | Proginoskes wrote:
bill wrote:
Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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.
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?
It looks like you made Kempe's mistake again.
Maybe not ... (See the note below.)
What note?
|
The one further along in my post.
| Quote: | 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 |
|
| Back to top |
|
|
a.khanm@gmail.com
science forum beginner
Joined: 02 Jul 2006
Posts: 14
|
| Posted: Sun Jul 09, 2006 1:08 pm Post subject:
Re: Four Color Theorem
|
|
|
What became of your proof of the four colour theorem? Usually people
tend to make the same old mistakes in proving the elusive theorem. Have
you found any errors yet or do you still think that your proof is
correct? Another person by the name of Ashay Dharwadker has claimed to
have proven the four colour theorem but his proof never made it to a
journal. What is the current status of your proof?
icahit@gmail.com wrote:
| Quote: | May be the answer to Bill's question is in my response in another group
at
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1150590683
Cahit
Proginoskes wrote:
a.khanm@gmail.com 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.
But a variation of induction is the backbone to reduction, a technique
that both Appel and Haken, and Robertson Sanders Seymour Thomas used to
find a proof.
--- Christopher Heckman |
|
|
| Back to top |
|
|
icahit@gmail.com
science forum beginner
Joined: 30 Sep 2005
Posts: 24
|
| Posted: Sun Jul 09, 2006 2:37 pm Post subject:
Re: Four Color Theorem
|
|
|
I must be careful when I give an answer to your question. Yes my proof
has not yet published but I think it has been passed through extensive
public review in the last two years. I have given several seminars in
the mathematical departments. The bottom line is that, no one yet and
I trust, no one will be given serious comments against its correctness.
It is true that in the final publisheable form I will have to add a few
things and delete some unneccessary figures and explain my spiral-chain
coloring algorithm in more detail, that is three-coloring of the spiral
segments and additional stuff on three-plus-one spiral chain coloring
and also adding may be a new section for the three-edge coloring of
(bridgeless) cubic planar graphs. I will going to present my proof in
the ICM2006, Madrid in August (
http://icm2006.org/v_f/web_UC.php?CodiSeccio=14&CodiTipusEvent=4sc&Ordenacio=ae&Format=UC4sc)
and based on the feedback I receive I am going to submit to a proper
journal. The only thing that I would say about Dharwadker's "proof" of
the 4CT is that there is no description of an coloring algorithm in his
proof.
Cahit
a.khanm@gmail.com wrote:
| Quote: | What became of your proof of the four colour theorem? Usually people
tend to make the same old mistakes in proving the elusive theorem. Have
you found any errors yet or do you still think that your proof is
correct? Another person by the name of Ashay Dharwadker has claimed to
have proven the four colour theorem but his proof never made it to a
journal. What is the current status of your proof?
icahit@gmail.com wrote:
May be the answer to Bill's question is in my response in another group
at
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1150590683
Cahit
Proginoskes wrote:
a.khanm@gmail.com 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.
But a variation of induction is the backbone to reduction, a technique
that both Appel and Haken, and Robertson Sanders Seymour Thomas used to
find a proof.
--- Christopher Heckman |
|
|
| Back to top |
|
|
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Sun Jul 09, 2006 10:10 pm Post subject:
Re: Four Color Theorem
|
|
|
Proginoskes wrote:
| Quote: | bill wrote:
a.khanm@gmail.com 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.
|..............| Internal chord
|--------------| External {B-C-B} Kempe chain
D <== A-----B-----C-----B------D ==> A
|--------------|--------------| External {A-C) & {C-D}
Kempe chains
The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D} chains
at common 'C' vertices.
It looks like you made Kempe's mistake again.
|
Exactly what was "Kempe's mistake"?
| Quote: |
--- Christopher Heckman |
|
|
| Back to top |
|
|
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Sun Jul 09, 2006 11:07 pm Post subject:
Re: Four Color Theorem
|
|
|
Proginoskes wrote:
| Quote: | bill wrote:
Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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".
| Quote: |
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.
----Bill J
| Quote: |
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 |
|
|
| Back to top |
|
|
bill
science forum Guru
Joined: 07 Sep 2005
Posts: 311
|
| Posted: Sun Jul 09, 2006 11:25 pm Post subject:
Re: Four Color Theorem
|
|
|
Proginoskes wrote:
| Quote: | bill wrote:
Proginoskes wrote:
bill wrote:
Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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.
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?
It looks like you made Kempe's mistake again.
Maybe not ... (See the note below.)
What note?
The one further along in my post.
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.
|
Specificially, the B-C-B chain is discussed in the sci.math thread;
"Vertex Coloring", April 19, 2006.
---Bill J
| Quote: |
--- Christopher Heckman |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Sun Jul 09, 2006 11:44 pm Post subject:
Re: Four Color Theorem
|
|
|
icahit@gmail.com wrote:
| Quote: | I must be careful when I give an answer to your question. Yes my proof
has not yet published but I think it has been passed through extensive
public review in the last two years. I have given several seminars in
the mathematical departments. The bottom line is that, no one yet and
I trust, no one will be given serious comments against its correctness.
|
I have my doubts, especially in Cahit's paper which "settles"
Steinberg's Conjecture. In particular, I derived the conditions under
which a "triangulated ring" is 3-colorable, and they do not match
Cahit's. That, and my increased workload during the academic year, have
kept me from looking at more of the problem.
It is my opinion that the proofs should be worked out in more detail.
--- Christopher Heckman
| Quote: | It is true that in the final publisheable form I will have to add a few
things and delete some unneccessary figures and explain my spiral-chain
coloring algorithm in more detail, that is three-coloring of the spiral
segments and additional stuff on three-plus-one spiral chain coloring
and also adding may be a new section for the three-edge coloring of
(bridgeless) cubic planar graphs. I will going to present my proof in
the ICM2006, Madrid in August (
http://icm2006.org/v_f/web_UC.php?CodiSeccio=14&CodiTipusEvent=4sc&Ordenacio=ae&Format=UC4sc)
and based on the feedback I receive I am going to submit to a proper
journal. The only thing that I would say about Dharwadker's "proof" of
the 4CT is that there is no description of an coloring algorithm in his
proof.
Cahit
a.khanm@gmail.com wrote:
What became of your proof of the four colour theorem? Usually people
tend to make the same old mistakes in proving the elusive theorem. Have
you found any errors yet or do you still think that your proof is
correct? Another person by the name of Ashay Dharwadker has claimed to
have proven the four colour theorem but his proof never made it to a
journal. What is the current status of your proof?
icahit@gmail.com wrote:
May be the answer to Bill's question is in my response in another group
at
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1150590683
Cahit
Proginoskes wrote:
a.khanm@gmail.com 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.
But a variation of induction is the backbone to reduction, a technique
that both Appel and Haken, and Robertson Sanders Seymour Thomas used to
find a proof.
--- Christopher Heckman |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Sun Jul 09, 2006 11:53 pm Post subject:
Re: Four Color Theorem
|
|
|
bill wrote:
| Quote: | Proginoskes wrote:
bill wrote:
a.khanm@gmail.com 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.
|..............| Internal chord
|--------------| External {B-C-B} Kempe chain
D <== A-----B-----C-----B------D ==> A
|--------------|--------------| External {A-C) & {C-D}
Kempe chains
The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D} chains
at common 'C' vertices.
It looks like you made Kempe's mistake again.
Exactly what was "Kempe's mistake"?
|
See http://www.stonehill.edu/compsci/LC/Four-Color/Four-color.htm . Two
graphs for which his proof fails are given by:
http://mathworld.wolfram.com/KittellGraph.html (23 vertices)
http://mathworld.wolfram.com/ErreraGraph.html (17 vertices)
--- Christopher Heckman |
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