Timeline for Is there easy proof for triangle-free two-coloring of planar graphs?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2016 at 12:00 | comment | added | domotorp | @Yaakov: Not at all, my motivation was that this paper uses only this corollary of the four color theorem, and so I wondered how difficult it is: arxiv.org/abs/1512.01953 | |
| Dec 2, 2016 at 11:15 | comment | added | Yaakov Baruch | Interesting. Well, the devil is in the "suitable" of course. Given coloring A for ONE triangle, there are 4 non-uniform colorings B (out of 8) such that AB would give a different color to each vertex. So "suitable" would mean a second coloring where for each triangle the coloring comes from one of those 4 possibilities. I have have no idea of course how to find globally this second coloring. But I thought something like that might have been the inspiration for your question. | |
| Dec 2, 2016 at 8:32 | comment | added | domotorp | @Yaakov: What would be the use of triangle-free? I was thinking about something related earlier, but my hopes were crushed by this answer: mathoverflow.net/questions/255409/… | |
| Dec 2, 2016 at 7:44 | comment | added | Yaakov Baruch | Perhaps it's silly to ask, but conversely, could there be any hope to prove the FCT by finding and merging 2 suitable such triangle-free colorings? | |
| Dec 2, 2016 at 0:28 | vote | accept | domotorp | ||
| Dec 1, 2016 at 22:06 | answer | added | Gjergji Zaimi | timeline score: 7 | |
| Dec 1, 2016 at 7:15 | answer | added | Brendan McKay | timeline score: 5 | |
| Nov 30, 2016 at 16:31 | history | asked | domotorp | CC BY-SA 3.0 |