Are there many self-complementary perfect graphs?

Question

A graph is perfect if it has no induced subgraph that is either an odd cycle (other than a triangle) or a complement thereof (note that the class of perfect graphs is closed under graph complement). A graph is self-complementary if it is isomorphic to its complement.

Question: How many self-complementary graphs are perfect? More precisly, which of the following cases are we in?:

1. most self-complementary graphs are perfect in the sense that one can classify or concisely characterize the self-complementary graphs that are not perfect.
2. most self-complementary graphs are not perfect in the sense that one can classify or concisely characterize the self-complementary perfect graphs.
3. There is no significant relation between the self-complementary graphs and the perfect graphs.

Note that the self-complementary perfect graphs are exactly the self-complementary graphs that do not contain odd induced cycles other than triangles.

Some examples of self-complementary perfect graphs are: the single vertex graph, the path of length three, and (I believe) $C_3\times C_3$. More examples can be found in this list of small self-complementary graphs.

So far I am not aware of a way to generate an infinite family of self-complementary graphs that are either perfect or not perfect.

Answer 1

I think this gives an infinite family of self-complementary perfect graphs:

$V = \{1,\ldots, 4n\}$

$E = \{\{i, j + 2n\} \mid i, j \in \{1, \ldots, 2n\} \wedge i \equiv j \mod 2 \} \cup \{\{i, j\} \mid i, j \in \{1, \ldots, 2n\}\wedge i \neq j\}$

Note that it is self-complementary (by mapping $i$ to $2n + i$ and $2n + i$ to $2n + 1 - i$ for all $i \in \{1 \ldots 2n\}$) and perfect (every cycle of length at least 5 must have three vertices in $\{1, \ldots, 2n\}$ which induce a triangle; similarly the graph contains no complement of a cycle of length at least 5 as an induced subgraph).

For a imperfect self-complementary graph, I think the following works:

$V = \{0,\{(i, j) \mid i \in \{1, 2\} \wedge j \in \{1,\ldots, 2n\}\}$

$E = \{\{0, (1, 1)\}, \{0, (1, 2n)\}\} \cup \{\{(1, i), (1, i+1)\}\mid i \in \{1, \ldots 2n - 1\}\}\cup \{\{0, (2, i)\} \mid i \in \{2, \ldots, 2n - 1\}\} \cup \{\{(2, i), (2, j)\}\mid i, j \in \{1, \ldots 2n\}\wedge |i - j| \neq 1\} \cup \{\{(1, i), (2, j)\}\mid i, j \in \{1, \ldots 2n\}\wedge i \equiv j \mod 2\}$

Note that it is self-complementary (by mapping $(1, i)$ to $(2, i)$ and $(2, i)$ to $(1, 2n + 1 - i)$) and imperfect ($\{0\} \cup \{(1, i) \mid i \in \{1, \ldots, 2n\}\}$ induces an odd cycle).