Approximate form equivalent to Seymour’s conjecture

Establish that for every ε > 0, every oriented graph G has a vertex v such that |N^+_2(G,v)| ≥ (1−ε) |N^+_1(G,v)|.

Background

Using lexicographic product constructions, the paper shows that if Seymour’s conjecture fails then one can build digraphs in which all vertices violate the inequality by a fixed multiplicative margin. Conversely, the authors note an equivalent approximate formulation: proving that for every ε > 0 there is always a vertex with |N+_2| ≥ (1−ε)|N+_1| would settle Seymour’s conjecture.

References

So Seymour's second neighbourhood conjecture is equivalent to the following conjecture: Let $\epsilon >0$ be arbitrary. Then every oriented graph $G$ has at least one vertex satisfying $|\NGi{2}{G}{v}| \geq (1-\epsilon) |\NGi{1}{G}{v}|$.

Seymour-tight orientations  (2603.29626 - Guo et al., 31 Mar 2026) in Subsection 3.1 (Putative counterexamples)