Timeline for How many colors do we need to avoid bichromatic triangles?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 1, 2016 at 16:39 | vote | accept | domotorp | ||
| Sep 1, 2016 at 9:33 | answer | added | David Conlon | timeline score: 9 | |
| Sep 1, 2016 at 5:32 | history | edited | domotorp | CC BY-SA 3.0 |
added update
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| Aug 31, 2016 at 21:13 | answer | added | Jan Kyncl | timeline score: 3 | |
| Aug 31, 2016 at 13:51 | answer | added | Ilya Bogdanov | timeline score: 3 | |
| Aug 31, 2016 at 13:25 | comment | added | Ilya Bogdanov | $biR_k(K_3)=k+2$ or $k+1$, depending on the parity of $k$. Indeed, a bichromatic $K_3$ exists iff there are two adjacent edges of the same color. | |
| Aug 31, 2016 at 13:15 | comment | added | David Roberson | Suppose I have a $2k$ coloring of the edges of $K_n$. Partition the edge colors into $k$ pairs and consider each as a single color. If $n \ge R_k(G)$, then there is a monochromatic $G$ in this coloring and therefore a bichromatic $G$ in the original coloring. Therefore you can actually get that $biR_{2k}(G) \le R_k(G)$. This is still an upper bound of course, but I thought it was worth pointing out. | |
| Aug 31, 2016 at 12:31 | history | asked | domotorp | CC BY-SA 3.0 |