Twin domination number of Tournaments

TL;DR

This paper investigates the twin domination number in orientations of complete graphs, proving exact values for specific cases and providing new bounds that disprove a longstanding conjecture for all sufficiently large n.

Contribution

The authors determine the exact twin domination number for K8 and K9, and establish new upper bounds for larger n, challenging previous conjectures.

Findings

Proved DOM^{*}(K_8)=DOM^{*}(K_9)=4

Established new upper bounds for DOM^{*}(K_n)

Disproved the conjecture that DOM^{*}(K_n)=ceil((n+1)/2) for n ≥ 8

Abstract

Let $D=(V,A)$ be a digraph. A subset $S$ of $V$ is called a twin dominating set of $D$ if for every vertex $v\in V-S$, there exists vertices $u_1,u_2 \in S$ such that $(v,u_1)$ and $(u_2,v)$ are arcs in $D$. The minimum cardinality of a twin dominating set in $D$ is called the twin domination number of $D$ and is denoted by $\gamma ^{*}(D)$. The upper orientable twin domination number of a graph $G$ is $DOM^{*}(G)=\max\{ \gamma ^{*}(D)|D \ \text{is an orientation of G} \}.$ It has been conjectured that for the complete graph $K_n$ with $n\geq 8$, $DOM^{*}(K_n)=\left\lceil \frac{n+1}{2}\right\rceil$. In this work we prove $DOM^{*}(K_8)= DOM^{*}(K_9)= 4$ and establish new upper bounds for $DOM^{*}(K_n)$, disproving the same above conjecture for all $n \geq 8$.

Equations

$$DO{M}^{\ast }(G)=\max \{{\gamma }^{\ast }(D)|D\text{is an orientation of }G\}.$$

$$\sum _{v\in V(T)}id(v)=\sum _{v\in V(T)}od(v)=\frac{n(n-1)}{2},$$

$$T_1$$

$$T_i$$

$$|S|$$

$$\le \lceil {\log }_2\left(od(u)id(u)\right)\rceil +2\le 2\lceil {\log }_2\frac{n-1}{2}\rceil +2=2\lceil {\log }_2(n-1)\rceil .$$

$$DOM^{*}(K_{n})=\left\{\begin{array}[]{ll}\left\lceil\frac{n+1}{2}\right\rceil&\textrm{if $1\leq n\leq 7$,}\\ 4&\textrm{if $n\in\{8,9\}$;}\end{array}\right.$$

$$DOM^{*}(K_{n})\leq\left\{\begin{array}[]{ll}\left\lfloor\frac{n}{2}\right\rfloor&\textrm{if $9<n\leq 21$,}\\ 2\left\lceil\log_{2}{(n-1)}\right\rceil&\textrm{if $21<n$.}\end{array}\right.$$

Full text

In [2], is defined the concept of upper orientable twin domination number of a graph $G$, $DOM^{*}(G)=\max\{\gamma^{*}(D)|D\ \text{is an orientation of G}\}.$ In [1], it is conjectured that for the complete graph $K_{n}$ with $n\geq 8$, $DOM^{*}(K_{n})=\left\lceil\frac{n+1}{2}\right\rceil$. In this work we prove $DOM^{*}(K_{8})=DOM^{*}(K_{9})=4$ and establish new upper bounds for $DOM^{*}(K_{n})$, disproving the same above conjecture for all $n\geq 8$.

Keywords: Twin domination number, twin dominating set, upper orientable twin domination number, tournaments.

AMS Subject Classification Numbers: 05C69; 05C20; 05C35.

1 Introduction

Let $D=(V,A)$ be a digraph. For any vertex $v\in V$, the sets $I_{D}(v)=\{u|(u,v)\in A\}$ and $O_{D}(v)=\{u|(v,u)\in A\}$ are called the inset and outset of $v$. The indegree and outdegree of $v$ are defined by $id_{D}(v)=|I_{D}(v)|$ and $od_{D}(v)=|O_{D}(v)|$. For any $S\subseteq V$, graph $\mathcal{G}_{D}[S]$ is the subgraph of $D$ induced by the set of vertices $S$. The bottom index is omitted, when the digraph is clear from the context. We say that a set of vertices $S\subseteq V$ out-dominates (in-dominates) $V$ if for every $v\in V-S$ there exists $u\in S$ such that $(v,u)\in A$ ($(u,v)\in A$). A tournament is an orientation of a complete graph.

Definition 1.1

Let $D=(V,A)$ be a digraph. A subset $S$ of $V$ is called a twin dominating set of $D$ if for every vertex $v\in V-S$, there exists vertices $u_{1},u_{2}\in S$ ($u_{1}$ and $u_{2}$ may be equal) such that $(v,u_{1})$ and $(u_{2},v)$ are arcs in $D$. The minimum cardinality of a twin dominating set in $D$ is called the twin domination number of $D$ and is denoted by $\gamma^{*}(D)$.

For different orientations $D_{1}$ and $D_{2}$ of a graph $G$, it is possible to have $\gamma^{*}(D_{1})\not=\gamma^{*}(D_{2})$. In Chartrand et al. [2], the authors defined the concept of upper orientable twin domination number of a graph $G$,

[TABLE]

Arumugam et al. [1] proved that for $1\leq n\leq 7$ and the complete graph $K_{n}$, $DOM^{*}(K_{n})=\left\lceil\frac{n+1}{2}\right\rceil$ and they conjectured that for $n\geq 8$, $DOM^{*}(K_{n})=\left\lceil\frac{n+1}{2}\right\rceil$. We prove in Section $2$ that for $n=8$, $DOM^{*}(K_{8})=4$ and in Section $3$ that $DOM^{*}(K_{9})=4$. Then in Section $4$ we expand these results by proving $DOM^{*}(K_{n})\leq\left\lfloor\frac{n}{2}\right\rfloor$ for all $n\geq 8$. In Section $5$ the upper bound of $2\left\lceil\log_{2}{(n-1)}\right\rceil$ is shown and the summary of all results presented.

2 $DOM^{*}(K_{8})=4$

In this section we prove the exact value of the upper orientable twin domination number of the graph $K_{8}$ to be equal to $4$, i.e. $DOM^{*}(K_{8})=4$. The following observation is a consequence of the results given in Arumugam et al. [1].

Observation 2.1

Let $T$ be a tournament of order $n\geq 3.$ From Theorem 2.6 of Arumugam et al. [1], if $T$ contains at least one vertex $u\in V(T)$ such that $id(u)=0$ or $od(u)=0$, then $\gamma^{*}(T)\leq\left\lceil\log_{2}{(n-1)}+1\right\rceil$. In the case of $n\geq 8$, $\gamma^{*}(T)\leq\left\lfloor\frac{n}{2}\right\rfloor$.

Above observation assures us that if a tournament of order $8$ contains a sink or source vertex then $DOM^{*}(K_{8})\leq 4$. In the next lemma we prove that this result holds also when a tournament contains a vertex of in- or outdegree $2$.

Lemma 2.1

Let $T$ be an orientation of $K_{8}$. If there exists $v\in V(T)$ such that $id(v)=2$ or $od(v)=2$, then $\gamma^{*}(T)\leq 4$.

Proof. Suppose there exists a vertex $v\in V(T)$ such that $id(v)=2$, $I(v)=\{i,i^{\prime}\}$ and $O(v)=\{o_{1},o^{\prime}_{1},o_{2},o^{\prime}_{2},z\}$. Without loss of generality we can assume that the arcs $(i,i^{\prime}),(o_{1}^{\prime},o_{1}),(o_{2}^{\prime},o_{2})\in A(T)$.

If the arcs $(z,i),(z,o_{1})$ or $(z,o_{2})$ are in $A(T)$, then $S=\{v,i,o_{1},o_{2}\}$ is a twin dominating set of $T$. So we can assume that the arcs $(i,z),(o_{1},z),(o_{2},z)\in A(T)$. See Figure 1. 2. 2.

If the arc $(o_{1}^{\prime},o_{2}^{\prime})\in A(T)$, then the set $S=\{v,i,z,o_{2}^{\prime}\}$ is a twin dominating set of $T$. If the arc $(o_{2}^{\prime},o_{1}^{\prime})\in A(T)$, then the set $S=\{v,i,z,o_{1}^{\prime}\}$ is a twin dominating set of $T$.

Therefore, if $id(v)=2$, $\gamma^{*}(T)\leq 4$. The case $od(v)=2$, is symmetric.

In the next considered case a tournament of order $8$ contains a vertex of in- or outdegree $1$.

Lemma 2.2

Let $T$ be an orientation of $K_{8}$. If there exists $v\in V(T)$ such that $id(v)=1$ or $od(v)=1$, then $\gamma^{*}(T)\leq 4$.

Proof. Let $v\in V(T)$ such that $id(v)=1$ with $I(v)=\{z\}$ and $O(v)=\{o_{1},o_{1}^{\prime},o_{2},\allowbreak o_{2}^{\prime},o_{3},o_{3}^{\prime}\}$. Without loss of generality we can assume that the arcs $(o_{1}^{\prime},o_{1}),$ $(o_{2}^{\prime},o_{2}),$ $(o_{3}^{\prime},o_{3})\in A(T)$.

If $(o_{1},z),(o_{2},z)$ or $(o_{3},z)$ are in $A(T)$, then $S=\{v,o_{1},o_{2},o_{3}\}$ is a twin dominating set of $T$. So we can assume that $\{v,o_{1},o_{2},o_{3}\}\subseteq O(z)$. 2. 2.

By Observation 2.1, we can assume that $id(z)\neq 0$. Assume, without lost of generality, that $(o_{3}^{\prime},z)\in A(T)$. If one of the arcs $(o_{3},o_{1})$ or $(o_{3},o_{2})$ are in $T$, then $S=\{v,z,o_{1},o_{2}\}$ is a twin dominating set of $T$. So let us assume that $(o_{1},o_{3}),(o_{2},o_{3})\in A(T)$. See Figure 2. 3. 3.

If $(o_{3},o_{1}^{\prime})$ and $(o_{3},o_{2}^{\prime})$ are arcs in $T$, then $od(o_{3})=2$ and from Lemma 2.1, $\gamma^{*}(T)\leq 4$. On the other hand, if $(o^{\prime}_{1},o_{3})$ (resp. $(o^{\prime}_{2},o_{3})$) is an arc in $T$, then $S=\{v,z,o_{2},o_{3}\}$ (resp. $S=\{v,z,o_{1},o_{3}\}$) is a twin dominating set of $T$.

The case $od(v)=1$ is symmetric.

Before we prove the final equality we make an observation about the lower bound of $4$ for $DOM^{*}(K_{n})$, for any $n\geq 6$.

Observation 2.2

Denote by $T^{6}_{n}$ the tournament of order $n+6$ of Figure 3. Arumugam et al. [1] proved that $\gamma^{*}(T^{6}_{0})=\gamma^{*}(T^{6}_{1})=4$. So for any $n\geq 0$, $\gamma^{*}(T^{6}_{n})=4$ and $4\leq DOM^{*}(K_{n+6})$ .

We close this section by presenting the main theorem, which summarizes previous results and considers the remaining case, when all vertices of a tournament of order $8$ have in- or outdegree equal to $3$.

Theorem 2.1

The upper orientable twin domination number of $K_{8}$, $DOM^{*}(K_{8})=4$.

Proof.

Let $T$ be an orientation of $K_{8}$. By Observation 2.1 and Lemmas 2.1–2.2, if there exist $v\in V(T)$ such that $id(v)\in\{0,1,2,5,6,7\}$, then $\gamma^{*}(T)\leq 4$. So, we can suppose that for every $v\in V(T)$, $id(v)=3$ or $id(v)=4$.

Let $v\in V(T)$ such that $O(v)=\{o_{1},o_{1}^{\prime},o_{2},o_{2}^{\prime}\}$ and $I(v)=\{i,i^{\prime},z\}$. Without loss of generality, we can suppose that the arcs $(i,i^{\prime}),(o_{1}^{\prime},o_{1}),(o_{2}^{\prime},o_{2})\in A(T)$.

If one of the arcs $(i,z),(o_{1},z)$ or $(o_{2},z)$ are in $A(T)$, then $S=\{v,i,o_{1},o_{2}\}$ is a twin dominating set of $T$. So, we can suppose that $O(z)=\{v,i,o_{1},o_{2}\}$ and $I(z)=\{i^{\prime},o_{1}^{\prime},o_{2}^{\prime}\}$.

If $(o_{1},o_{2})\in A(T)$, then $S=\{v,z,i,o_{2}\}$ is a twin dominating set of $T$. 2. 2.

If $(o_{2},o_{1})\in A(T)$, then $S=\{v,z,i,o_{1}\}$ is a twin dominating set of $T$.

The case when $id(v)=4$ is symmetric. Therefore, $DOM^{*}(K_{8})\leq 4$. By Observation 2.2, we can conclude that $DOM^{*}(K_{8})=4.$

3 $DOM^{*}(K_{9})=4$

In this section we prove the exact value of the upper orientable twin domination number of the graph $K_{9}$ to be equal to $4$, i.e., $DOM^{*}(K_{9})=4$. First, let us make a simple observation, that will be very useful in the rest of the section.

Observation 3.1

For every tournament $T$ of order $n$, since

[TABLE]

there exist vertices $u_{1},u_{2}\in V(T)$, such that $od(u_{1}),id(u_{2})\leq\left\lfloor\frac{n-1}{2}\right\rfloor$.

Similarly, as for the previous case of the number of vertices $8$, we show that our result holds if a graph contains at least one vertex with a specified in- or out-degree. Observation 2.1 assures us that if a tournament of order $9$ contains a sink or source vertex then $DOM^{*}(K_{9})\leq 4$. In the next lemma we prove that this result holds also when a tournament contains a vertex of in- or outdegree $2$.

Lemma 3.1

Let $T$ be an orientation of $K_{9}$. If there exists $v\in V(T)$ such that $id(v)=2$ or $od(v)=2$, then $\gamma^{*}(T)\leq 4$.

Proof. Suppose there exists a vertex $v\in V(T)$ such that $id_{T}(v)=2$. Denote the tournament induced by the outset of $v$ as $T_{1}=\mathcal{G}[O_{T}(v)]$. By Observation 3.1 there exists a vertex $v_{1}\in V(T_{1})$, such that $od_{T_{1}}(v_{1})\leq 2$ (because $|V(T_{1})|=6$). Let $T_{2}=\mathcal{G}[O_{T_{1}}(v_{1})]$. Let $v^{\prime}_{1}\in I_{T}(v)$ be a vertex such that $od_{\mathcal{G}[I_{T}(v)]}(v^{\prime}_{1})=1$, and let $v_{2}\in V(T_{2})$ be such that $id_{T_{2}}(v_{2})=1$ if $|V(T_{2})|=2$ or simply $\{v_{2}\}=V(T_{2})$ if $|V(T_{2})|=1$. Notice now that $S=\{v,v_{1},v^{\prime}_{1},v_{2}\}$ is a twin dominating set of $T$. Indeed,

• •

$v$ in-dominates $V(T_{1})\cup\{v\}$, $v^{\prime}_{1}$ in-dominates $I_{T}(v)$ and

• •

$v$ out-dominates $V(T)\backslash V(T_{1})$, $v_{1}$ out-dominates $V(T_{1})\backslash V(T_{2})$ and $v_{2}$ out-dominates $V(T_{2})$.

See Figure 4 for the illustration. Notice that if $V(T_{2})$ is the empty set, then $S=\{v,v_{1},v^{\prime}_{1}\}$ is a twin dominating set. Therefore, if $id_{T}(v)=2$, $\gamma^{*}(T)\leq 4$. The case $od_{T}(v)=2$ is symmetric.

In the next considered case a tournament of order $9$ contains a vertex of in- or outdegree $1$.

Lemma 3.2

Let $T$ be an orientation of $K_{9}$. If there exists $v\in V(T)$ such that $id(v)=1$ or $od(v)=1$, then $\gamma^{*}(T)\leq 4$.

Proof. Suppose there exists a vertex $v\in V(T)$ such that $id_{T}(v)=1$. We denote the tournament induced by the outset of $v$ as $T_{1}=\mathcal{G}[O_{T}(v)]$ and $\{v^{\prime}_{1}\}=I_{T}(v)$. By Observation 3.1 there exists a vertex $v_{1}\in T_{1}$ such that $od_{T_{1}}(v_{1})\leq 3$. Let $T_{2}=\mathcal{G}[O_{T_{1}}(v_{1})]$. Repeating this reasoning, we pick a vertex $v_{2}\in V(T_{2})$ such that $od_{T_{2}}(v_{2})\leq 1$ and let $T_{3}=\mathcal{G}[O_{T_{2}}(v_{2})]$. While $|V(T_{2})|\leq 3$ and $|V(T_{3})|\leq 1$ several cases have to be considered. If $|V(T_{2})|=0$, then $S=\{v,v_{1},v^{\prime}_{1}\}$ is a twin domination set of $T$. If $|V(T_{2})|\in\{1,2\}$ then $S=\{v,v_{1},v^{\prime}_{1},u\}$, where $u\in V(T_{2})$ such that $id_{T_{2}}(u)=1$ if $|V(T_{2})|=2$ or simply $\{u\}=V(T_{2})$ if $|V(T_{2})|=1$, is a twin domination set of $T$. Finally, if $|V(T_{2})|=3$ and $|V(T_{3})|=0$, then $S=\{v,v_{1},v^{\prime}_{1},v_{2}\}$ is a twin domination set of $T$. Therefore, assume $|V(T_{2})|=3$ and $|V(T_{3})|=1$ and let $\{v_{3}\}=V(T_{3})$.

If $(v_{1},v_{1}^{\prime}),(v_{2},v^{\prime}_{1})$ or $(v_{3},v_{1}^{\prime})$ are in $A(T)$, then $S=\{v,v_{1},v_{2},v_{3}\}$ is a twin domination set of $T$. So let us assume that $(v_{1}^{\prime},v_{1}),(v^{\prime}_{1},v_{2}),(v_{1}^{\prime},v_{3})\in A(T)$. 2. 2.

If $(s,v_{1}^{\prime})\in A(T)$, where $\{s\}=I_{T_{2}}(v_{2})$, then $S=\{v,v_{1},v^{\prime}_{1},v_{3}\}$ is a twin domination set of $T$. So let us assume that $(v_{1}^{\prime},s)\in A(T)$.

If $id_{T}(v_{1}^{\prime})\in\{0,2\}$, then from Observation 2.1 and Lemma 3.1 we have $\gamma^{*}(T)\leq 4$. Therefore, we consider two remaining cases.

Case 1: $id_{T}(v_{1}^{\prime})=1$. Let $c_{1},c_{2}$ and $c_{3}$ denote the vertices, which form an oriented cycle in $\mathcal{G}[I_{T_{1}}(v_{1})]$. Without loss of generality we assume $(v^{\prime}_{1},c_{1}),(c_{2},v^{\prime}_{1}),(v_{1}^{\prime},c_{3})\in A(T)$. It can be observed now, that $S=\{v^{\prime}_{1},v_{2},v_{3},c_{1}\}$ is a twin domination set of $T$. See Figure 5(a).

Case 2: $id_{T}(v_{1}^{\prime})=3$. We immediately obtain a twin dominating set of $T$ as $S=\{v,v^{\prime}_{1},v_{2},v_{3}\}$. See Figure 5(b).

Therefore, if $id_{T}(v)=1$, $\gamma^{*}(T)\leq 4$. The case $od_{T}(v)=1$ is symmetric.

We continue our reasoning by looking at a tournament of order $9$ that contains a vertex of in- or outdegree $3$.

Lemma 3.3

Let $T$ be an orientation of $K_{9}$. If there exists $v\in V(T)$ such that $id(v)=3$ or $od(v)=3$, then $\gamma^{*}(T)\leq 4$.

Proof. Suppose there exists a vertex $v\in V(T)$ such that $id_{T}(v)=3$. We denote the tournaments induced by the out- and insets of $v$ as $T_{1}=\mathcal{G}[O_{T}(v)]$ and $T^{\prime}_{1}=\mathcal{G}[I_{T}(v)]$ respectively. By Observation 3.1 there exist vertices $v_{1}\in T_{1}$ and $v_{1}^{\prime}\in T^{\prime}_{1}$, such that $od_{T_{1}}(v_{1})\leq 2$ and $id_{T^{\prime}_{1}}(v_{1}^{\prime})\leq 1$. Denote $T_{2}=\mathcal{G}[O_{T_{1}}(v_{1})]$ and $T^{\prime}_{2}=\mathcal{G}[I_{T^{\prime}_{1}}(v^{\prime}_{1})]$. While $|V(T_{2})|\leq 2$ and $|V(T^{\prime}_{2})|\leq 1$ several cases have to be considered. Notice first that if $|V(T_{2})|=|V(T^{\prime}_{2})|=0$, then $S=\{v,v_{1},v^{\prime}_{1}\}$ is a twin dominating set of $T$. On the other hand, if $|V(T_{2})|=0$ and $|V(T^{\prime}_{2})|\neq 0$ ($|V(T^{\prime}_{2})|=0$ and $|V(T_{2})|\neq 0$), then set $S=\{v,v_{1},v^{\prime}_{1},u\}$, where $u\in V(T^{\prime}_{2})$ ($u\in V(T_{2})$, respectively) is a twin dominating set of $T$. Therefore, there are two remaining cases.

Case 1: $|V(T_{2})|=|V(T^{\prime}_{2})|=1$. Let $\{v_{2}\}=V(T_{2})$, $\{v_{2}^{\prime}\}=V(T^{\prime}_{2})$ and $\{s\}=O_{T^{\prime}_{1}}(v^{\prime}_{1})$.

If $(v_{2},v_{2}^{\prime}),(v_{2},v^{\prime}_{1})$ or $(v_{1},v_{2}^{\prime})$ are in $A(T)$, then $S=\{v,v_{1},v_{1}^{\prime},v_{2}\}$ or $S=\{v,v_{1},v_{1}^{\prime},v_{2}^{\prime}\}$ are twin domination sets of $T$. So let us assume that $(v_{2}^{\prime},v_{2}),\allowbreak(v^{\prime}_{1},v_{2}),(v_{2}^{\prime},v_{1})\in A(T)$. See Figure 6(a). 2. 2.

If $(s,v_{1})\in A(T)$, then $od(v_{1})\leq 2$ and from previous lemmas $\gamma^{*}(T)\leq 4$. On the other hand, if $(v_{1},s)\in A(T)$, then $S=\{v,v_{1},v_{2},v_{2}^{\prime}\}$ is a twin domination set of $T$.

Case 2: $|V(T_{2})|=2$ and $|V(T^{\prime}_{2})|=1$. Let $\{v_{2}^{\prime}\}=V(T^{\prime}_{2})$, $\{s\}=O_{T^{\prime}_{1}}(v^{\prime}_{1})$ and $v_{2}\in V(T_{2})$, such that $id_{T_{2}}(v_{2})=1$.

If $(v_{2},v_{2}^{\prime})$ or $(v_{1},v_{2}^{\prime})$ are in $A(T)$, then $S=\{v,v_{1},v_{1}^{\prime},v_{2}\}$ is a twin domination set of $T$. So let us assume that $(v_{2}^{\prime},v_{2}),(v_{2}^{\prime},v_{1})\in A(T)$. 2. 2.

If $(v_{1},s),(v_{2},s)$ or $(v^{\prime}_{2},s)$ are in $A(T)$, then $S=\{v,v_{1},v_{2},v^{\prime}_{2}\}$ is a twin domination set of $T$. So let us assume that $(s,v_{1}),(s,v_{2}),(s,v^{\prime}_{2})\in A(T)$. See Figure 6(b). 3. 3.

If $(v^{\prime}_{1},v_{1})\in A(T)$, then $od_{T}(v_{1})=2$ and from previous lemma $\gamma^{*}(T)\leq 4$. On the other hand, if $(v_{1},v^{\prime}_{1})\in A(T)$, then $S=\{v,v_{1},v_{2},s\}$ is a twin domination set of $T$.

Therefore, if $id_{T}(v)=3$, $\gamma^{*}(T)\leq 4$. The case $od_{T}(v)=3$ is symmetric.

We close this section by presenting the main theorem, which summarizes previous results and considers the remaining case, when all vertices of a tournament of order $9$ have indegree equal to $4$.

Theorem 3.1

The upper orientable twin domination number of $K_{9}$, $DOM^{*}(K_{9})=4$.

Proof. Let $T$ be an orientation of $K_{9}$. By Observation 2.1 and Lemmas 3.1–3.3, if there exists $v\in V(T)$ such that $id_{T}(v)\in\{0,1,2,3,5,6,7,8\}$, then $\gamma^{*}(T)\leq 4$. So, we can suppose that for every $v\in V(T)$, $id_{T}(v)=4$.

Let $v\in V(T)$ be any vertex. We denote the tournaments induced by the out- and insets of $v$ as $T_{1}=\mathcal{G}[O_{T}(v)]$ and $T^{\prime}_{1}=\mathcal{G}[I_{T}(v)]$ respectively. By Observation 3.1 there exist vertices $v_{1}\in T_{1}$ and $v_{1}^{\prime}\in T^{\prime}_{1}$, such that $od_{T_{1}}(v_{1}),id_{T^{\prime}_{1}}(v_{1}^{\prime})\leq 1$. Repeating the reasoning from the proof of the previous lemma, we observe that if $od_{T_{1}}(v_{1})=0$ or $id_{T^{\prime}_{1}}(v_{1}^{\prime})=0$, then $\gamma^{*}(T)\leq 4$. Therefore, let $od_{T_{1}}(v_{1})=1$ and $id_{T^{\prime}_{1}}(v_{1}^{\prime})=1$ and we denote these vertices as $\{v_{2}\}=O_{T_{1}}(v_{1})$ and $\{v^{\prime}_{2}\}=I_{T^{\prime}_{1}}(v^{\prime}_{1})$ respectively. If $(v_{2},v_{2}^{\prime}),(v_{2},v^{\prime}_{1})$ or $(v_{1},v_{2}^{\prime})$ are in $A(T)$, then $S=\{v,v_{1},v_{1}^{\prime},v_{2}\}$ or $S=\{v,v_{1},v_{1}^{\prime},v_{2}^{\prime}\}$ are twin domination sets of $T$. So let us assume that $(v_{2}^{\prime},v_{2}),(v^{\prime}_{1},v_{2}),(v_{2}^{\prime},v_{1})\in A(T)$. Now, because $v_{2}$ (resp. $v_{2}^{\prime}$) has already four incoming (resp. outcoming) arcs, all of the rest arcs have to be outcoming (resp. incoming), which means that the set $S=\{v_{2},v^{\prime}_{2}\}$ is a twin dominating set of $T$. Therefore, $DOM^{*}(K_{9})\leq 4$. By Observation 2.2, we can conclude that $DOM^{*}(K_{9})=4.$

4 The Linear Upper Bound of $DOM^{*}(K_{n})$

In this section we prove that for every integer $n\geq 8$, the conjecture given in Arumugam et al. [1] is false.

Theorem 4.1

For every integer $n\geq 8$, the upper orientable twin domination number of $K_{n}$, $DOM^{*}(K_{n})\leq\left\lfloor\frac{n}{2}\right\rfloor$.

Proof. We prove first the theorem for even $n$.

Let $n=2k$, $k\geq 4$. We use induction on $k$.

If $k=4$, the theorem holds by Theorem 2.1.

Suppose the theorem is true for any tournament $T_{1}$ of order $2k$. Let $T_{2}$ be a tournament of order $2k+2$. Consider $v_{1},v_{2}\in V(T_{2}),\ (v_{1},v_{2})\in A(T_{2})$ and $T_{1}$ the subtournament of $T_{2}$ induced by $V(T_{1})=V(T_{2})-\{v_{1},v_{2}\}$.

By our induction hypothesis, there exist a twin domination set $S_{1}$ of $T_{1}$, such that $|S_{1}|\leq k$. If $|S_{1}|<k$, then $S_{2}=S_{1}\cup\{v_{1},v_{2}\}$ is a twin dominating set of $T_{2}$ with $|S_{2}|\leq k+1$. So, suppose $|S_{1}|=k.$

If for one vertex $v\in S_{1}$, $(v,v_{1})\in A(T_{2})$ or $(v_{2},v)\in A(T_{2})$, then the set $S_{2}=S_{1}\cup\{v_{2}\}$ or $S_{2}=S_{1}\cup\{v_{1}\}$, respectively, is a twin dominating set of $T_{2}$ with cardinality $|S_{2}|=k+1$. Therefore we can assume that $v_{1}$ is a source and $v_{2}$ a sink with respect to the set $S_{1}$. Notice that $\{v_{1},v_{2}\}$ is a twin dominating set of $S_{1}$. 2. 2.

If one of the vertices in $T_{2}$ is source or sink the proof is finished according to Observation 2.1. So there exists $o_{1},o_{2}\in V(T_{2})$ such that the arcs $(o_{1},v_{1}),(v_{2},o_{2})\in A(T_{2})$. If $o_{1}=o_{2}$, $v_{1}v_{2}o_{2}$ or $o_{1}v_{1}v_{2}$ is an oriented cycle in $T_{2}$, then $S_{2}=V(T_{2})-\{S_{1}\cup\{o_{i}\}\}$ with $i\in\{1,2\}$ is a twin dominating set of $T_{2}$ with $|S_{2}|=k+1$. Thus, the arcs $(o_{1},v_{2}),(v_{1},o_{2})\in A(T_{2})$. It is clear, that the set $S_{2}=V(T_{2})-S_{1}$ is a twin dominating set of $T_{2}$ with cardinality $|S_{2}|=k+2$. We will prove that $S_{2}$ it is not minimum. Let $v\in S_{2}-\{v_{1},v_{2},o_{1},o_{2}\}$.

1. (a)

If the arc $(o_{2},o_{1})\in A(T_{2})$, then $S_{2}-\{o_{2}\}$ is a twin dominating set of $T_{2}$ with cardinality $k+1$. So, we can suppose we have the arc $(o_{1},o_{2})\in A(T_{2})$. See Figure 7. 2. (b)

If the arc $(v,o_{1})$ or $(o_{2},v)$ is in $A(T_{2})$, then $S_{2}-\{o_{1}\}$ or $S_{2}-\{o_{2}\}$ are twin dominating set of $T_{2}$, respectively. So we can assume that in $T_{2}$, we have the arcs $(o_{1},v),(v,o_{2})$, which implies that $S_{2}-\{v\}$ is a twin dominating set of $T_{2}$ with cardinality $k+1$.

The reasoning for odd numbers, i.e., $n=2k+1$ for $k\geq 4$, is similar, as the theorem holds for $k=4$ by Theorem 3.1.

5 The Logarithmic Upper Bound of $DOM^{*}(K_{n})$

In the following section we prove the $O(\log_{2}(n))$ upper bound of $DOM^{*}(K_{n})$ for any $n\geq 4$.

Theorem 5.1

For every natural number $n\geq 4$, the upper orientable twin domination number of $K_{n}$, $DOM^{*}(K_{n})\leq 2\left\lceil\log_{2}{(n-1)}\right\rceil$.

Proof. Let $T$ be any orientation of $K_{n}$ and $u\in V(T)$ any vertex. Let $T_{1}$ be the subtournament of $T$ induced by $O_{T}(u)$. As $u$ in-dominates $T_{1}$ there has to be found an out-dominating set in $T_{1}$, which is done in the following recursive way. Let $u_{i}\in T_{i},\ i\in\mathbb{N}^{+}$ such that $od_{T_{i}}(u_{i})\leq\left\lfloor\frac{n_{i}-1}{2}\right\rfloor$ (the existence of such a vertex is guaranteed by Observation 3.1), where $n_{i}=|V(T_{i})|$, then a recursive equations are

[TABLE]

where $l$ is a first integer for which $|V(T_{l})|\leq 2$. Set $S_{1}=\{u_{i}|i=1,\ldots,l\}$ out-dominates $T_{1}$ and while the number of vertices at each step decreases twice $|S_{1}|\leq\left\lceil\log_{2}{od(u)}\right\rceil$. Similarly, we construct the in-dominating set $S_{2}$ for a subtournament of $T$ induced by $I_{T}(u)$, of the maximum cardinality $|S_{2}|\leq\left\lceil\log_{2}{id(u)}\right\rceil$. Set $S=S_{1}\cup S_{2}\cup\{u\}$ twin dominates $T$ and

[TABLE]

This logarithmic upper bound is tighter than the linear one for $n>21$, i.e., $2\left\lceil\log_{2}{(n-1)}\right\rceil<\left\lfloor\frac{n}{2}\right\rfloor$ for $n>21$. We summarize all results for $DOM^{*}(K_{n})$ from Arumugam et al. [1] and this paper in the following theorem.

Theorem 5.2

For $n\geq 1$,

[TABLE]

[TABLE]