Double Roman Domination in Digraphs

Abstract

Let D be a finite and simple digraph with vertex set V(D). A double Roman dominating function (DRDF) on a digraph D is a function $f:V(D)\rightarrow \{0,1,2,3\}$ satisfying the condition that if $f(v)=0$, then the vertex v must have at least two in-neighbors assigned 2 under f or one in-neighbor assigned 3, while if $f(v)=1$, then the vertex v must have at least one in-neighbor assigned 2 or 3. The weight of a DRDF f is the sum $\sum _{v\in V(D)}f(v)$. The double Roman domination number of a digraph D is the minimum weight of a DRDF on D. In this paper, we initiate the study of the double Roman domination of digraphs, and we give several relations between the double Roman domination number of a digraph and other domination parameters such as Roman domination number, k-domination number and signed domination number. Moreover, various bounds on the double Roman domination number of a digraph are presented, and a Nordhaus–Gaddum type inequality for the parameter is also given.

1 Introduction

Due to the diversity of its applications to both theoretical and practical problems, domination and its variants have become one of the important research topics in graph theory (see, for example, [2, 5, 6, 11, 13]). Our aim in this paper is to initiate the study of the double Roman domination in digraphs.

Throughout this paper, $D=(V,A)$ is a finite digraph with neither loops nor multiple arcs (but pairs of opposite arcs are allowed). For two vertices $u,v\in V(D)$, we use (u, v) to denote the arc with direction from u to v, and we also call v an out-neighbor of u and u an in-neighbor of v. For $v\in V(D)$, the out-neighborhood and in-neighborhood of v, denoted by $N^+_D(v)=N^+(v)$ and $N^-_D(v)=N^-(v)$, are the sets of out-neighbors and in-neighbors of v, respectively. The closed out-neighborhood and closed in-neighborhood of a vertex $v\in V(D)$ are the sets $N_D^+[v]=N^+[v]=N^+(v)\cup \{v\}$ and $N_D^-[v]=N^-[v]=N^-(v)\cup \{v\}$, respectively. In general, for a set $X\subseteq V(D)$, we denote $N^+_D(X)=\bigcup _{v\in X}N^+_D(v)$ and $N^+_D[X]=\bigcup _{v\in X}N^+_D[v]$. The out-degree and in-degree of a vertex $v\in V(D)$ are defined by $d^+_D(v)=d^+(v)=|N^+_D(v)|$ and $d^-_D(v)=d^-(v)=|N_D^-(v)|$, respectively. The maximum out-degree, maximum in-degree, minimum out-degree and minimum in-degree among the vertices of D are denoted by $\Delta ^+(D)=\Delta ^+$, $\Delta ^-(D)=\Delta ^-$, $\delta ^+(D)=\delta ^+$ and $\delta ^-(D)=\delta ^-$, respectively. For two vertices u and v of D, the distance d(u, v) from u to v is the length of a shortest directed u-v path in D. If D contains no directed u-v path, then $d(u,v)=\infty$. For a subdigraph H of D and $v\in V(D),$ the distance from H to v in D is $d(H,v)=\min \{d(u,v) : u\in V(H)\}$. Let $\overrightarrow{P}_n$ and $\overrightarrow{C}_n$ denote a directed path and a directed cycle of order n, respectively.

A rooted tree is a connected digraph with a vertex of in-degree 0, called the root, such that every vertex different from the root has in-degree 1. The height of a rooted tree T, denoted by h(T), is $\max \{d(r, v) : v\in V(T)\}$, where r is the root of T. A digraph D is contrafunctional if each vertex of D has in-degree 1. The complement of a digraph D is the digraph $\overline{D}$, where $V(\overline{D})=V(D)$ and $(u,v)\in A(\overline{D})$ if and only if $(u,v)\notin A(D)$.

A k-dominating set of a digraph D is a subset S of the vertex set of D such that every vertex not in S has at least k in-neighbors in S. The minimum cardinality of a k-dominating set of a digraph D is called the k-domination number of D and is denoted by $\gamma _k(D)$. A k-dominating set of D of cardinality $\gamma _k(D)$ is called a $\gamma _k(D)$-set. If $k=1$, then the k-dominating set is exactly the dominating set and we simply write $\gamma (D)$ for $\gamma _1(D)$, which was introduced by Fu [7] and now has been studied extensively (see, for example, [4, 8, 9]).

A signed dominating function (abbreviated SDF) on D is a function $f:V(D)\rightarrow \{-1,1\}$ such that $\sum _{x\in N^-[v]}f(x)\ge 1$ for each vertex $v\in V(D)$. The weight of an SDF f is $\omega (f)=\sum _{v\in V(D)}f(v)$. The signed domination number$\gamma _{S}(D)$ of a digraph D is the minimum weight of an SDF on D. An SDF on D with weight $\gamma _{S}(D)$ is called a $\gamma _{S}(D)$-function. The signed domination number of a digraph was introduced by Zelinka [16] and has been studied by several authors, for example, in Karami et al. [12], Volkmann [15] and elsewhere.

A Roman dominating function (abbreviated RDF) on a digraph D is a function $f: V(D)\rightarrow \{0,1,2\}$ satisfying the condition that every vertex v with $f(v)=0$ has an in-neighbor u with $f(u)=2$. The weight of an RDF f is the sum $\omega (f)=\sum _{v\in V(D)}f(v)$. The Roman domination number of a digraph D, denoted by $\gamma _R(D)$, is the minimum weight of an RDF on D. A Roman dominating function on D with weight $\gamma _R(D)$ is called a $\gamma _R(D)$-function. An RDF f on D can be represented by the ordered partition $(V_0,V_1,V_2)$, where $V_i=\{v\in V(D):f(v)=i\}$ for $i\in \{0,1,2\}$. The Roman domination of a digraph has been studied by Sheikholeslami and Volkmann [14].

Let G be a finite, simple and undirected graph with vertex set V(G). A double Roman dominating function (abbreviated DRDF) on a graph G is defined in [3] as a function $f:V(G)\rightarrow \{0,1,2,3\}$ having the property that if $f(v)=0$, then the vertex v must be adjacent to at least two vertices assigned 2 under f or one vertex assigned 3, while if $f(v)=1$, then the vertex v must be adjacent to at least one vertex assigned 2 or 3. The weight of a DRDF f is $\omega (f)=\sum _{v\in V(G)}f(v)$. The double Roman domination number$\gamma _{dR}(G)$ of a graph G is the minimum weight of a DRDF on G.

In this paper, motivated by the work in [3, 14], we initiate the study of the double Roman domination number of digraphs. A double Roman dominating function (abbreviated DRDF) on a digraph D is a function $f:V(D)\rightarrow \{0,1,2,3\}$ having the property that if $f(v)=0$, then the vertex v must have at least two in-neighbors assigned 2 under f or one in-neighbor assigned 3, while if $f(v)=1$, then the vertex v must have at least one in-neighbor assigned 2 or 3. The weight of a DRDF f is $\omega (f)=\sum _{v\in V(D)}f(v)$. The double Roman domination number$\gamma _{dR}(D)$ of a digraph D is the minimum weight of a DRDF on D. A $\gamma _{dR}(D)$-function is a DRDF on D with weight $\gamma _{dR}(D)$. A DRDF f on D can be represented by the ordered partition $(V_0,V_1,V_2,V_3)$, where $V_i=\{v\in V(D):f(v)=i\}$ for $i\in \{0,1,2,3\}$.

The rest of the paper is organized as follows. In the next section, we give two simple but useful properties of the double Roman domination number of a digraph. We then relate the double Roman domination number of digraphs to other domination parameters such as Roman domination number, k-domination number and signed domination number in Sect. 3. In Sect. 4, we establish lower and upper bounds on the double Roman domination number of a digraph in terms of its order, maximum out-degree and minimum in-degree. Finally, in Sect. 5 we present a Nordhaus–Gaddum result for the double Roman domination number of a digraph.

2 Preliminaries

In this section, we shall give two simple properties of the double Roman domination number of a digraph that will be useful in the next sections.

Proposition 1

For any digraph D, there exists a $\gamma _{dR}(D)$-function such that no vertex needs to be assigned the value 1.

Proof

Let f be a $\gamma _{dR}(D)$-function. Suppose that there exists some vertex v of D such that $f(v)=1$. Then by the definition of $\gamma _{dR}(D)$-function, we have that there exists a vertex $u\in N^-(v)$ such that either $f(u)=2$ or $f(u)=3$. If $f(u)=3$, then we define the function $f'$ by $f'(v)=0$ and $f'(x)=f(x)$ for each $x\in V(D)\backslash \{v\}$. Obviously, $f'$ is a DRDF of weight $\gamma _{dR}(D)-1$, a contradiction. If $f(u)=2$, then we define the function $f''$ by $f''(v)=0$, $f''(u)=3$ and $f''(x)=f(x)$ for each $x\in V(D)\backslash \{u,v\}$. Clearly, $f''$ is a DRDF of weight $\gamma _{dR}(D)$, implying that $f''$ is a $\gamma _{dR}(D)$-function. $\square$

By Proposition 1, it is reasonable to claim that $V_1=\emptyset$ for all double Roman dominating functions under consideration. In this case, any double Roman dominating function f on D can be represented by the ordered partition $(V_0,V_2,V_3)$, where $V_i=\{v\in V(D):f(v)=i\}$ for $i\in \{0,2,3\}$.

Proposition 2

Let D be a digraph and let $f=(V_0,V_1,V_2)$ be a $\gamma _{R}(D)$-function. Then $\gamma _{dR}(D)\le 2|V_1|+3|V_2|$.

Proof

We set $g(v)=0$ for each $v\in V_0$, $g(v)=2$ for each $v\in V_1$ and $g(v)=3$ for each $v\in V_2$. Then it is easy to see that g is a DRDF on D and hence $\gamma _{dR}(D)\le \omega (g)=2|V_1|+3|V_2|$. $\square$

3 Relations to Other Domination Parameters

In this section, we shall relate the double Roman domination number of digraphs to other domination parameters such as Roman domination number, k-domination number and signed domination number.

We first give some relations between the double Roman domination number and Roman domination number of digraphs.

Theorem 1

For any digraph D, $\gamma _{dR}(D)\le 2\gamma _{R}(D)$ with equality if and only if D is empty.

Proof

Let $f=(V_0,V_1,V_2)$ be a $\gamma _{R}(D)$-function such that $|V_1|$ is minimal. Then by Proposition 2,

$$\begin{aligned} \gamma _{dR}(D)\le 2|V_1|+3|V_2|=2\gamma _{R}(D)-|V_2|\le 2\gamma _{R}(D), \end{aligned}$$

(1)

establishing the desired result.

The sufficiency is trivial. To show the necessity, let $\gamma _{dR}(D)=2\gamma _{R}(D)$. Then we have equality throughout the inequality chain (1). Therefore, $|V_2|=0$ and hence by the definition of $\gamma _{R}(D)$-function, $|V_0|=0$. This implies that $V_1=V(D)$. If there exists some arc, say (u, v), of D, then we set $g(v)=0$, $g(u)=2$ and $g(x)=1$ for each $x\in V(D)\backslash \{u,v\}$. It is easy to see that g is an RDF on D with $\omega (g)=\gamma _{R}(D)$. This implies that g is a $\gamma _{R}(D)$-function. Moreover, $|x\in V(D):f(x)=1\}|-|x\in V(D):g(x)=1\}|=2$, contradicting the minimality of f. Therefore, D is empty. $\square$

Theorem 2

For any digraph D, $\gamma _{dR}(D)\ge \gamma _{R}(D)+1$.

Proof

Let $f=(V_0,V_2,V_3)$ be a $\gamma _{dR}(D)$-function. If $V_3\ne \emptyset$, then every vertex in $V_3$ can be reassigned the value 2 and the resulting function will be an RDF on D and hence

$$\begin{aligned} \gamma _{dR}(D)=2(|V_2|+|V_3|)+|V_3|\ge \gamma _{R}(D)+|V_3|\ge \gamma _{R}(D)+1. \end{aligned}$$

Suppose next that $V_3=\emptyset$. Then by the definition of $\gamma _{dR}(D)$-function, $V_2\ne \emptyset$, for otherwise, $V_0=V(D)$, a contradiction. Therefore, all vertices are assigned either the value 0 or the value 2, and all vertices in $V_0$ must have at least two in-neighbors in $V_2$. In this case one vertex in $V_2$ can be reassigned the value 1 and the resulting function will be an RDF on D and hence $\gamma _{dR}(D)=2|V_2|\ge \gamma _{R}(D)+1$. $\square$

Combining Theorems 1 and 2, we may obtain the following result immediately.

Corollary 1

For any nontrivial connected digraph D,

$$\begin{aligned} \gamma _{R}(D)+1\le \gamma _{dR}(D)\le 2\gamma _{R}(D)-1. \end{aligned}$$

We now establish a relation between the double Roman domination number and the domination number of digraphs.

Theorem 3

For any digraph D,

$$\begin{aligned} 2\gamma (D)\le \gamma _{dR}(D)\le 3\gamma (D). \end{aligned}$$

Moreover,

• (a) The left equality holds if and only if $\gamma (D)=\gamma _2(D).$

• (b) The right equality holds if and only if there exists a $\gamma _{dR}(D)$-function $(V_0,V_2,V_3)$ such that $V_2=\emptyset$.

Proof

Let $f=(V_0,V_2,V_3)$ be a $\gamma _{dR}(D)$-function and let S be a $\gamma (D)$-set. We set $f'(v)=3$ for each $v\in S$ and $f'(v)=0$ otherwise. Then it is easy to see that $f'$ is a DRDF on D and hence $\gamma _{dR}(D)\le 3|S|=3\gamma (D)$. On the other hand, by the definition of $\gamma _{dR}(D)$-function, $V_2\cup V_3$ is a dominating set of D and hence $\gamma (D)\le |V_2|+|V_3|$, implying that

$$\begin{aligned} \gamma _{dR}(D)=2|V_2|+3|V_3|\ge 2(|V_2|+|V_3|)\ge 2\gamma (D). \end{aligned}$$

(2)

(a) Suppose that $\gamma _{dR}(D)=2\gamma (D)$. Then we have equality throughout the inequality chain (2). This means that $V_3=\emptyset$ and hence $\gamma _{dR}(D)=2|V_2|$, implying that $\gamma (D)=|V_2|$. Therefore, by the definition of $\gamma _{dR}(D)$-function, we have that every vertex in $V(D)\backslash V_2$ must have at least two in-neighbors in $V_2$. Thus, $\gamma _2(D)\le |V_2|=\gamma (D)$. On the other hand, clearly $\gamma (D)\le \gamma _2(D).$ As a result, $\gamma (D)=\gamma _2(D).$

Conversely, suppose that $\gamma (D)=\gamma _2(D).$ Let $S'$ be a $\gamma _2(D)$-set. We set $g(v)=2$ for each $v\in S'$ and $g(v)=0$ otherwise. Then clearly g is a DRDF on D and hence $\gamma _{dR}(D)\le 2|S'|=2\gamma _2(D)=2\gamma (D)$. As proved previously, $\gamma _{dR}(D)\ge 2\gamma (D)$. Thus, $\gamma _{dR}(D)=2\gamma (D)$.

(b) Suppose that $\gamma _{dR}(D)=3\gamma (D)$. Set $g(v)=3$ for each $v\in S$ and $g(v)=0$ otherwise. Then clearly $\omega (g)=3|S|=3\gamma (D)=\gamma _{dR}(D)$, implying that g is a $\gamma _{dR}(D)$-function.

Conversely, suppose that $V_2=\emptyset$. Then $V_3$ is a dominating set of D and hence $|V_3|\ge \gamma (D)$. Thus, $\gamma _{dR}(D)=3|V_3|\ge 3\gamma (D)$. As proved earlier, $\gamma _{dR}(D)\le 3\gamma (D)$. Therefore, $\gamma _{dR}(D)=3\gamma (D)$. $\square$

Proposition 3

If D is a digraph, then $\gamma _{dR}(D)\le 2\gamma _2(D)$.

Proof

Let S be a $\gamma _2(D)$-set. Define the function $f(x)=2$ for $x\in S$ and $f(x)=0$ otherwise. Then it is easy to verify that f is a DRDF on D and hence $\gamma _{dR}(D)\le 2|S|=2\gamma _2(D)$. $\square$

If $K_n^*$ is the complete digraph of order $n\ge 2$, then $\gamma (K_n^*)=1$, $\gamma _R(K_n^*)=2$ and $\gamma _{dR}(K_n^*)=3$. Thus, Corollary 1 and the upper bound in Theorem 3 are sharp. In addition, let $u,v,x_1,x_2,\ldots ,x_{n-2}$ be the vertex set of the digraph H such that $(u,x_i),(v,x_i)\in A(H)$ for $1\le i\le n-2$. Then $\gamma (H)=2$, $\gamma _2(H)=2$ and $\gamma _{dR}(H)=4$. This example shows that the lower bound in Theorem 3 and Proposition 3 is sharp.

We end this section by relating the double Roman domination number to signed domination number of digraphs. To this end, we need a result due to Ahangar et al. [1].

Let G be a bipartite (undirected) graph with bipartition $(\mathcal {L}, \mathcal {R})$ (standing for “left” and “right”). A subset S of vertices in $\mathcal {R}$ is a left dominating set of G if every vertex of $\mathcal {L}$ is adjacent to a vertex in S. The left domination number, denoted by $\gamma _{\mathcal {L}}(G)$, is the minimum cardinality of a left dominating set of G. A left dominating set of G of cardinality $\gamma _{\mathcal {L}}(G)$ is called a $\gamma _{\mathcal {L}}(G)$-set. Let $\delta _{\mathcal {L}}(G)$ denote the minimum degree of a vertex of $\mathcal {L}$ in G. Ahangar et al. [1] established the following upper bound on the left domination number of a bipartite (undirected) graph in terms of its order.

Theorem 4

([1]) Let G be a bipartite (undirected) graph of order n with bipartition $(\mathcal {L}, \mathcal {R})$. If $\delta _{\mathcal {L}}(G)\ge 2$, then $\gamma _{\mathcal {L}}(G)\le n/3$.

Theorem 5

For any digraph D of order n,

$$\begin{aligned} \gamma _{dR}(D)\le \gamma _{S}(D)+4n/3. \end{aligned}$$

Proof

Let f be a $\gamma _{S}(D)$-function and let $\mathcal {L}$ and $\mathcal {R}$ denote the sets of those vertices in D which are assigned under f the values $-1$ and 1, respectively. Then $|\mathcal {L}|+|\mathcal {R}|=n$ and $\gamma _{S}(D)=\omega (f)=|\mathcal {R}|-|\mathcal {L}|$, implying that $2|\mathcal {R}|=n+\gamma _{S}(D)$.

If $\mathcal {L}=\emptyset$, that is, if $\mathcal {R}=V(D)$, then we set $g(x)=2$ for each $x\in V(D)$. Then it is easy to see that g is a DRDF on D, implying that

$$\begin{aligned} \gamma _{dR}(D)\le \omega (g)=2n=2|\mathcal {R}|=2\gamma _{S}(D)<\gamma _{S}(D)+4n/3. \end{aligned}$$

Hence we may assume that $\mathcal {L}\ne \emptyset$. Let $D'$ be the bipartite spanning subdigraph of D with bipartition $(\mathcal {L}, \mathcal {R})$, where $A(D')=\{(u,v)\in A(D):u\in \mathcal {R}\ \text {and}\ v\in \mathcal {L}\}$. Since f is a $\gamma _{S}(D)$-function, each vertex of $\mathcal {L}$ has at least 2 in-neighbors in $\mathcal {R}$ in $D'$ and hence $\delta ^-_{\mathcal {L}}(D')\ge 2$, where $\delta ^-_{\mathcal {L}}(D')=\min \{d^-_{D'}(v):v\in \mathcal {L}\}$. Let H be the (undirected) graph obtained from $D'$ by replacing any arc with an edge and let $\mathcal {R}_2$ be a $\gamma _{\mathcal {L}}(H)$-set. Then $\delta _{\mathcal {L}}(H)=\delta ^-_{\mathcal {L}}(D')\ge 2$ and hence by Theorem 4, $|\mathcal {R}_2|=\gamma _{\mathcal {L}}(H)\le n/3$. Moreover, since $\mathcal {R}_2$ is a $\gamma _{\mathcal {L}}(H)$-set, any vertex in $\mathcal {L}$ is adjacent to some vertex in $\mathcal {R}_2$ in H and hence any vertex in $\mathcal {L}$ has at least one in-neighbor in $\mathcal {R}_2$ in $D'$ and so in D. Let $\mathcal {R}_1=\mathcal {R}\backslash \mathcal {R}_2$. Set

$$\begin{aligned} g'(x)=\left\{ \begin{array}{ll} 0,&{}\text {if}\ x\in \mathcal {L},\\ 2,&{}\text {if}\ x\in \mathcal {R}_1,\\ 3,&{}\text {if}\ x\in \mathcal {R}_2. \end{array}\right. \end{aligned}$$

Then $g'$ is a DRDF on D and hence

$$\begin{aligned} \gamma _{dR}(D)\le&\,\,\omega (g')=2|\mathcal {R}_1|+3|\mathcal {R}_2|\\ =&\,\,2(|\mathcal {R}_1|+|\mathcal {R}_2|)+|\mathcal {R}_2|=2|\mathcal {R}|+|\mathcal {R}_2|\\ =&\,\,n+\gamma _{S}(D)+|\mathcal {R}_2|\\ \le&\,\,\gamma _{S}(D)+4n/3, \end{aligned}$$

which completes the proof. $\square$

4 Upper and lower bounds

Our aim in the section is to establish upper and lower bounds on the double Roman domination number of a digraph in term of its order, maximum out-degree and minimum in-degree.

We first present upper bounds on the double Roman domination number of digraphs.

Proposition 4

If D is a digraph of order n, then $\gamma _{dR}(D)\le 2n$ with equality if and only if D is empty.

Proof

Define the function f by $f(x)=2$ for each $x\in V(D)$. Then f is a DRDF on D and hence $\gamma _{dR}(D)\le 2n$. If D is empty, then $\gamma _{dR}(D)=2n$. Now assume that $\gamma _{dR}(D)=2n$, and suppose to the contrary that D contains an arc (u, v). Define the function $g(v)=0$, $g(u)=3$ and $g(x)=2$ for each $x\in V(D)\setminus \{u,v\}$. Then g is a DRDF on D of weight $2n-1$, a contradiction. $\square$

Theorem 6

Let D be a digraph of order $n\ge 2$ such that $|A(D)|\ge 1$. Then $\gamma _{dR}(D)\le 2n-1$ with equality if and only if D has exactly one nontrivial component of order 2 or one nontrivial component H of order 3 such that H is a directed path or a directed cycle.

Proof

Proposition 4 implies $\gamma _{dR}(D)\le 2n-1$. If D has exactly one nontrivial component of order 2 or one nontrivial component H of order 3 such that H is a directed path or a directed cycle, then it is easy to see that $\gamma _{dR}(D)=2n-1$. Conversely, assume that $\gamma _{dR}(D)=2n-1$. Suppose that D contains two arcs (u, v) and (w, z).

If $u\ne w,z$ and $v\ne w,z$, then define f by $f(v)=f(z)=0$, $f(u)=f(w)=3$ and $f(x)=2$ for each $x\in V(D)\backslash \{u,v,w,z\}$. Then f is a DRDF on D of weight $2n-2$, a contradiction. Therefore, D contains at most one nontrivial component H.

If $v=z$ and $u\ne w$, then define g by $g(v)=0$ and $g(x)=2$ otherwise. Then g is a DRDF on D of weight $2n-2$, a contradiction.

If $u=w$ and $v\ne z$, then define h by $h(u)=3$, $h(v)=h(z)=0$ and $h(x)=2$ otherwise. Then h is a DRDF on D of weight $2n-3$, a contradiction.

Using these observations, we deduce that $2\le |V(H)|\le 3$, and if $|V(H)|=3$, then H is a directed path or a directed cycle of order 3. $\square$

Proposition 5

Let T be a rooted tree with $h(T)=1$. Then $\gamma _{dR}(T)=3$.

Proof

Let r be the root of T. We set $f(r)=3$ and $f(u)=0$ for each $u\in V(D)\backslash \{r\}$. Then it is easy to see that f is a $\gamma _{dR}(D)$-function and hence $\gamma _{dR}(T)=\omega (f)=3$. $\square$

Theorem 7

Let $T\ncong \overrightarrow{P}_3$ be a rooted tree of order $n\ge 2$. Then

$$\begin{aligned} \gamma _{dR}(T)\le (5n-1)/3. \end{aligned}$$

Proof

We proceed by induction on n. If $n=2$, then by Proposition 5, $\gamma _{dR}(T)=3=(5n-1)/3.$ Hence we may assume that $n\ge 3$. If $h(T)=1$, then again by Proposition 5, $\gamma _{dR}(T)=3\le (5n-1)/3.$

Suppose next that $h(T)\ge 2$. Let r be the root of T; let x be a vertex of T such that $d(r,x)=h(T)-1$; let $T_1$ be the connected component of $T-x$ that contains the root r and let $T_2=T-T_1$. Note that $h(T_2)=1$. Therefore, by Proposition 5, $\gamma _{dR}(T_2)=3\le (5|V(T_2)|-1)/3$. If $|V(T_1)|=1$, then clearly $|V(T_2)|\ge 3$ since $T\ncong \overrightarrow{P}_3$ and $(V_0,V_2,V_3)$ is a DRDF on D, where $V_3=\{x\}$, $V_2=\{r\}$ and $V_0=V(D)\backslash \{r,x\}$, and hence $\gamma _{dR}(T)\le 3+2\le (5n-1)/3.$ Assume next that $|V(T_1)|\ge 2$. If $T_1\ncong \overrightarrow{P}_3$, then by the induction hypothesis, $\gamma _{dR}(T_1)\le (5|V(T_1)|-1)/3$ and hence

$$\begin{aligned} \gamma _{dR}(T)&\le \gamma _{dR}(T_1)+\gamma _{dR}(T_2)\\&\le (5|V(T_1)|-1)/3+(5|V(T_2)|-1)/3\\&<(5n-1)/3. \end{aligned}$$

If $T_1\cong \overrightarrow{P}_3$, then $\gamma _{dR}(T_1)=5=5|V(T_1)|/3$ and hence

$$\begin{aligned} \gamma _{dR}(T)\le&\,\gamma _{dR}(T_1)+\gamma _{dR}(T_2)\\ \le&\,5|V(T_1)|/3+(5|V(T_2)|-1)/3\\ =&\,(5n-1)/3, \end{aligned}$$

which completes our proof. $\square$

Harary et al. [10] showed that every connected contrafunctional digraph D has a unique directed cycle and the removal of any arc of the directed cycle results in a rooted tree T. Therefore, we have $\gamma _{dR}(D)\le \gamma _{dR}(T)$, which, together with Theorem 7, would yield the following result directly.

Corollary 2

Let D be a connected contrafunctional digraph of order n. Then $\gamma _{dR}(D)=5$ if $D\cong \overrightarrow{C}_3$, and $\gamma _{dR}(D)\le (5n-1)/3$ otherwise.

For a special class of contrafunctional digraphs, we will improve Corollary 2 slightly. For this purpose, we define the height of a connected contrafunctional digraph D, denoted by h(D), to be the maximum distance from its unique directed cycle C to all vertices of D, i.e., $h(D)=\max \{d(C, v) : v\in V(D)\}$. In particular, the height of a directed cycle is exactly equal to 0.

Theorem 8

Let D be a connected contrafunctional digraph of order n with $h(D)=1$. Then $\gamma _{dR}(D)\le 3n/2$.

Proof

Let C be the unique directed cycle of D, $v_{i_j}$ be the vertex set of C such that $v_{i_j}$ has at least one out-neighbor not in C for $1\le j\le t$, and let $V'$ be the set of out-neighbors of $v_{i_j}$ not in C for $1\le j\le t$. Define the function f by $f(v_{i_j})=3$ for $1\le j\le t$ and $f(x)=0$ for each $x\in V'$. We observe that $D'=D\setminus (\{v_{i_1},v_{i_2},\ldots ,v_{i_t}\}\cup V')$ is empty or consists of some directed paths. If $w_1w_2\ldots w_k$ is such a directed path of $D'$, then for $1\le i\le k$, we define $f(w_i)=0$ if i is odd and $f(w_i)=3$ if i is even. Altogether, it is easy to verify that f is a DRDF on D of weight $\omega (f)\le 3n/2$. Therefore, $\gamma _{dR}(D)\le \omega (f)\le 3n/2$. $\square$

Theorem 9

Let $D\ncong \overrightarrow{C}_3$ be a connected digraph of order $n\ge 3$ with $\delta ^-(D)\ge 1$. Then

$$\begin{aligned} \gamma _{dR}(D)\le (5n-1)/3. \end{aligned}$$

Proof

If $n=3$, then it is easy to see that $\gamma _{dR}(D)=3\le (5n-1)/3$ since $D\ncong \overrightarrow{C}_3$. Hence we may assume that $n\ge 4$. Since $\delta ^-(D)\ge 1$, we can choose an arbitrary incoming arc of v for each vertex v of D. Then all such arcs induce a spanning subdigraph H of D consisting of some connected components, say $H_1, H_2,\ldots , H_l$. Moreover, $H_i$ ($i\in \{1,2,\ldots ,l\}$) is a connected contrafunctional subdigraph of D since each vertex of $H_i$ has in-degree 1.

Firstly, we consider the case that H is not the disjoint union of copies of $\overrightarrow{C}_3$. Without loss of generality, assume that $H_1\ncong \overrightarrow{C}_3$. Then by Corollary 2, we have $\gamma _{dR}(H_1)\le (5|V(H_1)|-1)/3$ and $\gamma _{dR}(H_i)\le 5|V(H_i)|/3$ for each $i\in \{2,3,\ldots ,l\}$. Therefore,

$$\begin{aligned} \gamma _{dR}(D)\le&\,\gamma _{dR}(H)=\sum _{i=1}^l\gamma _{dR}(H_i)\\ \le&\,(5|V(H_1)|-1)/3+\sum _{i=2}^l5|V(H_i)|/3\\ =&\,(5n-1)/3. \end{aligned}$$

Next, we consider the case that H is the disjoint union of copies of $\overrightarrow{C}_3$. This implies that $H_i\cong \overrightarrow{C}_3$ for $i\in \{1,2,\ldots ,l\}$. Note that $n\ge 4$. Therefore, $l\ge 2$. Since D is connected but H is not, the arc set A(D) of D consists of A(H) and some arcs not in H. In addition, if we add some arc in $A(D)\backslash A(H)$ to H, then it is easy to verify that the resulting digraph has a strictly smaller double Roman domination number than that of H. Therefore, by Corollary 2, we have

$$\begin{aligned} \gamma _{dR}(D)\le&\,\gamma _{dR}(H)-1\\ =&\,\sum _{i=1}^l\gamma _{dR}(H_i)-1\\ =&\,\sum _{i=1}^l5|V(H_i)|/3-1\\ \le&\,(5n-1)/3, \end{aligned}$$

which completes our proof. $\square$

Theorem 10

For any digraph D of order n, $\gamma _{dR}(D)\le 2(n-\Delta ^+)+1$.

Proof

Let v be a vertex of out-degree $\Delta ^+$. Then it is easy to see that $f=(V_0,V_2,V_3)$ is a DRDF on D, where $V_0=N^+(v)$, $V_2=V(D)\backslash N^+[v]$ and $V_3=\{v\}$. Thus, $\gamma _{dR}(D)\le \omega (f)=3+2(n-d^+(v)-1)=2(n-\Delta ^+)+1$. $\square$

Theorem 11

For any digraph D of order n with $\delta ^-\ge 1$,

$$\begin{aligned} \gamma _{dR}(D)\le n\left\{ 3-3\left( \frac{3}{2(1+\delta ^-)}\right) ^{\frac{1}{\delta ^-}}+2\left( \frac{3}{2(1+\delta ^-)}\right) ^{\frac{1+\delta ^-}{\delta ^-}}\right\} . \end{aligned}$$

Proof

Given a digraph D and a real number p with $0\le p\le 1$, select a set X of vertices each of which is selected independently with probability p (with p to be defined later). Then the expected size of X is np since X admits the binomial distribution with parameters n and p. Let $Y=V(D)\backslash N^+_D[X]$. We set $f(v)=3$ for any $v\in X$, $f(v)=2$ for any $v\in Y$ and $f(v)=0$ otherwise. Then it is easy to see that f is a DRDF on D. Note that

$$\begin{aligned} P(v\in Y)=&\,P(v\in V(D)\backslash N^+_D[X])\\ =&\,(1-p)^{1+d^-(v)}\\ \le&\,(1-p)^{1+\delta ^-}. \end{aligned}$$

Thus,

$$\begin{aligned} \mathrm{\mathbf E}(\omega (f))\le 3np+2n(1-p)^{1+\delta ^-}, \end{aligned}$$

where E$(\omega (f))$ is the expected weight of f. It is not difficult to verify that the upper bound for E$(\omega (f))$ is minimum when $p=1-\left( \frac{3}{2(1+\delta ^-)}\right) ^{\frac{1}{\delta ^-}}$ and hence

$$\begin{aligned} \mathrm{\mathbf E}(\omega (f))\le n\left\{ 3-3\left( \frac{3}{2(1+\delta ^-)}\right) ^{\frac{1}{\delta ^-}}+2\left( \frac{3}{2(1+\delta ^-)}\right) ^{\frac{1+\delta ^-}{\delta ^-}}\right\} . \end{aligned}$$

This implies that there must be some DRDF on D with at most the above bound as its weight, which completes our proof. $\square$

We next give a lower bound on the double Roman domination number of a digraph.

Theorem 12

For any connected digraph D of order $n\ge 4$,

$$\begin{aligned} \gamma _{dR}(D)\ge \left\lceil \frac{6n+3}{2\Delta ^++3}\right\rceil . \end{aligned}$$

Proof

Let $f=(V_0,V_2,V_3)$ be a $\gamma _{dR}(D)$-function and let $n_0=|V_0|$. If $V_3=\emptyset$, then it is easy to see that $\gamma _{dR}(D)=2|V_2|=2(n-n_0)$. Moreover, by the definition of $\gamma _{dR}(D)$-function, each $v\in V_0$ must have at least two in-neighbors assigned 2 under f and hence $\sum _{u\in N^-(v)}f(u)\ge 4$. Thus, $\gamma _{dR}(D)=\omega (f)\ge n_0\cdot \frac{4}{\Delta ^+}$. Then it follows that

$$\begin{aligned} 2\gamma _{dR}(D)=4n-4n_0\ge 4n-\gamma _{dR}(D)\Delta ^+ \end{aligned}$$

and hence

$$\begin{aligned} \gamma _{dR}(D)(\Delta ^++2)\ge 4n, \end{aligned}$$

implying that $\gamma _{dR}(D)\ge \left\lceil \frac{4n}{\Delta ^++2}\right\rceil \ge \left\lceil \frac{6n+3}{2\Delta ^++3}\right\rceil .$

If $V_3\ne \emptyset$, then it is easy to see that $\gamma _{dR}(D)=2(|V_2|+|V_3|)+|V_3|\ge 2(n-n_0)+1$. Moreover, by the definition of $\gamma _{dR}(D)$-function, each $v\in V_0$ must have at least two in-neighbors assigned 2 under f or one in-neighbor assigned 3, and hence $\sum _{u\in N^-(v)}f(u)\ge 3$. Thus, $\gamma _{dR}(D)=\omega (f)\ge n_0\cdot \frac{3}{\Delta ^+}$. Then it follows that

$$\begin{aligned} 3\gamma _{dR}(D)\ge 6n-6n_0+3\ge 6n-2\gamma _{dR}(D)\Delta ^++3 \end{aligned}$$

and hence

$$\begin{aligned} \gamma _{dR}(D)(2\Delta ^++3)\ge 6n+3, \end{aligned}$$

implying that $\gamma _{dR}(D)\ge \left\lceil \frac{6n+3}{2\Delta ^++3}\right\rceil .$$\square$

The following result, derived from Theorem 12, shows that the upper bound in Theorem 10 and the lower bound in Theorem 12 are sharp.

Corollary 3

Let D be a connected digraph of order $n\ge 3$. Then $\gamma _{dR}(D)=3$ if and only if $\Delta ^+=n-1$.

Proof

Clearly, by the definition, $\gamma _{dR}(D)\ge 3$. Now let $\Delta ^+=n-1$, and let v be a vertex of out-degree $\Delta ^+$. Define the function f by $f(v)=3$ and $f(x)=0$ for $x\in V(D)\setminus \{v\}$. Then f is a DRDF on D of weight 3 and hence $\gamma _{dR}(D)\le 3$ and thus $\gamma _{dR}(D)=3$. Conversely, assume that $\gamma _{dR}(D)=3$. If $\Delta ^+\le n-2$, then Theorem 12 leads to the contradiction

$$\begin{aligned} \gamma _{dR}(D)\ge \left\lceil \frac{6n+3}{2\Delta ^++3}\right\rceil \ge \left\lceil \frac{6n+3}{2(n-2)+3}\right\rceil \ge 4, \end{aligned}$$

which completes our proof. $\square$

5 A Nordhaus–Gaddum type result

In this section, we derive a Nordhaus–Gaddum bound on the double Roman domination number of digraphs.

Theorem 13

For any digraph D of order $n\ge 4$,

$$\begin{aligned} \gamma _{dR}(D)+\gamma _{dR}(\overline{D})\le 2n+3. \end{aligned}$$

Proof

It is easy to see that $d_D^+(v)+d_{\overline{D}}^+(v)=n-1$ for any vertex $v\in V(D)$. This implies that $\Delta ^+(\overline{D})=n-1-\delta ^+(D)$. Then by Theorem 10, we have

$$\begin{aligned} \gamma _{dR}(D)+\gamma _{dR}(\overline{D})\le&\,\, (2n-2\Delta ^+(D)+1)+(2n-2\Delta ^+(\overline{D})+1) \nonumber \\ =&\,\,2n-2\Delta ^+(D)+2\delta ^+(D)+4 \nonumber \\ \le&\,\,2n+4. \end{aligned}$$

(3)

Suppose that $\gamma _{dR}(D)+\gamma _{dR}(\overline{D})=2n+4.$ Then we have equality throughout the inequality chain (3). This implies that $\Delta ^+(D)=\delta ^+(D)$. Let $k=\Delta ^+(D)=\delta ^+(D)$. Then $\Delta ^+(\overline{D})=\delta ^+(\overline{D})=n-1-k$. Without loss of generality, we may assume that $k\le (n-1)/2$, since our argument is symmetric in D and $\overline{D}$. Since equality holds, $\gamma _{dR}(D)=2(n-k)+1$ and $\gamma _{dR}(\overline{D})=2k+3$. Let $v\in V(D)$.

Claim 1

All of the out-neighbors of every vertex not in $N^+_D[v]$ are in $N^+_D[v]$.

Proof of Claim 1

If some vertex u outside $N^+_D[v]$ in D has at least one out-neighbor, say w, outside $N^+_D[v]$, then set $f(v)=3$, $f(w)=1$, $f(x)=0$ for $x\in N^+_D(v)$ and $f(x)=2$ otherwise. Clearly, f is a DRDF on D with weight $2(n-k)$, a contradiction to the fact that $\gamma _{dR}(D)=2(n-k)+1$. So, this claim is true.

Claim 2

For any vertex u outside $N^+_D[v]$, $(u,v)\in A(D)$.

Proof of Claim 2

Suppose, to the contrary, that there exists some vertex u outside $N^+_D[v]$ such that $(u,v)\notin A(D)$. Note that $\Delta ^+(D)=\delta ^+(D)=k$. Hence by Claim 1, $N^+_D(u)=N^+_D(v)$. We set $g(x)=0$ for any $x\in N^+_D(v)$ and $g(x)=2$ otherwise. Then g is a DRDF on D with weight $2(n-k)$, a contradiction. So, this claim is true.

Claim 3

There exists at most one vertex outside $N^+_D[v]$.

Proof of Claim 3

Suppose, to the contrary, that there exist at least two vertices, say u and w, outside $N^+_D[v]$. Then by Claim 2, we have that $(u,v),(w,v)\in A(D)$. Note that $\Delta ^+(D)=\delta ^+(D)=k$. Hence by Claim 1, $N^+_D(u)=N^+_D(w)$ or $|N^+_D(v)\backslash (N^+_D(u)\cap N^+_D(w))|=2$. Let $N^+_D(v)=\{v_1,v_2,\ldots ,v_k\}$.

If $N^+_D(u)=N^+_D(w)$, then we assume, without loss of generality, that $N^+_D(u)=N^+_D(w)=\{v,v_1,v_2,\ldots ,v_{k-1}\}$. We set $h(x)=0$ for any $x\in \{v,v_1,v_2,\ldots ,v_{k-1}\}$ and $h(x)=2$ otherwise. It is easy to see that h is a DRDF on D with weight $2(n-k)$, a contradiction. If $|N^+_D(v)\backslash (N^+_D(u)\cap N^+_D(w))|=2$, then we assume, without loss of generality, that $N^+_D(u)=\{v,v_1,v_2,\ldots ,v_{k-1}\}$ and $N^+_D(w)=\{v,v_2,v_3,\ldots ,v_{k}\}$. Set $h'(v_1)=h'(v_k)=1$, $h'(x)=0$ for any $x\in \{v,v_2,v_3,\ldots ,v_{k-1}\}$ and $h'(x)=2$ otherwise. It is easy to see that $h'$ is a DRDF on D with weight $2(n-k)$, a contradiction. So, this claim is true.

Thus, by Claim 3, we have $k=d_D^+(v)\ge n-2$ for any $v\in V(D)$. Together with our earlier assumptions, $k\le (n-1)/2$. Therefore, $n-2\le k\le (n-1)/2$ and hence $n\le 3$, a contradiction. This implies that $\gamma _{dR}(D)+\gamma _{dR}(\overline{D})\le 2n+3,$ which completes our proof. $\square$