Prev: Is This really a Proof that Standard Math is Inconsistent?
Next: From Cauchy to Hensel p-adics should have been Cauchy to Peano #323 should be #2 ; Correcting Math
From: Alan Smaill on 21 Jan 2010 04:52
Charlie-Boo <shymathguy(a)gmail.com> writes:

> One obvious case is when proving by induction. Any others?

When it makes the formulas being manipulated much shorter.

> C-B

--
Alan Smaill
From: Larry Hammick on 21 Jan 2010 17:04

"Bill Taylor" <w.taylor(a)math.canterbury.ac.nz> wrote in message
news:6760e09d-f308-48e1-b96c-f0b764df6bb0(a)a32g2000yqm.googlegroups.com...
> "Larry Hammick" <larryhamm...(a)telus.net> wrote:
>
>> Along those lines, I'm right now writing up a conjecture that
>> is considerably stronger than the four colour theorem, but technically
>> simper to state, because it applies to some graphs that are not planar as
>> well as all graphs that are planar.
>
> May we be permitted an advance preview of the statement of the
> conjecture?
>
> -- Wondering William

I'll have something coherent on the internet by the end of January. But
here's a quick sketch:

The edges of a planar graph can be partitioned into three subsets each of
which defines a "forest" graph (no circuits) see e.g. "Schnyder's
algorithm". Better, each of those forest graphs can be oriented, i.e. the
vertices, which are what we want to colour, can be given three forest /order
relations/ which moreover satisfy certain axioms relative to one another.
I've worked out a set of such axioms on a triple of forest orders which are
necessary and sufficient for the corresponding graph to be planar. But I've
also got a simpler and weaker set of such axioms and an expanded notion of
colouring, such that the corresponding graph seems to be colourable with X
colours in at least (X-3)^n ways, where n is the number of vertices. And
more -- I have a nontrivial theorem with proof about a graph defined by only
two forest orders, satisfying similar conditions relative to one another,
saying that it is X-colourable in at least (X-2)^n ways.

I'll try to state the 3-colour theorem in the Ascii-speak of Usenet. It
won't be easy.
Some def's needed. By a "relation" or "preorder relation" or the like on a
set S we mean a subset of SxS satisfying the familiar axioms. In our usage a
preorder relation contains the diagonal of SxS.
(A set may carry more than one preorder relation but any family of preorders
on any set has a least upper bound, when the set of preorders is ordered by
inclusion.)
Given a set S carrying an order P and a total order T (two subsets of SxS)
we say that P is "segmental" with respect to T if
-- P is contained in T
-- for any x in S the set of y in S such that xPy, is an interval of S with
respect to the total order T.
There are several equivalent definitions and it can be seen pretty easily
that if P is segmental in some total order then P is a forest order.
If P is segmental in T, then the set (T-P+(diagonal of S)) is a forest
order, and it is segmental in the opposite of T.)
Now for our colouring issue, it is convenient to adjoin a least element
("pole") to each of the forest orders in play. A planar graph, thus
augmented, will have
-- n vertices
-- 3n edges
-- 2n linearly independant "circuits" from one pole to another and distinct
pole.
We want to colour the edges with elements of a group G (it need not be
commutative) such that the sum of the values of the colourations has some
prescribed value (not necessarily 0 as in the original four colour theorem)
on each element of a /circuit basis/. This more general problem means that a
graph is not automatically reducible if two different edges connect the same
two points.
Now what graphs are we claiming are 3-colourable? We define the structure as
a set S with two forest orders U and V, with poles, such that:
-- there exists a total order T1 such that
---- U is segmental in T1
---- V is contained in the opposite of T1.
-- there exists a total order T2 such that
---- V is segmental in T2
---- U is segmental in T2.
(For the four-colour rather than three-colour statement, the axioms are just
like the previous, but for each /pair/ out of the 3 forest orders.)
Claim, which I can prove: For a group G of X>=2 elements, the edges of such
a structure can be coloured with non-identity elements of G such that the
product of those values, along any element of the afforementioned circuit
basis, have any prescribed values, and in at least (X-2)^n ways.

The conjecture goes over to the three-forest case easily, but I don't have a
proof. The conjecture has however passed a few tests.

Footnote:
The notion of segmentality is not altogether new. Take a formula that uses
brackets, e.g.
((x+y)*z)+(u+v).
We see a total order on the set of operands -- their order of appearance
from left to right. But the operations or brackets define a /forest order/
on those operands, and the forest order is /segmental/ with respect to the
total order. With respect to the opposite total order, from right to left,
we get the complimentary forest order. (Speaking loosely here.)

More coming as time permits. Diagrams need to be done, and TeX won't be
enough, so I'll resort to scans of hand-drawn diagrams, and embed them in a
pdf along with the Tex.

LH.


First  |  Prev  | 
Pages: 1 2
Prev: Is This really a Proof that Standard Math is Inconsistent?
Next: From Cauchy to Hensel p-adics should have been Cauchy to Peano #323 should be #2 ; Correcting Math