# Prove that a self-complementary graph has radius 2 and diameter 2 or 3 if G ~= bar G, then both graphs G and bar G are connected, and that, for any graph, if rad(G)>=3 then rad(bar G)<=2, and that if diam(G)>=3 then diam(bar G)<=3, should make the proof quite easier

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 $G\cong \overline{G}$, then both graphs G and $\overline{G}$are connected, and that, for any graph, if $rad\left(G\right)\ge 3$ then $rad\left(\overline{G}\right)\le 2$, and that if $diam\left(G\right)\ge 3$ then $diam\left(\overline{G}\right)\le 3$, should make the proof quite easier, but I don't know how to develop it. Could you help me? Thanks in advance!
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sweetwisdomgw
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.