The second out-neighborhood for local tournaments

Conjecture 1.1

(Seymour’s Second Neighborhood Conjecture (SSNC)) For any oriented graph D, there exists a vertex v in D such that d ⁺⁺(v) ≥ d ⁺ (v).

Theorem 1.2

[3] In any tournament T, there exists a Seymour vertex.

Theorem 1.3

[4] A tournament T with no vertex of out-degree zero has at least two Seymour vertices.

Conjecture 1.4

[11] (1) Every oriented graph D has a vertex x such that d ⁺⁺(x) ≥ d ⁻(x).

(2) Every oriented graph D has a vertex x such that d ⁺⁺(x) + d ⁺(x) ≥ 2d ⁻(x).

Theorem 1.5

[12] Every tournament has a Sullivan-1 vertex and a Sullivan-2 vertex. Every tournament with no vertex of in-degree zero has at least three Sullivan-1 vertices.

Theorem 1.6

[12] A tournament T has at least two Sullivan-2 vertices unless T ∈ T .

Corollary 1.7

A strong tournament T with at least three vertices has at least two Sullivan-2 vertices.

Theorem 2.1

[15] Let D be a connected locally semicomplete digraph. Then, exactly one of the following possibilities holds:

1. D is round decomposable with a unique round decomposition R[S ₁, S ₂,…,S _r ], where R is a round local tournament on r ≥ 2 vertices and S _i is a strong semicomplete digraph for each i ∈ {1, 2,…,r};

2. D is nonround decomposable and not semicomplete and it has the structure as described in Theorem 2.6;

3. D is a semicomplete digraph which is nonround decomposable.

Corollary 2.2

Let D be a connected local tournament. Then, exactly one of the following possibilities holds:

1. D is round decomposable with a unique round decomposition R[S ₁, S ₂,…,S _r ], where R is a round local tournament on r ≥ 2 vertices and S _i is a strong tournament for i ∈ {1, 2,…,r};

2. D is nonround decomposable and not a tournament and it has the structure as described in Theorem 2.6;

3. D is a tournament which is nonround decomposable.

Theorem 2.3

[17] Let D be a connected, but not strong locally semicomplete digraph. Then, the following holds for D.

1. If A and B are distinct strong components of D with at least one arc between them, then either A ⇒ B or B ⇒ A;

2. If A and B are strong components of D, such that A ⇒ B, then A and B are semicomplete digraphs;

3. The strong components of D can be ordered in a unique way D ₁, D ₂,…,D _p such that there is no arc from D _j to D _i for j > i, and D _i completely dominates D _i+1 for i ∈ {1, 2,…,p − 1}.

Theorem 2.4

[17] Let D be a connected, but not strong locally semicomplete digraph, and let D ₁, D ₂,…,D _p be the acyclic ordering of the strong components of D. Then, D can be decomposed into r ≥ 2 induced subdigraphs D 1 ′ , D 2 ′ , …, D r ′ as follows:

and

The subdigraphs D 1 ′ , D 1 ′ , … , D r ′ satisfy the properties below:

1. D i ′ consists of some strong components of D and is semicomplete for each i∈ {1,2,…,r};

2. D i + 1 ′ completely dominates the initial component of D i ′ , and there exists no arc from D i ′ to D i + 1 ′ for i ∈ {1, 2,…,r − 1};

3. If r ≥ 3, then there is no arc between D i ′ and D j ′ for i, j satisfying |j − i| ≥ 2.

Theorem 2.5

[15] If D is a round decomposable locally semicomplete digraph, then it has a unique round decomposition D = R[S ₁, S ₂,…,S _r ], where R is a round local tournament on r ≥ 2 vertices and each S _i is a strong semicomplete digraph.

Theorem 2.6

[15] Let D be a strong locally semicomplete digraph which is not semicomplete. Then, D is nonround decomposable if and only if the following conditions are satisfied:

1. There is a minimal separating set S such that D − S is not semicomplete, and for each such S, D〈S〉 is semicomplete and the semicomplete decomposition of D − S has exactly three components D 1 ′ , D 2 ′ , D 3 ′ ;

2. There are integers α, β, μ, ν with λ ≤ α ≤ β ≤ p − 1 and p + 1 ≤ μ ≤ ν ≤ p + q such that

or

where D ₁, D ₂,…,D _p and D _p+1, D _p+2,…,D _p+q are the acyclic orderings of the strong components of D − S and D〈S〉, respectively, and D _λ is the initial component of D 2 ′ .

Theorem 2.7

[15] Let D be a strong nonround decomposable locally semicomplete digraph and let S be a minimal separating set of D such that D − S is not semicomplete. Let D ₁, D ₂,…,D _p be the acyclic ordering of the strong components of D − S and D _p+1, D _p+2,…,D _p+q be the acyclic ordering of the strong components of D〈S〉. The following holds:

1. D _p ⇒ S ⇒ D ₁;

2. Suppose that there is an arc s → v from S to D 2 ′ with s ∈ V(D _i ) and v ∈ V(D _j ). Then, D i ∪ D i + 1 ∪ ⋯ ∪ D p + q ⇒ D 3 ′ ⇒ D λ ∪ ⋯ ∪ D j ;

3. D p + q ⇒ D 3 ′ and D _f ⇒ D _f+1 for f ∈ {1, 2,…,p + q} where subscripts are modulo p + q.

Lemma 3.1

Let D be a round decomposable local tournament and R[S ₁, S ₂,…,S _r ] be the unique round decomposition of D. Let D* = R[V ₁,V ₂,…,V _r ] and v _i ∈ V _i be arbitrary, where V _i is the vertex set of S _i for i ∈ {1, 2,…,r}. If N D ⁎ + ( v j ) = V j + 1 ∪ ⋯ ∪ V k , then d D ⁎ + + ( v j ) ≥ d D ⁎ + ( v k ) .

Proof

Let v ∈ N D ⁎ + ( v k ) . We claim that v ∉ N D ⁎ + ( v j ) . In fact, if v _j → v, then v _j , v, v _k are in the order of the round labeling of R. Then, v _j → v _j since v _k → v. Note that v _j → v _k . This contradicts the fact that D has no 2-cycle. So, v ∉ N D ⁎ + ( v j ) . Thus, v ∈ N D ⁎ + + ( v j ) , i.e., N D ⁎ + ( v k ) ⊆ N D ⁎ + + ( v j ) . Then, d D ⁎ + + ( v j ) ≥ d D ⁎ + ( v k ) .□

Lemma 3.2

Let D be a round decomposable local tournament and R[S ₁, S ₂,…,S _r ] be the unique round decomposition of D. Let D = D* = R[V ₁, V ₂,…,V _r ] and v _j ∈ V _j be arbitrary, where V _j is the vertex set of S _j for j ∈ {1, 2,…,r}. If there is a vertex v ∈ V _j such that v is a Sullivan-i vertex of S _j and a Sullivan-i vertex of D*, then v is a Sullivan-i vertex of D for i ∈ {1, 2}.

Proof

For the case when i = 1, since v is a Sullivan-1 vertex of S _j and a Sullivan-1 vertex of D*, we have d S j + + ( v ) ≥ d S j − ( v ) , d D ⁎ + + ( v ) ≥ d D ⁎ − ( v ) . Clearly

Thus, d D + + ( v ) ≥ d D − ( v ) and v is a Sullivan-1 vertex of D.

For the case when i = 2, it can be proved similarly.□

Theorem 3.3

Let D be a round decomposable local tournament. Then, D has a Sullivan-i vertex for i ∈ {1, 2}.

Proof

Let R[S ₁, S ₂,…,S _r ] be the unique round decomposition of D. Let D* = R[V ₁, V ₂,…,V _r ] and v _j ∈ V _j be arbitrary, where V _j is the vertex set of S _j for j ∈ {1, 2,…,r}. W.l.o.g., assume that v ₁ ∈ V ₁ is a vertex of D* with minimum out-degree, i.e., d D ⁎ + ( v 1 ) = δ + ( D ⁎ ) . Let N D ⁎ + ( v 1 ) = V 2 ∪ ⋯ ∪ V t . Since v 1 ↛ v t + 1 , we have N D ⁎ − ( v t + 1 ) ⊆ V 2 ∪ ⋯ ∪ V t = N D ⁎ + ( v 1 ) . Then,

Let N D ⁎ + ( v t + 1 ) = V t + 2 ∪ ⋯ ∪ V h . By Lemma 3.1, we have

Thus, d D ⁎ + + ( v t + 1 ) ≥ δ + ( D ⁎ ) ≥ d D ⁎ − ( v t + 1 ) and v _t+1 is a Sullivan-1 vertex of D*. Note that d D ⁎ + ( v t + 1 ) ≥ δ + ( D ⁎ ) . Then, d D ⁎ + + ( v t + 1 ) + d D ⁎ + ( v t + 1 ) ≥ 2 δ + ( D ⁎ ) ≥ 2 d D ⁎ − ( v t + 1 ) and v _t+1 is also a Sullivan-2 vertex of D*.

Clearly, all the vertices of V _t+1 are Sullivan-i vertices of D* for i ∈ {1, 2}. By Corollary 1.5, the tournament S _t+1 always has a Sullivan-1 vertex and a Sullivan-2 vertex, say v _t+1 and v t + 1 ′ , respectively. By Lemma 3.2, v _t+1 is a Sullivan-1 vertex of D and v t + 1 ′ is a Sullivan-2 vertex of D.□

Theorem 3.4

Let D be a connected, but not strong local tournament with no vertex of in-degree zero. Then, D has at least three Sullivan-1 vertices and two Sullivan-2 vertices.

Proof

Let D ₁, D ₂,…,D _p be the acyclic ordering of the strong components of D. Let v be a Sullivan-1 vertex of D ₁ and v′ be a Sullivan-2 vertex of D ₁, i.e.,

Clearly,

Thus, d D + + ( v ) ≥ d D − ( v ) and d D + + ( v ′ ) + d D + + ( v ′ ) ≥ 2 d D − ( v ′ ) , i.e., v is a Sullivan-1 vertex of D and v′ is a Sullivan-2 vertex of D. So, the Sullivan-i vertex of D ₁ is always the Sullivan-i vertex of D for i ∈ {1, 2}.

Since D has no vertex of in-degree zero, we see that D ₁ has no vertex of in-degree zero and D ₁ has at least three vertices. By Corollaries 1.5 and 1.7, D ₁ has at least three Sullivan-1 vertices and two Sullivan-2 vertices. Then, these three vertices (respectively, two vertices) are Sullivan-1 (respectively, Sullivan-2) vertices of D.□

Theorem 3.5

Let D be a strong round decomposable local tournament. Then, D has at least two Sullivan-i vertices for i ∈ {1, 2}.

Proof

Let D = R[S ₁, S ₂,…,S _r ] be the unique round decomposition. Let D* = R[V ₁, V ₂,…,V _r ] and v _j ∈ V _j be arbitrary, where V _j is the vertex set of S _j for j ∈ {1, 2,…,r}. W.l.o.g., assume that v ₁ ∈ V ₁ is a vertex of D* with minimum out-degree, i.e., d D ⁎ + ( v 1 ) = δ + ( D ⁎ ) . Let N D ⁎ + ( v 1 ) = V 2 ∪ ⋯ ∪ V t . According to the proof of Theorem 3.3, a Sullivan-i vertex of S _t+1 is a Sullivan-i vertex of D for i ∈ {1, 2}.□

Case 1

v _g ≠ v ₁.

We see that d D ⁎ + + ( v t + 2 ) ≥ d D ⁎ + ( v g ) ≥ δ + ( D ⁎ ) + 1 ≥ d D ⁎ − ( v t + 2 ) , and v _t+2 is a Sullivan-1 vertex of D*.

Note v t + 2 ∉ V 1 . Then, d D ⁎ + ( v t + 2 ) ≥ δ + ( D ⁎ ) + 1 ≥ d D ⁎ − ( v t + 2 ) . Thus, d D ⁎ + + ( v t + 2 ) + d D ⁎ + ( v t + 2 ) ≥ 2 d D ⁎ − ( v t + 2 ) and v _t+2 is a Sullivan-2 vertex of D*.

Clearly, all the vertices of V _t+2 are Sullivan-i vertices of D* for i ∈ {1,2}. By Corollary 1.5, the tournament S _t+2 has a Sullivan-1 vertex and a Sullivan-2 vertex, say v _t+2 and v t + 2 ′ , respectively. By Lemma 3.2, v _t+2 is another Sullivan-1 vertex of D and v t + 2 ′ is another Sullivan-2 vertex of D.

Case 2

v _g = v ₁.

Now N D ⁎ + ( v t + 2 ) = V t + 3 ∪ ⋯ ∪ V r ∪ V 1 .

First, we show that v ₂ is another Sullivan-1 vertex of D*. Recall v ₂ → v _t+2. Then, V ₂ ⇒ V ₃ ∪ V ₄ ⋯ ∪ V _t+2. Also, V _t+2 ⇒ V _t+3 ∪⋯∪ V _r ∪ V ₁. Then, v ∈ N ⁺(v ₂) ∪ N ⁺⁺(v ₂) for any v ∈ V(D* − V ₂). Since N D ⁎ − V 2 + ( v 2 ) ∩ N D ⁎ − V 2 − ( v 2 ) = ∅ , we have N D ⁎ − V 2 − ( v 2 ) ⊆ N D ⁎ − V 2 + + ( v 2 ) . Then, d D ⁎ − V 2 + + ( v 2 ) ≥ d D ⁎ − V 2 − ( v 2 ) . This means that all vertices of V ₂ are Sullivan-1 vertices of D*. By Corollary 1.5, the tournament S ₂ has a Sullivan-1 vertex, say also v ₂. By Lemma 3.2, then v ₂ is another Sullivan-1 vertex of D.

To find another Sullivan-2 vertex of D, we consider the following two cases:

If d D ⁎ + ( v t + 2 ) ≥ δ + ( D ⁎ ) + 2 , we have d D ⁎ + + ( v t + 2 ) ≥ d D ⁎ + ( v g ) = d D ⁎ + ( v 1 ) = δ + ( D ⁎ ) . Thus, d D ⁎ + + ( v t + 2 ) + d D ⁎ + ( v t + 2 ) ≥ δ + ( D ⁎ ) + δ + ( D ⁎ ) + 2 ≥ 2 ( δ + ( D ⁎ ) + 1 ) ≥ 2 d D ⁎ − ( v t + 2 ) , and hence, all vertices of V _t+2 are Sullivan-2 vertices of D*. By Corollary 1.5, the tournament S _t+2 has a Sullivan-2 vertex, say also v _t+2. By Lemma 3.2, v _t+2 is another Sullivan-2 vertex of D.

If d D ⁎ + ( v t + 2 ) ≤ δ + ( D ⁎ ) + 1 , we have d D ⁎ + ( v t + 2 ) = δ + ( D ⁎ ) + 1 since we note v _t+2 ∉ V ₁. Also note v ₂ → v _t+2. Then, N D ⁎ + ( v 2 ) ⊇ V 3 ∪ ⋯ ∪ V t + 2 , and hence, N D ⁎ − ( v 2 ) ⊆ V t + 3 ∪ ⋯ ∪ V r ∪ V 1 . So,

Let N D ⁎ + ( v 2 ) = V 3 ∪ ⋯ ∪ V k . By Lemma 3.1, we have d D ⁎ + + ( v 2 ) ≥ d D ⁎ + ( v k ) . Note v ₂ → v _k and v ₁ → v ₂. Then, v k ∉ V 1 since D has no 2-cycle. Note that

Then, d D ⁎ + + ( v 2 ) + d D ⁎ + ( v 2 ) ≥ 2 ( δ + ( D ⁎ ) + 1 ) ≥ 2 d D ⁎ − ( v 2 ) , and hence, all vertices of V ₂ are Sullivan-2 vertices of D*. By Corollary 1.5, the tournament S ₂ always has a Sullivan-2 vertex, say also v ₂. By Lemma 3.2, v ₂ is another Sullivan-2 vertex of D.

Corollary 3.6

1. Every round decomposable local tournament has a Sullivan-i vertex for i ∈ {1, 2}.

2. Every round decomposable local tournament with no vertex of in-degree zero has at least two Sullivan-i vertices for i ∈ {1, 2}.

Lemma 4.1

For any strong component D _j of D 3 ′ , d D − D j + ( D j ) ≥ | S | .

Proof

Let D j ⊆ D 3 ′ be arbitrary. Since N D − D j + ( D j ) is a separating set of D and D − N D − D j + ( D j ) is not a tournament, we have d D − D j + ( D j ) ≥ | S | by the choice of S.□

Lemma 4.2

If |X| < |S|, then S − B ⇒ A and S − B ⇒ D 3 ′ .

Proof

Note that X is also a separating set of D. By the choice of S, combining with |X| < |S|, we see that D − X = D 〈 V ( D 3 ′ ) ∪ A ∪ ( S − B ) 〉 is a tournament. So, any vertex of S − B is adjacent to the vertices of A and D 3 ′ . Since N S − B + ( A ) = ∅ , we have S − B ⇒ A. Also, an arc from D 3 ′ to S implies that D 3 ′ and D 1 ′ are adjacent, which contradicts Theorem 2.4 (c). So S − B ⇒ D 3 ′ .□

Lemma 4.3

If |X| < |S|, then for any v ∈ V(D′) = S − B,

1. X ⊆ N D − D ′ + ( v ) ∪ N D − D ′ + + ( v ) ;

2. N D − D ′ − ( v ) ⊆ N D − D ′ + + ( v ) ⊆ X ;

3. d D − D ′ + ( v ) ≥ d D − D ′ − ( v ) + 2 .

Proof

1. Recall that X = B ∪ ( V ( D 2 ′ ) − A ) ∪ V ( D p ) . Since |X| < |S|, we have |B| ≤ |X| −1 ≤ |S| − 2, and hence, |S − B| ≥ 2. By Lemma 4.2, for any v ∈ V(D′) = S − B, we have v ⇒ A ⇒ D 2 ′ − A and v ⇒ A ⇒ D _p . Since B = N S + ( A ) , for any y ∈ B, there exists a vertex x ∈ A such that x → y and hence v → x → y . So, any vertex of X either belongs to N D − D ′ + ( v ) or belongs to N D − D ′ + + ( v ) , i.e., X ⊆ N D − D ′ + ( v ) ∪ N D − D ′ + + ( v ) .

2. Recall that D′ = D〈S − B〉. By the definition of A, B, and X, we have V ( D − D ′ ) − X = A ∪ V ( D ′ 3 ) . For any v ∈ V(D′), by Lemma 4.2, we have v ⇒ A and v ⇒ D 3 ′ . Then,

   N D − D ′ + ( v ) ⊇ A ∪ V ( D 3 ′ ) = V ( D − D ′ ) − X

   and hence, N D − D ′ − ( v ) ⊆ X and N D − D ′ + + ( v ) ⊆ X . By (a), N D − D ′ − ( v ) ⊆ X ⊆ N D − D ′ + ( v ) ∪ N D − D ′ + + ( v ) . Since N D − D ′ + ( v ) ∩ N D − D ′ − ( v ) = ∅ , we have N D − D ′ − ( v ) ⊆ N D − D ′ + + ( v ) ⊆ X .

3. By Lemma 4.1, we have d D − D 1 + ( D 1 ) ≥ | S | . Then, | A ∪ V ( D 3 ′ ) | = | D 1 | + d D − D 1 + ( D 1 ) ≥ 1 + | S | ≥ | X | + 2 since |X| < |S|. Note that A ∪ V ( D 3 ′ ) ⊆ N D − D ′ + ( v ) and N D − D ′ − ( v ) ⊆ X . Then,

So, d D − D ′ + ( v ) ≥ d D − D ′ − ( v ) + 2 .□

Lemma 4.4

If |X| < |S|, then a Sullivan-i vertex of D′ is a Sullivan-i vertex of D for i ∈ {1, 2}.

Proof

By Lemma 4.3 (c), for any v ∈ V(D′), d D − D ′ + ( v ) ≥ d D − D ′ − ( v ) + 2 . Combining with Lemma 4.3 (b), we have d D − D ′ + + ( v ) + d D − D ′ + ( v ) ≥ 2 d D − D ′ − ( v ) + 2 . By Corollary 1.5, the tournament D′ always has a Sullivan-1 vertex and a Sullivan-2 vertex, say v′ and v″, respectively. Then, v′ is a Sullivan-1 vertex of D and v″ is a Sullivan-2 vertex of D.□

Lemma 4.5

If |X| ≥ |S|, then a Sullivan-i vertex of D ₁ is a Sullivan-i vertex of D for i ∈ {1, 2}.

Proof

Let v be a Sullivan-i vertex of D ₁ for i ∈ {1, 2}. We will prove that v is a Sullivan-i vertex of D for i ∈ {1, 2}.

Note that there is no arc from D ₁ to S. Otherwise D ₁ and D 1 ′ are adjacent which contradicts Theorem 2.4 (c). So, N D − D 1 + ( v ) = N D − S − D 1 + ( v ) . Combining with Theorem 2.3 (a), we have

Similarly, we have N D − D 1 − ( v ) = N D − D 1 − ( D 1 ) . By Lemma 4.1,

By the structure of D described in Theorems 2.6 and 2.7, we have N D − D 1 − ( D 1 ) ⊆ S and X ⊆ N D − D 1 + + ( D 1 ) = N D − D 1 + + ( v ) . Now

So, d D − D 1 + + ( v ) ≥ | S | ≥ d D − D 1 − ( v ) and d D − D 1 + + ( v ) + d D − D 1 + ( v ) ≥ 2 | S | ≥ 2 d D − D 1 − ( v ) . Since v is also a Sullivan-i vertex of D ₁, we see that v is a Sullivan-i vertex of D.□

Lemma 4.6

[18] Let D be a nonround decomposable locally semicomplete digraph. Then, D has a king.

Proposition 4.7

[12] Let D be an oriented graph. A king of D is a Sullivan-1 vertex.

Corollary 4.8

Let D be a nonround decomposable local tournament, which is not a tournament. Then, D has a Sullivan-1 vertex.

Theorem 4.9

Let D be a nonround decomposable local tournament, which is not a tournament. Then, D has at least two Sullivan-1 vertices.

Proof

Recall that A = N D 2 ′ + ( D 1 ) = V ( D λ ) ∪ V ( D λ + 1 ) ∪ ⋯ ∪ V ( D i ) , B = N S + ( A ) , X = N D − A + ( A ) = B ∪ ( V ( D 2 ′ ) − A ) ∪ V ( D p ) . The structure of D is illustrated in Figure 2.□

Case 1

There exists no arc between u and D _i+1.

We will prove that u is another Sullivan-1 vertex of D.

We claim that N D − D ′ − ( u ) ⊈ N D − D ′ + + ( u ) . By Lemma 4.3 (b), we have N D − D ′ − ( u ) ⊆ N D − D ′ + + ( u ) . We only need to prove that N D − D ′ − ( u ) ≠ N D − D ′ + + ( u ) . By Lemma 4.2, u ⇒ A ⇒ D _i+1. Combining with the fact that there exists no arc between u and D _i+1, we have D i + 1 ⊆ N D − D ′ + + ( u ) and D i + 1 ⊆ N D − D ′ − ( u ) . Then, N D − D ′ − ( u ) ≠ N D − D ′ + + ( u ) .