Timeline for Maximum chromatic number of a $k$-regular graph [closed]
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2019 at 15:22 | history | closed |
Yemon Choi Chris Godsil Ben Barber Jan-Christoph Schlage-Puchta Neil Hoffman |
Not suitable for this site | |
| Jan 16, 2019 at 22:44 | comment | added | Yemon Choi | I'm voting to close this question for the reasons pointed out ib @verret's comment | |
| Jan 16, 2019 at 20:18 | comment | added | Dominic van der Zypen | @verret I absolutely agree - my apologies | |
| Jan 16, 2019 at 20:05 | review | Close votes | |||
| Jan 17, 2019 at 15:22 | |||||
| Jan 16, 2019 at 19:52 | comment | added | verret | @Dominic: In the past 10 days, you've asked 11 questions and currently the average vote on them is lower than 1 positive vote. I think you should think a little bit more about your questions before posting them, or consider posting some of them on math.stackexchange.com. This is particularly true of your questions in graph theory (or perhaps I just know more about this area); many, if not most of them, admit examples/counterexamples of very small order that could be found by anyone who actually spent a few minutes thinking about it. | |
| Jan 16, 2019 at 17:29 | comment | added | Wojowu | I would also like to mention the upper bound can be easily shown with a greedy algorithm, no need to appeal to any theorems for just that. | |
| Jan 16, 2019 at 15:28 | vote | accept | Dominic van der Zypen | ||
| Jan 16, 2019 at 15:27 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
edited body
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| Jan 16, 2019 at 15:18 | comment | added | Gro-Tsen | Uh, the chromatic number is bounded by the maximal degree plus one, so $c_k\leq k+1$, but conversely the complete graph on $k+1$ vertices shows that $c_k\geq k+1$. What did I miss? | |
| Jan 16, 2019 at 15:13 | answer | added | Francesco Polizzi | timeline score: 2 | |
| Jan 16, 2019 at 14:49 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |