Abstract

A double Roman dominating function on a graph is a function satisfying the conditions that every vertex for which is adjacent to at least one vertex for which or two vertices and for which and every vertex for which is adjacent to at least one vertex for which . The weight of a double Roman dominating function is the value . The minimum weight of a double Roman dominating function on a graph is called the double Roman domination number of . A graph with is called a double Roman graph. In this paper, we study properties of double Roman domination in graphs. Moreover, we find a class of double Roman graphs and give characterizations of trees with for .

1. Introduction

In this paper, we shall only consider graphs without multiple edges or loops. Let be a graph, , and the neighborhood of in is denoted by . That is to say, . The closed neighborhood of in is defined as . The complementary graph of is denoted by . A vertex of degree one is called a leaf. A graph is trivial if it has a single vertex. The degree of a vertex is denoted by , i.e., . Denote by , , and the complete graph, path, and cycle on vertices, respectively. The maximum degree and the minimum degree of a graph are denoted by and , respectively. For a set , the graph induced by is denoted by . Let , and we denote by the graph obtained from by contracting the edge . For an edge , we denote by the graph obtained from by deleting .

A subset of the vertex set of a graph is a dominating set if every vertex not in has at least one neighbour in . The domination number is the minimum cardinality of a dominating set of .

The domination and its variations of graphs have attracted considerable attention [1, 2]. Many varieties of dominating sets are listed in the book Fundamentals of Domination in Graphs [3]. However, Roman domination and double Roman domination are not listed in this book. Roman domination and double Roman domination appear to be a new variety of interest [4–11].

A Roman dominating function (RDF) of a graph is a function such that every vertex for which is adjacent to at least one vertex for which . The weight of a Roman dominating function is the value . The minimum weight of a Roman dominating function on a graph is called the Roman domination number of . An RDF of with is called a function.

A double Roman dominating function (DRDF) on a graph is a function satisfying the condition that every vertex for which is adjacent to at least one vertex for which or two vertices and for which and every vertex for which is adjacent to at least one vertex for which . The weight of a double Roman dominating function is the value . The minimum weight of a double Roman dominating function on a graph is called the double Roman domination number of . A DRDF of with is called a function. We denote by the weight of a double Roman dominating function in , i.e., .

Beeler et al. [12] initiated the study of the double Roman domination in graphs. They showed that and defined a graph to be double Roman if . Moreover, they suggest to find double Roman graphs.

In this paper, we study properties of double Roman domination in graphs and show that the double Roman domination problem is NP-complete for bipartite graphs. Moreover, we find a class of double Roman graphs and give characterizations of trees with for .

2. Properties of Double Roman Domination

Proposition 1 (see [12]). In a double Roman dominating function of weight , no vertex needs to be assigned the value 1.

By Proposition 1, when we consider a function, we assume no vertex has been assigned the value 1.

Proposition 2 (see [12]).  (i)Let be a graph and be a function. Then, .(ii)For any graph G, with equality if and only if .

Proposition 3 (see [12]).  (i)For every graph , .(ii)If is any function, then .

The following result is immediate.

Proposition 4. For any graph , .

Proof. The desired inequality obviously holds if . In order to prove the proposition for , we introduce the discharging approach. Let be a function. The initial charge of every vertex is set to be . We apply the discharging procedure defined by applying the following rule.
For each vertex , we send charge to each adjacent vertex in . Then, the final charge of is satisfying with .
For each vertex , we send charge to each adjacent vertex in . Then, the final charge of is satisfying with .
For each vertex , by the definition of double Roman domination, has a neighbor assigned 3 or two vertices assigned 2. Due to the discharging rule above, if has a neighbor assigned 3, receives charge from . We have .
If has two vertices assigned 2, receives charge from . We have . Thus, . The proof is complete.

Proposition 5. Let be a graph. If and is a function,(i)then  (ii)if , then and there exists a vertex with degree .

Proof.  (i)By Proposition 3, we have . So we have .(ii)If , let and . We have the following claim.

Claim 1. is empty.

Proof. Otherwise, any vertex in assigned with 0 has at least two vertices assigned with 2. Let be a vertex in assigned with 2, we consider a function with , , and for any . Then, is an RDF of with weight . So , a contradiction.
Since is empty, we have as a vertex with degree . Now, we have and , and so the result holds.

Theorem 1. For every graph on vertices without isolated vertex, .

Proof. Clearly, we have . We select a subset of , where each vertex is selected with probability independently. Let , we consider a function with for , for , and for other vertex . Then, is a DRDF of . We have . When we consider the expectation, we also have . First, it is clear that . For each vertex with degree , if neither nor any neighbor is selected, then . So we have , and thus, . Consequently, . Let . By , the maximum of is given by at . Thus, .
If is a graph with some isolated vertices, then . Let be the set of isolated vertices of and let . Therefore, . Because all isolated vertices must be assigned 3, it is easy to prove that , where .

Proposition 6. If is a connected graph of order , then if and only if there exists a vertex of degree in .

Proof.   () Let be a function with minimum . Then, we have being independent. Together with Proposition 2, we have . Then, we have . If , we have , and thus . Since is independent, it is impossible. If , we have . Let , , and . Then, we have and so . () By Proposition 2 (ii), we have for a connected graph . Assume contains a vertex of degree in . Let , , and . Then, is a function and so . Hence, .Let be the family of connected graphs such that for any function , we have .

Proposition 7. Let , then(i) contains no strong support vertex(ii)if for an edge , then and .

Proof.  (i)Suppose be a strong support vertex, then there exist two leaves . Since , we have . Now consider the function with for any , , and for any . Then, is a DRDF of with fewer weight than , a contradiction.(ii)If for an edge , we have as also a DRDF of . So . Since for any graph , we have , so . Suppose to the contrary that there exists a function such that for a vertex . Then, is also a function, contradicting with . For the graph , the proof is similar.

Lemma 1. Let be a graph on vertices, then if and only if contains a complete bipartite graph as a subgraph and .

Proof.    If , then no vertex is assigned with 3 and thus we have two vertices assigned with 2 and the others 0. Also, each vertex 0 must be adjacent to both and . Therefore, contains a complete bipartite graph as a subgraph. Since , we have .  If contains a complete bipartite graph with partitions () as a subgraph and . Then, let for any and for . Then, is a DRDF of and so . Since contains no vertex with degree , we have .Note that if contains a complete bipartite graph as a subgraph. Thus, can be replaced with in the lemma.

Theorem 2. Let be a graph on vertices, then . Furthermore, equality holds in the upper bound if or is .

Proof. If is a graph on vertices, we have , and if , then has a vertex with degree . But its complement is neither a star nor a graph with (see the graph stated in Lemma 1). So we have and thus, . If is a star, then and so the lower bound is attainable.
Let be a vertex with maximum degree ; consider a function with , for any , and for . Then, is a DRDF of and so . Since , we have . Therefore, . It can be seen that if , then . Hence, is -regular for some . By symmetry, we may assume that . Then, if , we have and . Let . If for some , then is a DRDF of with fewer weight than , a contradiction. Therefore, each vertex not in has at least neighbors in . Analogously, each vertex in has at most 2 neighbors outside . Then, we have . Since , we have . If , then . This is impossible. If , then . If , we have , and so , a contradiction. If , we have , and so , a contradiction. If , we have , and so , a contradiction. If , then , and we have and , and so , a contradiction. Therefore, we conclude that . If , then and thus the upper bound is attainable.

3. Some Double Roman Graphs

The Cartesian product of graphs and is the graph with vertex set and whenever and , or and . The Cartesian product is commutative and associative, having the trivial graph as a unit (cf. [13]).

Let be a double Roman dominating function of , and we write for . When no confusion arise, we simply write as . We use to denote the value for . Let be the weight of , i.e., .

Theorem 3. Let . Then, the Cartesian product graphs are double Roman.

Proof. The lower bound follows from Proposition 4. Let , , , , , , , , , and . Then, and so is a DRDF of with weight and we have . Since we have is double Roman.

4. Trees with

Theorem 4. If is a tree, then if and only if is a star for .

Proof.    If is a star for , it is clear that and , and the theorem holds.  By Proposition 3, we have . If , by Proposition 5, we have is a star for some . If , then each vertex in is assigned 2 or 0. Since is a tree, has a least two leaves and . If are adjacent to the same vertex , then we can obtain a DRDF of with fewer weight by changing to 3 and to 0 and obtaining a contradiction. If are adjacent to different vertices, we consider a function with , , and for any . Then, is an RDF of with weight . So , a contradiction.For a positive integer , a wounded spider is a star with at most of its edges subdivided. In a wounded spider, a vertex of degree t will be called the head vertex, and the vertices at distance two from the head vertex will be the foot vertices.

Theorem 5. If is a tree, then if and only if is a wounded spider with only one foot or is obtained by adding an edge between two stars and for .

Proof.    If is a wounded spider with only one foot, it is clear that and , and the theorem holds. If is a tree obtained by adding an edge between two stars and for , then and , and the theorem holds.  By Proposition 3, we have . If , let and . Similar to the proof of Theorem 5, we have as empty. Otherwise, there exists a vertex in assigned with 2. Now, we change the function values of from to , respectively, and obtain an RDF of with weight , a contradiction. In this case, is a tree obtained by adding an edge between two stars and for . If , let and . Then, has at most one connected component. Otherwise, we can make a vertex in each connected component to change the function values including the vertex to obtain an RDF of with weight . Since contains no vertex assigned with 3, then is not a star with at least two leaves. We claim that the leaves of are at most two. Otherwise, we change the function values of and choose two leaves to obtain an RDF of with weight . Therefore, is a path on at least four vertices. In this case, we can obtain an RDF of with weight at most , a contradiction. Therefore, is a wounded spider with only one foot.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work was supported by Sichuan Science and Technology Program under grant 2018ZR0265, Sichuan Military and Civilian Integration Strategy Research Center under grant JMRH-1818, and Sichuan Provincial Department of Education (Key Project) under grant 18ZA0118.