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Two examples of such graphs with the minimum number of edges are two 4-cliques (each with 6 edges) or a cube. But why is 12 the minimum? Can this be proven?

EDIT: now that someone has disproven my cube example (not sure what I was thinking), I suspect it might be that all graphs with $n$ vertices where 2 out of any 3 vertices are adjacent consist of 2 cliques of size $\lfloor \frac{n}{2}\rfloor$ and $\lceil\frac{n}{2}\rceil$, respectively. Not sure if that is right, though.

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    $\begingroup$ Do you mean "2 out of any 3 vertices are adjacent..."? $\endgroup$
    – Dave
    Jun 30 '19 at 23:59
  • $\begingroup$ Yes, sorry - I edited the title $\endgroup$ Jul 1 '19 at 0:01
  • $\begingroup$ Well there are 56 ways to choose 3 vertices out of the 8. And for a given edge, there are 56*3=168 ways to select a group of three and decide which two vertices in that group it will join. I'm not sure what you mean, however. $\endgroup$ Jul 1 '19 at 0:21
  • $\begingroup$ A cube does have 3 vertices in which none are adjacent. $\endgroup$ Jul 1 '19 at 1:34
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If every vertex has at least $3$ incident edges, then that makes $24$ incident (vertex,edge) pairs. Since each edge contributes only $2$ such pairs, we have at least $12$ edges, as required.

So suppose, from now on, that some vertex $P$ doesn't have at least $3$ incident edges. If $P$ has only one incident edge or none, then there are $6$ or more vertices not adjacent to $P$. Every two of these $6$ must be adjacent, because any two that are not adjacent would, when combined with $P$, constitute three vertices with no adjacency. So the number of edges would be at least $\binom62=15$, which is more than enough.

There remains the case that $P$ has exactly two incident edges, say $PQ$ and $PR$. Among the $5$ vertices other than $P,Q,R$, every two must be adjacent, lest they and $P$ form three points with no adjacencies. So that gives us $\binom52=10$ edges among those $5$ vertices. Together with the edges $PQ$ and $PR$, we have $12$, as required.

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