Skip to main content
14 events
when toggle format what by license comment
Apr 12, 2021 at 17:55 answer added Louis D timeline score: 1
Apr 10, 2021 at 16:29 vote accept CommunityBot
Apr 10, 2021 at 10:14 answer added user153000 timeline score: 8
S Apr 9, 2021 at 19:37 history bounty ended CommunityBot
S Apr 9, 2021 at 19:37 history notice removed user153000
Apr 9, 2021 at 18:46 answer added RobPratt timeline score: 2
Apr 8, 2021 at 19:27 comment added RobPratt Optimization for $n \le 15$ and $\lfloor n/2 \rfloor \le k \le n-1$ yields empirical formula $\max \mathcal{I}_k(G) = (k+1)(n-k-1)$.
Apr 8, 2021 at 18:08 comment added Louis D Another observation is that you can assume $R=\emptyset$ since if there were a counterexample with $R\neq \emptyset$, then you could add all edges inside the set $R$ and all edges between the sets $A$ and $R$ which has at least as many pairs of vertices with degree difference at least $k$.
Apr 8, 2021 at 14:05 comment added Louis D A couple of trivial comments: (i) You can let $B:=\{v_i:\deg(v_i)\leq n-k-1\}$ since there can be no vertices of degree $n$. (ii) Since $\mathcal{I}_k(G)\leq |A||B|$, you can assume you are in the case where $n-k<\min(|A|, |B|)\leq \max(|A|, |B|)<k$.
S Apr 8, 2021 at 13:08 history bounty started CommunityBot
S Apr 8, 2021 at 13:08 history notice added user153000 Draw attention
Apr 7, 2021 at 16:56 history undeleted user153000
Apr 7, 2021 at 15:55 history deleted user153000 via Vote
Apr 6, 2021 at 12:29 history asked user153000 CC BY-SA 4.0