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Apr 12, 2017 at 19:51 comment added Tony Huynh I guess the minor version is not that interesting. You can just take the complete graph on $\kappa$ vertices. Since you can delete edges, you can make any graph on $\kappa$ vertices.
Apr 12, 2017 at 15:06 vote accept Dominic van der Zypen
Apr 12, 2017 at 14:39 answer added Will Brian timeline score: 11
Apr 12, 2017 at 14:31 comment added John Pardon Surely you want some more properties than you ask for. As stated, you can just take $G$ to be the disjoint union of all (one for each isomorphism class of) simple undirected graphs on $\kappa$ vertices.
Apr 12, 2017 at 14:31 answer added Tony Huynh timeline score: 6
Apr 12, 2017 at 14:25 history undeleted Dominic van der Zypen
Apr 12, 2017 at 14:25 history deleted Dominic van der Zypen via Vote
Apr 12, 2017 at 14:21 history asked Dominic van der Zypen CC BY-SA 3.0