Why is it not possible to recognize a self-complementary graph just by searching for a self-complementary graph on $8$ vertices?

Question

From the paper by Gibbs (1974) it is known that every self-complementary graph on $4n$ vertices has $n$ disjoint induced $P_4$ subgraphs. Let's call the collection of these subgraphs a $P_4$ cover.

Given a $P_4$ cover for a self-complementary graph, it appears that for every pair $\{A,B\}$ of disjoint $P_4$ in it, the subgraph induced on $V(A\cup B)$ must be a self-complementary graph on $8$ vertices. Therefore finding any induced self-complementary subgraph on $8$ vertices should reveal two of the disjoint induced $P_4$ in the given graph. Once you know one of the $P_4$ subgraphs in the $P_4$ cover, finding all other $P_4$ subgraphs is possible in $O(n^4)$ time.

Now, the potential flaw in this reasoning is that, for a given $P_4$ cover, the original self-complementary graph might contain an induced self-complementary subgraph on $8$ vertices, such that its vertices are not vertices of any pair of $P_4$ in the $P_4$ cover. But do graphs with this property actually exist?

Answer 1

Suppose a graph $G=$ $P_4\cdot P_4$. For any three disjoint induced $P_4$ subgraphs $A,B,C:$ $G[V(A)\cup V(B)]$ or $G[V(A)\cup V(C)]$ is not self-complementary.

This shows that the following statement is false:

Given a $P_4$ cover for a self-complementary graph, it appears that for every pair $\{A,B\}$ of disjoint $P_4$ in it, the subgraph induced on $V(A\cup B)$ must be a self-complementary graph on $8$ vertices.

Therefore this algorithm does not allow to decide if the graph is self-complementary.