4 ms·

Like I said, describing it is hard, but showing it is not See this: https://imgur.com/iU7oc8d https://imgur.com/iU7oc8d There's no way to color that with 4 colo

Like I said, describing it is hard, but showing it is not See this: https://imgur.com/iU7oc8d https://imgur.com/iU7oc8d There's no way to color that with 4 colors without a color hitting itself. Yes, it's contrived, but so are borders. Also, the 4-color theorem didn't limit itself to established borders, it claims to be for an arbitrary map. This image, is well, arbitrary and drawn up in paint in a few minutes, but it shows the point.

kr99x 8y ago
Nope. https://imgur.com/a/SkjlD https://imgur.com/a/SkjlD
guskel 8y ago
https://imgur.com/a/JE5Ov https://imgur.com/a/JE5Ov
benchaney 8y ago
You can do R G B Y B from the inside working out. The third and fifth rings (counting from the inside) don’t touch each other so, you can give them the same color.
Sniffnoy 8y ago
You're mistaken about that not being 4-colorable. Color the "rings", from outside to inside: red, blue, green, blue, yellow. Again, as mentioned in my earlier comment, I would suggest thinking of things / describing things in terms of plane graphs rather than "maps". It'll make everything easier. I might also suggest learning some of the relevant graph theory? It sounds like you're trying and failing to embed a K_5, or perhaps an arbitrary K_n, into the plane. That can't be done, and the proof that this is impossible is much easier than the full 4-color theorem. (There are also the 5-color and 6-color theorems, which again have much easier proofs than the 4-color theorem, and which are another reason you can't embed an arbitrary K_n in the plane.)
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tathougies 8y ago
/r/iamverysmart
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heroohwaitnot 8y ago
Oh man! Few things more hilarious than seeing condescending ignorance like hermitdev's get put down with impunity. In any case, the proofs are widely available and decades old: Appel, Kenneth; Haken, Wolfgang (1989), Every Planar Map is Four-Colorable, Contemporary Mathematics, 98, With the collaboration of J. Koch., Providence, RI: American Mathematical Society, doi:10.1090/conm/098, ISBN 0-8218-5103-9, MR 1025335 And: http://www.ams.org/notices/200811/tx081101382p.pdf http://www.ams.org/notices/200811/tx081101382p.pdf

kr99x 8y ago

Nope. https://imgur.com/a/SkjlD https://imgur.com/a/SkjlD

guskel 8y ago

https://imgur.com/a/JE5Ov https://imgur.com/a/JE5Ov

benchaney 8y ago

You can do R G B Y B from the inside working out. The third and fifth rings (counting from the inside) don’t touch each other so, you can give them the same color.

Sniffnoy 8y ago

You're mistaken about that not being 4-colorable. Color the "rings", from outside to inside: red, blue, green, blue, yellow. Again, as mentioned in my earlier comment, I would suggest thinking of things / describing things in terms of plane graphs rather than "maps". It'll make everything easier. I might also suggest learning some of the relevant graph theory? It sounds like you're trying and failing to embed a K_5, or perhaps an arbitrary K_n, into the plane. That can't be done, and the proof that this is impossible is much easier than the full 4-color theorem. (There are also the 5-color and 6-color theorems, which again have much easier proofs than the 4-color theorem, and which are another reason you can't embed an arbitrary K_n in the plane.)

deleted 8y ago

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tathougies 8y ago

/r/iamverysmart

deleted 8y ago

[deleted]

heroohwaitnot 8y ago

Oh man! Few things more hilarious than seeing condescending ignorance like hermitdev's get put down with impunity. In any case, the proofs are widely available and decades old: Appel, Kenneth; Haken, Wolfgang (1989), Every Planar Map is Four-Colorable, Contemporary Mathematics, 98, With the collaboration of J. Koch., Providence, RI: American Mathematical Society, doi:10.1090/conm/098, ISBN 0-8218-5103-9, MR 1025335 And: http://www.ams.org/notices/200811/tx081101382p.pdf http://www.ams.org/notices/200811/tx081101382p.pdf