Timeline for "n-partite n-clique" with added conditions
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 30, 2011 at 4:57 | vote | accept | Pawan Aurora | ||
| May 30, 2011 at 4:56 | comment | added | Pawan Aurora | You are right. I wish I could think like you. | |
| May 29, 2011 at 21:20 | comment | added | fedja | OK, you have your proof and I offered my counterexample. Now, we can either check your proof (which may be long and hard) or refute my $n=5$ construction (which should be easy if it is wrong). So, you claim that you can find a triangle in it or a vertex with a wrong set of neighbors. Do it and I'll retract my claim. | |
| May 29, 2011 at 18:39 | comment | added | Pawan Aurora | I am totally lost here. As long as every vertex has exactly one neighbor in each row and column except its own, we always have a $n$-clique. In fact every vertex is part of some $n$-clique. What am i missing here? | |
| May 29, 2011 at 14:20 | comment | added | fedja | Since when is -3=0 a nonresidue modulo 3? | |
| May 29, 2011 at 3:49 | comment | added | Pawan Aurora | Consider $n=3$. So we have the following edges $(V_{11},V_{22}),(V_{11},V_{33})$ as well as the edge $(V_{22},V_{33})$ apart from some other edges. Clearly we have a triangle (a $3$-clique here) among $V_{11},V_{22},V_{33}$. | |
| May 28, 2011 at 20:35 | history | answered | fedja | CC BY-SA 3.0 |