Tournaments and Even Graphs are Equinumerous

TL;DR

This paper proves a surprising conjecture that the number of non-isomorphic even graphs equals the number of non-isomorphic tournaments on the same number of vertices, using combinatorial counting methods.

Contribution

It establishes the equivalence in count between even graphs and tournaments, resolving a conjecture based on computational evidence with a rigorous proof.

Findings

Number of non-isomorphic even graphs equals the number of non-isomorphic tournaments.

The proof employs the Cauchy-Frobenius Theorem in combinatorial counting.

The result has implications for graph automorphism and symmetry studies.

Abstract

A graph is called odd if there is an orientation of its edges and an automorphism that reverses the sense of an odd number of its edges, and even otherwise. Pontus von Br\"omssen (n\'e Andersson) showed that the existence of such an automorphism is independent of the orientation, and considered the question of counting pairwise non-isomorphic even graphs. Based on computational evidence, he made the rather surprising conjecture that the number of pairwise non-isomorphic even graphs on $n$ vertices is equal to the number of pairwise non-isomorphic tournaments on $n$ vertices. We prove this conjecture using a counting argument with several applications of the Cauchy-Frobenius Theorem.