Constructing self-complementary graphs

How does one go about systematically constructing a self-complementary graph, on say 8 vertices?

[Added: Maybe everyone else knows this already, but I had to look up my guess to be sure it was correct: a self-complementary graph is a simple graph which is isomorphic to its complement. --PLC]

Solution 1:

Here's a nice little algorithm for constructing a self-complementary graph from a self-complementary graph $H$ with $4k$ or $4k+1$ vertices, $k = 1, 2, ...$ (e.g., from a self-complementary graph with $4$ vertices, one can construct a self-complementary graph with $8$ vertices; from $5$ vertices, construct one with $9$ vertices).

See this PDF on constructing self-complementary graphs.

Solution 2:

Take a complete graph with vertex set $V$ and edge set $E={V\choose2}$. Let $\alpha$ be any permutation of $V$ in which the length of each cycle is a multiple of $4$, except for at most one $1$-cycle. (Of course, such permutations exist if and only if $|V|\equiv 0$ or $1\pmod 4$.)

Let $\beta$ be the permutation of $E$ induced by $\alpha$. Observe that $\beta$ contains only cycles of even length. Color the edges in each cycle alternately black and white. The graph consisting of the black edges is self-complementary.

Example. To construct self-complementary graphs of order $5$, take $V=\{a,b,c,d,e\}$ and let $\alpha=(a\;b\;c\;d)(e)$ so that $\beta=(ab\;bc\;cd\;ad)(ac\;bd)(ae\;be\;\;ce\;de)\;$. If we choose the edges $ab,cd$ and $ac$ (i.e. "color them black") we get a $4$-point path $P_4$. Now we can choose the edges $be,de$ obtaining the self-complementary graph $C_5$, or else we can choose $ae,ce$ obtaining the other self-complementary graph of order $5$, the one that looks like the letter A.

Exercise. Construct all of the self-complementary graphs of order $8$.