Timeline for How many "different" colorings (excluding exchanges) exist for a given map (graph)?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S Nov 29, 2017 at 4:53 | history | suggested | jeq | CC BY-SA 3.0 |
Added OP's image to post.
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| Nov 29, 2017 at 3:57 | review | Suggested edits | |||
| S Nov 29, 2017 at 4:53 | |||||
| Dec 22, 2012 at 5:24 | answer | added | Thomas Zaslavsky | timeline score: 1 | |
| Jun 24, 2011 at 14:03 | answer | added | Emil Jeřábek | timeline score: 4 | |
| Jun 24, 2011 at 13:47 | answer | added | Matt Brin | timeline score: 3 | |
| Apr 14, 2011 at 6:38 | vote | accept | Mario Stefanutti | ||
| Apr 13, 2011 at 15:16 | vote | accept | Mario Stefanutti | ||
| Apr 13, 2011 at 15:16 | |||||
| Apr 13, 2011 at 15:15 | vote | accept | Mario Stefanutti | ||
| Apr 13, 2011 at 15:16 | |||||
| Apr 13, 2011 at 15:15 | vote | accept | Mario Stefanutti | ||
| Apr 13, 2011 at 15:15 | |||||
| Apr 12, 2011 at 15:02 | comment | added | Mario Stefanutti | @all: Thanks for the info. Yes, it is the problem I'm facing. But to get the "number of colorings" the only method I found is to compute the chromatic polynomial, which is known only for few graphs and it is hard to find for more complex cases. Do you know of papers that directly approach the computation of the "number of colorings without exchanges of colors"? I've implemented a brute force algorithm to color a given map with four colors. I'll try to extend it to find all possible colorings manually ... excluding exchanges. youtube.com/user/mariostefanutti#p/u/2/YmYGFxtj2es | |
| Apr 11, 2011 at 12:46 | history | edited | Gerry Myerson |
added arXiv tag
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| Apr 11, 2011 at 11:16 | answer | added | Christian Blatter | timeline score: 7 | |
| Apr 10, 2011 at 18:11 | comment | added | Thierry Zell | Let me see if I understand you correctly: you are looking at maps that are 4-colorable, but not 3-colorable. So given any coloring, you can simply move the colors around in exactly 4! ways (since you have to use all 4 colors) and get essentially the same coloring. So isn't the number you're looking for simply the number of colorings divided by 4! ? | |
| Apr 10, 2011 at 18:10 | answer | added | Siva | timeline score: 2 | |
| Apr 10, 2011 at 17:07 | history | asked | Mario Stefanutti | CC BY-SA 3.0 |