Prove that a self-complementary graph has radius $2$ and diameter $2$ or $3$.
I think that is one of the well-known properties of self-complementary graphs, but I am having some troubles trying to prove it. The facts that, if $G\cong \bar G$, then both graphs $G$ and $\bar G$ are connected, and that, for any graph, if $rad(G)\geq3$ then $rad(\bar G)\leq 2$, and that if $diam(G)\geq3$ then $diam(\bar G)\leq3$, should make the proof quite easier, but I don't know how to develop it. Could you help me? Thanks in advance!
Best Answer
The properties you list directly imply that a self-complementary graph cannot have radius $3$ or higher, and it cannot have diameter $4$ or higher.
The only thing that remains to show is that it can't have radius $1$ (the fact that the diameter can't be $1$ follows as a corollary). And (excluding the trivial example of a singleton), that can't happen either, as radius $1$ means there is a node which is connected to all other nodes, meaning the complement is not connected.