proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member

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Topic:	Science > Physics
User:	""
Date:	11 Feb 2006 12:59:48 PM
Object:	proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member

I need to post this to sci.physics also because the best example is
that of physics where if the Periodic Table of Elements had no smallest
atom, would atoms exist? If there was no smallest Element, would any
element exist? Obviously not.
But my surprize in all of this after making a brief search in the
literature is that it appears as if this idea is brand new and that I
just discovered it yesterday. It appears that no-one in mathematics or
logic or physics has discovered and used this idea before. That if a
category of "something" has no smallest member then the category does
not exist.
I do not know if this technique has a name already. I do not know if
anyone has already used it before. Logically it goes like this: if
something has no smallest member then that something does not exist.
The case example is the positive integers or counting-numbers. If there
is no smallest positive integer then there are no positive integers at
all. And I believe it is a twist on the Principle of Mathematical
Induction, whereas Math Induction gives all the positive integers, but
that a reversal would say no positive integer exists.
So I think my idea is brand new and never before used in all of
mathematical history until now.
Ordinary life experience examples would be that if no smallest
refrigerator exists, then no refrigerator exists. If there is no
smallest airplane, then there are no airplanes at all.
It reminds me of a reversal of Mathematical Induction.
So I have to give this technique a name and I call it "Smallest member
Existor" and state the technique as "If no smallest member of a
category
exists then the category is nonexistant". To have existence a
category must possess the quality of having a smallest member.
Now I keep in mind that the Reals and negative combined with positive
integers has no smallest member as these are infinite sets. What I am
talking about above are all finite sets and using the positive integers
only as assignment of number.
For humanity to exist, it must possess the smallest member which being
that of 1 human. If not 1 human exists, then no human exists.
I would be awfully surprized if this technique is brand new to logic,
mathematics, but it maybe the case, because anyone knowing of this
logic
would have been able to easily see it is a engine for the proof of both
Kepler Packing and 4 Color Mapping in a half page proof.
Proof of Kepler Packing: Suppose false then there exists a sphere that
has 13 kissing points, in contradiction to regular-polyhedra which can
have only have vertices of 4,6,8, 12, 20.
You see, the proof of KPP relies on the fact that there cannot be a
sphere with 13 kissing points which is the Smallest Member for a denser
KPP and because of the nonexistence of a 13 kissing point sphere-cell
there
cannot be a 14 kissing point or higher.
Proof of 4 Color Mapping: Suppose false then there exists a cell of 5
regions which are mutually adjacent and require 5 colors. This is the
smallest
cell that requires 5 colors. Contradiction to Moebius theorem saying 4
mutual-adjacencies is the maximum.
You see, if there can not be a smallest cell that requires 5 colors
then there are no cells that require 5 colors.
The objection raised against this proof is that Nevada is a cell that
has 6 adjacencies (of which no more than 3 mutual-adjacencies).
They are Nevada, California, Oregon, Idaho, Utah, Arizona. But does
it matter how many adjacencies any example proffers forth, whether it
is 6 or 7 or 8 or 2001 adjacencies, if the Smallest 5 coloring is
5 Mutual-Adjacency and none of those can
exist, then a larger cell requiring 5 colors cannot exist. If the
smallest 5 color cell
does not exist and cannot exist then
does any cell requiring 5 colors exists? The answer is no. If there is
no Smallest Positive Integer, no smallest Counting Number, are there
any
Counting Numbers? The answer again is no.
A requirement for existence is a Smallest member. If a finite set has
only one
member then it serves as the smallest member and the material object in
question exists, but if a finite set has no smallest member then
whatever the material object that the set refers to does not exist.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	16 Feb 2006 04:44:34 AM

wrote:

Thank Universe for physical constructivism.
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	12 Feb 2006 12:23:42 AM

wrote:

There is a smallest element. But the set of atoms in the universe may
be empty, so the "well-ordering" argument doesn't work here.

But my surprize in all of this after making a brief search in the
literature is that it appears as if this idea is brand new and that I
just discovered it yesterday. It appears that no-one in mathematics or
logic or physics has discovered and used this idea before.

Gee, I posted a reply saying that this idea is probably at least 100
years old. You obviously didn't read it.

I do not know if this technique has a name already. I do not know if
anyone has already used it before. Logically it goes like this: if
something has no smallest member then that something does not exist. [...]

Yes, it's called the "minimal counterexample" technique.

So I think my idea is brand new and never before used in all of
mathematical history until now.

Only someone ignorant of mathematical history would say this.

It reminds me of a reversal of Mathematical Induction.

It is; it's induction without base cases.

So I have to give this technique a name and I call it "Smallest member
Existor" and state the technique as "If no smallest member of a
category
exists then the category is nonexistant". To have existence a
category must possess the quality of having a smallest member.

Now I keep in mind that the Reals and negative combined with positive
integers has no smallest member as these are infinite sets.

And also because the usual ordering of the reals and integers is not a
well-ordering. The integers can be well-ordered by rearranging them:
0 < -1 < 1 < -2 < 2 < -3 < 3 < ...
The Axiom of Choice states that any set can be well-ordered, but no one
has provided a concrete ordering of the real numbers which is a
well-ordering. There is a current thread about this in sci.math.

I would be awfully surprized if this technique is brand new to logic,
mathematics, but it maybe the case, because anyone knowing of this
logic
would have been able to easily see it is a engine for the proof of both
Kepler Packing and 4 Color Mapping in a half page proof.

Robertson, Sanders, Seymour, and Thomas definitely used it in their new
4CT proof. Appel and Haken most likely used it.

Proof of 4 Color Mapping: Suppose false then there exists a cell of 5
regions which are mutually adjacent and require 5 colors.

You're still making the same mistake you've made the whole past month
.... This result does NOT follow, even if you have a minimal
counterexample. 5 colors does not require 5 mutual adjacencies. I've
posted counterexamples in the thread "Why AP will never proof the 4CT"
(yes, that's a typo).

You see, if there can not be a smallest cell that requires 5 colors
then there are no cells that require 5 colors.
The objection raised against this proof is that Nevada is a cell that
has 6 adjacencies (of which no more than 3 mutual-adjacencies).

This "cell" requires 4 colors without having 4 mutual adjacencies. This
map can be turned into an example of a map requiring 5 colors without
having 5 mutual adjacencies. AP is putting his fingers in his ears,
burying his head in the sand, and singing ATOMPLUTONIUM at the top of
his voice; in other words, he's supressing this information.

They are Nevada, California, Oregon, Idaho, Utah, Arizona. But does
it matter how many adjacencies any example proffers forth, whether it
is 6 or 7 or 8 or 2001 adjacencies, if the Smallest 5 coloring is
5 Mutual-Adjacency and none of those can
exist, then a larger cell requiring 5 colors cannot exist.

And a counterexample to the 4CT must be a planar map; it has to be
possible to draw it in the plane. Since the 5 mutual-adjacency
configuration is non-planar, it cannot be a counterexample to the 4CT.
So the only thing you learn about a minimal counterexample is that
there are NOT 5 mutually adjacent regions

A requirement for existence is a Smallest member.

Yes, but the Smallest member must be a member of the set. The 5-m-a
configuration isn't.

If a finite set has only one
member then it serves as the smallest member and the material object in
question exists, but if a finite set has no smallest member then
whatever the material object that the set refers to does not exist.

Every finite set can be well-ordered, doofus. You didn't read my post.
--- Christopher Heckman
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	12 Feb 2006 02:03:34 PM

Chris Heckman wrote:
(1) You seem to think that I'm ignoring the concept of adjacency in the
statement of the 4CT. I'm not; I'm just not explicitly stating it. (The
same is true about the "minimum counterexample" concept, which is part
of lots of proofs.) The definition of a 4-coloring is:
DEFINITION. A 4-coloring of a map M is a function c from the countries
of M to the set {1,2,3,4} such that c assigns different colors to
ADJACENT countries.
so adjacency does show up, after all. (In fact, anyone who knows what a
coloring is automatically knows that adjacency is involved, so it
doesn't need to be stated.) Note that two countries with different
colors are not necessarily adjacent.
A.P. writes:
Diagnostics of 4 Color Mapping and formal Proof outline:
What Chris has mentioned above that Appel & Haken and R, S, S, T assume
but never address in any proof of 4 Color Mapping is why there is no
short one page proof of 4 CM. When you ignore the central issue of
coloring which is ADJACENCY, you are bound to not get it right, and
bound to never have a short concise proof.
Outline of my Formal Proof of 4 Color Mapping: It will take more than a
paragraph, and more than a 1/2 page, I am afraid to say. Because I need
to address definitions which up to this moment I never addressed.
Cell: a cell in Kepler Packing KPP and a cell in 4 Color Mapping takes
each individual sphere (KPP) or region (4CM) and takes that sphere or
region as the center of the cell for which it asks how many surrounding
spheres are kissing point spheres (KPP) and how many surrounding
regions are adjacent (4CM).
Diagnostics:
In KPP density is a dependent variable of kissing point number and in
4CM, coloring is a dependent variable of adjacency number. Both proofs
are virtually identical in method and where I can substitute crosswise
the concepts, such as density is coloring and kissing point number is
adjacency number. I do not need this fact in the proof of either KPP or
4CM, but I state this as a aid in understanding and teaching guide to
the proof.
The KPP density is pi/3sqrt2 or 74.048,,,percent and achieved by having
every cell possess 12 kissing points. So a proof would say suppose
false then there exists at least one cell that has 13 kissing points.
Contradiction to regular-polyhedra theorem that says the spacing of
regular polyhedra can come in only the forms of 4,6,8,12,20 vertices.
Now, what about 14 kissing points or 15, or 20 and the answer is that
13 is the smallest number of kissing points and that you cannot have a
14 kissing point cell without traversing past a 13 kissing point cell.
The 4CM, suppose false then there exists a cell of adjacencies that
requires 5 colors. Obviously such a cell must have at least 5 regions
and 5 adjacencies as per the definition of mapp-coloring. So what is
the smallest cell that requires 5 colors:
(1) Smallest cell is 5 Mutually Adjacent regions
(2) Second smallest cell is 4 Mutually Adjacent regions and 2 nonmutual
adjacent regions
Let us review 4 Coloring cells:
(i) Smallest 4 Coloring cell is 4 Mutually Adjacent regions
(ii) Second smallest 4 coloring cell is 3 mutually adjacent regions
with 2 nonmutual adjacent
(iii) third smallest 4 coloring cell is 3 mutually adjacent regions
with 3 nonmutual adjacencies
Proof of 4 Color Mapping: suppose false, then there exists a mapping of
5 mutually adjacent regions which requires 5 colors. Contradiction to
Moebius theorem of 4 maximum mutual adjacencies. And, since all other 5
colorings require more adjacencies than the smallest which is 5 mutual
adjacencies are disposed with 4 colors required.
If you do not allow the existence of a 5 mutual adjacency in the plane
which is the smallest 5 coloring to exist, then more complex 5 coloring
all require a greater number of adjacencies. The second smallest
5-coloring requires 6 adjacencies.
Since the 5 mutual adjacencies is the smallest cell and it cannot exist
then the second smallest, third smallest, fourth smallest etc etc,
since they all are increasing in number of adjacencies, also do not
exist. Using the Smallest Member Existor Mathematical Induction
principle (what Chris calls Minimum Counterexample)
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	13 Feb 2006 12:04:17 AM

wrote:

[...]
Proof of 4 Color Mapping: suppose false, then there exists a mapping of
5 mutually adjacent regions which requires 5 colors. Contradiction to
Moebius theorem of 4 maximum mutual adjacencies. And, since all other 5
colorings require more adjacencies than the smallest which is 5 mutual
adjacencies are disposed with 4 colors required.

If you do not allow the existence of a 5 mutual adjacency in the plane
which is the smallest 5 coloring to exist, then more complex 5 coloring
all require a greater number of adjacencies. The second smallest
5-coloring requires 6 adjacencies.

Since the 5 mutual adjacencies is the smallest cell and it cannot exist
then the second smallest, third smallest, fourth smallest etc etc,
since they all are increasing in number of adjacencies, also do not
exist. Using the Smallest Member Existor Mathematical Induction
principle (what Chris calls Minimum Counterexample)

Okay. We're going to look at the structure of this proof, without the
details. The 4CT can be reduced to a statement of the form
4CT: If M is an element of S, then M has property C.
(S is the set of all (planar) maps, and C is the property that the
mentioned map can be colored with 4 or fewer colors.)
AP's "proof", in its structural form, is:
Assume the 4CT is false. Then there is a minimal counterexample A.
There is also a specific counterexample B. Since B has fewer
regions
(AP calls these "adjacencies") than A, A is not a minimal
counterexample. Contradiction, therefore the 4CT is true.
Here, A is the counterexample which has at least 6 regions. B is the
configuration which has 5 mutually adjacent regions.
There is no problem with the structure of the proof. The error is that
B is NOT a counterexample to the 4CT, since a counterexample M to the
4CT must satisfy two conditions:
(1) M is an element of S, and
(2) M does not have property C,
and the configuration B is NOT an element of S. Therefore, it cannot
satisfy both (1) and (2), therefore it is not a counterexample to the
4CT. So AP's "proof" becomes:
Assume the 4CT is false. Then there is a minimal counterexample A.
A in S and does not have property C. There also exists a smaller B
which does not have property C. This contradicts the minimality of
A.
Therefore, the 4CT is true.
This outline does not provide a valid proof, because it is never shown
that B is in S.
To see that this argument is invalid, let's make some changes::
S: now the set of even integers >= 4
C: being the sum of two primes
B: 3
The proof becomes a "proof" of the Goldbach Conjecture (GC):
Assume GC is false. Then there is a minimal counterexample A.
Thus A is even and >= 4 and cannot be written as the sum of two
primes. There also exists a number, 3, which cannot be written as
the sum of two primes. This contradicts the minimality of A.
Therefore,
the GC is true.
Congratulations, AP, you've found a "valid" proof of GC as well!
--- Christopher Heckman
P.S. That last paragraph, "Congratulations ...", is meant to be read
with irony.
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	13 Feb 2006 02:13:26 AM

Chris Heckman wrote:
There is no problem with the structure of the proof. The error is that
B is NOT a counterexample to the 4CT, since a counterexample M to the
4CT must satisfy two conditions:
A.P. writes:
This is criticism without understanding of what I wrote. I used 4
coloring to make easier the explanation of 5 coloring. Perhaps I should
not have talked about both 4 coloring and 5 coloring for it seemed to
have confused Chris.
So let me explain the proof structure a different way using only 5
coloring, and perhaps it may make the entire argument even more clear.
In the Kepler Packing Problem KPP proof we assumed false and a Reductio
Ad Absurdum and that brought up a cell of 13 kissing points and the
smallest such cell counter to the KPP density.
So in this Reductio Ad Absurdum of KPP the "smallest member of
reverse-Math-Induction" (others call it minimal counterexamples) is 13
kissing points and the entire list goes like this:
13 kissing point cell, 14 kissing point cell, 15 kissing point cell,
..... ad infinitum
Now the 13 kissing point cell cannot exist because there is no 13
vertices regular-polyhedra.
So if the smallest member of the counterexample cannot exist for KPP,
then neither does the 14 kissing point cell and higher exist for KPP.
Hence the proof.
Now with 4CM proof, assume false in a Reductio Ad Absurdum RAA argument
which delivers unto us the smallest cell counter to the 4CM and this
smallest cell is a 5 mutually-adjacent requiring 5 colors.
And here is the list of cells starting with the smallest that satisfies
this RAA proof of 4 CM.
5 mutually adjacent cell, 6 adjacent cell consisting of 4
mutual-adjacencies and 2 nonmutual, 7 adjacent cell consisting of 4
mutual adjacencies and 3 nonmutual, 8 adjacency cell consisting of 4
mutual adjacencies and 4 nonmutual, etc etc ad infinitum.
Now apply the "smallest member of reverse-Math-Induction" (others call
it minimal counterexamples). Since the smallest cell of 5 mutually
adjacent cannot exist because of Moebius theorem which says 4 mutual
adjacencies is the maximum. Then neither can any of the other cells
requiring 5 colors exist. Hence the proof of 4 CM.
Example: if no smallest human exists then no humans exist. The list of
humans goes like this: 1 human being, 2 human beings, 3 human beings ,
ad infinitum. So if the smallest member is nonexistant; that is not one
single human exists, then every other member of that list is
nonexistant also.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	13 Feb 2006 02:08:59 PM

Last night I wrote:
5 mutually adjacent cell, 6 adjacent cell consisting of 4
mutual-adjacencies and 2 nonmutual, 7 adjacent cell consisting of 4
mutual adjacencies and 3 nonmutual, 8 adjacency cell consisting of 4
mutual adjacencies and 4 nonmutual, etc etc ad infinitum.
Today I further write:
Now I am going to anticipate the next mistake by Chris Heckman over my
one page proof of 4 Color Mapping. I anticipate this is where he is
going to have flagging, flailing, and falsing comments.
In the above well-ordered list of the smallest counterexamples can be
reduced even further to this list:
(i) 5 mutually-adjacent
(ii) all other 5 adjacent regions
(iii) 6 adjacent regions
(iv) 7 adjacent regions
(v) 8 adjacent regions
ad infinitum
Now Chris is likely to think that there can be a 6 adjacent region
which requires 5 colors and for which the Moebius theorem would not
touch. This would be an error of logic for Chris because if you have a
6 adjacent region that is requiring of 5 colors can be reduced to a 5
mutually adjacent region by combining two of the regions.
Example: Nevada is surrounded by Ca,Or,Id,Ut,Ar and is not 4 mutual
adjacency, but it is easy to combine these states and force it to be 4
mutual adjacency.
So, if anyone crabs and bitches and complains about the proof I have
given, saying that ho ho ho, what if a 8 adjacency cell requires 5
colors, well, they are not thinking well enough, because that can be
reduced, easily, to a 5 mutual adjacency cell.
So if a 5 mutual adjacency cell is NonExistable, then a cell of higher
adjacency will never require 5 colors either because those adjacencies
can be topologically formed into a 5 mutual adjacency.
And likewise the logic transfers to the Kepler Packing so that if
someone crabs complains and bitches that , ho ho ho, what about a cell
of 14 kissing points or 20 kissing points and then I come back with the
answer that if you cannot have a cell of 13 kissing points you cannot
ever traverse further into one of 14 or 20 kissing points.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "ldb"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	13 Feb 2006 02:29:52 PM

"if you have a
6 adjacent region that is requiring of 5 colors can be reduced to a 5
mutually adjacent region by combining two of the regions.
Example: Nevada is surrounded by Ca,Or,Id,Ut,Ar and is not 4 mutual
adjacency, but it is easy to combine these states and force it to be 4
mutual adjacency."
No one believes proofs by example. Prove that you can do it, always. If
you can, you win. Just a little FYI, you are actually getting pretty
close the actual proof now. The only problem is, proving that
X-adjacents can be reduced takes up several thousand pages.
Good luck with that.
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	13 Feb 2006 05:31:43 PM

wrote:

No; this is not a counterexample, because this "configuration" is not a
map. A counterexample ot the 4CT must BOTH (1) be a map, and (2)
require at least 5 colors. You're making the same mistake over and
over, repetitiously.

(ii) all other 5 adjacent regions

These are not counterexamples, because if you one of these, it is both
a map and 4-colorable.

(iii) 6 adjacent regions

There could be counterexamples here, but it was proven long ago (100
years ago?) that there aren't, and without a computer.

(iv) 7 adjacent regions
(v) 8 adjacent regions
ad infinitum

Now Chris is likely to think that there can be a 6 adjacent region
which requires 5 colors and for which the Moebius theorem would not
touch. This would be an error of logic for Chris because if you have a
6 adjacent region that is requiring of 5 colors can be reduced to a 5
mutually adjacent region by combining two of the regions.

Example: Nevada is surrounded by Ca,Or,Id,Ut,Ar and is not 4 mutual
adjacency, but it is easy to combine these states and force it to be 4
mutual adjacency.

Yes, but combining states can also change the number of colors
necessary to color it, so you may go from a map requiring 5 colors to
one requiring 4, and then you get stuck because you can't do any more
reduction at that point.
Starting with Nevada and its neighbors, if you combine any two
neighboring states, you get a map that only requires 3 colors.
However, What you're doing sounds reasonable ... In fact, you seem to
be on the road to proving Hadwiger's Conjecture in addition to the 4CT.

So, if anyone crabs and bitches and complains about the proof I have
given, saying that ho ho ho, what if a 8 adjacency cell requires 5
colors, well, they are not thinking well enough, because that can be
reduced, easily, to a 5 mutual adjacency cell.

I object only to the word "easily". There could potentially be many
many cases, even though you could do it for one particular
configuration.
--- Christopher Heckman
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	14 Feb 2006 03:49:25 AM

Chris Heckman wrote:
No; this is not a counterexample, because this "configuration" is not a
map. A counterexample ot the 4CT must BOTH (1) be a map, and (2)
require at least 5 colors. You're making the same mistake over and
over, repetitiously.
(snips)
However, What you're doing sounds reasonable ... In fact, you seem to
be on the road to proving Hadwiger's Conjecture in addition to the 4CT.
A.P. writes:
Well both of us share the blame for the error of terminology. You said
it is called "Minimal Counterexamples". I want to call it Reversal
Mathematical Induction. So using your terminology I started calling the
well-orderings as counterexamples. So there is a mix-up here in
terminology, but the logic of my proof is steadfast.
As to the comment of Hadwiger. That is indeed what this proof has
become. However, Hadwiger did not go the final distance, and it seems
bizarre as to why Appel, Haken, R,S,S, T never arrived at this point.
The point I speak of is the full use of the Moebius theorem that no 5
mutual adjacency can exist in the plane. Now, take every cell of the
plane and count the adjacency of every cell. Every cell that has a 4
adjacency can be reduced to a 4 mutual adjacency. Every cell that has a
5 adjacency can be reduced to a 4 mutual adjacency, ad infinitum.
Now take a 5 mutual adjacency cell which is nonexistent by the Moebius
theorem. Take a 5 adjacency cell and it must reduce to a 5 mutual
adjacency cell. Take a 6 adjacency cell and it must reduce to a 5
mutual adjacency cell, ad infinitum.
Thus the 4 CM becomes simply the mechanical idea that because 5 mutual
adjacency cell cannot exist by Moebius theorem and all other adjacency
cells of 5 or greater must reduce to 5 mutual adjacencies, impossible
via Moebius theorem, and hence 4 CM.
So, Chris, why has Hadwiger, Appel, Haken, R, S, S, T and Heckman all
failed to realize the full extent of the Moebius theorem? When you know
a 5 mutual adjacency is nonexistent and you combine that fact with the
realization that all adjacencies of cells of 5 or greater reduce to
that of the condition of 5 mutual adjacency and thus the proof of 4CM.
Why did all these people have such thick blinders on for so many years?
I mean the logic above is rather elementary.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	15 Feb 2006 01:10:29 AM

wrote:

Chris Heckman wrote:

However, What you're doing sounds reasonable ... In fact, you seem to
be on the road to proving Hadwiger's Conjecture in addition to the 4CT.
[...]
[now AP:]
Now, take every cell of the
plane and count the adjacency of every cell. Every cell that has a 4
adjacency can be reduced to a 4 mutual adjacency.

This is true. Basically, if the consecutive neighbors of the cell C are
C1, C2, C3, ..., Cn
(n >= 3). (Note that Cn is adjacent to C1.) You can get a 4 mutual
adjacency using the "supercells" C, C1, C2, and C3 union C4 union ...
Cn. This is the main part of a proof of Hadwiger's Conjecture with N=4,
and all you need is one cell with at least 3 neighbors.

Every cell that has a
5 adjacency can be reduced to a 4 mutual adjacency, ad infinitum.

Yes, but you can't reduce all of them at once.

Now take a 5 mutual adjacency cell which is nonexistent by the Moebius
theorem. Take a 5 adjacency cell and it must reduce to a 5 mutual
adjacency cell.

That's not true; if you have a country with 5 neighbors, you cannot get
a 5 mutual adjacency cell out of it. If you have a country C with
consecutive neighbors C1, C2, ..., C5, you need to establish the
existence of "chains" from (say) C1 to C3 and from C2 to C4, which do
not have any countries in common. There's no guarantee that C1 and C3
are adjacent (they might, but they might not be), so you may need to
use extra countries.
I have a feeling that Kempe's "proof" (which I haven't seen in its
entirety but know via Kempe chains) follows along these lines. (It also
uses the fact that any map has at least one country with at most 5
neighbors --- This is a consequence of Euler's Polyhedron Formula:
V - E + F = 2.)

Take a 6 adjacency cell and it must reduce to a 5
mutual adjacency cell, ad infinitum.

Once again, if you look at a country with 6 neighbors, there just
aren't enough guaranteed adjacencies to get a 5 mutual adjacency cell.

Thus the 4 CM becomes simply the mechanical idea that because 5 mutual
adjacency cell cannot exist by Moebius theorem and all other adjacency
cells of 5 or greater must reduce to 5 mutual adjacencies, impossible
via Moebius theorem, and hence 4 CM.

So, Chris, why has Hadwiger, Appel, Haken, R, S, S, T and Heckman all
failed to realize the full extent of the Moebius theorem? When you know
a 5 mutual adjacency is nonexistent and you combine that fact with the
realization that all adjacencies of cells of 5 or greater reduce to
that of the condition of 5 mutual adjacency and thus the proof of 4CM.

It's a good idea, but the details aren't necessarily easy.
Robertson, Seymour, and Thomas went on to prove Hadwiger's Conjecture
for N=6, so I'm pretty sure that they knew about the approach.

Why did all these people have such thick blinders on for so many years?
I mean the logic above is rather elementary.

"The devil's in the details."
--- Christopher Heckman
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	16 Feb 2006 01:05:38 AM

Proginoskes wrote:

a_plutonium@hotmail.com wrote:

Actually, no, you don't; there's no guarantee that C1 and C2 are
adjacent. And if you look at the node which C, C1, and C2 are all
incident with, and consider the rest of the countries incident with
that node, that won't necessarily work, either. The string of countries
from C1 to C2, C2 to C3, etc., may share countries, which destroys the
4 mutual adjacency.
--- Christopher Heckman
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	14 Feb 2006 12:38:40 PM

wrote:

Before I stick my foot-in-my-mouth, I need to know what you mean by "4
adjacency"?
What is a "4 mutual adjacency"?
If a cell is bordered by more than three other cells. it is possible
that no group of four cells is mutually adjacent!

Now take a 5 mutual adjacency cell which is nonexistent by the Moebius
theorem. Take a 5 adjacency cell and it must reduce to a 5 mutual
adjacency cell. Take a 6 adjacency cell and it must reduce to a 5
mutual adjacency cell, ad infinitum.

Thus the 4 CM becomes simply the mechanical idea that because 5 mutual
adjacency cell cannot exist by Moebius theorem and all other adjacency
cells of 5 or greater must reduce to 5 mutual adjacencies, impossible
via Moebius theorem, and hence 4 CM.

So, Chris, why has Hadwiger, Appel, Haken, R, S, S, T and Heckman all
failed to realize the full extent of the Moebius theorem? When you know
a 5 mutual adjacency is nonexistent and you combine that fact with the
realization that all adjacencies of cells of 5 or greater reduce to
that of the condition of 5 mutual adjacency and thus the proof of 4CM.

Why did all these people have such thick blinders on for so many years?
I mean the logic above is rather elementary.

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	14 Feb 2006 01:34:31 PM

Bill wrote:
Before I stick my foot-in-my-mouth, I need to know what you mean by "4
adjacency"?
What is a "4 mutual adjacency"?
If a cell is bordered by more than three other cells. it is possible
that no group of four cells is mutually adjacent!
A.P. writes:
You are extremely correct in that the slippery slopes of proving the
4CM theorem hinges on the concept of cell. That is why I coupled the
proof of Kepler Packing KPP with that of 4CM so as to borrow from one
another in cases of danger of thinking and flow of logic.
Bill yet does not have the full concept of "cell" in mind. Without the
concept of cell, it was nay impossible for Appel, Haken, R, S, S, T to
prove 4 CM in such a short and concise proof.
The concept of "cell" is similar in both KPP and 4CM. In KPP a cell is
defined as a sphere (circle in 2nd dimension) as to how many
surrounding spheres possess a kissing point. So the cell that has 12
kissing points is called a closed-pack-lattice. Mind you, there are 13
spheres in total involved in this cell. The beauty of the cell concept
is that we can point to any sphere in 3rd dimension and then ask what
the kissing point tally is for that sphere and that is the *cell* of
that sphere.
We never concern ourselves with the how one cell is related to other
cells. The cell concept concerns itself only with any given sphere x
and how many kissing points surround that given sphere x.
Now with 4CM, the cell concept is given any region, call it y, how many
adjacencies does y possess. You see, here again we never bother with
how cells relate to other cells, cell adjacency is not defined. Given a
region y, we ask what the adjacency number for that region y is. So we
count and tabulate how many adjacencies each region of the plane
possess.
Some region may possess 1 adjacency, an island country surrounded by
the ocean. And this would be a cell of 1 adjacency. Another region may
possess 2 adjacencies and this would be a cell of 2 adjacencies.
Now we get to the tricky part of
^
/ /\ \
/ / \ \
/2 / \3 \
/ / 4 \ \
/----------------------\
1
-------------------------|
where the region #4 is a cell of 3 adjacencies and overall is a 4
mutual adjacency. This is the maximum mutual adjacency in the plane as
per Moebius theorem.
All other cells in the plane of greater number of adjacencies are
reducable to this 4 mutual adjacency.
All other cells that are 5 adjacencies or greater than 5 adjacencies
are reducable to 4 mutual adjacency. None of these 5 adjacency cells
are reducable to 5 mutual adjacencies as per Moebius theorem. Hence the
proof of 4 Color Mapping theorem.
You see, Bill, the reason Appel, Haken, R,S,S, T were unable to prove
4CM in a short concise proof is that they failed to align the concept
of Cell with the theorem of Moebius that 4 adjacency is the maximum.
If there exists a mapping that requires 5 colors is equivalent to
saying that the Moebius theorem is false and that a 5 mutual adjacency
exists.
So it is important to get the concept of Cell perfect in mind for
without the concept of cell, you are flailing and failing in the bushes
and weeds and woods as were Appel, Haken, R, S, S, T.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "ldb"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	14 Feb 2006 04:01:55 PM

"All other cells that are 5 adjacencies or greater than 5 adjacencies
are reducable to 4 mutual adjacency."
We are still waiting for your proof of that claim. It's easy to do 1
page proofs when you don't prove the hard parts.
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	15 Feb 2006 01:32:36 AM

ldb wrote:

"All other cells that are 5 adjacencies or greater than 5 adjacencies
are reducable to 4 mutual adjacency."

We are still waiting for your proof of that claim. It's easy to do 1
page proofs when you don't prove the hard parts.

It's not quite true, but if you make assumptions which are reasonable
to make when proving the 4CT, it is true.
(I'll use standard graph theory terminology here.)
If you are going to color a planar graph G, you can assume that every
face is a triangle. That means two "consecutive" neighbors of any
vertex are themselves adjacent.
Suppose you have a vertex v of degree >= 3. Then the graph G(v)
consisting of v and its neighbors (and all the adjacencies in G) has K4
as a minor.
Proof: Let the neighbors of v be, in order, u(1), u(2), ..., u(d),
where d is the degree of v. Then u(i) is adjacent to u(i+1) for all i,
v is adjacent to u(i) for all i, u(1) is adjacent to u(d), and d is at
least 3. If you contract the edges u(3)u(4), u(4)u(5), ..., u(d-1)u(d),
the resulting graph is K4. QED.
--- Christopher Heckman
.

User: "bill"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	18 Feb 2006 02:45:44 PM

wrote:

Bill wrote:
Before I stick my foot-in-my-mouth, I need to know what you mean by "4
adjacency"?
What is a "4 mutual adjacency"?

If a cell is bordered by more than three other cells. it is possible
that no group of four cells is mutually adjacent!

A.P. writes:

You are extremely correct in that the slippery slopes of proving the
4CM theorem hinges on the concept of cell. That is why I coupled the
proof of Kepler Packing KPP with that of 4CM so as to borrow from one
another in cases of danger of thinking and flow of logic.

Bill yet does not have the full concept of "cell" in mind. Without the
concept of cell, it was nay impossible for Appel, Haken, R, S, S, T to
prove 4 CM in such a short and concise proof.

The concept of "cell" is similar in both KPP and 4CM. In KPP a cell is
defined as a sphere (circle in 2nd dimension) as to how many
surrounding spheres possess a kissing point. So the cell that has 12
kissing points is called a closed-pack-lattice. Mind you, there are 13
spheres in total involved in this cell. The beauty of the cell concept
is that we can point to any sphere in 3rd dimension and then ask what
the kissing point tally is for that sphere and that is the *cell* of
that sphere.

We never concern ourselves with the how one cell is related to other
cells. The cell concept concerns itself only with any given sphere x
and how many kissing points surround that given sphere x.

Now with 4CM, the cell concept is given any region, call it y, how many
adjacencies does y possess. You see, here again we never bother with
how cells relate to other cells, cell adjacency is not defined. Given a
region y, we ask what the adjacency number for that region y is. So we
count and tabulate how many adjacencies each region of the plane
possess.

Some region may possess 1 adjacency, an island country surrounded by
the ocean. And this would be a cell of 1 adjacency. Another region may
possess 2 adjacencies and this would be a cell of 2 adjacencies.

Now we get to the tricky part of

^
/ /\ \
/ / \ \
/2 / \3 \
/ / 4 \ \
/----------------------\
1
-------------------------|

where the region #4 is a cell of 3 adjacencies and overall is a 4
mutual adjacency. This is the maximum mutual adjacency in the plane as
per Moebius theorem.

All other cells in the plane of greater number of adjacencies are
reducable to this 4 mutual adjacency.

All other cells that are 5 adjacencies or greater than 5 adjacencies
are reducable to 4 mutual adjacency. None of these 5 adjacency cells
are reducable to 5 mutual adjacencies as per Moebius theorem. Hence the
proof of 4 Color Mapping theorem.

How do you reduce a cell to mutual adjaciencies?

You see, Bill, the reason Appel, Haken, R,S,S, T were unable to prove
4CM in a short concise proof is that they failed to align the concept
of Cell with the theorem of Moebius that 4 adjacency is the maximum.

If there exists a mapping that requires 5 colors is equivalent to
saying that the Moebius theorem is false and that a 5 mutual adjacency
exists.

So it is important to get the concept of Cell perfect in mind for
without the concept of cell, you are flailing and failing in the bushes
and weeds and woods as were Appel, Haken, R, S, S, T.

If one region is adjacent to four regions each of which has a different
color then that group of regions is 5-C. It is intuitively obvious
that the adjacent regions cannot be forced to require four colors by
the other vertices and edges in the graph; but this has not yet been
specifically proven!

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	15 Feb 2006 01:31:37 AM

Earlier today I wrote:
All other cells in the plane of greater number of adjacencies are
reducable to this 4 mutual adjacency.
A.P. writes:
I need to define one more concept, and then I think I am mostly done.
The concept of "Reducable". I need to call it "Topologically
Reducable".
The idea is basically that given any cell of a number of adjacencies
larger than 3 adjacencies is topologically-reducable to that of 4
Mutual Adjacency.
Chris Heckman called them superstates. But I do not like that
terminology. I wonder if Moebius in proving his theorem used an idea of
Reducable. I wonder if Moebius was the father of the subject of
topology and that his theorem is the beginnings of topology. Whether he
used some form of reducability.
The central idea of "Topological Reducibility" is that a given region
call it #1
forms a cell that has x number of adjacencies where x is greater than 2
And where you combine those adjacencies such that it is a triangle
forming 4 mutual adjacencies.
^
/ /\ \
/ / \ \
/2 / \3 \
/ / 1 \ \
/----------------------\
4
-------------------------|
Perhaps it is a more generalized Moebius theorem that this Topological
Reduction bespeaks of. And thus is a automatic proof of 4CM, since all
cells topologically reduce to 4 mutual adjacencies and none can be 5
mutual adjacencies.
I did say the Moebius theorem was equivalent to the 4CM.
Now in Kepler Packing KPP, there is no call for topological
reducability because all the spheres are uniform in size. There is no
topology maneuvers in KPP. But in 4CM it is all topology because the
Moebius theorem is purely topology.
Now in topology, all closed figures are topologically-equivalent to the
triangle and it is the triangle-cell above of the 4 Mutual Adjacency
that is the most simple representation of the 4 Mutual Adjacency cell.
So what I am saying is that given any cell of greater than 2
adjacencies is able to be topologically bent, shaped and combined to
form a 4 mutual adjacency cell.
Perhaps - A - Proof: given a region that forms a cell that has greater
than 2 adjacencies, then that cell can be reformed into a triangle and
where those adjacencies are bent into the shape of #2, #3,#4 above and
combined.
So, if I can be granted the concept of Topological Reduction, and then
combine that with the cell concept and then applying Moebius theorem
and lastly applying Reverse Mathematical Induction (minimal
counterexample). Ends as a nice solid proof of 4CM and where 4 and only
4 paragraphs are necessary and sufficient, eh.
P.S. I would like to tie into Moebius, for I cannot imagine how he
could have proven his theorem without the idea of Topological-Reduction
surfacing within his proof of that theorem. If he used the idea, then I
borrow it. If he did not use the idea, then I have a stronger theorem
that 4 mutual adjacencies is the maximum adjacencies. Stronger because
it has an additional concept of Topological Reduction.
Anyone familar with how Moebius proved his theorem and whether he used
topological-reduction or whether he implied it.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	15 Feb 2006 02:29:14 AM

Okay, here is the outline of the final formal proof of 4 Color Mapping
theorem, which I estimate is about one page in length:
Define the concepts of "cell" and "topological reduction".
Proof Outline: Suppose false, then there exists a cell in the plane
that requires 5 colors.
This cell has 4 or more adjacencies. From the Moebius theorem 4 mutual
adjacencies is the maximum. From topological-reduction all cells are
reducible to a 4 mutual adjacency cell. So the cell requiring 5 colors,
is also topologically-reducible not only to a 4 mutual adjacent cell
but to a 5 mutual adjacent cell. All 5 colorable cells topologically
reduce to a 5 mutual adjacent cell. Using
Reversal-Mathematical-Induction (minimal counterexample), no cell
requiring 5 colors exists because the 5 mutual adjacent cell does not
exist via Moebius theorem. Hence every cell in the plane requires only
4 colors and since cells cover the entire plane, hence 4CM.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	16 Feb 2006 01:18:17 AM

wrote:

Good so far. The terminology makes me cringe, but I think my AP-Graph
Theory dictionary is correct.

All 5 colorable cells topologically reduce to a 5 mutual adjacent cell.

This is a huge step. Details will be necessary.

Using
Reversal-Mathematical-Induction (minimal counterexample), no cell
requiring 5 colors exists because the 5 mutual adjacent cell does not
exist via Moebius theorem. Hence every cell in the plane requires only
4 colors and since cells cover the entire plane, hence 4CM.

I'm reminded of a cartoon in Sidney Harris's _What's So Funny About
Science?_, where two mathematicians are standing next to a blackboard.
On the board are a group of equations, the phrase "Then a miracle
occurs", then some more equations. One of the mathematicians is
pointing to the phrase and saying, "I think you should be more explicit
here in step two."
--- Christopher Heckman
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	16 Feb 2006 03:36:27 AM

Chris Heckman wrote:

Define the concepts of "cell" and "topological reduction".
Proof Outline: Suppose false, then there exists a cell in the plane
that requires 5 colors.
This cell has 4 or more adjacencies. From the Moebius theorem 4 mutual
adjacencies is the maximum. From topological-reduction all cells are
reducible to a 4 mutual adjacency cell. So the cell requiring 5 colors,
is also topologically-reducible not only to a 4 mutual adjacent cell
but to a 5 mutual adjacent cell.

Good so far. The terminology makes me cringe, but I think my AP-Graph
Theory dictionary is correct.

All 5 colorable cells topologically reduce to a 5 mutual adjacent cell.

This is a huge step. Details will be necessary.

A.P. writes:
I should not use the word "colorable" for that sounds like something
from an A,H,R,S,S,T proof. So that sentence is confusing.
I should have said " Every cell that requires 5 colors is
topologically-reducable to a cell that is 5 mutually adjacent".
We can topologically-reduce all polygons to a triangle and all
irregular closed curves to a triangle. We can topologically reduce all
cells to a 1 adjacency cell; all 2 adjaceny cells to 3 mutual
adjacencies. We can topologically reduce all 3 adjacency cells to 4
mutual adjacencies.
So, when we suppose there exists a cell requiring 5 colors, we can
topologically reduce that cell to 5 mutual adjacencies.
The reason I need that in the proof is to eliminate all cases of cells
that require 5 colors. And the way to do that is to reduce all cells
requiring 5 colors to that of 5 mutual adjacency cell.
We know that no cell exists that requires 5 colors, but in this
Reductio Ad Absurdum method, we infer such a cell exists. We do not
know if this cell is 5 mutual adjacency or some hybrid adjacencies. So
we want to have under control all cells that require 5 colors, and this
is the beauty of topological-reduction in that all such cells requiring
5 colors reduces to 5 mutual adjacencies. They all reduce to 4
mutual-adjacencies, or 3 mutual-adjacencies if we want to reduce them
even further, to a 2 mutual adjacencies.
Just as a 20 sided polygon can be reduced topologically to a 12 sided
polygon or to a 8 sided polygon or a 4 sided polygon or to a triangle.
You see, the simple fact of the Moebius theorem does most all the work
of proving 4 color mapping. So why stop at applying Moebius theorem to
just 5 mutual adjacencies and extend the Moebius theorem in proving
that all cells requiring 5 colors with all sorts of adjacencies, are
all vanquished simultaneously and thus no messy numerous cases to
solve.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	17 Feb 2006 01:17:50 AM

wrote:

[...]
I should have said " Every cell that requires 5 colors is
topologically-reducable to a cell that is 5 mutually adjacent".

That's what I thought you mean. Even so, it's something that needs to
be proven.

We can topologically-reduce all polygons to a triangle and all
irregular closed curves to a triangle.

Even so, you wouldn't want to do this for the proof of the 4CT. To get
a contradiction, you need to end up with 5 mutually adjacent countries,
and this requires each of those countries to have at least 4 neighbors.

We can topologically reduce all
cells to a 1 adjacency cell; all 2 adjaceny cells to 3 mutual
adjacencies. We can topologically reduce all 3 adjacency cells to 4
mutual adjacencies.

No, you can't. If you have a country C which is adjacent to C1 and C2,
but C1 is not adjacent to C2, how do you reduce C, C1, and C2 to a 3
mutual adjacency?

You see, the simple fact of the Moebius theorem does most all the work
of proving 4 color mapping. So why stop at applying Moebius theorem to
just 5 mutual adjacencies and extend the Moebius theorem in proving
that all cells requiring 5 colors with all sorts of adjacencies, are
all vanquished simultaneously and thus no messy numerous cases to
solve.

I mentioned how you can do this, by defining "supercountries": A
supercountry consists of the union of at least one country, provided
it's in one piece. Also, two supercountries are not allowed to contain
a country in the original map.
If you can topologically reduce a map to 5-mutually adjacent countries,
then you can also find five supercountries which are all adjacent to
each other. Then you can use Moebius's result directly.
--- Christopher Heckman
.

User: "a_plutonium"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same techniqueof Smallest Member	17 Feb 2006 12:23:26 PM

Proginoskes wrote:

a_plutonium@hotmail.com wrote:

[...]
I should have said " Every cell that requires 5 colors is
topologically-reducable to a cell that is 5 mutually adjacent".

That's what I thought you mean. Even so, it's something that needs to
be proven.

My proof of this goes to 3rd dimensional space where I use the very
finest example of regular-adjacency versus mutual-adjacency. And which
offers a modern day proof of Moebius theorem. Call it the Finest Example
Method.
The finest example of 4 mutual-Adjacency is the triangle 3 sides as
mutual adjacent regions with the 4th as the interior landlocked region.
The finest example of infinite mutual adjacency is the bundled and
twisted wiring where the central wire is bigger than the outer wires.
Now to prove that any cell requiring 5 colors is topologically-reducible
to 5 mutual adjacency is to use this wiring example of 3rd dimension
where all adjacencies of 5 or larger is topologically-reducible to 5
mutual-adjacency and this is easily done.

We can topologically-reduce all polygons to a triangle and all
irregular closed curves to a triangle.

We can topologically reduce all
cells to a 1 adjacency cell; all 2 adjaceny cells to 3 mutual
adjacencies. We can topologically reduce all 3 adjacency cells to 4
mutual adjacencies.

No, you can't. If you have a country C which is adjacent to C1 and C2,
but C1 is not adjacent to C2, how do you reduce C, C1, and C2 to a 3
mutual adjacency?

Okay, I was typing too fast with too many examples.

I mentioned how you can do this, by defining "supercountries": A
supercountry consists of the union of at least one country, provided
it's in one piece. Also, two supercountries are not allowed to contain
a country in the original map.

If you can topologically reduce a map to 5-mutually adjacent countries,
then you can also find five supercountries which are all adjacent to
each other. Then you can use Moebius's result directly.

--- Christopher Heckman

I am going to start to offer a new modern day proof of Moebius theorem
using the idea that the finest example of 4 mutual adjacency is a
triangle with landlocked interior region and the finest example in 3rd
dimension is a electrical bundling of wires.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	15 Feb 2006 01:26:24 AM

When you say "A & H and R, S, S, T didn't know about the concept of a
cell", you are arguing without any evidence. In fact, if you read their
proofs, or even an outline, you run across the idea of a
"configuration", which uses your concept of a "cell" (which they
actually called the "neighborhood"). They just couldn't use this
concept in an easy manner.
a_plutonium@hotmail.com wrote:

A, H, R, S, S, and T all know of a "cell" as a "neighborhood", and the
"adjacency" as "degree".

The concept of "cell" is similar in both KPP and 4CM. In KPP a cell is
defined as a sphere (circle in 2nd dimension) as to how many
surrounding spheres possess a kissing point. So the cell that has 12
kissing points is called a closed-pack-lattice. Mind you, there are 13
spheres in total involved in this cell. The beauty of the cell concept
is that we can point to any sphere in 3rd dimension and then ask what
the kissing point tally is for that sphere and that is the *cell* of
that sphere.

We never concern ourselves with the how one cell is related to other
cells. The cell concept concerns itself only with any given sphere x
and how many kissing points surround that given sphere x.

Now with 4CM, the cell concept is given any region, call it y, how many
adjacencies does y possess. You see, here again we never bother with
how cells relate to other cells, cell adjacency is not defined. Given a
region y, we ask what the adjacency number for that region y is. So we
count and tabulate how many adjacencies each region of the plane
possess.

Some region may possess 1 adjacency, an island country surrounded by
the ocean. And this would be a cell of 1 adjacency. Another region may
possess 2 adjacencies and this would be a cell of 2 adjacencies.

Now we get to the tricky part of

^
/ /\ \
/ / \ \
/2 / \3 \
/ / 4 \ \
/----------------------\
1
-------------------------|

where the region #4 is a cell of 3 adjacencies and overall is a 4
mutual adjacency. This is the maximum mutual adjacency in the plane as
per Moebius theorem.

All other cells in the plane of greater number of adjacencies are
reducable to this 4 mutual adjacency.

You need a cell with at least 3 adjacencies (4 countries) to get a 4
mutual adjacency. If every country is adjacent to 1 or 2 other
countries, then you cannot get a 4 mutual adjacency.

All other cells that are 5 adjacencies or greater than 5 adjacencies
are reducable to 4 mutual adjacency. None of these 5 adjacency cells
are reducable to 5 mutual adjacencies as per Moebius theorem. Hence the
proof of 4 Color Mapping theorem.

Kempe's proof, anyway. I may post it separately, converted to
Archimedes Plutonium's terminology.

Not true; they were familiar with the concept. (It is part of the
definition of a "configuration".)

If there exists a mapping that requires 5 colors is equivalent to
saying that the Moebius theorem is false and that a 5 mutual adjacency
exists.

And you've proven Hadwiger's Conjecture in the process.
--- Christopher Heckman
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	15 Feb 2006 02:00:41 AM

Chris Heckman wrote:
When you say "A & H and R, S, S, T didn't know about the concept of a
cell", you are arguing without any evidence. In fact, if you read their
proofs, or even an outline, you run across the idea of a
"configuration", which uses your concept of a "cell" (which they
actually called the "neighborhood"). They just couldn't use this
concept in an easy manner.
A.P. writes:
Thanks for that information, Chris, but I think A,H,R,S,S,T were
handicapped too much in that they never had the luxury of using another
mathematics proof as a instruction manual on how to prove 4CM. I am
speaking of Kepler Packing. Rarely in the history of mathematics or
sciences has one used another problem to guide oneself through a
different problem. In KPP the terminology is lattice and I cannot use
lattice in 4CM. Neighborhood is okay for 4CM but falls short and
confusing in KPP, but cell serves both accurately. And the term
"degree" tends to obfuscate rather than clear up anything. Instead of
filling a treatise with unnessary terms like "degree" should have just
kept the term "number of adjacencies".
But I wonder if A,H,R,S,S,T used a concept of "topological-reduction".
Where given any cell of 3 or more adjacencies can be combined, bent,
shaped into a triangle of 4 mutual-adjacencies. Chris, I see nothing in
topology that prevents such a action from being realized. That every
cell of 3 or more adjacencies can be reformed and manufactured into a
triangle cell of 4 mutual adjacencies, provided we can combine regions.
Did A,H, R,S,S, T ever mention or discuss such a topological reforming
of regions? I call it "topological-reduction"
I wonder if Moebius used a form of this concept or implied it in his
theorem?
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	16 Feb 2006 01:10:21 AM

wrote:

[...]
But I wonder if A,H,R,S,S,T used a concept of "topological-reduction".
Where given any cell of 3 or more adjacencies can be combined, bent,
shaped into a triangle of 4 mutual-adjacencies.

If so, yes, it's called a "graph minor" (or sometimes a "topological
graph minor"). Robertson and Seymour have written a 12-part paper about
graph minors, so I'm sure they're aware of the concept. 8-)

I wonder if Moebius used a form of this concept or implied it in his
theorem?

I suspect not, but it was easily adapted. (If you do a "topological
reduction" on a map and get 5 adjacent regions, then that map is said
to have a "K5 minor". This takes us to the idea of Hadwiger's
Conjecture as well.)
--- Christopher Heckman
.

User: ""
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	16 Feb 2006 03:47:37 AM

Chris Heckman wrote:
I suspect not, but it was easily adapted. (If you do a "topological
reduction" on a map and get 5 adjacent regions, then that map is said
to have a "K5 minor". This takes us to the idea of Hadwiger's
Conjecture as well.)
A.P. writes:
Then I think the idea of Topological-Reduction creates an extension
unto the Moebius theorem. Perhaps topology was not extant during the
life of Moebius and that his theorem was derived geometrically. So that
this concept of Topological-Reduction extends the Moebius theorem and
makes it even more general.
It seems to me that the Moebius theorem of 4 mutual adjacencies is the
maximum mutual adjacencies borrows from the Jordan Curve theorem, yet
the Moebius theorem seems to precede JCT by about 50 years.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	17 Feb 2006 01:30:07 AM

wrote:

No; actually it shows that Moebius _did_ know about topology.

So that
this concept of Topological-Reduction extends the Moebius theorem and
makes it even more general.

According to
http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Topology_in_mathemat=
ics.html
, topology
] Perhaps the first work which deserves to be considered as the
beginnings
] of topology is due to Euler. In 1736 Euler published a paper on the
solution
] of the K=F6nigsberg bridge problem [...]
This result is also said to be the start of Graph Theory.

It seems to me that the Moebius theorem of 4 mutual adjacencies is the
maximum mutual adjacencies borrows from the Jordan Curve theorem, yet
the Moebius theorem seems to precede JCT by about 50 years.

The usual proof of Mobius's proof is through the use of Euler's formula
for polyhedra:
v - e + f =3D 2
In the "map" with 5 mutual adjacencies, f =3D 5, e =3D C(5,2) =3D 10, and v
<=3D 2/3 * 10.
(Thus v <=3D 6.) Substituting into this equation, we get
2 =3D v - e + f <=3D 6 - 10 + 5 =3D 1.
Of course, Euler's polyhedron formula may have been proved using the
JCT.
--- Christopher Heckman
.

User: "Proginoskes"
Title: Re: proving Kepler Packing and 4 Color Mapping using the same technique of Smallest Member	17 Feb 2006 01:37:54 AM

Proginoskes wrote:

[...]
The usual proof of Mobius's proof is through the use of Euler's formula
for polyhedra:

v - e + f = 2

In the "map" with 5 mutual adjacencies, f = 5, e = C(5,2) = 10, and
v <= 2/3 * 10.
(Thus v <= 6.) Substituting into this equation, we get

2 = v - e + f <= 6 - 10 + 5 = 1.

Of course, Euler's polyhedron formula may have been proved using the
JCT.

No, it couldn't have. The Geometry Junkyard --- at
http://www.ics.uci.edu/~eppstein/junkyard/euler/ --- links to 19 proofs
of Euler's formula, even though some of them (like "Induction on
Faces") uses it. (BTW, Euler didn't prove this formula; he only
conjectured that it was true.)
--- Christopher Heckman
.

User: "a_plutonium"
Title: modern day proof of Moebius theorem using Smallest Example Re: provingKepler Packing and 4 Color Mapping	17 Feb 2006 12:40:41 PM

Proginoskes wrote:

I would have been happier if Moebius tried proving his theorem using the
idea of triangles as the smallest unit of 4 mutual adjacency. But I am
happy that Moebius used Euler polyhedra for it is exactly that idea that
proves the Kepler Packing Problem where no regular-polyhedra of 13
vertices can exist.
But now I am going to start in offering a proof of the Moebius theorem,
provided it does not take too much of my time.
I am going to use what I call the Finest Example, or Bare Minimum
Example Method.
The finest example of the 4 mutual adjacency in 2nd dimension is the
triangle with its interior landlocked region.
The finest example of 5 mutual adjacency in 3rd dimension is the bundled
electrical wiring with the central wire as thicker in diameter from the
wires adjacent. So to achieve mutual adjacency we twist the outlying
wires around the central wire and in one of those twists we achieve
mutual adjacency no matter what number we desire (in our case we want 5
mutual adjacency). To achieve regular-adjacency we simply add parallel
wires in the opposite direction to the central wire of that of the
twisted wires. So for mutual adjacency we twist the wires but for
regular-adjacency we have the wires running parallel to the central wire
in the opposite direction of the twisted bundle. In this manner we
achieve any hybrid that we desire-- e.g. say we wanted 78 mutual
adjacencies with 39 regular-adjacencies and thus do that wiring.
Now for a proof of Moebius theorem that in 2nd dimension
4-mutual-adjacencies is the maximum. Note that the Euclidean plane has
two coordinate axes. The smallest closed figure in the 2nd dimension
plane is the triangle. We use the axes as lines. In 3rd dimension we
cannot close the 3rd dimension using the axes as lines and so the number
of mutual adjacencies is infinite.
Now we use the Smallest Example of mutual adjacency in 2nd dimension
which is the triangle of its landlocked interior region. And the
Smallest Example of the mutual adjacency in 3rd dimension which is the
twisted wires bundle.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: ""
Title: Re: modern day proof of Moebius theorem using Smallest Example Re: proving Kepler Packing and 4 Color Mapping	17 Feb 2006 01:03:47 PM

Make one stipulation to the electrical wiring example, in that the
wires are not round but having flat surfaces like miniature polygons so
that the adjacencies are between flat surfaces and not
pointwise-adjacencies.
A.P.
.