Seymour's Second Neighborhood Conjecture
Prove that for every oriented directed graph G = (V, E) with no loops and no pairs of opposite arcs, there exists a vertex v in V such that the size of its second out-neighborhood N++(v) is at least the size of its first out-neighborhood N+(v), i.e., |N++(v)| ≥ |N+(v)|. Here, N+(v) denotes the set of out-neighbors of v, and N++(v) denotes the set of vertices reachable from v by a directed path of length two.
References
Conjecture 1.1. (Seymour's Second Neighborhood Conjecture). For every oriented graph G = (V, E), there exists a verter v E V such that |N++(v)| ≥ |N+(v)|, where Nt and N++ represent the first and second neighborhoods of v respectively.
The following is known as Seymour's second neighbourhood conjecture. Every orientation $G$ contains at least one vertex $v$ such that $|\NGi{2}{G}{v}| \geq |\NGi{1}{G}{v}|$.