user568995

@wizzwizz4 Did you fully understand the two examples of 4 coloring planar graphs?

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How can I get some up votes?

OK

Your post

Why do I not divide the rational exponents to get $x^{20}y^{16/7}$ from simplifying $\frac{\left(x^5y^{8/3}\right)^{1/4}}{x^{1/16}y^{7/24}}$?

shouldn't I bve able to see math formulas

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I'm in stackexchange

Mathematical formulas posted by others are not rendered from TeX to Print

my page is not converting TeX to readable print. What can I do?

Consider it slave labour.

@ Otaku, Yes High school students may have a quired math project to do. I think they would like this.

OK I understand the generaly accepted def,

It can still be peer reviewed by the broeder community outside academia.

It depends on what you mean by peer?

What does publishing mean?

Huh?

We can just work out a few examples. viXra will publish it.

If you want we can publish papers togethr

We can put them to work doing real research.

Even High school students could do this.

There is a program available for testing with a set of instructions to see if it works . I think a ton of examples will elucidate the correctness or lack therof the algorithm.

I find that a little challenging, better to break the algorithm into well defined components and deal with them separately. I don't know the capabilities of computer proof assistants, But at this point I'm more concerned about the efficacy and correctness of the algorithm in finding proper colorings. Once that is establish and understood it will be time to attempt a proof.

You know when youve done them correctly whem propertys A,B are satisfied. very easy to check

@wizzwizz4 The intermediate step would be fiinding a generalized alternating chain ( a couple of chains of 2 or 3 length appear at some points in the examples). the chains could be more complicated but should be easy to find even tho they might have self intersections and. This is the magic of property B. However I now believe it might much easier than I had originaly thought. because it might be possible to eliminate all branching chains, with simple chains.

@wizz its only half an algorithm as only the main steps are given, but intermediate steps are missing, but I think they can be easily supplied. I think it might be in fact very easy.

Hi

Wizzwizz, are you AI?

I should say that each edge is conflict free wrt the checkers on that edge.

Were trying to prove the four color theorem

this is true for all colors.

Actually we say that all edges are unconflicted wrt any color because There is no edge with a ( lets use red) red checker on it and a red checker on both ends.color say Red

This means that edge 1-6 is unconflicted wrt. green

And a Green checker is placed on edge 1-6

on figure 3 the green checker on vertex 6 is removed

this indicates that the green checker on vertex 1 is pressuring the green checker on vertex 6

arrow from vertex 1 to vertex 6

OK let me fingd the graph now

on fig 2 there is an arrow from

This is the situation on fig 1 of the second paper I showed you

Ok Start off with no checkers on any edge, but checkers of all 4 colors on each vertex

its not hard at all.

I understand the algorithm, the idea of it

there is a program you can download which will run the instructions

Yes but before a program we can learn much by just implementing a sety of instructions

That should give us a ghood idea

I think so . I think the thing to do first is a lot of examples as is done in the papers

look at properties A and B. They are always maintained at each majoe itteration

The algorithm itself is interesting irregardless of proof