Timeline for Quasi-Ramsey Graphs
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2011 at 4:15 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Dec 9, 2011 at 17:33 | comment | added | Aaron Meyerowitz | It is not a complete answer but I thought it relevant. There is at least one 16 vertex graph that is $K_4$ free (including the complement) , the Paley graph with a deleted vertex. Are there any others? If so, they are all Quasi-Ramsey. I agree that it is not easy to find one. If in fact there are no others then note that the Payley graph has only 2 15 vertex sugraphs so look for a 15 graph other than one of those which is $K_4$ free any of those would be Quasi-Ramsey. | |
| Dec 9, 2011 at 8:54 | comment | added | Zello | Your answer gives no explication. The 17 vertex Paley graph is obviously not quasiRamsey. And generally, a Ramsey graph contains no quasiRamsey subgraph. How to find an irregular 16 vertex quasiRamsey graph? | |
| Dec 9, 2011 at 7:59 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 398 characters in body
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| Dec 9, 2011 at 7:15 | comment | added | Zello | I have corrected the minor fault in my question. Itself and its complement has no essential difference if you see it as a two-colored graph. | |
| Dec 9, 2011 at 6:56 | comment | added | Yemon Choi | Zhipeng, it seems that Aaron has answered one of the questions you asked. If that's not the question you wanted to ask, then perhaps you should edit the question? | |
| Dec 9, 2011 at 6:38 | comment | added | Zello | Yes, I know, but this is not essential improvement. | |
| Dec 9, 2011 at 6:35 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |