6
$\begingroup$

It can be easily shown that if a graph is self-complementary and regular then the number of vertices, $n$, is equal to $4k +1$ for some $k \in \mathbb{Z}$.

But, how to we prove (prove by constructing) that there is a self-complementary regular graph for $n = 4k +1$

$\endgroup$
3
$\begingroup$

Here is a source for a construction:

S.B. Rao, On regular and strongly-regular self-complementary graphs. Discrete Mathematics 54 (1985), pp. 73–82.

See Theorem 2.3. This solution seems to answer a more specific question, so it's likely that there is a simpler answer to your question, which I would be interested in.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.