Actually, if you just change the extension, the website does not even allow the file to be uploaded. Therefore, I have to actually save it as a different type of file to display it as a different filetype here.[/font]
[font=fixedsys]On the other hand, grav, thank you very much for your advice: I learned many new things. However, I did not learn much about the [IMG] tag (see, I know now how to keep it from trying to convert that tag into an image). I had done in the past exactly what the instructions outlined. So I'm still going along with the filetype case. Apparently, gif works with the [IMG] tag, but gif loses quality. So which other images work for the tag without losing quality? 
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My signature to identify my posts until I find a small enough avatar.

What do you mean by that? GIF is a lossless compression, isn't it?

 Remember, we went through this? Well, just in case, this post is where I demonstrated the loss of quality. 
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My signature to identify my posts until I find a small enough avatar.

GIF only loses quality if you have more than 256 colours AND you are using a standard colour palette. When you save your picture, use the option to have a customized palette, and it will be fine.

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"It's turtles all the way down."

Here is a copy of Dragon's PNG file, saved as a GIF:

 Well, first of all, the image that I posted here way back was saved as a GIF with only 16 colours and a custom palette, and so it still did not work. And my computer classifies hhEb09'1's picture as a 256-colours BMP image. Really odd, because were that true, it would show up as a link. So I investigated the case if that is not true. I open his image, click "Save As GIF", and it does not change. I open my image, do the same thing, and it doesn't. I save his file as 16-colour, then PNG, and JPG, and it still works. I open a new file of mine, and it doesn't. I try different ways of saving with my file, and it does not help. I try modifying your file to see if there's something magical about it; as long as I only pour colours, it works. But add something else, and voila.  
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My signature to identify my posts until I find a small enough avatar that I really like.

Attached Images

untitled1.GIF (1.4 KB, 96 views)

*Bump*

I recall that Dragon was supposed to have a "proof" of the theorem published by July....

Does anyone else recall that?

Pete

__________________
PJE

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

He mentioned it last in this post, but I'm not sure he said where it would be published. He hasn't post to BAUT since his last post to this thread.

grav, how's your proof going?

I was just reading Courant and Robbins and thought this'd be a good place to display that proof of the five-color theorem, since I don't remember doing it before. They go about it in a roundabout way, but here's the essentials.

Any map can be drawn on a sphere (just take the outside region that surrounds the whole map and shrink its border to a point--a coloring of the 2D map will color the spherical one, and vice versa). Applying the algorithm that I mentioned earlier, results in a polyhedron where 3 edges meet at each vertex (this is called regular), but each edge is shared by two vertices, so the number of edges E is 3/2 times the number of vertices V (in other words, 3V-2E=0). If all the faces have six or more edges then E is greater than 6 times F divided by 2 (in other words, 3F-E<=0), or

0 >= 3F - E
0 >= 3F - E + (3V - 2E)
0 >= 3F - 3E + 3V
0 >= F - E + V

But that contradicts Euler's famous relationship 2 = F - E + V, so we know that at least one region of a regular map must have less than six other regions touching it. (I ran across this part of the proof years ago in D'Arcy Thompson's On Growth and Form. Applied to the illustration (webpage) of the radiolarian Aulonia hexagona Hkl., it says that there must be a pentagon or two mixed up in all those hexagons!)

Take one of the regions, R, that has five or less other regions touching it. If two of the surrounding regions do not touch each other (this must be so, for four or five), remove the associated edges, otherwise just remove one edge. This results in a regular map that has either 1 or 2 less regions. If that smaller (fewer regions) map can be colored with five colors, then so can the larger, just by adding back in the boundaries, and coloring R with whatever color does not touch it. So, we can reduce the size of the map until we only have to use five colors, then we can add back the edges until the original map is five-colored.

Obviously, the proof of the four-color theorem is a long ways from this simple proof!

PS: over two years ago, peter eldergill and Grey linked to webpages that had explanations of the above proof.

That's interesting. I would have to look it over some more to get the full jest of it, though. Euler's relationship is fascinating to me. It's one of the first things that got me thinking in three dimensions. I never thought about starting with a two dimensional map and wrapping that around, though, making each area a face. It's easier for me to think about if I don't wrap the border around to a point, though, but another face, which would be the outside of the map originally, then, although the point can be opened up anyway, using the algorithm you provided earlier in the thread. The way you stated that with the spherical map, I can now see how Euler's relationship can be shown to be true with what you were describing.

If we start with a single area and build up from that, such as a hexagon, then the border will have an equal number of vertices and edges, and two faces, inside and outside of the shape, since the outside can be wrapped around into another face in the manner you described above. This provides us with the relationship 2 = F - E + V, although kind of after the fact, since we really just have F=2 and E=V, but we can still show thereon that the relationship holds regardless. We can then add another face that connects to the original one, brought out from any two of the vertices of the original face, with any number of vertices and edges of its own, always gaining one more edge than vertices in the process, plus one extra face, so the relationship stays the same. If we continue to add faces this way, the same thing will occur each time, so the relationship remains constant throughout the process. Also, if we use the algorithm you demonstrated earlier in the thread to make the map regular, then we add one face, along with an equal number of edges and vertices, but subtract the original vertice, so the relationship stays the same. It therefore works out for vertices with any number of edges stemming from them.

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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."

Since any polyhedron has a vertex with at least three edges radiating from it (other wise, it would be just a ring, wouldn't it?), you can subtract faces, edges, vertices, until you arrive at a tetrahedron. It's easy to show that Euler's formula works for a tetrahedron, and then your approach shows that it works for the original too.

The last step, going from five color theorem to four color theorem, took a long time!

Oh, so that would be the proof for the four color theorem right there, then! Is that what you mean? If we can start with any polyhedron and reduce it to a tetrahedron by subtracting faces, then we should be able to start with a tetrahedron and add faces to produce any polyhedron. So we have four faces on the tetrahedron originally, each with a different color since they are all joined, which would also show one needs at least four colors to begin with. Then we open up any vertice to create another face. The vertice will join three faces, each necessarily with a different color, so we use the fourth color for the new face each time. If we just keep opening up new faces in the same positions relative to the others in the original map we are trying to color, after it is made regular, plus one face that will act as the outside beyond the border once the polyhedron is unwrapped, then we should be able to reproduce any regular map, already four-colored, in this way, shouldn't we? After unwrapping the polyhedron, we would have the map we made regular, and then we would just make it irregular again by subtracting areas until we regain the original map, already four-colored. Does that sound like it might work out to you?

__________________
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."

Darn, I thought this was a thread about comic books!

__________________
"If you think the LHC will create black holes, you might as well believe Hobbits are at the bottom of your garden."- Dr. Mike Inglis
Rovers forever! - ToSeek
"Carl Sagan sent a message to ET,
Neil Armstrong walked in the Sea of Tranquility
Steve Squyers built Spirit and Opportunity
Dan Haylen upchucked in zero gravity." -Brent Simon, The Space Camp Song

Comic books have theorems now? Whoa, I guess I gotta get back into them

Actually, what I described above is too simple. It would just produce a bunch of triangular areas. We would also have to open up some of the edges from one of the vertices, splitting the original edge, and producing an extra edge for one of the areas, along with creating another face and vertice, though. We could also stretch vertices out into another edge and vertice as well, creating a border between two of the original faces, and an extra edge each. That sort of thing. But those might affect the four coloring, I'm not sure.

Basically, though, I'm thinking that if we can make the original map regular, then twist it around into a polyhedron, and reduce that to a tetrahedron by subtracting faces, then whatever precise method we used to do that can then be exactly reversed, four-coloring each face as we add them back in. In fact, now that I think about it some more as I'm writing this, we could probably by-step the 3D part of that altogether, and might not need to make it regular either, actually, but just reduce the original map by subtracting faces until we get three adjoining areas and the outside of the border, which would represent the same thing as the tetrahedron anyway, and then just build that back up again, four-coloring as we go, right? But that might be different than that algorithm you mentioned after all, however, since adding faces to make it regular and subtracting the same ones again without affecting the four-coloring is not the same as subtracting areas and then adding them back in. No more than three colors could meet at any vertice as we build it back up so we can give the fourth color to the new face each 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."

Since we cannot subtract and then add areas back in the same way according to that algorithm for making maps regular, I'm not sure about getting the method I described above to work out, but it gave me an idea. I can still use the algorithm for that two-color method I came up with earlier in the thread. It's a simple matter to show that if we can two-color a map in that way, it can be four colored. The only problem was coming up with a method to two-color it to begin with. Well, I'm thinking that once the map is made regular, we don't need to stop there. We can keep adding areas by opening up the triple vertices at will. So maybe we could just two-color the map with a simple spiral or something. If we get to an area that thwarts us somehow, we can just open up an extra area to keep us on track. Then, after the map has been two-colored and then four-colored, we can close those extra areas, along with the ones we added to make the map regular.

__________________
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."

You know it is impossible to two color some maps?

Maybe I need to go back over some old posts to see what you mean here.

Some may. But comic book-realted stuff is often described as "four colour".

__________________
"If you think the LHC will create black holes, you might as well believe Hobbits are at the bottom of your garden."- Dr. Mike Inglis
Rovers forever! - ToSeek
"Carl Sagan sent a message to ET,
Neil Armstrong walked in the Sea of Tranquility
Steve Squyers built Spirit and Opportunity
Dan Haylen upchucked in zero gravity." -Brent Simon, The Space Camp Song