You know, I'm probably going to end up cheating with this. It would just take too long to apply colors in every way imaginable to various maps to find out which rules must be applied under given circumstances. I made a program once that played the game 'connect four' against itself by starting out with random moves when it switched to each side. After each game, points were subtracted from whatever moves that led to a loss for the one side according to the circumstances involved, and added for the moves for the side that won, but placed into the same memory. After a few hundred such games, I played against it myself and ended up losing seven out of ten games.

I might try the same thing with this, although it might take a while. I can make random areas on the screen and have the computer fill them in, randomly at first, and then according to its program as points for certain moves are added or subtracted, depending on whether it fills in the entire area or not. I will probably have to start with simple areas for it and then get more complex as it learns which ways work best. If it gets to the point where it can fill in all of the areas every time, then I can just "pick its brain" to see how it is doing it.

If I do that, though, would it be considered cheating, or would I still get credit for it?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

OK, sorry, I am too sleepy for now, but when I say "your rules work", I don't mean that they are not mandatory: I just mean that they are not sufficient to provide a full colouration of a graph. The pictures in the post mentioned above showed that these rules were sufficient to colour that one graph, in this order:[/font]
1. Start out with three colours.
2. If an area has a region definitely coloured adjacent to it, that area cannot be the colour of the adjacent area.
3. If an are only has one possibility for a colouration, it is that possibility.
4. If an area has two possibilities and another area has those two possibilities and a third one, colour the other area with that third colour.
However, these are not sufficient for all graphs. So I showed you how to show that certain rules are sufficient for a graph. Your job now is to check your new rules against the fc.png graph to make sure that they work BEFORE posting them here, and showing that they do work when you post them.
Now here is something else: you don't have to show all the possible colourations for EVERY area: if you notice what I did, I just picked one territory and showed all the possibilities for it, and when I could do no more, I picked another area and showed all the possibilities for it, etc. It is a rule that if one area at a time is followed, that is sufficient to show that the same would work for any combination of areas in the same circumstances one area at a time.
Also, the only reason the current four-colour proof does not get credited by some is because there is no way to extract it from the computer in a way that would be followable by people.[font=Verdana]
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My signature to identify my posts until I find a small enough avatar.

Okay, good. I think we are on the same page now. I tend to work backwards with these sort of things, though. What I want to do at this point is to compile a list of all mandatory rules that must be applied no matter what, and then later to find an algorithm for choosing colors for areas in such a way that the least number of rules may be applied, which is the same thing you're saying, but I'm kind of coming at it from the other direction, I think. In other words, instead of using an algorithm at all for this, we could just color the territories randomly and see where we get into trouble each time, find the rule that would prohibit coloring that way depending on the circumstances, and then apply that rule and any others we have already found with each run until there are none left to keep us from coloring an entire map in randomly, except for wherever those mandatory rules must be applied, however many there turn out to be. And the rules wouldn't just be something we come up with that just happens to work out for a particular situation or something, but it must be absolute, a rule for what we definitely cannot do, so that it can be proven to be so, as with the five rules so far, and this would also help when coming up with a proof for the theorem.

That's true. If I make a program for it, I would have to set it up in a way that when I go through it afterwards, I would be able to pinpoint exactly what it is doing each time. It won't be easy.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

No offense, but you can also work on expressing a rule as briefly as you can. Did you see how concisely I explained yours in my previous post? See if you can do the same with all of your five rules now. Attempting to do so without looking at my rules would help you do the same for any new rules you come up with. By the way, the rules I named do not work by themselves for the fc graph: more rules are needed; and since fc failed the Kempe's original 1890 proof, it certainly is a good map to experiment with for any future rules. In fact I'm fairly sure that even all the rules you currenty have can fail one way or another trying to colour fc. So go ahead and try to see how much failures you can obtain, and if you ever think up of a set of rules that you can't fail colouring fc with, you can post them here (concisely, once again) so that I could try, and if I can't, then we both can work on eliminating redundant rules.
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My signature to identify my posts until I find a small enough avatar.

Neither. The computer program itself would prove nothing. Picking its brains to see how it works will just give you a long list of rules like list you're compiling now. You'd still have to attempt to prove that your list is sufficient before anyone's even going to bother looking at it.

PS There's nothing wrong with using a computer to help you find a tentative set of such rules. The use of a computer in the successfull proof of the theorem was different - they already proved that there were only so many cases to check, then programmed a computer to systematically check all those cases.

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There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

Actually, I think I've changed my mind about that, after thinking about it for a while. I might do it eventually, just for kicks, but coming up with a program that can learn well enough and in such a way that I can go through it afterwards to see what its doing would be quite a chore, and may take weeks or months. I can pretty much tell what goes wrong when I get stuck, and find the mandatory rule that applies accordingly, so I guess I'll just program in the rules as I find them and let the computer run through various maps randomly while running through the rules after coloring each area. If it gets stuck, I'll look through the map and see why, and I'll have a new rule. I can then see if there might be an algorithm that reduces them. I'll definitely need to write a program for this much, at least, so I can quickly run through all possible combinations for a particular map, which would take a tremendous amount of time otherwise. And if anyone else comes up with an algorithm they think works, I could also check that for them.

I'm thinking the number of mandatory rules, which can be run in any order without affecting the others, and would be applied with or without an algorithm if that particular situation arises, would be limited. This is the case if considering only the possibilities for three bordering areas, two of which have the same possibilities for their colors, or at least, that's what I see so far. All of the rules I've come up with include one area with two others that border it, although directly or indirectly through a chain or otherwise, that have the same two possibilities for their colors, so I may be able to eventually prove that all possibilities are accounted for, and if I can do that, I would be just a jump away from a proof for the theorem itself.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Well good luck with the final little jump

__________________
There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

Not quite. Only the ones that you called "common sense"--because those are the ones that we can easily prove that a violation of the rule would result in a failure. The other rules, you do not know that they work. They may or may not.Add "use only four colors" to your list I would bet that if you could prove that the number of rules was limited, you would have accomplished most of the effort needed to prove the theorem.Well, as I say, I think that is where most of the work is.

You're probably .001% finished.

This is probably quite similar to what the guys who originally solved this thing did, and it looks like the right way to approach it. (The accepted proof is down to about 600 "rules" that must be met. I have a feeling that your rule set is going to get very big, and quickly.)

Even if you accept that, it's probably still worth playing with, though. What if you could find a rule that reduced some of those 600 cases into a single case? You'd have reduced the current proof in size, which is pretty significant.

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"It's turtles all the way down."

It would probably help to obtain the set of 600 rules and then look at whether you're going the correct direction.

Grav, I'm sure you probably understand the problem better now than you ever thought you would, don't you (or better than you cared to know it! HA!)

Honestly, I haven't read most of the posts, but I find the problem very intersting.

Pete

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PJE

There's so much I don't know about astrophysics. I wish I had read that book by that wheelchair guy.

Okay, I have finally written a program for that 'fc' map. I only put in those first two mandatory rules to start with for the trial run, though. It only works a little better than about three-quarters of the time for just those two. The program can color the entire map four or five times per second, very much faster than I was doing, so that's pretty good, right? Well, as it turns out, it's still not quite good enough. I've been waiting half the day for the program to run through all of the possible ways for doing it, and now I realize why it's taking so long. With only twenty-five areas to fill in with one of four different colors, there are about a million billion combinations. Considering also the orders that they can be colored in one by one in any particular order drives it up to about possible 10^40 combinations for color and order of coloring of each area. Now, once the rules are applied to the algorithm I've been using, there still seems to be about 1.2 billion combinations possible. Figuring four or five done per second, it should only take about 8.5 years to run the program all of the way through, and adding more rules would require it to take that much longer. I'm definitely going to need a more efficient algorithm. Any suggestions?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Maybe start with Appel and Haken?

I thought the extra rules beyound the first two were to reduce the number of times when you had to make a random choice. Each extra rule's flavour was to take some more definite action before having to resort to arbitrarily chosing a perimeter territory and arbitrarily chosing a color for it. If so then adding the extra rules should reduce rather increase the number of cases to test for each map.

But all that's rather moot when you've still got infinite maps to check

__________________
There are 10 types of people in the world. Those who understand ternary, those who don't, and those waiting for a bus.
If logic doesn't work, then surely it does.

Thanks. I generally like to try to work things through on my own, and only resort to checking out other people's methods at times when I've gone as far as I can go, or get stuck. I guess this would be the time.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

I meant that adding more rules would take up more program time to go through each one, making it take even longer to run, but now that you mention it, it would make for less arbitrary choices, so less overall run-throughs, whereby it might take roughly the same time or possibly even less, you're right. But I'm thinking more along the lines of a starting point that limits the number of areas we can begin with. For example, if we start with the inner most perimeter, by subsequently eliminating all outer perimeters until we get down to the last few areas, then there is only one area in the 'fc' map, anyway, we can use to color in our first area, and then we can add two of the adjoining areas to that.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."