Domination related parameters in rooted product graphs
Abstract
A set of vertices of a graph is a dominating set in if every vertex outside of is adjacent to at least one vertex belonging to . A domination parameter of is related to those sets of vertices of a graph satisfying some domination property together with other conditions on the vertices of . Here, we investigate several domination related parameters in rooted product graphs.
Keywords: Domination; Roman domination; domination related parameters; rooted product graphs.
AMS Subject Classification Numbers: 05C12; 05C76.
1 Introduction
Domination in graph constitutes a very important area in graph theory [9]. An enormous quantity of researches about domination in graphs have been developed in the last years. Nevertheless, there are still several open problems and incoming researches about that. One interesting question in this area is related to the study of domination related parameters in product graphs. For instance, the Vizing’s conjecture [20, 21], is one of the most popular open problems about domination in product graphs. The Vizing’s conjecture states that the domination number of Cartesian product graphs is greater than or equal to the product of the domination numbers of the factor graphs. Moreover, several kind of domination related parameters have been studied in the last years. Some of the most remarkable examples are the following ones. The domination number of direct product graphs was studied in [3, 11, 16]. The total domination number of direct product graphs was studied in [5]. The upper domination number of Cartesian product graphs was studied in [2]. The independence domination number of Kronecker product graphs was studied in [10]. Several domination related parameters of corona product graphs and the conjunction of two graphs were studied in [8] and [22], respectively. The Roman domination number of lexicographic product graphs was studied in [12]. According to the quantity of works devoted to the study of domination related parameters in product graphs it is noted that not only Vizing’s conjecture is an interesting topic related to domination in product graphs. In this sense, in this paper we pretend to contribute with the study of some domination related parameters for the case of rooted product graphs.
We begin by establishing the principal terminology and notation which we will use throughout the article. Hereafter represents an undirected finite graph without loops and multiple edges with set of vertices and set of edges . The order of is and the size (If there is no ambiguity we will use only and ). We denote two adjacent vertices by and in this case we say that is an edge of or . For a nonempty set and a vertex , denotes the set of neighbors that has in : and the degree of in is denoted by In the case we will use only , which is also called the open neighborhood of a vertex , and to denote the degree of in . The close neighborhood of a vertex is . The minimum and maximum degrees of are denoted by and , respectively. The subgraph induced by is denoted by and the complement of the set in is denoted by . The distance between two vertices of is denoted by (or if there is no ambiguity).
The set of vertices is a dominating set of if for every vertex it is satisfied that . The minimum cardinality of any dominating set of is the domination number of and it is denoted by . A set is a set if it is a dominating set and . Throughout the article we follow the terminology and notation of [9].
Given a graph of order and a graph with root vertex , the rooted product is defined as the graph obtained from and by taking one copy of and copies of and identifying the vertex of with the vertex in the copy of for every [7]. If or is the singleton graph, then is equal to or , respectively. In this sense, to obtain the rooted product , hereafter we will only consider graphs and of orders greater than or equal to two. Figure 1 shows the case of the rooted product graph . Hereafter, we will denote by the set of vertices of and by the copy of in .
It is clear that the value of every parameter of the rooted product graph depends on the root of the graph . In the present article we present some results related to some domination parameters in rooted product graphs.
2 Domination number
We begin with the following remark which will be useful into proving next results.
Lemma 1.
Let be a graph of order and let be any graph with root and at least two vertices. If does not belong to any set or belongs to every set, then
Proof.
If is a dominating set of minimum cardinality in , , then it is clear that is a dominating set in . Thus . Suppose does not belong to any set. Let be a set and let for every . Notice that the set dominates all the vertices of except maybe the root which could be dominated by other vertex not in .
If for some , then is a dominating set in . So . Moreover, since does not belong to any set, it is satisfied that . If , then we have that , a contradiction. On the other side, if for some , then is a dominating set in . So . Therefore, for every and we obtain that .
On the other hand, let us suppose belongs to every set. Thus, dominates at least one vertex in which is not dominated by any other vertex in every set and, as a consequence, . As above, denotes a set and for every . If for some , then either is not a dominating set in (which is a contradiction) or . On the contrary, if , then is a dominating set in and . Therefore, we have that and the proof is complete. ∎
Theorem 2.
Let be a graph of order . Then for any graph with root and at least two vertices,
Proof.
It is clear that and also, from Lemma 1, there are rooted product graphs such that . Now, let us suppose that . Let be the set of vertices of and let , , be the set of vertices of the copy of in . Hence, if is a set, then there exists such that . Notice that the set dominates all the vertices in excluding . If , then the set is a dominating set in and , which is a contradiction. So, for every .
Let be the number of copies of in which the vertex of is not dominated by (i.e., is dominated by a vertex of belonging to other copy , with ). On the contrary, let be the number of copies of in which the vertex of is dominated by or . Note that the vertices of satisfying the above property form a dominating set in and, as a consequence, . Since , we have that . Also, notice that if the vertex of is dominated by or , then is a dominating set in . So, for every copy in which the vertex of is dominated by or . Thus we have the following.
On the other side, let be a set. Since , there exists at least one copy of such that , which implies . Since dominates all the vertices of except maybe the root , we have that if , then is a dominating set in , which is a contradiction. So, . Now, as for every , we obtain that . So, is a set. Let us denote by , , the set of vertices of in each copy of .
Let be a set and let . Since dominates the vertices of for every and dominates the vertices of , we obtain that is a dominating set in . Thus
Therefore, we obtain that and the result follows. ∎
3 Roman domination number
Roman domination number was defined by Stewart in [18] and studied further by some researchers, for instance in [4]. Given a graph , a map is a Roman dominating function for if for every vertex with , there exists a vertex such that . The weight of a Roman dominating function is given by . The minimum weight of a Roman dominating function on is called the Roman domination number of and it is denoted by . A function is a function in a graph if it is a Roman dominating function and .
Let be a Roman dominating function on and let , and be the sets of vertices of induced by , where . Frequently, a Roman dominating function is represented by the sets , and , and it is common to denote . It is clear that for any Roman dominating function on the graph of order we have that and . The following lemmas will be useful into proving other results in this section.
Lemma 3.
[4] For any graph , .
Lemma 4.
Let be a graph and let be a function. Then for every ,

If , then .

If , then .

If , then .
Proof.
Let be a function. By making we have that is a Roman dominating function in . Thus
(1) 
Now, if , then it is clear that and (i) is proved.
Moreover, if , then is a Roman dominating function in . Thus . Therefore, by (1) we obtain (ii).
On the other hand, if , then is a Roman dominating function in . Thus
Therefore, (iii) is proved. ∎
Lemma 5.
Let be a graph. If for every function is satisfied that , then
Proof.
From Lemma 4 (i) we have that . If , then there exists a function such that , which leads to . If is a function in such that for every , , we have that and , then is a Roman dominating function in . Thus, . So, and we have that is a function such that , which is a contradiction. Therefore, . ∎
The Roman domination number of rooted product graphs is studied at next.
Theorem 6.
Let be a graph of order . Then for any graph with root and at least two vertices,
Proof.
It is clear that . Let be the set of vertices of for every and let be a function. Now, for every and every , let . Let . We consider the following cases.
Case 1: . If , then is a Roman dominating function in . On the contrary, if , then is adjacent to some vertex , with and, again is a Roman dominating function in . So, . By Lemma 4 (i) we have that . Thus .
Case 2: . Hence, it is clear that is a Roman dominating function in . So, . By Lemma 4 (ii) we have that . Thus .
Case 3: . Thus is a Roman dominating function in . So, .
Now, let be the set of vertices of and let be the set of vertices of such that for every vertex is satisfied that . So, every vertex is dominated by some vertex in , with . As a consequence, is a dominating set and . Since , it is satisfied that equals at most the numbers of copies of such that (those copies satisfying Case 1). Thus we have the following,
Therefore the lower bound is proved. ∎
As the following proposition shows, the above bounds are tight.
Theorem 7.
Let be a graph of order and let be a graph with root and at least two vertices. Then,

If for every function is satisfied that , then

If there exist two functions and such that and , then
Proof.
Let be a function and let be the set of vertices of , . Now, for every , let . From Theorem 6 we have that . If , then there exists such that . So is not a Roman dominating function in . If or , then every vertex in is adjacent to a vertex in and, as a consequence, is a Roman dominating function in , which is a contradiction. So and is a function. Since for every function, by Lemma 5 we have that and this is a contradiction. Therefore, and (i) is proved.
To prove (ii), for every we consider two functions and such that and , and let be a set. Now, we define a function in in the following way.

For every vertex belonging to a copy of such that the root we make (notice that ).

For every vertex , except the corresponding root, belonging to a copy of such that the root , we make .

For every root of every copy satisfying the conditions of the above item we make (note that these vertices are adjacent to a vertex of for which ).
Since every vertex not in , with , is adjacent to a vertex such that and also, every vertex of , with , is adjacent to a vertex with , we obtain that is a Roman dominating function in . Thus
Therefore, (ii) follows by Theorem 6. ∎
On the other hand, we can see that there are rooted product graphs for which the bounds of Theorem 6 are not achieved.
Theorem 8.
Let be a graph of order and let be a graph with root and at least two vertices. If for every function is satisfied that , then
Proof.
Let be a function and let be a function. Now, let us define a function in such that if , then Otherwise, . Since for every function, it is satisfied that every vertex of with is adjacent to a vertex in with . Thus is a Roman dominating function in and we have that
On the other hand, let , , be the set of vertices of the copy of in and let be the set of vertices of . Now, let be a function and for every let . Since the root of satisfies that for every function , we have the following cases.
Case 1: If there exists such that , then is a Roman dominating function in , but it is not a function. Thus , which leads to
Case 2: If there exists such that , then is a Roman dominating function in and is a Roman dominating function in . Thus, by Lemma 4 (ii), it is satisfied that
Case 3: If there exists such that , then we have one of the following possibilities:

is not a Roman dominating function in . So, should be adjacent to a vertex , , of such that . Moreover, is a Roman dominating function in and by Lemma 4 (ii) it is satisfied that .

is a Roman dominating function in . Since for every function , we have that . Let be a function. Now, by taking a function on , such that if , then and if , then , we obtain that is a Roman dominating function for and the weight of is given by
and this is a contradiction.
As a consequence, we obtain that if , then is not a Roman dominating function in . So, every vertex of for which is adjacent to a vertex , , of such that and it is satisfied that the function