The second out-neighbourhood for local tournaments

TL;DR

This paper proves Sullivan's conjectures regarding out-neighbourhoods in local tournaments, showing that certain degree inequalities hold for vertices in these graphs, with specific conditions and multiple vertices satisfying the conjectures.

Contribution

It establishes the validity of Sullivan's conjectures for local tournaments and identifies conditions under which multiple vertices satisfy these degree inequalities.

Findings

Sullivan's conjecture (1) holds for local tournaments with no in-degree zero.

Sullivan's conjecture (2) holds or a stronger inequality is satisfied in local tournaments.

At least two vertices satisfy the conjecture (1) in local tournaments without in-degree zero.

Abstract

Sullivan stated the conjectures: (1) every oriented graph $D$ has a vertex $x$ such that $d^{++}(x)\geq d^{-}(x)$; (2) every oriented graph $D$ has a vertex $x$ such that $d^{++}(x)+d^{+}(x)\geq 2d^{-}(x)$. In this paper, we prove that these conjectures hold for local tournaments. In particular, for a local tournament $D$, we prove that $D$ has at least two vertices satisfying $(1)$ if $D$ has no vertex of in-degree zero. And, for a local tournament $D$, we prove that either there exist two vertices satisfying $(2)$ or there exists a vertex $v$ satisfying $d^{++}(v)+d^{+}(v)\geq 2d^{-}(v)+2$ if $D$ has no vertex of in-degree zero.

Equations

$$N_H^{++}(x)=\bigcup _{u\in N_H^{+}(x)}{N}_H^{+}(u)-N_H^{+}(x)$$

$$N_H^{+}(H^{\prime })=\bigcup _{x\in V(H^{\prime })}{N}_H^{+}(x)-V(H^{\prime }),N_H^{-}(H^{\prime })=\bigcup _{x\in V(H^{\prime })}{N}_H^{-}(x)-V(H^{\prime }).$$

$$D_1^{\prime }=D_p,{\lambda }_1=p\text{mbox},$$

$${\lambda }_{i+1}=\min \{j|N^{+}(D_j)\cap V(D_i^{\prime })\ne \varnothing \}\text{mbox}foreachi\in \{1,2,\dots ,r-1\}\text{mbox},$$

$$\text{mbox}and{D}_{i+1}^{\prime }=D⟨V(D_{{\lambda }_{i+1}})\cup V(D_{{\lambda }_{i+1}+1})\cup \dots \cup V(D_{{\lambda }_i-1})⟩\text{mbox}.$$

$$N^{-}(D_{\alpha })\cap V(D_{\mu })\ne \varnothing and{N}^{+}(D_{\alpha })\cap V(D_{\nu })\ne \varnothing \text{mbox},$$

$$\text{mbox}or{N}^{-}(D_{\mu })\cap V(D_{\alpha })\ne \varnothing and{N}^{+}(D_{\mu })\cap V(D_{\beta })\ne \varnothing \text{mbox},$$

$$d_D^{++}(v)=d_{S_j}^{++}(v)+d_{D^{\ast }}^{++}(v),d_D^{-}(v)=d_{S_j}^{-}(v)\cup d_{D^{\ast }}^{-}(v)\text{mbox}.$$

$$d_{D^{\ast }}^{-}(v_{t+1})\le d_{D^{\ast }}^{+}(v_1)={\delta }^{+}(D^{\ast })\text{mbox}.$$

$$d_{D^{\ast }}^{++}(v_{t+1})\ge d_{D^{\ast }}^{+}(v_h)\ge {\delta }^{+}(D^{\ast })\text{mbox}.$$

$$d_{D_1}^{++}(v)\ge d_{D_1}^{-}(v),d_{D_1}^{++}(v^{\prime })+d_{D_1}^{+}(v^{\prime })\ge 2d_{D_1}^{-}(v^{\prime })\text{mbox}.$$

$$d_D^{++}(v)\ge d_{D_1}^{++}(v),d_D^{-}(v)=d_{D_1}^{-}(v)\text{mbox}.$$

$$d_D^{++}(v^{\prime })\ge d_{D_1}^{++}(v^{\prime }),d_D^{+}(v^{\prime })\ge d_{D_1}^{+}(v^{\prime }),d_D^{-}(v)=d_{D_1}^{-}(v^{\prime })\text{mbox}.$$

$$d_{D^{\ast }}^{-}(v_{t+2})=|V_2\cup V_3\cup \dots \cup V_t\cup V_{t+1}|=d_{D^{\ast }}^{+}(v_1)+1={\delta }^{+}(D^{\ast })+1\text{mbox}.$$

$$d_{D^{\ast }}^{++}(v_{t+2})\ge d_{D^{\ast }}^{+}(v_g)\text{mbox}.$$

$$d_{D^{\ast }}^{-}(v_2)\le |V_{t+3}\cup \dots \cup V_r\cup V_1|=d_{D^{\ast }}^{+}(v_{t+2})={\delta }^{+}(D^{\ast })+1\text{mbox}.$$

$$d_{D^{\ast }}^{++}(v_2)\ge d_{D^{\ast }}^{+}(v_k)\ge {\delta }^{+}(D^{\ast })+1,d_{D^{\ast }}^{+}(v_2)\ge {\delta }^{+}(D^{\ast })+1\text{mbox}.$$

$$A=N_{D_2^{\prime }}^{+}(D_1)=V(D_{\lambda })\cup V(D_{\lambda +1})\cup \dots \cup V(D_i),B=N_S^{+}(A)\text{mbox},$$

$$X=N_{D-A}^{+}(A)=B\cup (V(D_2^{\prime })-A)\cup V(D_p)\text{mbox},$$

$$D^{\prime}=D\langle S-B\rangle\mbox{,\,\,when $|X|\leq|S|$}.$$

$$N_{D-D^{\prime }}^{+}(v)\supseteq A\cup V(D_3^{\prime })=V(D-D^{\prime })-X$$

$$|X|+2\le |A\cup V(D_3^{\prime })|\le d_{D-D^{\prime }}^{+}(v),d_{D-D^{\prime }}^{-}(v)\le |X|.$$

$$N_{D-D_1}^{+}(v)=N_{D-S-D_1}^{+}(v)=N_{D-S-D_1}^{+}(D_1)=N_{D-D_1}^{+}(D_1).$$

$$d_{D-D_1}^{+}(v)=d_{D-D_1}^{+}(D_1)\ge |S|.$$

$$d_{D-D_1}^{++}(v)\ge |X|\ge |S|,d_{D-D_1}^{-}(v)\le |S|\text{mbox}.$$

$$N_{D-D^{\prime }}^{+}(u)\supseteq A\cup V(D_3^{\prime })=V(D-D^{\prime })-X$$

$$d_{D-D^{\prime }}^{-}(u)\le d_{D-D^{\prime }}^{++}(u)\le |X|.$$

$$d_{D-D^{\prime }}^{+}(u)\ge |A\cup V(D_3^{\prime })|\ge 1+|S|\ge 1+|X|.$$

$$A^{\prime }=N_{D_2^{\prime }}^{+}(D_2)=V(D_{\lambda })\cup V(D_{\lambda +1})\cup \dots \cup V(D_{i^{\prime }}),B^{\prime }=N_S^{+}(A^{\prime })\text{mbox},$$

$$X^{\prime }=N_{D-A^{\prime }}^{+}(A^{\prime })=B^{\prime }\cup (V(D_2^{\prime })-A^{\prime })\cup V(D_p)\text{mbox}.$$

Full text

Keywords: local tournaments; the second out-neighbourhood; Sullivan’s conjectures; round decomposable; non-round decomposable

1 Introduction

In this paper, we consider finite digraphs without loops and multiple arcs. The main source for terminology and notation is [1].

Let $D$ be a digraph. We denote the vertex set and the arc set of $D$ by $V(D)$ and $A(D)$, respectively. For a vertex subset $X$, we denote the subdigraph of $D$ induced by $X$ (respectively, $D-X$) by $D\langle X\rangle$ (respectively, $D\langle V(D)-X\rangle$). For convenience, we write $D-X$ instead of $D\langle V(D)-X\rangle$. In addition $D-x=D-\{x\}$. And if $X$ is a subdigraph, we write $D-X$ instead of $D-V(X)$.

Let $x,y$ be distinct vertices of $D$. If there is an arc from $x$ to $y$, we say that $x$ dominates $y$ and denote it by $x\rightarrow y$ and call $y$ (respectively, $x$) an out-neighbour (respectively, an in-neighbour) of $x$ (respectively, $y$). If $V_{1}$ and $V_{2}$ are disjoint subsets of vertices of $D$ such that there is no arc from $V_{2}$ to $V_{1}$ and $a\rightarrow b$ for all $a\in V_{1}$ and $b\in V_{2}$, then we say that $V_{1}$ completely dominates $V_{2}$ and denote it by $V_{1}\Rightarrow V_{2}$. We will use the same notation when $V_{1}$ or $V_{2}$ is subdigraphs of $D$. In particular, if $V_{1}$ contains only one vertex $v$, denote it by $v\Rightarrow V_{2}$.

For a subdigraph or simply a vertex subset $H$ of $D$ (possibly, $H=D$), we let $N^{+}_{H}(x)$ (respectively, $N^{-}_{H}(x)$) denote the set of out-neighbours (respectively, in-neighbours) of $x$ in $H$ and call it the out-neighbourhood (respectively, in-neighbourhood) of $x$ in $H$. Furthermore, $d^{+}_{H}(x)=|N^{+}_{H}(x)|$ (respectively, $d^{-}_{H}(x)=|N^{-}_{H}(x)|)$ is called the out-degree (respectively, in-degree) of $x$ in $H$. Let

[TABLE]

which is called the second out-neighbourhood of $x$ in $H$. Furthermore, $d^{++}_{H}(x)=|N^{++}_{H}(x)|$. We will omit the subscript $H$ if the digraph is known from the context. For a pair of vertex disjoint subdigraphs $H$ and $H^{\prime}$, we define

[TABLE]

A vertex $x$ is a 2-king (for short, a king) of $D$, if for $y\in V(D)-x$, there exists an $(x,y)$-path of length at most 2.

A digraph $D$ is strong if, for every pair $x$, $y$ of distinct vertices, $D$ contains a path from $x$ to $y$ and a path from $y$ to $x$. A strong component of a digraph $D$ is a maximal induced subdigraph of $D$ which is strong. If $D_{1},D_{2},\ldots,D_{t}$ are the strong components of $D$, then clearly $V(D_{1})\cup V(D_{2})\cup\ldots\cup V(D_{t})=V(D)$ (note that a digraph with only one vertex is strong). Moreover, we must have $V(D_{i})\cap V(D_{j})=\emptyset$ for every $i\neq j$. The strong components of $D$ can be labelled $D_{1},D_{2},\ldots,D_{t}$ such that there is no arc from $D_{j}$ to $D_{i}$ unless $j<i$. We call such an ordering an acyclic ordering of the strong components of $D$.

For a vertex subset $S$ of strong digraph $D$, $S$ is called a separating set of $D$ if $D-S$ is not strong. A separating set $S$ of $D$ is minimal if for any proper subset $S^{\prime}$ of $S$, the subdigraph $D-S^{\prime}$ is strong.

A digraph $R$ on $n$ vertices is round if we can label its vertices $v_{1},v_{2},\ldots,v_{n}$ so that for each $i$, we have $N^{+}(v_{i})=\{v_{i+1},v_{i+2},\ldots,v_{i+d^{+}(v_{i})}\}$ and $N^{-}(v_{i})=\{v_{i-d^{-}(v_{i})},\ldots,v_{i-2},v_{i-1}\}$(all subscripts are taken modulo $n$). We will refer to the ordering $v_{1},v_{2},\ldots,v_{n}$ as a round labelling of $R$.

A digraph $D$ is semicomplete if, for every pair $x$, $y$ of distinct vertices in $D$, either $x$ dominates $y$ or $y$ dominates $x$ (or both). Tournaments are semicomplete digraphs with no 2-cycle.

A digraph $D$ with no 2-cycle is an oriented graph.

In 1990, Seymour [2] proposed the following conjecture which is one of the most interesting and challenging open questions concerning oriented graphs.

Conjecture 1.1**.**

(Seymour’s Second Neighbourhood Conjecture(SSNC)) For any oriented graph $D$, there exists a vertex $v$ in $D$ such that $d^{++}(v)\geq d^{+}(v)$.

We call such a vertex $v$ satisfying Conjecture 1.1 a Seymour vertex. The first non-trivial result for SSNC was obtained by Fisher [3] who proved Dean’s conjecture [2], which is SSNC restricted to tournaments. Fisher used Farkas’ Lemma and averaging arguments.

Theorem 1.2**.**

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

A more elementary proof of SSNC for tournaments was given by Havet and Thomassé [4] who introduced a median order approach. Their proof also yields the following stronger result.

Theorem 1.3**.**

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

Fidler and Yuster [5] further developed the median order approach and proved that SSNC holds for oriented graphs $D$ with minimum degree $|V(D)|-2$, tournaments minus a star and tournaments minus the arc set of a subtournament. The median order approach was also used by Ghazal [6] who proved a weighted version of SSNC for tournaments missing a generalized star. Kaneko and Locke [7] proved SSNC for oriented graphs with minimum out-degree at most $6$. Cohn, Godbole, Wright, Harkness and Zhang [8] proved SSNC for random oriented graphs with probability $p<\frac{1}{2}-\delta$. Gutin and Li [9] proved SSNC for extended tournaments and quasi-transitive oriented graphs.

Another approach to SSNC is to determine the maximum value $\gamma$ such that in every oriented graph $D$, there exists a vertex $x$ such that $d^{+}(x)\leq\gamma d^{++}(x)$. SSNC asserts that $\gamma=1$. Chen, Shen and Yuster [10] proved that $\gamma\geq r$ where $r=0.657298\ldots$ is the unique real root of $2x^{3}+x^{2}-1=0$. Furthermore, they improves this bound to $0.67815\ldots$ mentioned in the end of the article [10].

Sullivan [11] stated the following “compromise conjectures” on SSNC, where $d^{-}(v)$ is used instead of or together with $d^{+}(v)$.

Conjecture 1.4**.**

[11]** (1) Every oriented graph $D$ has a vertex $x$ such that $d^{++}(x)\geq d^{-}(x)$.

(2) Every oriented graph $D$ has a vertex $x$ such that $d^{++}(x)+d^{+}(x)\geq 2d^{-}(x)$.

For convenience, a vertex $x$ in $D$ satisfying Conjecture 1.4 ($i$) is called a Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$.

Li and Sheng [12] [13], proved Sullivan’s Conjectures for tournaments, extended tournaments, quasi-transitive oriented graphs as well as bipartite tournaments. For tournaments, they obtained the following results:

Corollary 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 $T\in\mathcal{T}$.

$\mathcal{T}$ is a special class of tournaments. $T\in\mathcal{T}$ if $T$ is a tournament consisting of exactly two strong components $T_{1}$ and $T_{2}$ such that $T_{1}$ dominates $T_{2}$, $T_{1}$ is a single vertex $v$ and $T_{2}$ is a tournament satisfying that $d_{T_{2}}^{+}(x)\leq d_{T_{2}}^{-}(x)+1$ for any $x\in V(T_{2})$. It is easy to check that $v$ is the unique Sullivan-2 vertex of $T$.

From Theorem 1.6, we obtain immediately the following result:

Corollary 1.7**.**

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

A digraph $D$ is locally semicomplete if $D\langle N^{+}(x)\rangle$ and $D\langle N^{-}(x)\rangle$ are both semicomplete for every vertex $x$ of $D$. Specifically, every round digraph is locally semicomplete [15]. A local tournament is a locally semicomplete digraph with no 2-cycle.

Let $D$ be a digraph with vertex set $\{v_{1},v_{2},\ldots,v_{n}\}$, and let $G_{1},G_{2},\ldots,G_{n}$ be digraphs which are pairwise vertex disjoint. The composition $D[G_{1},G_{2},\ldots,G_{n}]$ is the digraph $L$ with vertex set $V(G_{1})\cup V(G_{2})\cup\ldots\cup V(G_{n})$ and arc set $(\cup^{n}_{i=1}A(G_{i}))\cup\{g_{i}g_{j}|g_{i}\in V(G_{i}),g_{j}\in V(G_{j}),v_{i}v_{j}\in A(D)\}$. If $D=H[V_{1},V_{2},\ldots,\\ V_{n}]$ and none of the digraphs $V_{1},V_{2},\ldots,V_{n}$ has an arc, then $D$ is an extension of $H$.

A digraph $D$ is round decomposable if there exists a round local tournament $R$ on $r\geq 2$ vertices such that $D=R[S_{1},S_{2},\ldots,S_{r}]$, where each $S_{i}$ is a strong semicomplete digraph. We call $R[S_{1},S_{2},\ldots,S_{r}]$ a round decomposition of $D$. Clearly, a round decomposable digraph is locally semicomplete.

Locally semicomplete digraphs were introduced in 1990 by Bang-Jensen [14]. The following theorem, due to Bang-Jensen, Guo, Gutin and Volkmann, stated a full classification of locally semicomplete digraphs.

Theorem 1.8**.**

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

$(a)$* $D$ is round decomposable with a unique round decomposition $R[S_{1},S_{2},\ldots,\\ S_{r}]$, where $R$ is a round local tournament on $r\geq 2$ vertices and $S_{i}$ is a strong semicomplete digraph for each $i\in\{1,2,\ldots,r\}$;*

$(b)$* $D$ is non-round decomposable and not semicomplete and it has the structure as described in Theorem 2.4;*

$(c)$* $D$ is a semicomplete digraph which is non-round decomposable.*

If $D$ is restricted to a local tournament, we have the following result:

Corollary 1.9**.**

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

$(a)$* $D$ is round decomposable with a unique round decomposition $R[S_{1},S_{2},\ldots,\\ S_{r}]$, where $R$ is a round local tournament on $r\geq 2$ vertices and $S_{i}$ is a strong tournament for $i\in\{1,2,\ldots,r\}$;*

$(b)$* $D$ is non-round decomposable and not a tournament and it has the structure as described in Theorem 2.4;*

$(c)$* $D$ is a tournament which is non-round decomposable.*

In [16], we investigate SSNC for local tournaments. In this paper, we discuss Sullivan’s Conjectures for local tournaments. In Section 2, we introduce the structure of a local tournament. In Section 3 and Section 4, we investigate the Sullivan-$i$ vertex in a round decomposable local tournament and a non-round decomposable local tournament, respectively, for $i\in\{1,2\}$.

2 The structure of a local tournament

In this section, all theorems are on the structure of locally semicomplete digraphs. Clearly, these theorems also hold if the digraph is restricted to a local tournament.

Theorem 2.1**.**

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

$(a)$* If $A$ and $B$ are distinct strong components of $D$ with at least one arc between them, then either $A\Rightarrow B$ or $B\Rightarrow A$.*

$(b)$* If $A$ and $B$ are strong components of $D$, such that $A\Rightarrow B$, then $A$ and $B$ are semicomplete digraphs.*

$(c)$* The strong components of $D$ can be ordered in a unique way $D_{1},D_{2},\ldots,D_{p}$ such that there is no arc from $D_{j}$ to $D_{i}$ for $j>i$, and $D_{i}$ completely dominates $D_{i+1}$ for $i\in\{1,2,\ldots,p-1\}$.*

A kind of the decomposition of non-strong locally semicomplete digraphs described in [17] is the following.

Theorem 2.2**.**

[17]** Let $D$ be a connected, but not strong locally semicomplete digraph, and let $D_{1},D_{2},\ldots,D_{p}$ be the acyclic ordering of the strong components of $D$. Then $D$ can be decomposed into $r\geq 2$ induced subdigraphs $D^{\prime}_{1},D^{\prime}_{2},\ldots,D^{\prime}_{r}$ as follows:

[TABLE]

[TABLE]

[TABLE]

The subdigraphs $D^{\prime}_{1},D^{\prime}_{2},\ldots,D^{\prime}_{r}$ satisfy the properties below:

$(a)$* $D^{\prime}_{i}$ consists of some strong components of $D$ and is semicomplete for each $i\in\{1,2,\ldots,r\}$;*

$(b)$* $D^{\prime}_{i+1}$ completely dominates the initial component of $D^{\prime}_{i}$ and there exists no arc from $D^{\prime}_{i}$ to $D^{\prime}_{i+1}$ for $i\in\{1,2,\ldots,r-1\}$;*

$(c)$* If $r\geq 3$, then there is no arc between $D^{\prime}_{i}$ and $D^{\prime}_{j}$ for $i,\,j$ satisfying $|j-i|\geq 2$.*

The unique sequence $D^{\prime}_{1},D^{\prime}_{2},\ldots,D^{\prime}_{r}$ defined in Theorem 2.2 will be referred to as the $semicomplete$ $decomposition$ of $D$.

Theorem 2.3**.**

[15]** If $D$ is a round decomposable locally semicompete digraph, then it has a unique round decomposition $D=R[S_{1},S_{2},\ldots,S_{r}]$, where $R$ is a round local tournament on $r\geq 2$ vertices and each $S_{i}$ is a strong semicomplete digraph.

Theorem 2.4**.**

[15]** Let $D$ be a strong locally semicomplete digraph which is not semicomplete. Then $D$ is non-round decomposable if and only if the following conditions are satisfied:

$(a)$* There is a minimal separating set $S$ such that $D-S$ is not semicomplete, and for each such $S$, $D\langle S\rangle$ is semicomplete and the semicomplete decomposition of $D-S$ has exactly three components $D^{\prime}_{1}$, $D^{\prime}_{2}$, $D^{\prime}_{3}$;*

$(b)$* There are integers $\alpha,\beta,\mu,\nu$ with $\lambda\leq\alpha\leq\beta\leq p-1$ and $p+1\leq\mu\leq\nu\leq p+q$ such that*

[TABLE]

[TABLE]

where $D_{1},D_{2},\ldots,D_{p}$ and $D_{p+1},D_{p+2},\ldots,D_{p+q}$ are the acyclic orderings of the strong components of $D-S$ and $D\langle S\rangle$, respectively, and $D_{\lambda}$ is the initial component of $D_{2}^{{}^{\prime}}$.

By Theorem 2.4, $D$ is always strong if $D$ is a non-round decomposable locally semicomplete digraph. An example of a non-round decomposable locally semicomplete digraph is shown in Figure 2.

Theorem 2.5**.**

[15]** Let $D$ be a strong non-round decomposable locally semicomplete digraph and let S be a minimal separating set of $D$ such that $D-S$ is not semicomplete. Let $D_{1},D_{2},\ldots,D_{p}$ be the acyclic ordering of the strong components of $D-S$ and $D_{p+1},D_{p+2},\ldots,D_{p+q}$ be the acyclic ordering of the strong components of $D\langle S\rangle$. The following holds:

$(a)$* $D_{p}\Rightarrow S\Rightarrow D_{1}$.*

$(b)$* Suppose that there is an arc $s\rightarrow v$ from $S$ to $D^{\prime}_{2}$ with $s\in V(D_{i})$ and $v\in V(D_{j})$. Then $D_{i}\cup D_{i+1}\cup\ldots\cup D_{p+q}\Rightarrow D^{\prime}_{3}\Rightarrow D_{\lambda}\cup\ldots\cup D_{j}$.*

$(c)$* $D_{p+q}\Rightarrow D^{\prime}_{3}$ and $D_{f}\Rightarrow D_{f+1}$ for $f\in\{1,2,\ldots,p+q\}$ where subscripts are modulo $p+q$.*

3 In a round decomposable local tournament

In this section, $D$ is always a round decomposable local tournament and let the unique round decomposition of $D$ be $R[S_{1},S_{2},\ldots,S_{r}]$, where $R$ is a round local tournament on $r\geq 2$ vertices and each $S_{i}$ is a strong tournament.

We begin with a useful observation.

Lemma 3.1**.**

Let $D$ be a round decomposable local tournament and $R[S_{1},S_{2},\ldots,\\ S_{r}]$ be the unique round decomposition of $D$. Let $D^{*}=R[V_{1},V_{2},\ldots,V_{r}]$ and $v_{i}\in V_{i}$ be arbitrary, where $V_{i}$ is the vertex set of $S_{i}$ for $i\in\{1,2,\ldots,r\}$. If $N_{D^{*}}^{+}(v_{j})=V_{j+1}\cup\ldots\cup V_{k}$, then $d_{D^{*}}^{++}(v_{j})\geq d_{D^{*}}^{+}(v_{k})$.

Proof. Let $v\in N_{D^{*}}^{+}(v_{k})$. We claim that $v\notin N_{D^{*}}^{+}(v_{j})$. In fact, if $v_{j}\rightarrow v$, then $v_{j}$, $v$, $v_{k}$ are in the order of the round labelling of $R$. Then $v_{k}\rightarrow v_{j}$ since $v_{k}\rightarrow v$. Note that $v_{j}\rightarrow v_{k}$. This contradicts the fact that $D$ has no 2-cycle. So $v\notin N_{D^{*}}^{+}(v_{j})$. Thus $v\in N_{D^{*}}^{++}(v_{j})$, i.e., $N_{D^{*}}^{+}(v_{k})\subseteq N_{D^{*}}^{++}(v_{j})$. Then $d^{++}_{D^{*}}(v_{j})\geq d^{+}_{D^{*}}(v_{k})$.∎

First, we consider the existence of a Sullivan-$i$ vertex in $D$ for $i\in\{1,2\}$.

Lemma 3.2**.**

Let $D$ be a round decomposable local tournament and $R[S_{1},S_{2},\ldots,\\ S_{r}]$ be the unique round decomposition of $D$. Let $D^{*}=R[V_{1},V_{2},\ldots,V_{r}]$ and $v_{j}\in V_{j}$ be arbitrary, where $V_{j}$ is the vertex set of $S_{j}$ for $j\in\{1,2,\ldots,r\}$. If there is a vertex $v\in V_{j}$ 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 $i\in\{1,2\}$.

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 $d^{++}_{S_{j}}(v)\geq d^{-}_{S_{j}}(v),\,\,d^{++}_{D^{*}}(v)\geq d^{-}_{D^{*}}(v)$. Clearly

[TABLE]

Thus $d^{++}_{D}(v)\geq d^{-}_{D}(v)$ 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 $i\in\{1,2$}.

Proof. Let $R[S_{1},S_{2},\ldots,S_{r}]$ be the unique round decomposition of $D$. Let $D^{*}=R[V_{1},V_{2},\ldots,V_{r}]$ and $v_{j}\in V_{j}$ be arbitrary, where $V_{j}$ is the vertex set of $S_{j}$ for $j\in\{1,2,\ldots,r\}$. W.l.o.g., assume that $v_{1}\in V_{1}$ is a vertex of $D^{*}$ with minimum out-degree, i.e., $d_{D^{*}}^{+}(v_{1})=\delta^{+}(D^{*})$. Let $N_{D^{*}}^{+}(v_{1})=V_{2}\cup\ldots\cup V_{t}$. Since $v_{1}\nrightarrow v_{t+1}$, we have $N_{D^{*}}^{-}(v_{t+1})\subseteq V_{2}\cup\ldots\cup V_{t}=N_{D^{*}}^{+}(v_{1})$. Then

[TABLE]

Let $N_{D^{*}}^{+}(v_{t+1})=V_{t+2}\cup\ldots\cup V_{h}$. By Lemma 3.1, we have

[TABLE]

Thus $d_{D^{*}}^{++}(v_{t+1})\geq\delta^{+}(D^{*})\geq d_{D^{*}}^{-}(v_{t+1})$ and $v_{t+1}$ is a Sullivan-1 vertex of $D^{*}$. Note that $d_{D^{*}}^{+}(v_{t+1})\geq\delta^{+}(D^{*})$. Then $d_{D^{*}}^{++}(v_{t+1})+d_{D^{*}}^{+}(v_{t+1})\geq 2\delta^{+}(D^{*})\geq 2d_{D^{*}}^{-}(v_{t+1})$ 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 $i\in\{1,2\}$. 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 $v^{\prime}_{t+1}$, respectively. By Lemma 3.2, $v_{t+1}$ is a Sullivan-1 vertex of $D$ and $v^{\prime}_{t+1}$ is a Sullivan-2 vertex of $D$.∎

Next, we consider the number of Sullivan-$i$ vertices in a connected round decomposable local tournament with no vertex of in-degree zero for $i\in\{1,2\}$. Note that every non-strong local tournament is round decomposable. We consider two cases: $(1)$ a connected, but not strong local tournament with no vertex of in-degree zero; $(2)$ a strong round decomposable local tournament.

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 $D_{1},D_{2},\ldots,D_{p}$ be the acyclic ordering of the strong components of $D$. Let $v$ be a Sullivan-1 vertex of $D_{1}$ and $v^{\prime}$ be a Sullivan-2 vertex of $D_{1}$, i.e.,

[TABLE]

Clearly

[TABLE]

[TABLE]

Thus $d^{++}_{D}(v)\geq d^{-}_{D}(v)$ and $d^{++}_{D}(v^{\prime})+d^{++}_{D}(v^{\prime})\geq 2d^{-}_{D}(v^{\prime})$, i.e., $v$ is a Sullivan-1 vertex of $D$ and $v^{\prime}$ is a Sullivan-2 vertex of $D$. So the Sullivan-$i$ vertex of $D_{1}$ is always the Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$.

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 Corollary 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 $i\in\{1,2\}$.

Proof. Let $D=R[S_{1},S_{2},\ldots,S_{r}]$ be the unique round decomposition. Let $D^{*}=R[V_{1},V_{2},\ldots,V_{r}]$ and $v_{j}\in V_{j}$ be arbitrary, where $V_{j}$ is the vertex set of $S_{j}$ for $j\in\{1,2,\ldots,r\}$. W.l.o.g., assume that $v_{1}\in V_{1}$ is a vertex of $D^{*}$ with minimum out-degree, i.e., $d_{D^{*}}^{+}(v_{1})=\delta^{+}(D^{*})$. Let $N_{D^{*}}^{+}(v_{1})=V_{2}\cup\ldots\cup V_{t}$. According to the proof of Theorem 3.3, a Sullivan-$i$ vertex of $S_{t+1}$ is a Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$.

For the case when $|V_{t+1}|\geq 2$, by Corollary 1.5 and 1.7, the strong tournament $S_{t+1}$ has at least three Sullivan-1 vertices and two Sullivan-2 vertices. Then these three vertices (respectively, two vertices) are the Sullivan-1 (respectively, Sullivan-2) vertices of $D$.

For the case when $|V_{t+1}|=1$ and there exists $v_{h}\notin V_{1}$ such that $d_{D^{*}}^{+}(v_{h})=\delta^{+}(D^{*})$, we can repeat the proof of Theorem 3.3 and obtain a different $``v_{t+1}$”. Now the so-called $``v_{t+1}$” is another Sullivan-$i$ vertex for $i\in\{1,2\}$.

Now we consider the case when $|V_{t+1}|=1$ and there exists no $v_{h}\notin V_{1}$ such that $d_{D^{*}}^{+}(v_{h})=\delta^{+}(D^{*})$. Then $d^{+}_{D^{*}}(v_{j})>\delta^{+}(D^{*})\geq 1$ for any $j\neq 1$. According to the proof of Theorem 3.3, the only vertex $v_{t+1}$ of $V_{t+1}$ is a Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$. It is sufficient to find another Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$.

We claim that $v_{2}\rightarrow v_{t+2}$ and $v_{t+2}\notin V_{1}$. In fact, we have $v_{t+2}$ must be in the set $N_{D^{*}}^{+}(v_{2})$ since $d^{+}_{D^{*}}(v_{2})>d^{+}_{D^{*}}(v_{1})$, $|V_{2}|\geq|V_{t+1}|=1$ and $N^{+}_{D^{*}}(v_{1})=V_{2}\cup V_{3}\cup\ldots\cup V_{t}$, $N_{D^{*}}^{+}(v_{2})\supseteq V_{3}\cup\ldots\cup V_{t}\cup V_{t+1}$. So $v_{2}\rightarrow v_{t+2}$. Note that $v_{1}\rightarrow v_{2}$. Furthermore, $v_{t+2}\notin V_{1}$ since $D$ has no 2-cycle.

Note $v_{2}\rightarrow v_{t+2}$ and $v_{1}\nrightarrow v_{t+2}$. Then $N^{-}_{D^{*}}(v_{t+2})=V_{2}\cup V_{3}\cup\ldots\cup V_{t}\cup V_{t+1}$. Since $|V_{t+1}|=1$, we have

[TABLE]

Let $N_{D^{*}}^{+}(v_{t+2})=V_{t+3}\cup\ldots\cup V_{g}$. By Lemma 3.1, we have

[TABLE]

Case 1. $v_{g}\neq v_{1}$.

We see that $d^{++}_{D^{*}}(v_{t+2})\geq d^{+}_{D^{*}}(v_{g})\geq\delta^{+}(D^{*})+1\geq d_{D^{*}}^{-}(v_{t+2})$, and $v_{t+2}$ is a Sullivan-1 vertex of $D^{*}$.

Note $v_{t+2}\notin V_{1}$. Then $d_{D^{*}}^{+}(v_{t+2})\geq\delta^{+}(D^{*})+1\geq d_{D^{*}}^{-}(v_{t+2})$. Thus $d_{D^{*}}^{++}(v_{t+2})+d_{D^{*}}^{+}(v_{t+2})\geq 2d_{D^{*}}^{-}(v_{t+2})$ 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 $i\in\{1,2\}$. By Corollary 1.5, the tournament $S_{t+2}$ has a Sullivan-1 vertex and a Sullivan-2 vertex, say $v_{t+2}$ and $v^{\prime}_{t+2}$, respectively. By Lemma 3.2, $v_{t+2}$ is another Sullivan-1 vertex of $D$ and $v^{\prime}_{t+2}$ is another Sullivan-2 vertex of $D$.

Case 2. $v_{g}=v_{1}$.

Now $N_{D^{*}}^{+}(v_{t+2})=V_{t+3}\cup\ldots\cup V_{r}\cup V_{1}$.

First we show that $v_{2}$ is another Sullivan-1 vertex of $D^{*}$. Recall $v_{2}\rightarrow v_{t+2}$. Then $V_{2}\Rightarrow V_{3}\cup V_{4}\cup\ldots\cup V_{t+2}$. Also $V_{t+2}\Rightarrow V_{t+3}\cup\ldots\cup V_{r}\cup V_{1}$. Then $v\in N^{+}(v_{2})\cup N^{++}(v_{2})$ for any $v\in V(D^{*}-V_{2})$. Since $N_{D^{*}-V_{2}}^{+}(v_{2})\cap N_{D^{*}-V_{2}}^{-}(v_{2})=\emptyset$, we have $N_{D^{*}-V_{2}}^{-}(v_{2})\subseteq N_{D^{*}-V_{2}}^{++}(v_{2})$. Then $d_{D^{*}-V_{2}}^{++}(v_{2})\geq d_{D^{*}-V_{2}}^{-}(v_{2})$. 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 $d_{D^{*}}^{+}(v_{t+2})\geq\delta^{+}(D^{*})+2$, we have $d^{++}_{D^{*}}(v_{t+2})\geq d^{+}_{D^{*}}(v_{g})=d^{+}_{D^{*}}(v_{1})=\delta^{+}(D^{*})$. Thus $d_{D^{*}}^{++}(v_{t+2})+d_{D^{*}}^{+}(v_{t+2})\geq\delta^{+}(D^{*})+\delta^{+}(D^{*})+2\geq 2(\delta^{+}(D^{*})+1)\geq 2d_{D^{*}}^{-}(v_{t+2})$ 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 $d_{D^{*}}^{+}(v_{t+2})\leq\delta^{+}(D^{*})+1$, we have $d_{D^{*}}^{+}(v_{t+2})=\delta^{+}(D^{*})+1$ since we note $v_{t+2}\notin V_{1}$. Also note $v_{2}\rightarrow v_{t+2}$. Then $N_{D^{*}}^{+}(v_{2})\supseteq V_{3}\cup\ldots\cup V_{t+2}$ and hence $N_{D^{*}}^{-}(v_{2})\subseteq V_{t+3}\cup\ldots\cup V_{r}\cup V_{1}$. So

[TABLE]

Let $N_{D^{*}}^{+}(v_{2})=V_{3}\cup\ldots\cup V_{k}$. By Lemma 3.1, we have $d^{++}_{D^{*}}(v_{2})\geq d^{+}_{D^{*}}(v_{k})$. Note $v_{2}\rightarrow v_{k}$ and $v_{1}\rightarrow v_{2}$. Then $v_{k}\notin V_{1}$ since $D$ has no 2-cycle. Note that

[TABLE]

Then $d_{D^{*}}^{++}(v_{2})+d_{D^{*}}^{+}(v_{2})\geq 2(\delta^{+}(D^{*})+1)\geq 2d_{D^{*}}^{-}(v_{2})$ 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**.**

$(a)$* Every round decomposable local tournament has a Sullivan-$i$ vertex for $i\in\{1,2\}$.*

$(b)$* Every round decomposable local tournament with no vertex of in-degree zero has at least two Sullivan-i vertices for $i\in\{1,2\}$.*

Two examples of round local tournament, which have exactly two Sullivan-$i$ vertices for $i\in\{1,2\}$, are illustrated in Figure 1.

4 In a non-round decomposable local tournament

In this section, $D$ is always a non-round decomposable local tournament, which is not a tournament. We also assume that $S$ is chosen with minimum cardinality among all minimal separating sets of $D$ satisfying that $D-S$ is not a tournament, $D_{1},D_{2},\ldots,D_{p}$ is the acyclic ordering of the strong components of $D-S$, $D_{p+1},D_{p+2},\ldots,D_{p+q}$ is the acyclic ordering of the strong components of $D\langle S\rangle$, $D^{\prime}_{1},D^{\prime}_{2},D^{\prime}_{3}$ is the semicomplete decomposition of $D-S$ and $D_{\lambda}$ is the initial component of $D^{\prime}_{2}$. Clearly, in a local tournament $D$, the subdigraphs $S$, $D^{\prime}_{1}$, $D^{\prime}_{2}$ and $D^{\prime}_{3}$ are all tournaments. See Figure 2. Let

[TABLE]

[TABLE]

[TABLE]

Lemma 4.1**.**

For any strong component $D_{j}$ of $D^{\prime}_{3}$, $d_{D-D_{j}}^{+}(D_{j})\geq|S|$.

Proof. Let $D_{j}\subseteq D^{\prime}_{3}$ 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 $d_{D-D_{j}}^{+}(D_{j})\geq|S|$ by the choice of $S$.∎

Lemma 4.2**.**

If $|X|<|S|$, then $S-B\Rightarrow A$ and $S-B\Rightarrow D^{\prime}_{3}$.

Proof. Note that $X$ is also a separating set of $D$. By the choice of $S$, combining with $|X|<|S|$, we see that $D-X=D\langle V(D^{\prime}_{3})\cup A\cup(S-B)\rangle$ is a tournament. So any vertex of $S-B$ is adjacent to the vertices of $A$ and $D^{\prime}_{3}$. Since $N_{S-B}^{+}(A)=\emptyset$, we have $S-B\Rightarrow A$. Also, an arc from $D^{\prime}_{3}$ to $S$ implies that $D^{\prime}_{3}$ and $D^{\prime}_{1}$ are adjacent, which contradicts Theorem 2.2 ($c$). So $S-B\Rightarrow D^{\prime}_{3}$.∎

Lemma 4.3**.**

If $|X|<|S|$, then for any $v\in V(D^{\prime})=S-B$,

$(a)$* $X\subseteq N_{D-D^{\prime}}^{+}(v)\cup N_{D-D^{\prime}}^{++}(v)$.*

$(b)$* $N^{-}_{D-D^{\prime}}(v)\subseteq N^{++}_{D-D^{\prime}}(v)\subseteq X$.*

$(c)$* $d_{D-D^{\prime}}^{+}(v)\geq d_{D-D^{\prime}}^{-}(v)+2$.*

Proof. $(a)$ Recall that $X=B\cup(V(D^{\prime}_{2})-A)\cup V(D_{p})$. Since $|X|<|S|$, we have $|B|\leq|X|-1\leq|S|-2$ and hence $|S-B|\geq 2$. By Lemma 4.2, for any $v\in V(D^{\prime})=S-B$, we have $v\Rightarrow A\Rightarrow D^{\prime}_{2}-A$ and $v\Rightarrow A\Rightarrow D_{p}$. Since $B=N_{S}^{+}(A)$, for any $y\in B$, there exists a vertex $x\in A$ such that $x\rightarrow y$ and hence $v\rightarrow x\rightarrow y$. So any vertex of $X$ either belongs to $N_{D-D^{\prime}}^{+}(v)$ or belongs to $N_{D-D^{\prime}}^{++}(v)$, i.e., $X\subseteq N_{D-D^{\prime}}^{+}(v)\cup N_{D-D^{\prime}}^{++}(v)$.

$(b)$ Recall that $D^{\prime}=D\langle S-B\rangle$. By the definition of $A$, $B$ and $X$, we have $V(D-D^{\prime})-X=A\cup V(D^{\prime}_{3})$. For any $v\in V(D^{\prime})$, by Lemma 4.2, we have $v\Rightarrow A$ and $v\Rightarrow D^{\prime}_{3}$. Then

[TABLE]

and hence $N^{-}_{D-D^{\prime}}(v)\subseteq X$ and $N^{++}_{D-D^{\prime}}(v)\subseteq X.$ By $(a)$, $N^{-}_{D-D^{\prime}}(v)\subseteq X\subseteq N_{D-D^{\prime}}^{+}(v)\cup N_{D-D^{\prime}}^{++}(v)$. Since $N_{D-D^{\prime}}^{+}(v)\cap N_{D-D^{\prime}}^{-}(v)=\emptyset$, we have $N^{-}_{D-D^{\prime}}(v)\subseteq N^{++}_{D-D^{\prime}}(v)\subseteq X$.

$(c)$ By Lemma 4.1, we have $d^{+}_{D-D_{1}}(D_{1})\geq|S|$. Then $|A\cup V(D^{\prime}_{3})|=|D_{1}|+d^{+}_{D-D_{1}}(D_{1})\geq 1+|S|\geq|X|+2$ since $|X|<|S|$. Note that $A\cup V(D^{\prime}_{3})\subseteq N^{+}_{D-D^{\prime}}(v)$ and $N^{-}_{D-D^{\prime}}(v)\subseteq X$. Then

[TABLE]

So $d_{D-D^{\prime}}^{+}(v)\geq d_{D-D^{\prime}}^{-}(v)+2.$ ∎

Lemma 4.4**.**

If $|X|<|S|$, then a Sullivan-i vertex of $D^{\prime}$ is a Sullivan-i vertex of $D$ for $i\in\{1,2\}$.

Proof. By Lemma 4.3 $(c)$, for any $v\in V(D^{\prime})$, $d^{+}_{D-D^{\prime}}(v)\geq d^{-}_{D-D^{\prime}}(v)+2$. Combining with Lemma 4.3 $(b)$, we have $d^{++}_{D-D^{\prime}}(v)+d^{+}_{D-D^{\prime}}(v)\geq 2d^{-}_{D-D^{\prime}}(v)+2$. By Corollary 1.5, the tournament $D^{\prime}$ always has a Sullivan-1 vertex and a Sullivan-2 vertex, say $v^{\prime}$ and $v^{\prime\prime}$, respectively. Then $v^{\prime}$ is a Sullivan-1 vertex of $D$ and $v^{\prime\prime}$ is a Sullivan-2 vertex of $D$.∎

Lemma 4.5**.**

If $|X|\geq|S|$, then a Sullivan-$i$ vertex of $D_{1}$ is a Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$.

Proof. Let $v$ be a Sullivan-$i$ vertex of $D_{1}$ for $i\in\{1,2\}$. We will prove that $v$ is a Sullivan-$i$ vertex of $D$ for $i\in\{1,2\}$.

Note that there is no arc from $D_{1}$ to $S$. Otherwise $D_{1}$ and $D^{\prime}_{1}$ are adjacent which contradicts Theorem 2.2 ($c$). So, $N_{D-D_{1}}^{+}(v)=N_{D-S-D_{1}}^{+}(v)$. Combining with Theorem 2.1 ($a$), we have

[TABLE]

Similarly, we have $N_{D-D_{1}}^{-}(v)=N_{D-D_{1}}^{-}(D_{1})$. By Lemma 4.1,

[TABLE]

By the structure of $D$ described in Theorems 2.4 and 2.5, we have $N_{D-D_{1}}^{-}(D_{1})\subseteq S$ and $X\subseteq N_{D-D_{1}}^{++}(D_{1})=N_{D-D_{1}}^{++}(v)$. Now

[TABLE]

So $d^{++}_{D-D_{1}}(v)\geq|S|\geq d^{-}_{D-D_{1}}(v)$ and $d^{++}_{D-D_{1}}(v)+d^{+}_{D-D_{1}}(v)\geq 2|S|\geq 2d^{-}_{D-D_{1}}(v)$. Since $v$ is also a Sullivan-$i$ vertex of $D_{1}$, we see that $v$ is a Sullivan-$i$ vertex of $D$.∎

Now, we consider the existence of a Sullivan-1 vertex in $D$ and the number of Sullivan-1 vertices of $D$. In fact, the existence can be directly obtained from the following two results, which were proved by Wang, Yang and Wang [18], Li and Sheng [12], respectively.

Lemma 4.6**.**

[18]** Let $D$ be a non-round 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 non-round decomposable local tournament, which is not a tournament. Then $D$ has a Sullivan-1 vertex.

We consider primarily the number of Sullivan-1 vertices.

Theorem 4.9**.**

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

Proof. Recall that $A=N_{D^{\prime}_{2}}^{+}(D_{1})=V(D_{\lambda})\cup V(D_{\lambda+1})\cup\ldots\cup V(D_{i}),\,\,B=N_{S}^{+}(A),\,\,X=N^{+}_{D-A}(A)=B\cup(V(D^{\prime}_{2})-A)\cup V(D_{p})$. The structure of $D$ is illustrated in Figure 2.

For the case when $|X|\geq|S|$, let $v$ be a Sullivan-1 vertex of $D_{1}$. By Lemma 4.5, $v$ is a Sullivan-1 vertex of $D$. By the proof of Lemma 4.6 (See reference [18]), there exists a king either belonging to $D_{2}^{\prime}$ or belonging to $S$ in $D$, say $v^{\prime}$. By Proposition 4.7, $v^{\prime}$ is a Sullivan-1 vertex of $D$. Clearly, $v^{\prime}\neq v$. Then $v^{\prime}$ is another Sullivan-1 vertex of $D$.

For the case when $|X|<|S|$, note that $|B|\leq|S|-2$. Recall that $D^{\prime}=D\langle S-B\rangle$. Let $v$ be a Sullivan-1 vertex of $D^{\prime}$. By Lemma 4.4, $v$ is Sullivan-1 vertex of $D$. Next we will find another Sullivan-1 vertex of $D$.

If $A=D^{\prime}_{2}$, we will show that a king of $D_{p}$, say $v^{\prime}$, is another Sullivan-1 vertex of $D$. By Theorem 2.5 ($a$), we have $D_{p}\Rightarrow S$. Combining with Lemma 4.2, $D_{p}\Rightarrow D^{\prime}\Rightarrow A=D^{\prime}_{2}$ and $D_{p}\Rightarrow D^{\prime}\Rightarrow D^{\prime}_{3}$. Then $v^{\prime}$ is a king of $D$. Clearly $v^{\prime}\neq v$. By Proposition 4.7, $v^{\prime}$ is another Sullivan-1 vertex of $D$.

Now $A\varsubsetneqq D^{\prime}_{2}$. Let $D^{\prime\prime}=D^{\prime}-v$ and $u$ be a Sullivan-1 vertex of $D^{\prime\prime}$. Assume $u\in V(D_{j})$, where $D_{j}$ is a strong component of $S$. We consider the following two cases.

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 $N^{-}_{D-D^{\prime}}(u)\varsubsetneqq N^{++}_{D-D^{\prime}}(u)$. By Lemma 4.3 (b), we have $N^{-}_{D-D^{\prime}}(u)\\ \subseteq N^{++}_{D-D^{\prime}}(u)$. We only need to prove that $N^{-}_{D-D^{\prime}}(u)\neq N^{++}_{D-D^{\prime}}(u)$. By Lemma 4.2, $u\Rightarrow A\Rightarrow D_{i+1}$. Combining with the fact that there exists no arc between $u$ and $D_{i+1}$, we have $D_{i+1}\subseteq N^{++}_{D-D^{\prime}}(u)$ and $D_{i+1}\nsubseteq N^{-}_{D-D^{\prime}}(u)$. Then $N^{-}_{D-D^{\prime}}(u)\neq N^{++}_{D-D^{\prime}}(u)$.

Now $d^{++}_{D-D^{\prime\prime}}(u)\geq d^{++}_{D-D^{\prime}}(u)\geq d^{-}_{D-D^{\prime}}(u)+1\geq d^{-}_{D-D^{\prime\prime}}(u)$ since $N^{-}_{D-D^{\prime}}(u)\varsubsetneqq N^{++}_{D-D^{\prime}}(u)$. Clearly $u\neq v$. Since $u$ is also a Sullivan-1 vertex of $D^{\prime\prime}$, we see that $u$ is another Sullivan-1 vertex of $D$.

Case 2. There exists at least one arc between $u$ and $D_{i+1}$.

Let $g$ be a king of $D_{i+1}$. We will show that $g$ is another Sullivan-1 vertex.

We claim that $D_{i+1}\Rightarrow D_{j}$. Since $A$ is a separating set, we see that $D_{i+1},\ldots,D_{p},D_{p+1},\ldots,D_{p+q},D_{1},\ldots,D_{\lambda-1}$ is the acyclic ordering of the strong components of $D-A$. Since $D_{i+1}$ and $D_{j}$ are distinct strong components of $D-A$, by Theorem 2.1 ($a$), we only need to prove that there exists at least one arc from $D_{i+1}$ to $D_{j}$. Now there exists at least one arc between $u$ and $D_{i+1}$ and hence there exists one arc between $D_{j}$ and $D_{i+1}$. However, by Theorem 2.5 ($b$), an arc from $D_{j}$ to $D_{i+1}$ implies that $D_{1}\Rightarrow D_{i+1}$ which contradicts the definition of $A$. Then there exists no arc from $D_{j}$ to $D_{i+1}$ and hence there exists at least one arc from $D_{i+1}$ to $D_{j}$.

Note that $g\rightarrow u$ since $g\in V(D_{i+1})$ and $u\in V(D_{j})$. By Lemma 4.2, $u\Rightarrow A$ and $u\Rightarrow D^{\prime}_{3}$. Hence $g\rightarrow u\Rightarrow A\,\,\,\mbox{and}\,\,\,g\rightarrow u\Rightarrow D^{\prime}_{3}.$ By the structure of $D$ described in Theorems 2.4 and 2.5, we see that $g\Rightarrow D^{\prime}_{2}-A-D_{i+1}$ and $g\Rightarrow D_{p}\Rightarrow S$. So $g$ is a king of $D$. Clearly $g\neq v$. By Proposition 4.7, $g$ is another Sullivan-1 vertex of $D$.∎

Next, we consider the existence of a Sullivan-2 vertex and the number of Sullivan-2 vertices in $D$.

Lemma 4.10**.**

If $|X|=|S|$, then $D$ has at least two Sullivan-2 vertices of $D$.

Proof. Let $v$ be a Sullivan-2 vertex of $D_{1}$. By Lemma 4.5, $v$ is a Sullivan-2 of $D$. To find another Sullivan-2 vertex of $D$, we consider the following two cases.

For the case when $|X|=|S|$ and $D-X$ is not a tournament, let $X$ be a minimal separating set of $D$ instead of $S$. By Lemma 4.4 and Lemma 4.5, there exists a Sullivan-2 vertex in $D$, say $u$. We can check $u\in S-B$ or $u\in X$ due to the new separating set $X$. Then $u\neq v$ since $v\in V(D_{1})$. Thus $u$ is another Sullivan-2 vertex of $D$.

For the case when $|X|=|S|$ and $D-X$ is a tournament, note that $|B|\leq|X|-1=|S|-1$ since $X=B\cup(V(D^{\prime}_{2})-A)\cup V(D_{p})$. Recall that $D^{\prime}=D\langle S-B\rangle$. Let $u$ be a Sullivan-2 vertex of the tournament $D^{\prime}$. We will show that $u$ is another Sullivan-2 vertex of $D$.

We claim $D^{\prime}\Rightarrow A$ and $D^{\prime}\Rightarrow D^{\prime}_{3}$. Since $D-X=D\langle V(D^{\prime}_{3})\cup A\cup V(D^{\prime})\rangle$ is a tournament, we see that any vertex of $D^{\prime}$ is adjacent to the vertices of $A$ and $D^{\prime}_{3}$. Since $N_{D^{\prime}}^{+}(A)=\emptyset$, we have $D^{\prime}\Rightarrow A$. Also, an arc from $D^{\prime}_{3}$ to $S$ implies that $D^{\prime}_{3}$ and $D^{\prime}_{1}$ are adjacent, which contradicts Theorem 2.2 ($c$). So $D^{\prime}\Rightarrow D^{\prime}_{3}$.

By the definition of $A$, $B$ and $X$, we have $V(D-D^{\prime})-X=A\cup V(D^{\prime}_{3})$. Combining with $u\Rightarrow A$ and $u\Rightarrow D^{\prime}_{3}$, we see that

[TABLE]

and hence $N^{-}_{D-D^{\prime}}(u)\subseteq X,\,\,\,\,N^{++}_{D-D^{\prime}}(u)\subseteq X.$

Also any vertex of $X=B\cup(V(D^{\prime}_{2})-A)\cup V(D_{p})$ either belongs to $N_{D-D^{\prime}}^{+}(u)$ or belongs to $N_{D-D^{\prime}}^{++}(u)$ since $u\Rightarrow A\Rightarrow D_{2}^{{}^{\prime}}-A$, $u\Rightarrow A\Rightarrow D_{p}$ and for any $g\in B$, there exists a vertex $h\in A$ such that $h\rightarrow g$. So $X\subseteq N_{D-D^{\prime}}^{+}(u)\cup N_{D-D^{\prime}}^{++}(u)$. Since $N_{D-D^{\prime}}^{+}(u)\cap N_{D-D^{\prime}}^{-}(u)=\emptyset$, we have $N^{-}_{D-D^{\prime}}(u)\subseteq N^{++}_{D-D^{\prime}}(u)\subseteq X$ and hence

[TABLE]

By Lemma 4.1, $d^{+}_{D-D_{1}}(D_{1})\geq|S|$. Then $|A\cup V(D^{\prime}_{3})|\geq|D_{1}|+d^{+}_{D-D_{1}}(D_{1})\geq 1+|S|.$ Combining with $A\cup V(D^{\prime}_{3})\subseteq N^{+}_{D-D^{\prime}}(u)$, we have

[TABLE]

So $d^{++}_{D-D^{\prime}}(u)+d^{+}_{D-D^{\prime}}(u)\geq 2d^{-}_{D-D^{\prime}}(u)+1$. Clearly, $u\neq v$ since $u\in S-B$ and $v\in V(D_{1})$. The fact that $u$ is a Sullivan-2 vertex of $D^{\prime}$ implies that $u$ is another Sullivan-2 vertex of $D$.∎

Theorem 4.11**.**

Let $D$ be a non-round decomposable local tournament, which is not a tournament. Then either $D$ has at least two Sullivan-2 vertices or $D$ has a Sullivan-2 vertex $v$ satisfying $d^{++}(v)+d^{+}(v)\geq 2d^{-}(v)+2$.

Proof. Recall that $A=N_{D^{\prime}_{2}}^{+}(D_{1})=V(D_{\lambda})\cup V(D_{\lambda+1})\cup\ldots\cup V(D_{i})$, $B=N_{S}^{+}(A)$ and $X=B\cup(V(D^{\prime}_{2})-A)\cup V(D_{p})$. We consider the following cases.

Case 1. $|X|<|S|$.

Note that $|B|\leq|S|-2$. Recall that $D^{\prime}=D\langle S-B\rangle$. Let $v$ be a Sullivan-2 vertex of $D^{\prime}$. By Lemma 4.4, $v$ is a Sullivan-2 vertex of $D$. Let $D^{\prime\prime}=D^{\prime}-v$ and $u$ be a Sullivan-2 vertex of the tournament $D^{\prime\prime}$. We will prove that $u$ is another Sullivan-2 vertex of $D$.

By Lemma 4.3 $(b)$ and $(c)$, we see that $d^{++}_{D-D^{\prime}}(u)+d^{+}_{D-D^{\prime}}(u)\geq 2d^{-}_{D-D^{\prime}}(u)+2$ and hence $d^{++}_{D-D^{\prime\prime}}(u)+d^{+}_{D-D^{\prime\prime}}(u)\geq 2d^{-}_{D-D^{\prime\prime}}(u)$. Since $u$ is also a Sullivan-2 vertex of $D^{\prime\prime}$, we have $d^{++}_{D^{\prime\prime}}(u)+d^{+}_{D^{\prime\prime}}(u)\geq 2d^{-}_{D^{\prime\prime}}(u)$. Clearly $u\neq v$. Thus $d^{++}_{D}(u)+d^{+}_{D}(u)\geq 2d^{-}_{D}(u)$ and $u$ is another Sullivan-2 vertex of $D$.

Case 2. $|X|=|S|$.

By Lemma 4.10, $D$ has at least two Sullivan-2 vertices.

Case 3. $|X|>|S|$ and $|D_{1}|\geq 2$.

By Lemma 4.5, a Sullivan-2 vertex of $D_{1}$ is a Sullivan-2 vertex of $D$. Since $D_{1}$ is strong and $|D_{1}|\geq 2$, we see that $D_{1}$ has at least three vertices. By Corollary 1.7, $D_{1}$ has at least two Sullivan-2 vertices. These two vertices are Sullivan-2 vertices of $D$.

Case 4. $|X|>|S|$, $|D_{1}|=1$ and $D_{2}\subseteq D^{\prime}_{2}$.

By Lemma 4.5, the only vertex $v$ of $D_{1}$ is a Sullivan-2 vertex of $D$. Note that $A=N^{+}_{D-D_{1}}(D_{1})$. By Lemma 4.1, $d_{D-D_{1}}^{+}(D_{1})=|A|\geq|S|$. We consider the following two cases. In the first case, we can find two Sullivan-2 vertices in $D$. In the second case, we can find a vertex $v$ satisfying $d^{++}(v)+d^{+}(v)\geq 2d^{-}(v)+2$ in $D$.

If $|A|=|S|$, let $A$ be a minimal separating set of $D$ instead of $S$. By Lemma 4.4 and Lemma 4.5, there exists a Sullivan-2 vertex in $D$, say $u$. We can check $u\in A$ or $u\in V(D_{i+1})$ due to the new separating set $A$. Then $u\neq v$ since $v\in V(D_{1})$. Thus $u$ is another Sullivan-2 vertex of $D$.

If $|A|>|S|$, we see that $N_{D-D_{1}}^{+}(v)=A$, $X\subseteq N_{D-D_{1}}^{++}(v)$ and $N_{D-D_{1}}^{-}(v)\subseteq S$. Then $d_{D-D_{1}}^{+}(v)=|A|\geq|S|+1$, $d_{D-D_{1}}^{++}(v)\geq|X|\geq|S|+1$ and $d_{D-D_{1}}^{-}(v)\leq|S|$ and hence $d_{D-D_{1}}^{++}(v)+d_{D-D_{1}}^{+}(v)\geq 2d_{D-D_{1}}^{-}(v)+2$. So $v$ is the desired vertex.

Case 5. $|X|>|S|$, $|D_{1}|=1$ and $D_{2}\subseteq D^{\prime}_{3}$.

By Lemma 4.5, the only vertex $v$ of $D_{1}$ is a Sullivan-2 vertex of $D$. Let

[TABLE]

[TABLE]

The structure of $D$ is illustrated in Figure 3. By Lemma 4.1, we have

[TABLE]

Let $C=N_{D-D_{2}}^{+}(D_{2})=(D_{3}\cup\ldots\cup D_{\lambda-1})\cup A^{\prime}$. Then $|C|\geq|S|.$

If $|X^{\prime}|<|S|$, let $D^{\prime\prime}=D\langle S-B^{\prime}\rangle$ and $u$ be a Sullivan-2 vertex of $D^{\prime\prime}$. Note that $u\neq v$ since $v\in V(D_{1})$. Similarly to the proof of Lemma 4.4, we can show that $u$ is another Sullivan-2 vertices of $D$.

If $|X^{\prime}|=|S|$, similarly to the proof of Lemma 4.10, we can prove that $D$ has two Sullivan-2 vertices.

If $|C|=|S|$, let $C$ be a minimal separating set of $D$ instead of $S$. By Lemma 4.4 and Lemma 4.5, there exists a Sullivan-2 vertex in $D$, say $u$. We can check $u\in C$ or $u\in V(D_{i^{\prime}+1})$ due to the new separating set $C$. Then $u\neq v$ since $v\in V(D_{1})$. Then $u$ is another Sullivan-2 vertex of $D$.

If $|C|>|S|$ and $|X^{\prime}|>|S|$, let $u$ be a Sullivan-2 vertex of $D_{2}$. We will show that $u$ is another Sullivan-2 vertex of $D$. By the structure of $D$ described in Theorems 2.4 and 2.5, we have $N_{D-D_{2}}^{-}(u)\subseteq S\cup D_{1}$, $N_{D-D_{2}}^{+}(u)=C$, $X^{\prime}\subseteq N_{D-D_{2}}^{++}(u)$. Then $d_{D-D_{2}}^{-}(u)\leq|S|+|D_{1}|=|S|+1$, $d_{D-D_{2}}^{+}(u)=|C|\geq|S|+1,d_{D-D_{2}}^{++}(u)\geq|X^{\prime}|\geq|S|+1.$ So $d_{D-D_{2}}^{++}(u)+d_{D-D_{2}}^{+}(u)\geq 2(|S|+1)\geq 2d_{D-D_{2}}^{-}(u)$. Clearly $u\neq v$. Since $u$ is also a Sullivan-2 vertex of $D_{2}$, we see that $u$ is another Sullivan-2 vertex of $D$.

In any case, we find either two Sullivan-2 vertices or a vertex $v$ satisfying $d^{++}(v)+d^{+}(v)\geq 2d^{-}(v)+2$ in $D$. The proof of the theorem is complete.∎

Corollary 4.12**.**

$(a)$* Every non-round decomposable local tournament which is not a tournament has a Sullivan-$i$ vertex for $i\in\{1,2\}$.*

$(b)$* Every non-round decomposable local tournament which is not a tournament has at least two Sullivan-1 vertices.*

$(c)$* Every non-round decomposable local tournament which is not a tournament either has at least two Sullivan-2 vertices or has a Sullivan-2 vertex $v$ satisfying $d^{++}(v)+d^{+}(v)\geq 2d^{-}(v)+2$.*

5 Conclusion

According to a full classification of local tournaments in Corollary 1.9, Corollary 1.5, Theorems 1.6, Corollary 3.6 and Corollary 4.12 imply the following theorem.

Theorem 5.1**.**

Let $D$ be a local tournament. Then the following holds for $D$.

$(a)$* Every local tournament $D$ has a Sullivan-$i$ vertex for $i\in\{1,2\}$.*

$(b)$* Every local tournament $D$ with no vertex of in-degree zero has at least two Sullivan-1 vertices.*

$(c)$* Every local tournament $D$ with no vertex of in-degree zero either has at least two Sullivan-2 vertices or has a Sullivan-2 vertex $v$ satisfying $d^{++}(v)+d^{+}(v)\geq 2d^{-}(v)+2$.*