Vertex–Edge Roman {2}-Domination

Abstract

A vertex–edge Roman $\left\{2\right\}$-dominating function on a graph $G=(V,E)$ is a function $f:V⟶\{0,1,2\}$ satisfying that, for every edge $uv\in E$ with $f\left(v\right)=f\left(u\right)=0$, ${\sum }_{w\in N\left(v\right)\cup N\left(u\right)}f\left(w\right)\ge 2$. The weight of the function f is the sum ${\sum }_{a\in V}f\left(a\right)$. The vertex–edge Roman $\left\{2\right\}$-domination number of G, denoted by ${\gamma }_{veR2}\left(G\right)$, is the minimum weight of a vertex–edge Roman $\left\{2\right\}$-dominating function on G. In this work, we begin the study of vertex–edge Roman $\left\{2\right\}$-domination. We determine the exact vertex–edge Roman $\left\{2\right\}$-domination number for cycles and paths, and we provide a tight lower bound and a tight upper bound for the vertex–edge Roman $\left\{2\right\}$-domination number of trees. In addition, we prove that the decision problem associated with vertex–edge Roman $\left\{2\right\}$-domination is NP-complete for bipartite graphs.

1. Introduction and Terminology

There are many articles in the literature investigating variables of either vertex or edge domination, while there are few papers in the literature investigating vertex-dominating edges. In many situations in real life, we need to dominate edges using vertices; for example, in the context of highway safety, if highways are considered as edges, and intersections (or towns) as vertices, the aim is to place services—such as road safety vehicles—at vertices with minimum cost; that is, we want to use as few road safety vehicles as possible. This highlights the usefulness of dominating edges using vertices or, as it is called in the literature, vertex–edge domination.

In this work, we introduce a variant of vertex–edge domination called vertex–edge Roman $\left\{2\right\}$-domination, which involves adding an extra condition to vertex–edge domination. In the highway safety example, this additional condition enhances the reliability of highway safety. Vertex–edge Roman $\left\{2\right\}$-domination relaxes the condition of vertex–edge Roman domination (the definition of this notion will be given later); thus, the vertex–edge Roman $\left\{2\right\}$-domination number lies between the vertex–edge Roman domination number and vertex–edge domination number (the definition of this notion will be given later). This may improve the understanding of the previous two vertex–edge domination variants.

All graphs $G=(V,E)$ in this paper are finite, undirected, and simple (i.e., they contain no parallel edges and no loops). Let $v\in V\left(G\right).$ The set of all vertices adjacent to v in G is denoted by $N_G\left(v\right)$, which is called the open neighborhood of v. The closed neighborhood of v in G is the set $N_G\left[v\right]:=N_G\left(v\right)\cup \left\{v\right\}$. If G is clear from the context, we write $N\left(v\right)$ and $N\left[v\right]$ instead of $N_G\left(v\right)$ and $N_G\left[v\right]$, respectively. The degree$d\left(v\right)$ of a vertex $v\in V\left(G\right)$ is $\left|N\right(v\left)\right|.$ A vertex $v\in V\left(G\right)$ is called a leaf if $d\left(v\right)=1$. A vertex $u\in V\left(G\right)$ is called a support vertex if u is adjacent to a leaf; u is called a strong support vertex if u is adjacent to at least two leaves. We denote the set of leaves in a graph G by $L\left(G\right)$. We denote the number of leaves and the number of support vertices in G by $l\left(G\right)$ and $s\left(G\right)$, respectively. A path with n vertices is denoted by $P_n$. A cycle with n vertices is denoted by $C_n$. A star graph with n vertices ($n-1$ leaves and one non-leaf vertex) is denoted by $S_n$. A double star graph is a connected graph that contains exactly two non-leaf vertices; a double star graph can be obtained from two star graphs by adding an edge between their centers. Let G be a connected graph. The distance between two vertices $u,v\in V\left(G\right)$, denoted by ${dist}_G(u,v)$, is the length of a shortest path between u and v in G. The diameter of G is $diam\left(G\right):=max\left\{{dist}_G(u,v)\right|(u,v)\in V\times V\}.$ A path in G is called a diametral path if its length is equal to $diam\left(G\right)$, and it is a shortest path between its ends. A tree that has a vertex r identified as the root is called a rooted tree. The successors of a vertex v in a rooted tree are the neighbors of v that are not in the unique path between v and the root r. In some of the literature, the successors of v are called the children of v. If T is a rooted tree and $v\in V\left(T\right)$, then $T_v$ denotes the subtree of T that is induced by v and its successors.

Let $G=(V,E)$ be a graph. A set $D\subseteq V$ is called a vertex–edge dominating set of G if, for every edge $uv\in E$, ($N\left(u\right)\cup N\left(v\right))\cap D\ne \varnothing$. The minimum cardinality of a vertex–edge dominating set of G is called the vertex–edge domination number of G, and it is denoted by ${\gamma }_{ve}\left(G\right).$ Vertex–edge domination was introduced in [1].

Let $f:V⟶B$ be a function, where $B\subseteq \mathbb{Z}$. The weight of the function f is $w\left(f\right):={\sum }_{a\in V}f\left(a\right)$. Let $V_b^f=\{a\in V\mid f\left(a\right)=b\}$, where $b\in B$. If H is a subgraph of G, then $f\left(H\right):={\sum }_{h\in V\left(H\right)}f\left(h\right)$.

A function $f:V⟶\{0,1,2\}$ is called a Roman dominating function on $G=(V,E)$ if every vertex $v\in V$ with $f\left(v\right)=0$ is adjacent to a vertex u with $f\left(u\right)=2$. The Roman domination number of G is the minimum weight of a Roman dominating function on G. Roman domination was motivated by a defensive strategy used to protect the Roman Empire by Constantine the Great, who was born in 272 and died in 337 BC. The strategy was then studied as a mathematical concept in graph theory in [2,3,4,5,6].

A function $f:V⟶\{0,1,2\}$ is called a vertex–edge Roman dominating function, abbreviated as veRDF, on $G=(V,E)$ if, for every edge $uv\in E$ with $f\left(v\right)=f\left(u\right)=0$, $(N\left(u\right)\cup N\left(v\right))\cap V_2^f\ne \varnothing$. The vertex–edge Roman domination number of G is ${\gamma }_{veR}\left(G\right):=min\{w\left(f\right)\mid f\mathrm{is}\mathrm{a}\mathrm{veRDF}\mathrm{on}G\}.$ Vertex–edge Roman domination was introduced in [7] and investigated further in [8]. A variant of vertex–edge Roman domination, called vertex–edge perfect Roman domination, was introduced and investigated in [9].

A function $f:V⟶\{0,1,2\}$ is called a Roman $\left\{2\right\}$-dominating function if, for every $v\in V$ with $f\left(v\right)=0$, ${\sum }_{u\in N\left(v\right)}f\left(u\right)\ge 2.$ The minimum weight of a Roman $\left\{2\right\}$-dominating function of G is the Roman $\left\{2\right\}$-domination number of G, and it is denoted by ${\gamma }_{R2}\left(G\right)$. Roman $\left\{2\right\}$-domination was introduced in [10]. For a recent survey of this notion, refer to [11].

In this work, we begin the study of vertex–edge Roman $\left\{2\right\}$-domination. A function $f:V⟶\{0,1,2\}$ is called a vertex–edge Roman $\left\{2\right\}$-dominating function, abbreviated as veR2DF, on $G=(V,E)$ if, (★) for every edge $uv\in E$ with $f\left(v\right)=f\left(u\right)=0$, ${\sum }_{w\in N\left(v\right)\cup N\left(u\right)}f\left(w\right)\ge 2$. We say the edge $uv\in E$ is f-dominated if either $max\left\{f\right(v),f(u\left)\right\}\ge 1$ or (★) holds. The vertex–edge Roman $\left\{2\right\}$-domination number of G is ${\gamma }_{veR2}\left(G\right):=min\{w\left(f\right)\mid f\mathit{is}\mathit{a}\mathit{veR2DF}\mathit{on}G\}.$ It is not difficult to see that the vertex–edge Roman $\left\{2\right\}$-domination number is bounded above by the Roman $\left\{2\right\}$-domination number and bounded below by the vertex–edge domination number; that is, for any graph G, ${\gamma }_{ve}\left(G\right)\le {\gamma }_{veR2}\left(G\right)\le {\gamma }_{R2}\left(G\right).$

Observe that, for every connected graph G with $\left|V\right(G\left)\right|\ge 2$, we have that ${\gamma }_{veR2}\left(G\right)\le \left|V\left(G\right)\right|-1$, as one can label a vertex with 0 and label each of the remaining vertices with 1 to obtain a veR2DF on G. Clearly, every veRDF is veR2DF. On the other hand, if f is a veR2DF on G, then the function $\theta :V⟶\{0,1,2\}$, defined as $V_0^{\theta }=V_0^f$, $V_1^{\theta }=\varnothing ,$$V_2^{\theta }=V_1^f\cup V_2^f$, is a veRDF on G with $w\left(\theta \right)\le 2w\left(f\right)$; hence, we have the following result.

Proposition 1. 

Let G be a graph. Then, ${\displaystyle \frac{1}{2}}{\gamma }_{veR}\left(G\right)\le {\gamma }_{veR2}\left(G\right)\le {\gamma }_{veR}\left(G\right)$.

The next result follows directly from the definition of Roman $\left\{2\right\}$-domination.

Proposition 2. 

For any graph G, ${\gamma }_{veR2}\left(G\right)\le {\gamma }_{R2}\left(G\right)$.

The following result gives the exact value of the ${\gamma }_{veR2}\left(G\right)$ of cycles.

Theorem 1. 

Let $n\ge 4.$ Then, ${\gamma }_{veR2}\left(C_n\right)=\lceil {\displaystyle \frac{n}{3}}\rceil$.

Proof. 

First, we show that ${\gamma }_{veR2}\left(C_n\right)\le \lceil {\displaystyle \frac{n}{3}}\rceil$ for any $n\ge 4$. Let $n\ge 4$, and $C_n=v_1{v}_2\dots v_n{v}_1$. Define a veR2DF f on $C_n$ such that $f\left(v_i\right)=1$ if $i\equiv 1(mod3)$ and $f\left(v_i\right)=0$ if $i\equiv 0,2(mod3)$. Clearly, $w\left(f\right)=\lceil {\displaystyle \frac{n}{3}}\rceil$; so, ${\gamma }_{veR2}\left(C_n\right)\le \lceil {\displaystyle \frac{n}{3}}\rceil$ for any $n\ge 4$. Now, we prove by induction on n that ${\gamma }_{veR2}\left(C_n\right)=\lceil {\displaystyle \frac{n}{3}}\rceil$ for any $n\ge 4$. It is not difficult to see that if $n\in \{4,5,6\}$, then ${\gamma }_{veR2}\left(C_n\right)=2=\lceil {\displaystyle \frac{n}{3}}\rceil$, and ${\gamma }_{veR2}\left(C_7\right)=3=\lceil {\displaystyle \frac{7}{3}}\rceil$. This has proven the base step. Let $C_n$ be a cycle of order $n\ge 8$, and assume that the statement holds for all $n^{\prime }$, where $4\le n^{\prime }<n$. Assume that there exists a veR2DF f on $C_n$ such that $w\left(f\right)<\lceil {\displaystyle \frac{n}{3}}\rceil$. Choose two distinct vertices $u,v\in V\left(C_n\right)$ such that $f\left(u\right),f\left(v\right)\ge 1$ and $dist(u,v)$ is as small as possible. From the definition of a vertex–edge Roman $\left\{2\right\}$-dominating function, we must have $dist(u,v)\le 4$. If $dist(u,v)\le 3$, contract the shortest path between u and v to a single vertex w, and let the new formed cycle be $C_{n^{\prime }}$. Observe that $n^{\prime }\in \{n-1,n-2,n-3\}$; so, $n^{\prime }\ge 5$. Define a veR2DF $f^{\prime }$ on $C_{n^{\prime }}$ such that $f^{\prime }\left(w\right)=max\{f\left(u\right),f\left(v\right)\}$ and $f^{\prime }\left(x\right)=f\left(x\right)$ for every $x\in V\left(C_{n^{\prime }}\right)$. Then, $w\left(f^{\prime }\right)\le w\left(f\right)-1<\lceil {\displaystyle \frac{n}{3}}\rceil -1=\lceil {\displaystyle \frac{n-3}{3}}\rceil \le \lceil {\displaystyle \frac{n^{\prime }}{3}}\rceil$. This contradicts the hypothesis; hence, the statement holds. Now, assume that $dist(u,v)=4$; so, we must have $f\left(u\right)=f\left(v\right)=2$. Contract the shortest path between u and v to a single vertex w, and let the new formed cycle be $C_{n^{\prime }}$. Observe that $n^{\prime }=n-4$; so, $n^{\prime }\ge 4$. Define a veR2DF $f^{\prime }$ on $C_{n^{\prime }}$ such that $f^{\prime }\left(w\right)=2$ and $f^{\prime }\left(x\right)=f\left(x\right)$ for every $x\in V\left(C_{n^{\prime }}\right)$. Thus, $w\left(f^{\prime }\right)=w\left(f\right)-2<\lceil {\displaystyle \frac{n}{3}}\rceil -2=\lceil {\displaystyle \frac{n-6}{3}}\rceil \le \lceil {\displaystyle \frac{n^{\prime }}{3}}\rceil$. This contradicts the hypothesis; hence, the statement holds. □

2. Paths and Trees

Paths and trees are important classes of graphs due to their applications. For example, paths represent connections and routes within a network, and trees are used to model hierarchical data and relationships. In this section, we provide the exact value of the vertex–edge Roman $\left\{2\right\}$-domination number of paths, and we give a tight lower bound and a tight upper bound for the vertex–edge Roman $\left\{2\right\}$-domination number of trees.

Theorem 2. 

Let $n\ge 2.$ Then, ${\gamma }_{veR2}\left(P_n\right)=\lceil {\displaystyle \frac{n}{3}}\rceil$.

Proof. 

First, we show that ${\gamma }_{veR2}\left(P_n\right)\le \lceil {\displaystyle \frac{n}{3}}\rceil$. Let $P_n=v_1{v}_2\dots v_n.$ If $n\equiv 0(mod3),$ define a veR2DF f on $P_n$ such that $f\left(v_i\right)=1$ if $i\equiv 2(mod3),$ and $f\left(v_i\right)=0$ otherwise. Clearly, $w\left(f\right)={\displaystyle \frac{n}{3}}.$ If $n\equiv 1,2(mod3),$ define a veR2DF f on $P_n$ such that $f\left(v_i\right)=1$ if $i\equiv 1(mod3),$ and $f\left(v_i\right)=0$ otherwise. Clearly, $w\left(f\right)=\lceil {\displaystyle \frac{n}{3}}\rceil .$ Thus, ${\gamma }_{veR2}\left(P_n\right)\le \lceil {\displaystyle \frac{n}{3}}\rceil .$

Now, we show by induction on n that ${\gamma }_{veR2}\left(P_n\right)=\lceil {\displaystyle \frac{n}{3}}\rceil .$ If $n=2$, then ${\gamma }_{veR2}\left(P_2\right)=1=\lceil {\displaystyle \frac{2}{3}}\rceil$; if $n=3$, then ${\gamma }_{veR2}\left(P_3\right)=1=\lceil {\displaystyle \frac{3}{3}}\rceil$; if $n=4$, then ${\gamma }_{veR2}\left(P_4\right)=2=\lceil {\displaystyle \frac{4}{3}}\rceil$. This establishes the base step. Let $P_n$ be a path of order $n\ge 5$, and assume that ${\gamma }_{veR2}\left(P_k\right)=\lceil {\displaystyle \frac{k}{3}}\rceil$, where $2\le k<n$. Assume, for the sake of contradiction, that there exists a veR2DF f on $P_n$ with $w\left(f\right)<\lceil {\displaystyle \frac{n}{3}}\rceil$. We check the possibilities of the summation $f\left(v_n\right)+f\left(v_{n-1}\right)+f\left(v_{n-2}\right).$ Clearly, $f\left(v_n\right)+f\left(v_{n-1}\right)+f\left(v_{n-2}\right)\ge 1$.

Case 1.

$f\left(v_n\right)+f\left(v_{n-1}\right)+f\left(v_{n-2}\right)=1$; then, either $f\left(v_n\right)=1, f\left(v_{n-1}\right)=f\left(v_{n-2}\right)=0$ or $f\left(v_{n-1}\right)=1, f\left(v_n\right)=f\left(v_{n-2}\right)=0$. If we are in the first sub-case (i.e., $f\left(v_n\right)=1, f\left(v_{n-1}\right)=f\left(v_{n-2}\right)=0$), then we must have $f\left(v_{n-3}\right)\ge 1$; if we are in the second sub-case, then either $f\left(v_{n-3}\right)\ge 1$ or $f\left(v_{n-4}\right)\ge 1$. By deleting the vertices $v_n,v_{n-1},v_{n-2}$, we obtain the path $P_{n-3}$, and the restriction of f on $V\left(P_{n-3}\right)$ gives a veR2DF $f^{\prime }$ on $P_{n-3}$ such that

$$w\left(f^{\prime }\right)=w\left(f\right)-1<\lceil {\displaystyle \frac{n}{3}}\rceil -1=\lceil {\displaystyle \frac{n-3}{3}}\rceil ,$$

contradicting the hypothesis.

Case 2.

$f\left(v_n\right)+f\left(v_{n-1}\right)+f\left(v_{n-2}\right)\ge 2.$ Assume that either $f\left(v_{n-3}\right)\ge 1$ or $f\left(v_{n-4}\right)\ge 1$. By deleting $v_n,v_{n-1},v_{n-2}$, we obtain the path $P_{n-3}$, and the restriction of f on $V\left(P_{n-3}\right)$ gives a veR2DF $f^{\prime }$ on $P_{n-3}$ such that

$$w\left(f^{\prime }\right)\le w\left(f\right)-2<\lceil {\displaystyle \frac{n}{3}}\rceil -2<\lceil {\displaystyle \frac{n-3}{3}}\rceil ,$$

contradicting the hypothesis. So, we may assume that $f\left(v_{n-3}\right)=f\left(v_{n-4}\right)=0$. By deleting $v_n,v_{n-1},v_{n-2}$, we obtain the path $P_{n-3}$. Define a veR2DF $f^{\prime }$ on $P_{n-3}$ such that $f^{\prime }\left(v_{n-4}\right)=1$, and $f^{\prime }\left(v_i\right)=f\left(v_i\right)$ otherwise. Then,

$$w\left(f^{\prime }\right)\le w\left(f\right)-1<\lceil {\displaystyle \frac{n}{3}}\rceil -1=\lceil {\displaystyle \frac{n-3}{3}}\rceil ,$$

contradicting the hypothesis.

Hence, the statement holds. □

We need a proposition and a lemma before providing a tight lower bound for the vertex–edge Roman $\left\{2\right\}$-domination number of trees.

Proposition 3. 

Let G be a graph of order $n\ge 3$ and θ a veR2DF on G such that $w\left(\theta \right)={\gamma }_{veR2}\left(G\right)$. Then, $L\left(G\right)\cap V_2^{\theta }=\varnothing .$

Proof. 

Assume that G contains a leaf v with $\theta \left(v\right)=2$. Let w be the support vertex adjacent to v. Define a veR2DF ${\theta }^{\prime }$ on G such that ${\theta }^{\prime }\left(v\right)=0, {\theta }^{\prime }\left(w\right)=1,$ and ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for all $x\in V\left(G\right)\setminus \{v,w\}$. Clearly, $w\left({\theta }^{\prime }\right)<w\left(\theta \right)$, which contradicts the hypothesis. □

Lemma 1. 

Let G be a graph of order $n\ge 3$. Then, there exists a veR2DF θ on G such that $w\left(\theta \right)={\gamma }_{veR2}\left(G\right)$, and $L\left(G\right)\subseteq V_0^{\theta }$.

Proof. 

Let f be a veR2DF on G that witnesses ${\gamma }_{veR2}\left(G\right)$. By Proposition 3, $L\left(G\right)\cap V_2^{\theta }=\varnothing .$ If $L\left(G\right)\subseteq V_0^f$, we are finished. Assume that $L\left(G\right)\cap V_1^f\ne \varnothing$. Since $w\left(f\right)={\gamma }_{veR2}\left(G\right),$ then every $v\in L\left(G\right)\cap V_1^f$ is adjacent to a support vertex $w_v$ with $f\left(w_v\right)=0.$ Define a veR2DF $\theta$ on G such that $\theta \left(v\right)=0$ and $\theta \left(w_v\right)=1$ for every $v\in L\left(G\right)\cap V_1^f$, and $\theta \left(x\right)=f\left(x\right)$ otherwise. Clearly, $w\left(\theta \right)\le w\left(f\right)$ and, as $w\left(f\right)={\gamma }_{veR2}\left(G\right)$, it follows that $w\left(\theta \right)={\gamma }_{veR2}\left(G\right)$. □

Theorem 3. 

Let T be a tree of order $n\ge 4$ with l leaves and s support vertices. If T is not a star graph, then ${\gamma }_{veR2}\left(T\right)\ge \lceil {\displaystyle \frac{n-l-s+4}{3}}\rceil$.

Proof. 

We proceed by induction on n. If $diam\left(T\right)=3$ (this includes the case $n=4$), then T is a double star graph. It is clear that ${\gamma }_{veR2}\left(T\right)=2=\lceil {\displaystyle \frac{n-l-s+4}{3}}\rceil$. This establishes the base step.

Assume that $diam\left(T\right)\ge 4$, and assume that the statement holds for every tree $T^{\prime }$ with $4\le |T^{\prime }|<|T|.$ Let $P:=x_1{x}_2\dots x_d$, $d\ge 5$ be a diametral path in T. Among all possible candidates of the diametral path, choose P such that $d\left(x_2\right)$ is the maximum possible. Let $\theta$ be a veR2DF on T.

Assume that T contains a strong support vertex v. Let u and w be two leaves adjacent to v. Let $T^{\prime }$ be the tree obtained from T by deleting w; then, $l\left(T^{\prime }\right)=l-1$, $s\left(T^{\prime }\right)=s$, and $|T^{\prime }|=n-1$. Define a veR2DF ${\theta }^{\prime }$ on $T^{\prime }$ as follows. If $\theta \left(w\right)=0$ or $\theta \left(v\right)\ge 1,$ let ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for every $x\in V\left(T^{\prime }\right)$; otherwise, let ${\theta }^{\prime }\left(v\right)=\theta \left(w\right)$ and ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for every $x\in V\left(T^{\prime }\right)\setminus \left\{v\right\}.$ From the induction hypothesis, we have

$$\begin{array}{cc}\hfill w\left(\theta \right)\ge & w\left({\theta }^{\prime }\right)\hfill \\ \hfill \ge & \lceil {\displaystyle \frac{|T^{\prime }|-s\left(T^{\prime }\right)-l\left(T^{\prime }\right)+4}{3}}\rceil \hfill \\ \hfill =& \lceil {\displaystyle \frac{n-1-s-l+1+4}{3}}\rceil \hfill \\ \hfill =& \lceil {\displaystyle \frac{n-s-l+4}{3}}\rceil ;\hfill \end{array}$$

so, the statement holds.

We may assume that T does not contain a strong support vertex. Then, we must have $d\left(x_2\right)=2$. Assume that $x_3$ is a support vertex; then, $x_3$ is adjacent to a leaf $\alpha$. Let $T^{\prime }$ be the tree obtained from T by deleting $\alpha$. Then, $l\left(T^{\prime }\right)=l-1$, $s\left(T^{\prime }\right)=s-1$, and $|T^{\prime }|=n-1$. Define a veR2DF ${\theta }^{\prime }$ on $T^{\prime }$ as follows. If $\theta \left(\alpha \right)=0$ or $\theta \left(x_3\right)\ge 1$, set ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for every $x\in V\left(T^{\prime }\right)$; otherwise, set ${\theta }^{\prime }\left(x_3\right)=\theta \left(\alpha \right)$ and ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for every $x\in V\left(T^{\prime }\right)\setminus \left\{\alpha \right\}.$ From the induction hypothesis, we have

$$\begin{array}{cc}\hfill w\left(\theta \right)\ge & w\left({\theta }^{\prime }\right)\hfill \\ \hfill \ge & \lceil {\displaystyle \frac{|T^{\prime }|-s\left(T^{\prime }\right)-l\left(T^{\prime }\right)+4}{3}}\rceil \hfill \\ \hfill =& \lceil {\displaystyle \frac{n-1-s+1-l+1+4}{3}}\rceil \hfill \\ \hfill =& \lceil {\displaystyle \frac{n-s-l+5}{3}}\rceil \hfill \\ \hfill \ge & \lceil {\displaystyle \frac{n-s-l+4}{3}}\rceil ;\hfill \end{array}$$

thus, the statement holds. We may assume that $x_3$ is not a support vertex. Assume that $d\left(x_3\right)\ge 3$. Then, $x_3$ is adjacent to a vertex $\beta$ that is different than $x_2$ and $x_4$, and $\beta$ is adjacent to a leaf $\alpha$. Let $T^{\prime }$ be the tree obtained from T by deleting $\alpha$ and $\beta$. Then, $l\left(T^{\prime }\right)=l-1$, and $s\left(T^{\prime }\right)=s-1$. Define a veR2DF ${\theta }^{\prime }$ on $T^{\prime }$ as follows. Let ${\theta }^{\prime }\left(x_3\right)=max\{\theta \left(x_3\right),\theta \left(\alpha \right),\theta \left(\beta \right)\}$. Observe that ${\theta }^{\prime }\left(x_3\right)\ge 1$. Let ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for all $x\in V\left(T^{\prime }\right)\setminus \left\{\alpha \right\}$. From the induction hypothesis, we have

$$\begin{array}{cc}\hfill w\left(\theta \right)\ge & w\left({\theta }^{\prime }\right)\hfill \\ \hfill \ge & \lceil {\displaystyle \frac{|T^{\prime }|-s\left(T^{\prime }\right)-l\left(T^{\prime }\right)+4}{3}}\rceil \hfill \\ \hfill =& \lceil {\displaystyle \frac{n-2-(s-1)-(l-1)+4}{3}}\rceil \hfill \\ \hfill =& \lceil {\displaystyle \frac{n-s-l+4}{3}}\rceil ;\hfill \end{array}$$

thus, the statement holds. We may assume that $d\left(x_3\right)=2$. If T is a path, the statement holds by Theorem 2. Assume that T is not a path; so, T has a leaf $\alpha$ different from $x_1$ and $x_d$. Observe that $\alpha$ is not adjacent to $x_{d-1}$, as T does not have a strong support vertex. Thus, the tree $T^{\prime }:=T-\{x_1,x_2,x_3\}$ has at least four vertices. Observe that $\theta \left(x_1\right)+\theta \left(x_2\right)+\theta \left(x_3\right)\ge 1$, and if $\theta \left(x_1\right)+\theta \left(x_2\right)+\theta \left(x_3\right)=1$, then either $\theta \left(x_1\right)=1, \theta \left(x_2\right)=\theta \left(x_3\right)=0,$ or $\theta \left(x_2\right)=1, \theta \left(x_1\right)=\theta \left(x_3\right)=0$. Define a veR2DF ${\theta }^{\prime }$ on $T^{\prime }$ as follows. If $\theta \left(x_1\right)+\theta \left(x_2\right)+\theta \left(x_3\right)=2$, let ${\theta }^{\prime }\left(x_4\right)=max\{1,\theta \left(x_4\right)\}$, and let ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for all $x\in V\left(T^{\prime }\right)\setminus \left\{x_4\right\}$. If $\theta \left(x_1\right)+\theta \left(x_2\right)+\theta \left(x_3\right)=1$, let ${\theta }^{\prime }\left(x\right)=\theta \left(x\right)$ for all $x\in V\left(T^{\prime }\right).$ From the induction hypothesis, we have

$$\begin{array}{cc}\hfill w\left(\theta \right)\ge & w\left({\theta }^{\prime }\right)+1\hfill \\ \hfill \ge & \lceil {\displaystyle \frac{|T^{\prime }|-s\left(T^{\prime }\right)-l\left(T^{\prime }\right)+4}{3}}\rceil +1\hfill \\ \hfill \ge & \lceil {\displaystyle \frac{n-3-s-l+4}{3}}\rceil +1\hfill \\ \hfill =& \lceil {\displaystyle \frac{n-s-l+4}{3}}\rceil ;\hfill \end{array}$$

thus, the statement holds. □

The bound is tight, as ${\gamma }_{veR2}\left(P_n\right)=\lceil {\displaystyle \frac{n-s-l+4}{3}}\rceil$, $n\ge 4$ (by Theorem 2).

We need to define some rooted trees before moving to the next theorem. Let $T_1:=P_2$, $T_2:=P_3=wxy$, $T_3:=P_4=vwxy$, $T_4:=P_5$, and let the middle vertex be labeled w. Let $T_5$ denote any tree obtained from the star graph $S_n,n\ge 3$ by subdividing each edge once, and denote the vertex in the center by w. Denote the non-leaf vertex in $S_4$ by w; let $T_6$ be the tree obtained from $S_4$ by subdividing two edges once. Let $T_7$ be any tree obtained from $T_5$ by adding a pendant edge to w. Let $T_8:=wxyz$. All the previous trees are rooted at w (see Figure 1).

Theorem 4. 

If T is a tree of order $n\ge 1$, then ${\gamma }_{veR2}\left(T\right)\le \lceil {\displaystyle \frac{3n}{7}}\rceil$.

Proof. 

We proceed by induction on n. If $n\in \{1,2\}$, then $T\in \{P_1,P_2\}$, and it is clear that ${\gamma }_{veR2}\left(T\right)\le 1=\lceil {\displaystyle \frac{3n}{7}}\rceil .$ If $diam\left(T\right)=2$ (this includes the case $n=3),$ then T is a star graph, and ${\gamma }_{veR2}\left(T\right)=1<\lceil {\displaystyle \frac{3n}{7}}\rceil .$ If $diam\left(T\right)=3$, then T is a double star graph, and $n\ge 4$. It is clear that ${\gamma }_{veR2}\left(T\right)=2\le \lceil {\displaystyle \frac{3n}{7}}\rceil .$ If $diam\left(T\right)=4$, there exists a vertex $v\in V\left(T\right)$ such that, for every $u\in V\left(T\right),$$dist(v,u)\le 2$. Define a veR2DF $\theta$ on T, such that $\theta \left(v\right)=2$, and $\theta \left(x\right)=0$, for all $x\in V\left(T\right)\setminus \left\{v\right\}.$ So, ${\gamma }_{veR2}\left(T\right)=2<\lceil {\displaystyle \frac{3n}{7}}\rceil$. This establishes the base step. We may assume that T is a tree with $diam\left(T\right)\ge 5$, and if $T^{\prime }$ is a tree with $1\le |T^{\prime }|<|T|$, then ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3|T^{\prime }|}{7}}\rceil .$ Throughout the proof, we use the symbol ${\theta }^{\prime }$ to denote a fixed veR2DF on $T^{\prime }$ that witnesses ${\gamma }_{veR2}\left(T^{\prime }\right)$ and satisfies the condition $L\left(T^{\prime }\right)\subseteq V_0^{{\theta }^{\prime }}$ (see Lemma 1).

Claim 1. 

If T contains a strong support vertex v, then the statement holds.

Proof. 

Let u and w be two leaves adjacent to v. Let $T^{\prime }$ be the tree obtained from T by deleting u. From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-1)}{7}}\rceil .$ Let ${\theta }^{\prime }$ be a veR2DF on $T^{\prime }$ with minimum weight and $L\left(T^{\prime }\right)\subseteq V_0^{{\theta }^{\prime }}$ (see Lemma 1); so, $w\left({\theta }^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-1)}{7}}\rceil .$ Observe that ${\theta }^{\prime }\left(w\right)=0$ as $w\in L\left(T^{\prime }\right)$, so either ${\theta }^{\prime }\left(v\right)\ge 1$ or ${\sum }_{a\in N_{T^{\prime }}\left(v\right)}{\theta }^{\prime }\left(a\right)\ge 2$. Define a veR2DF $\theta$ on T such that $\theta \left(u\right)=0$ and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus \left\{u\right\}$. Clearly, $w\left(\theta \right)=w\left({\theta }^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-1)}{7}}\rceil \le \lceil {\displaystyle \frac{3n}{7}}\rceil$, and we are finished. □

We may assume that T does not have a strong support vertex. Let $P:=x_1{x}_2\dots x_d$, $d\ge 6$ be a diametral path in T. Choose the diametral path P such that $d\left(x_3\right)$ is the maximum possible. Observe that $d\left(x_2\right)=2$ as T does not have a strong support vertex. Let $x_d$ be the root of the tree $T.$

Claim 2. 

If$|T_{x_3}|\ge 5$, then the statement holds.

Proof. 

Let $T^{\prime }$ be the tree obtained from T by deleting $x_3$ and all its successors (i.e., $T^{\prime }$ is the tree obtained from T by deleting $V\left(T_{x_3}\right)$). From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil .$ Let ${\theta }^{\prime }$ be a veR2DF on $T^{\prime }$ with minimum weight; so, $w\left({\theta }^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil .$ Define a veR2DF $\theta$ on T such that $\theta \left(x_3\right)=2$ and $\theta \left(x\right)=0$ for all the successors of $x_3$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_3}\right)$. Clearly, $w\left(\theta \right)=w\left({\theta }^{\prime }\right)+2\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil +2\le \lceil {\displaystyle \frac{3n}{7}}\rceil$, and we are finished. □

We may assume that $|T_{x_3}|\le 4$; so, either $d\left(x_3\right)=2$ or $d\left(x_3\right)=3$.

Claim 3. 

If $d\left(x_3\right)=3$, then the statement holds.

Proof. 

The neighbors of $x_3$ are $x_2$, $x_4$, and a vertex v. The vertex v must be a leaf as $|T_{x_3}|\le 4$. Assume that $d\left(x_4\right)=2$. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_4}\right)$. From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil .$ Define a veR2DF $\theta$ on T such that $\theta \left(x_3\right)=2$, $\theta \left(x_4\right)=0$, and $\theta \left(x\right)=0$ for all the successors x of $x_3$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_4}\right)$. Clearly, $w\left(\theta \right)=w\left({\theta }^{\prime }\right)+2\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil +2\le \lceil {\displaystyle \frac{3n}{7}}\rceil$, and we are finished.

We may assume now that $d\left(x_4\right)\ge 3$. Assume that $x_4$ is adjacent to a leaf w. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_3}\right)\cup \left\{w\right\}.$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil .$ Define a veR2DF $\theta$ on T such that $\theta \left(x_3\right)=2$, $\theta \left(w\right)=0$, and $\theta \left(x\right)=0$ for all the successors x of $x_3$. Thus, $w\left(\theta \right)=w\left({\theta }^{\prime }\right)+2\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil +2\le \lceil {\displaystyle \frac{3n}{7}}\rceil$, and we are finished. We may assume that $x_4$ is not adjacent to a leaf. Then, for every $w\in N\left(x_4\right)\setminus \left\{x_5\right\}$, $T_w=T_i$ for some $i\in \{1,2,3,4\}$. Let $a_i:=\left|\right\{w\in N\left(x_4\right)\setminus \left\{x_5\right\}|T_w=T_i\left\}\right|$, where $i\in \{1,2,3,4\}$ (as an example, the graph H in Figure 2, $T_{x_3}=T_3,$$T_a=T_b=T_1,$ and $T_c=T_2$; so, $a_1=2,$ $a_2=1$, $a_3=1$, and $a_4=0$). Observe that $a_3\ge 1$ as $x_3\in N\left(x_4\right)$, and $T_{x_3}=T_3.$ Observe also that $a_1+a_2+a_3+a_4\ge 2$ as $d\left(x_4\right)\ge 3$. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_4}\right)$. From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-(2a_1+3a_2+4a_3+5a_4))}{7}}\rceil .$ Assume that $a_1\ne 0$. Define a veR2DF $\theta$ on T such that $\theta \left(x_2\right)=1, \theta \left(x_4\right)=2$, and $\theta \left(w\right)=2$, if $T_w=T_4$, $\theta \left(x\right)=1$, if $T_w\in \{T_2,T_3\}$, $\theta \left(x\right)=0$, for all other successors of $x_4$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_4}\right)$. Thus,

$$\begin{array}{cc}\hfill w\left(\theta \right)=& w\left({\theta }^{\prime }\right)+2+a_2+a_3+2a_4\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3(n-(1+2a_1+3a_2+4a_3+5a_4))+7(2+a_2+a_3+2a_4)}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n-6a_1-2a_2-5a_3-a_4+11}{7}}\rceil \hfill \\ \hfill \underset{(\ast )}{\le }& \lceil {\displaystyle \frac{3n-2a_2-a_4}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n}{7}}\rceil .\hfill \end{array}$$

Inequality $(\ast )$ comes from the fact that $a_1\ge 1$ and $a_3\ge 1$.

Assume that $a_1=0$. Define a veR2DF $\theta$ on T such that $\theta \left(x_2\right)=1, \theta \left(x_4\right)=1$, and $\theta \left(w\right)=2$, if $T_w=T_4$, $\theta \left(x\right)=1$, if $T_w\in \{T_2,T_3\}$, $\theta \left(x\right)=0$, for all other successors of $x_4$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_4}\right)$. Thus,

$$\begin{array}{cc}\hfill w\left(\theta \right)=& w\left({\theta }^{\prime }\right)+1+a_2+a_3+2a_4\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3(n-(1+3a_2+4a_3+5a_4))+7(1+a_2+a_3+2a_4)}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n-2a_2-5a_3-a_4+4}{7}}\rceil \hfill \\ \hfill \underset{(\ast )}{\le }& \lceil {\displaystyle \frac{3n-2a_2-a_4-1}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n}{7}}\rceil .\hfill \end{array}$$

Inequality $(\ast )$ comes from the fact that $a_3\ge 1$. □

We may assume that $d\left(x_3\right)=2$.

Claim 4. 

If$d\left(x_4\right)\ge 3,$then the statement holds.

Proof. 

Assume that $x_4$ is adjacent to a leaf w. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_3}\right).$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-3)}{7}}\rceil .$ Observe that ${\theta }^{\prime }\left(w\right)=0$, and, if ${\theta }^{\prime }\left(x_4\right)=0,$ then ${\sum }_{v\in N\left(x_4\right)}{\theta }^{\prime }\left(v\right)\ge 2$. Define a veR2DF $\theta$ on T such that $\theta \left(x_1\right)=\theta \left(x_3\right)=0$, $\theta \left(x_2\right)=1$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_3}\right).$ Thus,

$$\begin{array}{cc}\hfill w\left(\theta \right)=& w\left({\theta }^{\prime }\right)+1\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3(n-3)}{7}}\rceil +1\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n-9+7}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n-2}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n}{7}}\rceil ,\hfill \end{array}$$

and we are finished. So, we may assume that $x_4$ is not adjacent to a leaf.

Assume that $x_4$ is adjacent to a support vertex w. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_3}\right).$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-3)}{7}}\rceil .$ Observe that, if ${\theta }^{\prime }\left(x_4\right)=0$, then we must have ${\theta }^{\prime }\left(w\right)\ge 1$. Define a veR2DF $\theta$ on T such that $\theta \left(x_1\right)=\theta \left(x_3\right)=0$, $\theta \left(x_2\right)=1$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_3}\right).$ Thus, $w\left(\theta \right)=w\left({\theta }^{\prime }\right)+1\le \lceil {\displaystyle \frac{3(n-3)}{7}}\rceil +1\le \lceil {\displaystyle \frac{3n}{7}}\rceil ,$ and we are finished. So, we may assume that $x_4$ is not adjacent to a support vertex.

Let $v\in N\left(x_4\right)\setminus \left\{x_5\right\}$. Since v is neither a leaf nor a support vertex, we must have $diam\left(T_v\right)\ge 2$. From the way that $x_3$ was chosen, we must have $d\left(v\right)=2$. So, $T_v=P_3$; see Figure 3.

Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_4}\right).$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-(3a+1\left)\right)}{7}}\rceil$, where $a=d\left(x_4\right)-1$. Define a veR2DF $\theta$ on T such that $\theta \left(x_4\right)=1,$$\theta \left(x\right)=1$ for every support vertex in $T_{x_4}$, $\theta \left(x\right)=0$ for the remaining vertices in $T_{x_4}$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_4}\right).$ Thus,

$$\begin{array}{cc}\hfill w\left(\theta \right)=& w\left({\theta }^{\prime }\right)+a+1\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3(n-(3a+1\left)\right)}{7}}\rceil +a+1\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n-9a-3+7a+7}{7}}\rceil \hfill \\ \hfill \underset{(\ast )}{\le }& \lceil {\displaystyle \frac{3n-2a+4}{7}}\rceil \hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n}{7}}\rceil ,\hfill \end{array}$$

where inequality $(\ast )$ comes from the assumption $d\left(x_4\right)\ge 3$ (so $a\ge 2)$. Thus, the statement holds. □

Claim 5. 

If$d\left(x_4\right)=2$, then the statement holds.

Proof. 

Assume that $d\left(x_5\right)=2$. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_5}\right).$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil .$ Define a veR2DF $\theta$ on T such that $\theta \left(x_2\right)=\theta \left(x_5\right)=1,$ $\theta \left(x_1\right)=\theta \left(x_3\right)=\theta \left(x_4\right)=0$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_5}\right).$ Thus, $w\left(\theta \right)=w\left({\theta }^{\prime }\right)+2\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil +2\le \lceil {\displaystyle \frac{3n}{7}}\rceil ,$ and we are finished. So, we may assume that $d\left(x_5\right)\ge 3$.

Assume that $x_5$ is adjacent to a leaf w. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_4}\right)\cup \left\{w\right\}.$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil .$ Define a veR2DF $\theta$ on T such that $\theta \left(x_2\right)=1, \theta \left(x_5\right)=max\{1,{\theta }^{\prime }\left(x_5\right)\},$ $\theta \left(x_1\right)=\theta \left(x_3\right)=\theta \left(x_4\right)=\theta \left(w\right)=0$, and $\theta \left(x\right)={\theta }^{\prime }\left(x\right)$ for all $x\in V\left(T\right)\setminus V\left(T_{x_5}\right).$ Thus, $w\left(\theta \right)\le w\left({\theta }^{\prime }\right)+2\le \lceil {\displaystyle \frac{3(n-5)}{7}}\rceil +2\le \lceil {\displaystyle \frac{3n}{7}}\rceil ,$ and we are finished. So, we may assume that $x_5$ is not adjacent to a leaf.

For every $w\in N\left(x_5\right)\setminus \left\{x_6\right\}$, $T_w=T_i$ for some $i\in \{1,2,3,4,5,6,7,8\}$. Let $a_i:=\left|\right\{w\in N\left(x_5\right)\setminus \left\{x_6\right\}|T_w=T_i\left\}\right|$, where $i\in \{1,2,3,4,5,6,7,8\}$. Observe that $a_8\ge 1$ as $T_{x_4}=T_8$, and ${\sum }_{i\in \{1,2,3,4,5,6,7,8\}}{a}_i\ge 2$ as $d\left(x_5\right)\ge 3$. Let $T^{\prime }$ be the tree obtained from T by deleting $V\left(T_{x_5}\right).$ From the induction hypothesis, ${\gamma }_{veR2}\left(T^{\prime }\right)\le \lceil {\displaystyle \frac{3(n-r)}{7}}\rceil ,$ where $r=1+2a_1+3a_2+4(a_3+a_8)+5(a_4+a_5+a_6+a_7)$. If $a_1\ge 1$, define a veR2DF $\theta$ on T such that $\theta \left(x_2\right)=1,$ $\theta \left(x_5\right)=2$, $\theta \left(y\right)=1$, if $y\in V\left(T_2\right)\cup V\left(T_8\right)$, $\theta \left(x\right)=1$, if $x\in V\left(T_3\right)$ (see Figure 1), $\theta \left(w\right)=2$, if $T_w\in \{T_4,T_5,T_6,T_7\}$, $\theta \left(v\right)=0$ for the remaining vertices in $V\left(T_{x_5}\right)$, and $\theta \left(v\right)={\theta }^{\prime }\left(v\right)$ for every $v\in V\left(T\right)\setminus V\left(T_{x_5}\right).$ Thus,

$$\begin{array}{cc}\hfill w\left(\theta \right)=& w\left({\theta }^{\prime }\right)+2+a_2+a_3+a_8+2(a_4+a_5+a_6+a_7)\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3(n-r)}{7}}\rceil +2+a_2+a_3+a_8+2(a_4+a_5+a_6+a_7)\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n+11-6a_1-2a_2-5a_3-a_4-a_5-a_6-a_7-5a_8}{7}}\rceil \hfill \\ \hfill \underset{(\ast )}{\le }& \lceil {\displaystyle \frac{3n}{7}}\rceil ,\hfill \end{array}$$

where inequality $(\ast )$ comes from the fact that $a_1,a_8\ge 1$. If $a_1=0$, define a veR2DF $\theta$ on T such that $\theta \left(x_2\right)=1,$ $\theta \left(x_5\right)=1$, $\theta \left(y\right)=1$ if $y\in V\left(T_2\right)\cup V\left(T_8\right)$, $\theta \left(x\right)=1$ if $x\in V\left(T_3\right)$, $\theta \left(w\right)=2$ if $T_w\in \{T_4,T_5,T_6,T_7\}$, $\theta \left(v\right)=0$ for the remaining vertices in $V\left(T_{x_5}\right)$, and $\theta \left(v\right)={\theta }^{\prime }\left(v\right)$ for every $v\in V\left(T\right)\setminus V\left(T_{x_5}\right).$ Thus,

$$\begin{array}{cc}\hfill w\left(\theta \right)=& w\left({\theta }^{\prime }\right)+1+a_2+a_3+a_8+2(a_4+a_5+a_6+a_7)\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3(n-r)}{7}}\rceil +1+a_2+a_3+a_8+2(a_4+a_5+a_6+a_7)\hfill \\ \hfill \le & \lceil {\displaystyle \frac{3n+4-2a_2-5a_3-a_4-a_5-a_6-a_7-5a_8}{7}}\rceil \hfill \\ \hfill \underset{(\ast )}{\le }& \lceil {\displaystyle \frac{3n}{7}}\rceil ,\hfill \end{array}$$

where inequality $(\ast )$ comes from the fact $a_8\ge 1$. Thus, the statement holds. □

This completes the proof of the theorem.

The bound in Theorem 4 is tight as the following theorem indicates.

Theorem 5. 

Let n be a positive integer such that $n\equiv 0(mod7)$. Then, there exists a tree T of order n such that ${\gamma }_{veR2}\left(T\right)=\lceil {\displaystyle \frac{3n}{7}}\rceil ={\displaystyle \frac{3n}{7}}$.

Proof. 

As $n\equiv 0(mod7)$, there exists a positive integer m such that $n=7m$. For every $i\in \{1,\dots ,m\},$ let $P_{i,7}:=a_i{b}_i{c}_i{d}_i{x}_i{y}_i{z}_i$; so, $P_{i,7}$ is a path with seven vertices. Let $T^m$ be the tree obtained from the disjoint union of $P_{i,7}$, $i\in \{1,\dots ,m\},$ by adding the set of edges $\left\{c_1{c}_j\right|2\le j\le m\}$. The tree $T^5$ is given in Figure 4. From Lemma 1, there exists a veR2DF $\theta$ on $T^m$ such that $w\left(\theta \right)={\gamma }_{veR2}\left(T^m\right)$, and $L\left(T^m\right)\subseteq V_0^{\theta }$. This means that $\theta \left(z_i\right)=\theta \left(a_i\right)=0$ for every $i\in \{1,\dots ,m\}$; so, either $\theta \left(y_i\right)=1$ or $\theta \left(x_i\right)=2$. Similarly, either $\theta \left(b_i\right)=1$ or $\theta \left(c_i\right)=2$. If $\theta \left(y_i\right)=1$ and $\theta \left(b_i\right)=1$, then we must have $\theta \left(c_i\right)+\theta \left(d_i\right)+\theta \left(x_i\right)\ge 1$. So, $\theta \left(P_{i,7}\right)\ge 3$ for every $i\in \{1,\dots ,m\}$. Thus, $w\left(\theta \right)\ge 3m={\displaystyle \frac{3n}{7}}$. From Theorem 4, we must have $w\left(\theta \right)={\displaystyle \frac{3n}{7}}$, as required. □

3. Complexity

Graph complexity is crucial for understanding and analyzing various real-world systems represented as graphs. Understanding graph complexity helps in designing efficient algorithms and classifying the difficulty of computational problems. In this section, we show that vertex–edge Roman $\left\{2\right\}$-domination is NP-complete even for bipartite graphs.

Consider the following decision problem.

veR2D

• Instance: A graph $G=(V,E)$ and $k\in {\mathbb{Z}}^{+}$ such that $k\le \left|V\right|$.
• Question: Is there a veR2DF f with $w\left(f\right)\le k$?

We transform an NP-complete problem, called Exact 3-Cover (X3C) [12], to veR2D.

X3C

• Instance: A set A of cardinality $3q$ and a collection $\mathcal{B}$ of 3-element subsets of A.
• Question: Does $\mathcal{B}$ contain a sub-collection ${\mathcal{B}}^{\prime }$ such that every $a\in A$ is in exactly one element in ${\mathcal{B}}^{\prime }$?

Theorem 6. 

veR2D is NP-complete for bipartite graphs.

Proof. 

It is clear that veR2D is in the NP class. Let $(A,\mathcal{B})$, where $A=\{a_1,\dots ,a_{3q}\}$ and $\mathcal{B}=\{B_1,\dots ,B_t\}$ is an instance of X3C. For every $i\in \left[3q\right]$, let $P_i:=x_i{y}_i{z}_i{a}_i$. Let $\mathcal{P}$ be the disjoint union of $P_i$’s, $i\in \left[3q\right]$. For every $i\in \left[t\right]$, let $Q_i:=b_i{c}_i{d}_i$. Let $\mathcal{Q}$ be the disjoint union of $Q_i$’s, $i\in \left[t\right]$. Let G be the graph obtained from the disjoint union of $\mathcal{P}$ and $\mathcal{Q}$ by adding the edges $a_i{b}_j$ if $a_i\in B_j$. Clearly, G is a bipartite graph; see Figure 5. Set $k=4q+t$.

Assume that $(A,\mathcal{B})$ has an exact 3-cover ${\mathcal{B}}^{\prime }$. Define a veR2DF f as follows. For every $i\in \left[3q\right]$, let $f\left(x_i\right)=f\left(z_i\right)=f\left(a_i\right)=0$, and $f\left(y_i\right)=1$; for every $i\in \left[t\right]$, let $f\left(c_i\right)=1, f\left(d_i\right)=0$ $f\left(b_i\right)=1$ if $B_i\in {\mathcal{B}}^{\prime }$, and $f\left(b_i\right)=0$ if $B_i\notin {\mathcal{B}}^{\prime }$. As ${\mathcal{B}}^{\prime }$ is an exact 3-cover, the edges $z_i{a}_i^{\prime }s$ and the edges $a_i{b}_j^{\prime }s$, where $i\in \left[3q\right]$ and $j\in \left[t\right],$ are f-dominated. The remaining edges are obviously f-dominated; so, f is a veR2DF with $w\left(f\right)=k$.

Conversely, assume that G admits a veR2DF f with $w\left(f\right)\le k$. Clearly, $f\left(P_i\right)\ge 1$ for every $i\in \left[3q\right]$; so, $f\left(\mathcal{P}\right)\ge 3q$, and if $f\left(P_i\right)=1$, then $f\left(x_i\right)=f\left(z_i\right)=f\left(a_i\right)=0$, and $f\left(y_i\right)=1$. Clearly, $f\left(Q_i\right)\ge 1$ for every $i\in \left[t\right]$; so, $f\left(\mathcal{Q}\right)\ge t$, and if $f\left(Q_i\right)=1$ for some $i\in \left[t\right]$, then $f\left(b_i\right)=0$. Let $r=\left|\right\{i\in \left[3q\right]\mid f\left(P_i\right)\ge 2\left\}\right|$, and $s=\left|\right\{i\in \left[t\right]\mid f\left(b_i\right)\ge 1\left\}\right|$. Then, $w\left(f\right)\ge 3q+r+t+s$. As $w\left(f\right)\le 4q+t$, $r+s\le q$. On the other hand, $s\ge {\displaystyle \frac{3q-r}{3}}$. So, $r=0$ and $s=q$. Hence, the set $\{B_i\mid f\left(b_i\right)\ge 1\}$ is an exact 3-cover. □

4. Conclusions

In this work, we introduced a variant of domination called vertex–edge Roman $\left\{2\right\}$-domination. The majority of the articles considering domination sets have studied vertex–vertex domination or edge–edge domination; however, in this work, vertices dominate edges. This type of domination can be applied to any real-life network where monitoring edges by nodes is required, for example, in the context of monitoring highway safety. In this domination, each vertex is assigned an integer between 0 and 2 such that, if there is an edge with endpoints each assigned 0, then the sum of the vertex weights on the neighborhood of those two endpoints is greater than or equal to 2. This new notion is related to a vertex–vertex domination called Roman $\left\{2\right\}$-domination (also called Italian domination). Vertex–edge Roman $\left\{2\right\}$-domination relaxes the condition of Roman $\left\{2\right\}$-domination; thus, the vertex–edge Roman $\left\{2\right\}$-domination number of a graph G is a lower bound for the Roman $\left\{2\right\}$-domination number of G. This new domination also adds an extra condition to a domination introduced in 1986, called vertex–edge domination; therefore, the vertex–edge Roman $\left\{2\right\}$-domination number of a graph G is an upper bound for the vertex–edge domination number of G. Vertex–edge Roman $\left\{2\right\}$-domination also relaxes the condition of vertex–edge Roman domination; as such, the vertex–edge Roman $\left\{2\right\}$-domination number of a graph G is a lower bound of the vertex–edge Roman domination number. In real-life contexts, adding an extra condition to a domination variant usually results in better service but with a higher cost; in contrast, relaxing the conditions of a domination variant reduces the cost and reduces the quality of service as well. Thus, vertex–edge Roman $\left\{2\right\}$-domination balances the quality of service and the cost between vertex–edge domination and vertex–edge Roman domination. In [13], the authors proved that, for any tree T of order $n\ge 3$, ${\gamma }_{ve}\left(T\right)\le {\displaystyle \frac{n}{3}}$; furthermore, they proved that this bound is tight. In our work, we showed that ${\gamma }_{ve}\left(T\right)\le \lceil {\displaystyle \frac{3n}{7}}\rceil$ and this bound is tight; thus, as mentioned earlier, vertex–edge Roman $\left\{2\right\}$-domination costs more than vertex–edge domination. In [14], the authors proved that, if T is a tree of order n with l leaves and s support vertices and $dim\left(T\right)\ge 3$, then ${\gamma }_{veR}\left(T\right)\ge {\displaystyle \frac{n-l-s+3}{2}}$, and this bound is tight. In our work, we proved that if T is a tree of order $n\ge 4$ and $dim\left(T\right)\ge 3$, then ${\gamma }_{veR2}\left(T\right)\ge \lceil {\displaystyle \frac{n-l-s+4}{3}}\rceil$, and this bound is tight. Therefore, the cost of vertex–edge Roman $\left\{2\right\}$-domination is lower than vertex–edge Roman domination.

In this work, we determined the exact vertex–edge Roman $\left\{2\right\}$-domination number for cycles and paths and provided tight lower and upper bounds for the vertex–edge Roman $\left\{2\right\}$-domination number of trees. In addition, we proved that the decision problem associated with vertex–edge Roman $\left\{2\right\}$-domination is NP-complete for bipartite graphs.

In future work, we aim to characterize graphs G such that ${\gamma }_{veR2}\left(G\right)={\gamma }_{veR}\left(G\right)$, as well as characterize graphs such that ${\gamma }_{veR2}\left(G\right)={\gamma }_{R2}\left(G\right)$. We also aim to investigate the vertex–edge Roman $\left\{2\right\}$-domination number for other classes of graphs, for example, cubic graphs.