# Number of non-isomorphic 6-regular graphs on 9 vertices (why find the complement)?

To clarify, the question has been asked on this site. However, I have a follow up question about the answer (but could not comment - since I am new to this forum).

The question: Count the number of non-isomorphic 6-regular graphs on 9 vertices.

The answer can be found here: Count the number of non-isomorphic 6-regular graphs on 9 vertices.

To summarize, the number of simple graphs with that property is the same as it's complement. I am wondering why is that the case.

Thus, if we take a set of representatives from each isomorphism class of $2$-regular $9$-vertex graphs and take their complements, we get non-isomorphic $6$-regular $9$-vertex graphs. Likewise, if we take a set of representatives from each isomorphism class of $6$-regular $9$-vertex graphs and take their complements, we get non-isomorphic $2$-regular $9$-vertex graphs.
This has the consequence that the set of non-isomorphic (or unlabelled) 6-regular 9-vertex graph $$\{\text{unlabelled 6-regular 9-vertex graph } G\}$$ is equal to the set of non-isomorphic 2-regular 9-vertex graph after replacing each graph by its complement $$\{G^c:\text{unlabelled 2-regular 9-vertex graphs } G\}.$$