On Seymour's and Sullivan's Second Neighbourhood Conjectures

For a vertex $x$ of a digraph, $d^+(x)$ ($d^-(x)$, resp.) is the number of vertices at distance 1 from (to, resp.) $x$ and $d^{++}(x)$ is the number of vertices at distance 2 from $x$. In 1995, Seymour conjectured that for any oriented graph $D$ there exists a vertex $x$ such that $d^+(x)\leq d^{++}(x)$. In 2006, Sullivan conjectured that there exists a vertex $x$ in $D$ such that $d^-(x)\leq d^{++}(x)$. We give a sufficient condition in terms of the number of transitive triangles for an oriented graph to satisfy Sullivan's conjecture. In particular, this implies that Sullivan's conjecture holds for all orientations of planar graphs and of triangle-free graphs. An oriented graph $D$ is an oriented split graph if the vertices of $D$ can be partitioned into vertex sets $X$ and $Y$ such that $X$ is an independent set and $Y$ induces a tournament. We also show that the two conjectures hold for some families of oriented split graphs, in particular, when $Y$ induces a regular or an almost regular tournament.


Introduction
A directed graph D = (V, A) is an oriented graph if xy ∈ A implies that yx ∈ A; we also say that D is an orientation of the underlying graph G = (V, E), where xy ∈ E if either xy ∈ A or yx ∈ A.
For a vertex u of a directed graph D = (V, A), let N − D (u) = {v ∈ V : vu ∈ A} and N + D (u) = {v ∈ V : uv ∈ A}.Also, N ++ D (u) = {v ∈ V : uw, wv ∈ A for some w ∈ A} \ N + D (u).In what follows we will often omit the subscript D in N − D (u), N + D (u), etc. when D is known from the context.In 1978, Caccetta and Häggkvist [2] introduced the following famous conjecture.
Conjecture 1.For any integer r > 0, every digraph with n vertices and the minimum out-degree at least n r has a cycle with length at most r.It is trivial when r = 2.When r ≥ 3, this conjecture remains open and for r = 3 it is one of the most well-known problems in graph theory.Seymour (see, e.g., [3]) proposed the following conjecture which would imply the special case of Conjecture 1 when the minimum in-and out-degree are both at least n/3.Seymour's conjecture turned out to be very difficult and was confirmed only for tournaments and other restricted classes of digraphs, see, e.g., [4,5,6,7].Note that Seymour's conjecture cannot be extended to all directed graphs as every complete directed graph with at least two vertices (a digraph obtained from a complete graph by replacing every edge xy by arcs xy, yx) does not have a Seymour vertex.
Sullivan [12] proposed the following variation of Seymour's conjecture.Note that this conjecture also implies the same special case of Conjecture 1 just as Seymour's conjecture does.Also note that the two conjectures coincide for Euler oriented graphs.Thus, it is possible that in general the two conjectures are of somewhat "equal difficulty." It seems that not much is known on Sullivan's conjecture (we have not found any publications with nontrivial results on the conjecture).In this paper, we present several results on Sullivan's conjecture proving the conjecture for tournaments, planar oriented graphs, some families of oriented split graphs, and almost all oriented graphs.We also prove that Seymour's conjecture holds for some families of oriented split graphs (defined in the next section).
Let us conclude this section with two simple results on Sullivan's conjecture and a brief discussion on the content of the other sections.
A tournament is an orientation of a complete graph.We say that a vertex It is well-known and it is easy to prove that every tournament has a 2-king, see, e.g., [11,1] for a proof.Note that N − (v) ⊆ N ++ (v) for every 2-king v of V and hence the following holds.Proposition 4. Every 2-king v in an oriented graph is a Sullivan vertex.In particular, every tournament has a Sullivan vertex.
As one can see the proof of Sullivan's conjecture for tournaments is much easier than Seymour's conjecture for tournaments [4,7].
For a real p with 0 < p < 1, let D(n, p) denote random oriented graphs with n vertices in which the probability of having an arc between a pair of vertices equals p.Let Q be a property of oriented graphs and let D Q (n, p) denote random oriented graphs in D(n, p) which satisfy property Q.We say that almost all oriented graphs have property Q if lim n→∞ D Q (n, p)/D(n, p) = 1 for each 0 < p < 1. (Our definition of almost all oriented graphs having property Q is a slight extension of the usual definition where only p = 1/2 is considered.)By Proposition 4, to show that almost all oriented graphs have a Sullivan vertex, it suffices to prove that almost all oriented graphs have a 2-king.

Proposition 5. Almost all oriented graphs have a 2-king.
The proof is quite simple and it is placed in Appendix.The rest of the paper is organised as follows: In the next section, we will introduce additional terminology and notation.In Section 3, we will prove that all planar oriented graphs satisfy Conjecture 2 and Conjecture 3 by counting the number of transitive triangles.In Section 4, we will prove that some families of oriented split graphs satisfy Conjecture 2 and Conjecture 3. Finally, in Section 5 we discuss open problems.

Additional Terminology and Notation
Let D = (V, A) be a digraph and let X ⊆ V.The subgraph of D induced by X is denoted by D Note that every source is a Sullivan vertex.The vertices in N + (x) (N − (x), respectively) are out-neighbours (in-neighbours, respectively) of x.Similarly to N ++ (u) we define N −− (u) = {v ∈ V : vw, wu ∈ A for some w ∈ A} \ N − (u).
To simplify some notation we use the following: When X and/or Y are singletons, we do not use brackets, e.g., x → y if X = {x} and Y = {y}.If x → y, the x dominates y and y is dominated by x.Thus, in particular, a vertex x dominates all its out-neighbours.
An oriented graph D is a oriented split graph if V (D) can be partitioned into two sets X and Y such that X is an independent set and Y induces a tournament.We will denote a oriented split graph D by D = (X, Y, A), where the order in the union matters.If for every x ∈ X and y ∈ Y either x → y or y → x then D = (X, Y, A) is a complete oriented split graph.Note that a complete oriented split graph is a multipartite tournament in which all but one partite set are singletons.
A tournament

Transitive Triangles and Planar Oriented Graphs
An orientation of a K 3 is a transitive triangle if it has a source.We denote the number of transitive triangles in a digraph D = (V, A) by tt(D).Observe that tt(D) ≤ |A|(n − 2)/3.The number of transitive triangles with source u in a digraph D is denoted by tt u (D).Proof.Suppose that D = (V, A) has no Sullivan vertex.Observe that for any u ∈ V , we have where tt u is the number of transitive triangles with u as a source, w u is the number of arcs from N + (u) to N ++ (u).Summing up this equation over all vertices, we have . Therefore, we have The above and u∈V d + (u) = |A| imply |A| ≤ tt(D).Thus, if tt(D) < |A|, then D has a Sullivan vertex.
Thus, every digraph with the number of transitive triangles less than the number of arcs has a Sullivan vertex.In particular, we have the following corollaries, of which the first two are immediate.Proof.Consider a planar embedding G of the underlying graph of D. Let n, m, f be the number of vertices, edges and faces of G, respectively.Suppose that n ≥ 3.If G is connected, then by Euler's formula, we have m − f = n − 2 > 0. Thus, m > f and this inequality can be extended to the case when G is not connected.Hence, tt(D) ≤ f < m = |A|.Therefore, by Theorem 6, we are done.
Note that Seymour's conjecture has been verified for digraph with the minimum out-degree at most 6 [8].This implies Seymour's conjecture holds for planar oriented graphs.In fact, a planar graph with n vertices has at most 3n − 6 edges.Thus, the minimum out-degree of planar oriented graphs is at most (3n − 6)/n < 3.

Oriented Split Graphs
We start from the following simple but useful lemma, whose simple proof is omitted.
Lemma 10.For every oriented split graph D = (X, Y, A), we have the following: y).We will consider three classes of oriented split graphs in the corresponding subsections.

Complete oriented split graphs
Note that a vertex of maximum out-degree in a tournament T is a 2-king of T .It has been shown in [5] that Seymour's second neighbourhood conjecture holds for oriented split digraphs with only one vertex in the independent set and for complete oriented split graphs.We show that Sullivan's conjecture also holds for these oriented graphs.
. Then, we can observe that y dominates all the vertices in N + [x] and therefore y is a 2-king of T .
Note that we may always assume that an oriented graph D is source-free since every source in D is a Sullivan vertex.
then y is also a 2-king of D and we are done.Otherwise, by Lemma 11, y is dominated by a 2-king , u dominates all the vertices in N + Y (v) which implies that the in-neighbours of u in Y are also in its second out-neighbourhood.Thus, d − Y (u) ≤ d ++ Y (u) and u is the required vertex since all in-neighbours of u are contained in Y .

Oriented Split Graphs with a Regular Tournament
Recall that a vertex of the maximum out-degree in a tournament is a 2-king.As a result, all vertices of a regular tournament are 2-kings.Note that (C1) and (C2) in the following theorem imply that Seymour's and Sullivan's conjectures hold for oriented split graphs with a regular tournament.
Proof.We first prove part (A).Let x ∈ X be arbitrary and let First assume that C = ∅.Note that all vertices in C per definition dominate all vertices in A. As T is eulerian there are equally many arcs entering C as leaving C in T , which imples that we must have |B| ≥ |A| (as otherwise more arcs would leave C than enter C in T ), or equivalently . This implies part (A1), as either C = ∅ or C = ∅.Part (A2) can be proved analogously.
We now prove part (B1).If for any x ∈ X, d ++ Y (x) < d + (x), then by part (A1), we have as desired.Part (B2) can be proved analogously.We now prove part (C1).By part (B1) we only need to consider the case when d −− Y (x) ≥ d − (x) for all x ∈ X.Note that every vertex y ∈ Y satisfies as T is a regular tournament and therefore every vertex is a 2-king in T .The following now holds due to Lemma 10.
This implies that for some y ∈ Y we must have d ++ (y) ≥ d + (y).Part (C2) can be proved by using similar arguments with (B2).

Oriented Split Graphs with an Almost Regular Tournament
Let T be an almost regular tournament.By definition, V (T ) can be partitioned into two sets V + T and V − T such that for every Proof.The first part of the proposition follows from the fact that in every digraph H, the sum of out-degrees equals the sum of in-degrees equals the number of arcs in H. Now we prove the formula for d and all vertices in N + (v) and also u ∈ V + T .However, either u or u ′ has out-degree d+1 (depending on the direction of the arc between u and u ′ ), a contradiction.Proof.Suppose to the contrary that there is no Sullivan vertex, i.e., for any u ∈ V (D), d ++ (u) < d − (u) and in particular d − (u) > 0. For any vertex x ∈ X, we may also assume that d + (x) > 0 as otherwise, since x is not contained in the in-neighbourhood of any other vertex, the resulting digraph obtained by deleting x still has no Sullivan vertex and then we can consider this digraph instead of D.
For any x ∈ X, since x is not a Sullivan vertex, we must have By Observation 1, we have Equality in (2) holds if and only if Otherwise, by a similar argument to the one for (2), we can obtain with equality if and only if 2) and ( 4), we have with equality if and only if equalities in (1), ( 2) and ( 4) hold.Note that if x attains equality in (1) and (2) then If X 1 = ∅, we will show the following properties for each x ∈ X 1 .

Proof of (A4):
, then by (A2) and (A3), we have that By (A0)-(A4), we have that for any x ∈ X 1 , Y can be partitioned into five disjoint sets, say Proof of (A5): We only need show that for any pair of vertices x, y ∈ X 1 N + (y) ⊆ N + (x) (N − (y) ⊆ N − (x) can be proved by a similar argument).Suppose that N + (y) ⊆ N + (x), then there exists an out-neighbour of y, z ∈ N ++ Y (x) ∪ C x .Note that N + (y) ∩ C x = ∅ since if not and w ∈ N + (y) ∩ C x , then by (A4), y → w → x, a contradiction to (A1).Thus, z ∈ N ++ Y (x).By (A2), y → z → C x .In addition, since N + (y) ∩ C x = ∅, C x ⊆ N ++ Y (y).Again by (A2), N ++ Y (y) → C y and therefore C x → C y .In particular, C x ∩ C y = ∅.But, by (A2), N ++ Y (x) → C x → N + (x).Thus, C y ⊆ N + (x).By (A4), x → C y → y also assume that d − (x) > 0 for otherwise x is not contained in the first or second out-neighbourhood of any other vertex, and therefore we can delete it and consider the resulting split digraph.Let First we claim that C x = ∅.Otherwise, by Observation 1, we have We have d ++ Y (x) = d + (x) − 1 since x is not a Seymour vertex, then all the vertices in C x with out-degree d.Note that N + X (A x ) = ∅, otherwise, x is a Seymour vertex.Choose a vertex y of out-degree d − 1 in A x (the number of vertices with out-degree d − 1 guarantees the existence of such y), we have y is a Seymour vertex by the arguments above.
So in the following we always assume that for any x ∈ X, C x = ∅.As a result, for any x ∈ X, since x is not a Seymour vertex and Now, we consider a vertex in y ∈ Y .Since y is not a Seymour vertex, where the last inequality follows from Proposition 17.
. We now partition X into two sets X a := {x ∈ X : C ′ x = ∅} and X b := X \ X a .For any x ∈ X a , since By arguments similar to those for (7), for any x ∈ X b , we have with equality if and only if We want to get a lower and a upper bound for |X 1 | in order to achieve a contradiction.We first try to get a upper bound.Now, let x * ∈ X 1 be a vertex with the minimum in-degree in X 1 , we partition X 1 into two sets X 11 := {x ∈ X 1 : x ∈ N ++ (x * ) ∩ X} and X 12 := X 1 \ X 11 .By (8), we have We now claim that there exists a vertex u ∈ N −− (x * ) ∩ N + (x * ) such that X 12 ⊆ N ++ X (u) (we postpone its proof to the end of the proof of the theorem in order not to break the flow of the proof).Therefore, by (9) we have where the last inequality holds because of N + X (u) ⊆ N ++ X (x * ).Again applying (8), we can obtain By ( 12) and (13), we have Now, we try to get the lower bound of |X 1 |.Let t be the number of vertices in Y which has out-degree d − 1 in T and is a 2-king of T .Using Proposition 17, the assumption that all vertices are not Seymour vertex and Lemma 10, we can get In fact, suppose there is a vertex y Now we show that u is the required vertex.Suppose it is not, then there exists a vertex x ′ ∈ X 12 \N ++ (u).In particular, x ′ ∈ N ++ (u) ∪ N ++ (x * ), i.e., N − (x ′ ) ∩ (N + (u) ∪ N + (x * )) = ∅.Thus we have But by the minimality of d − (x * ), a contradiction.This completes the proof.

Discussion
We proved Seymour's and Sullivan's conjectures for special classes of oriented graphs.Our results and those of other authors show that both conjectures are very difficult despite their simple formulations.In particular, we were unable to verify either conjecture for all oriented split graphs.As oriented split graphs are a subfamily of multipartite tournaments, which are orientations of complete multipartite graphs, also verifying that the conjectures hold for multipartite tournaments remains an open problem.

Theorem 6 .
Let D = (V, A) be an oriented graph.If tt(D) < |A|, then D has a Sullivan vertex.

Corollary 7 .Corollary 9 .
If D does not contain any transitive triangle, then D has a Sullivan vertex.Corollary 8.If D is an orientation of a triangle-free graph, then D has a Sullivan vertex.Every planar oriented graph D = (V, A) has a Sullivan vertex.

Theorem 14 .
Let D = (X, Y, A(D)) be an oriented split graph, where Y induces a regular tournament T in D. Let d + (x) = d + D (x) and d − (x) = d − D (x).Then (A) For every x ∈ X the following two statements hold.

For
any subset S of V (D), since d + (S) − d − (S) = v∈S (d + (v) − d − (v)), we have the following observation.Observation 1.Let D = (V, A) be an almost regular tournament with order 2d and S ⊆ V (D), then |d + (S) − d − (S)| ≤ |S|.Furthermore, d + (S) = d − (S) + |S| (d − (S) = d + (S) + |S|, respectively) if and only if S ⊆ V + T (S ⊆ V − T , respectively).Now we are ready to prove the first main result of this subsection.Theorem 16.Let D = (X, Y, A(D)) be a split digraph, where Y induces an almost regular tournament T with 2d vertices.Then D has a Sullivan vertex.