1 Introduction
The concept of dominating set and its variations are wellstudied topics in graph theory. There are thousands of papers on domination in graphs and several wellknown surveys and books on this topic such as [13, 12, 6].
The dominating set problem is a classic NPcomplete graph problem [16], however, there exist polynomial time algorithms for some graph classes such as trees, interval graphs, and graph with bounded treewidth [5, 2, 17].
Recently, there has been extensive research dealing with enumeration algorithms and combinatorial lower and upper bounds of minimal dominating sets, both on general graphs and on special classes of graphs [10]. In [9], Fomin et al. gave an algorithm with time complexity of to enumerate all of the minimal dominating sets in graphs with vertices and shown that the number of minimal dominating sets in such a graph is at most . They established a lower and an upper bound for the maximum number of minimal dominating sets in graphs. This gap has been narrowed on some wellknown graph classes, such as chordal graphs [7], trees [14] and cobipartite graphs [8]. Tight bounds have been obtained for some graph classes such as cographs and split graphs [7, 8]. Also, it is shown that the interval graphs and trees on vertices have at most minimal dominating sets [10].
Let denotes the collection of all dominating sets of and denotes the collection of all dominating sets of with size where . It is obvious that . The problem of minimum dominating set is to find the smallest such that and for all . Such smallest is called the minimum domination number of and is denoted by , or simply when there is no chance of confusions. Any set is called a set of .
In this paper, we propose a new concept, domination cover number, to evaluate a number of proposed dominating set like in a graph as follows:
where the index in refer to a special dominating set of . We concentrate on investigating maximum ( or minimum) of domination cover numbers. Maximum domination cover number, for sets, defined as
and minimum domination cover number, for sets, defined as
As an application of maximum domination cover number, consider we want to use minimum number of protective cameras such that maximum overlapping in the protected places occur.
The rest of the paper is organized as follows: In Section 2, we introduce some essential definitions and notations. In Section 3, we propose the concept of domination cover number and obtain some bounds for it. In Section 4, we show some results on the domination cover number in the product of graphs. Next, we investigate some algorithmic approaches in various classes of graphs in Section 5. Finally, a conclusion is drawn in last section.
2 Notations and Definitions
Let be a simple graph. The open neighborhood or simply the neighborhood of a vertex is the set of all vertices adjacent to and is denoted by , i.e. . The closed neighborhood of a vertex is defined as . For a subset of vertices , the open and close neighborhoods of are defined as
and
respectively. Note that when there is no chance of confusion, we may drop the subscript .
Let be a subset of and . Then, the private neighborhood of with respect to , denoted by , is the set . Each vertex in is called a private neighbor of with respect to .
A set is called a total dominating set of if every vertex in is adjacent to at least one vertex in . The concept of total domination is first introduced by Cockayne, Dawes, and Hedetniemi in 1980 in [6]. The minimum cardinality among all the total dominating sets of a graph is denoted by and each total dominating set of size is called a set.
A dominating set is called an efficient dominating set if every vertex is dominated by exactly one vertex in , and every vertex is not dominated by other vertices in . Such a dominating set is called an efficient open dominating set, if for every vertex , we have .
A subset in a graph is a total dominating set if, for every vertex , . The minimum cardinality of a total dominating set of is called the total domination number, denoted by .
For any connected graph , a vertex is called a cutvertex if is not connected, where is exactly the graph with vertex and all of its incident edges removed. A maximal connected subgraph of without any cutvertices is called a block of . Moreover, a graph is called a block graph if its blocks are complete subgraphs, or equivalently cliques, and the intersection of any two blocks is either empty or a cut vertex.
The lexicographic product of graphs and is denoted by , and is a graph with vertex set , where a vertex is adjacent to a vertex if either or and . Let . Then, the induced subgraph by is called the layer of with respect to the vertex and is denoted by . It is clear that is isomorphic to .
3 Domination Cover Number
In this section, we give formal definitions for the domination cover number and related problems.
Definition 3.1 (Covering Number of a Set of Vertices).
Let be a graph and . Then, the covering number of , denoted by , is defined as .
Remark 3.2.
Note that the covering number of a set of the vertices can be defined equivalently as .
Since a dominating set of a graph is a subset of vertices, then we can calculate the covering number of , i.e.
This parameter is called the domination cover number of the dominating set with respect to .
In this paper, we tend to solve the following problem which is referred to as the maximum domination covering of sets, denoted by . This problem is defined as
where is the collection of all sets.
Remark 3.3.
It is worth noting that the can be similarly defined as
It is clear that the domination number is unique, however, there may exist several dominating sets of size . The main purpose of introducing the domination cover number is to find a set with the maximum or minimum covering number.
Remark 3.4.
Note that the problem of finding maximum (minimum) domination cover number is extensible to other types of dominations as well.
Definition 3.5.
(Total Domination Cover Number) Let be a graph and be a total dominating set. The total domination cover number of with respect to is defined as
Let be a subgraph of and . Then, the domination cover number of with respect to the set equals
In the following, we investigate Domination cover number for some graphs.
Theorem 3.6.
([3], page 371) For any path and cycle , we have
Theorem 3.7.
Let . The domination cover number with respect to satisfies:

if , then ,

if , then ,

if , then .
Proof.
Let be a path of length . In the case that , the path has a unique set, i.e. is the only dominating set of in this case. It is obvious to see that the degree of each vertex in this set is two, therefore we have . In the case of , one of the sets is . It is easy to see this set is the minimum dominating set with the minimum covering number and it has vertices of degree two and two vertices of degree one. Therefore the lower bound is established. Also, we have another dominating set of size where none of its vertices has degree one which concludes the upper bound . In the case of , the upper and lower bounds are obtained similar to the previous case. ∎
There exist many different graph classes such as Barbell graphs and Book graphs that , e.g. see Figure 1.
Also, there are many graphs with unique domination cover number. For example, let be a graph with dominating set that satisfies the following condition for every vertex
Then, the set is the unique dominating set for [11]. Therefore, the graph has a unique domination cover number.
3.1 Bounds for the Domination Cover Number
In this section, we provide some bounds for domination cover number of graphs. For convenience, we assume that is of order .
Lemma 3.8.
Let , then we have
Proof.
It is clear that if is an efficient dominating set, then for every vertex , we have and for , we have , so this bound is concluded. In the case that is not an efficient dominating set, we have two cases to consider. In the first case, there exists at least one vertex such that . So,
In the second case, there exists at least one vertex such that . Therefor as in the first case, we have . ∎
Corollary 3.9.
If has an efficient dominating set, then
Theorem 3.10.
Let , then the domination cover number of is bounded as
(1) 
and these bounds are sharp, in the sense that there exist graphs satisfying the equalities for infinitely many values of .
Proof.
The lower bound is trivial, by the fact that and Lemma 3.8. Since is minimum dominating set, then every vertex has at least one private neighbor, otherwise if has not a private neighbor, we can remove from and find a smaller dominating set which is a contradiction. If has more than one private neighbor, e.g. , the vertex does not change the value of . So, we assume every vertex in has exactly one private neighbor. With this assumption, the maximum domination cover can be obtained when the vertices in construct a complete graph and every also have a private neighbor. Therefor
Since , so the upper bound is established.
Next, we prove that these bounds are sharp. Let be a graph with vertices which is constructed from base by adding an extra vertex adjacent to each vertex in . This graph is called a corona graph. It is clear that , so if the dominating set contains onedegree vertices, then and if the dominating set contains vertices of , then . ∎
A graph is called free, if does not contain an induced subgraph . In the next theorem, we establish bounds for domination cover number of free graphs. These graphs are also known as cographs.
Theorem 3.11.
Let be a free graph with n vertices and . Then, we have
Proof.
The lower bound is resulted by Lemma 3.8. In [4, 15], it is shown that in free graphs we have , i.e. domination number is equal to total [1,2] domination number of graph. So every vertex is dominated at most twice. But according the definition of total  domination every can be connected to at most one other vertex in , So we reduce the size dominating set from .
∎
4 Domination Cover Number in Graph Products
In this section, we investigate domination cover number for the lexicographic product of graphs and .
Theorem 4.1.
The minimum and maximum domination cover number of lexicographic product of and are as follows:
and
where , , and .
To prove Theorem 4.1, we need the following lemmas.
Lemma 4.2.
Let be a dominating set for with minimum cardinality and dominates all the vertices in . Then, the set is a dominating set for .
Proof.
It is sufficient to show the following statements:

The set dominates all the vertices in .

There exist no dominating set for of cardinality less than .
Since every vertex in is dominated by a vertex in , then it is easy to see that every vertex in is dominated by a vertex of such that . In other words every vertex such that is dominated by .
Next, we show that has minimum cardinality. Suppose be a set for and let . It is easy to see that is a dominating set for and . Therefore, has the minimum cardinality. ∎
Lemma 4.3.
Let be a set for and . Then, for each vertex the set is a set for .
Proof.
We will show that is a dominating set for with the minimum size among all dominating sets. Since is a total dominating set for , then for every , there exists a vertex such that . Therefore, each vertex is adjacent by a dominating vertex where .
Assume that is a dominating set for of cardinality less than and let and is not a total dominating set. So, there exists such that none of its neighbors are in . It means that there is a vertex such that there is no vertex where . Therefore, the vertex can be dominated by the vertex where . For every vertex in which is not total dominated, there exist at least two vertices in . Now, it is enough to select one of the vertices in and instead of other vertices, we select where . ∎
Lemma 4.4.
Let be a set for , be a set for with cardinality 2 and . Then, is a set for .
Proof.
Lemma 4.5.
Let be a set for such that . Then, the set is an independent dominating set.
Proof.
We prove this lemma by contradiction. Let there exist at least two vertices such that . For constructing total dominating set , it is enough to put vertices and in the set and for the rest of the vertices in like , we put and one of its neighbor in . As a result, we obtain a total dominating set for such that it has cardinality of at most which is a contradiction. ∎
Let , for every we define and be the degrees of in and in , respectively.
Lemma 4.6.
Let be a set for , and . Then the domination cover number constructed by is
Proof.
Let the set be the degrees for vertices of the dominating set . So,
∎
Lemma 4.7.
Let be a set for and either or . Then,
Proof.
Lemma 4.8.
Let be a set for , and . Then,
where .
Proof.
If be a set, the proof is similar to the proof of Lemma 4.7. otherwise, we consider the case where the set is a set. The set contains the degrees of vertices in . So
∎
Now, we have all of requirement to do the proof of Theorem 4.1.
Proof.
We have just proved that . The proof for is similar. According to the definition of domination cover number, we have
There are three cases to consider:
Case 1:
In this case, we have:
Case 2: Either
Case 3:
In this case, we have two types of dominating sets for

is a set for . Then, where is a set for which is similar to case 2.

is a set for and is a set for . In this case, is a set for and domination cover number of this set is
In this case, is the minimum value of the above types. ∎
5 Finding Domination Cover Number in Some Graphs
In this section, we use the dynamic programing approach to find the domination cover number of certain classes of graphs.
5.1 Domination Cover Number in Trees
5.1.1 Definitions
We first choose an arbitrary vertex of and consider it as a root. So, from now, we can think of as a rooted tree. For each vertex of , we define the following notation :

The set consists of all children of .

denotes the subtree of rooted at .

is the size of the smallest dominating set of which contains . This is welldefined because the set of all vertices of is a set of that contains .

is the size of the smallest dominating set of which does not contain . If no such set exists, we define to be .

() is the size of the maximum (minimum) domination cover number of which contains the vertex . This is welldefined because the set of all vertices of is a dominating set for that contains . We first initialize to zero.

() is the size of the maximum (minimum) domination cover number of that does not contain . This is welldefined because the set of all vertices of is a dominating set for that does not contains . The initialization value of this parameter is zero.
We are going to devise a linear time algorithm based on a bottomup dynamic programming technique to calculate the above notation for all of the vertices. To calculate these values for each vertex, we assume that similar values have been already computed for all of the descendants of the current vertex. This is a valid assumption since we process the vertices of in postorder. For convenience, we use black and white colorings for the vertices to denote whether they are in the domination or not, respectively.
5.2 Values for leaves
A vertex is called a leaf if it has no children. If is a leaf of , then consists only of and therefore, we set and , since there exists no dominating set for that does not contain .
5.3 Calculating when is not a leaf
Since we are concerned with when is not a leaf, we have the assumption that is white. Therefore, at least one of its children must be black. Moreover, has no effect on how one colors the subtrees for . We consider all of the possible valid bicolorings and choose one of the yielding numbers of black vertices with at least value of as follows:
At least one of the children of must be in the dominating set. So, we construct the set which consists of all children of like , which satisfy and is assumed as the set of all of children satisfying . Now, we solve the following equation
(2) 
Note that in the above equation, at least one of the children of must be chosen. If A is empty, then we select a vertex from B which has the maximum value of . Moreover, if is empty, we select the vertex in such that, is minimum.
The value of is calculated as
(3) 
and similarly, the value of is calculated as
(4) 
5.4 Calculating when is not a leaf
In this case, the vertex is black and all of the children of have a black neighbor and there is no restriction about the color of its children, i.e. each child of can be either white or black. The value of is calculated as
(5) 
and to find the value of , we consider all of the selected children of in the above equation as the set , i.e.
(6) 
Similarly, for we have
(7) 
The time complexity of calculating all of these equations is .
Remark 5.1.
With little changes in the above algorithm, we can find a dominating set for a subtree of with maximum or minimum domination cover number.
5.5 Block graphs
Our algorithm works on a treelike decomposition structure, named refined cuttree of a block graph [1]. Let be a block graph with blocks and the set of cutvertices . The cuttree of , denoted by is defined as and . It is shown in [1] that the cuttree of a block graph can be constructed in linear time by the depthfirstsearch method. For any block of , define where .
5.6 Domination Cover Number in Block Graphs
Let be a block graph and be its corresponding cuttree. We set the following notation:

as the set of all block nodes of ,

as the set of all cutvertices of ,

is the size of the smallest dominating set of that contains node ,

is the size of the smallest dominating set of that does not contain any vertices of ,

is the size of the smallest dominating set of that does not contain any vertices of but all of vertices in block which is constructed by vertices in are dominated,

,, and are defined similar to the ones at the beginning of Section 5.1.
Initialization
Let be a leaf. It is clear that is a block node of and has some vertices which are not cutvertices and the degree of all of them is the same. We initialize if a noncutvertex of is selected, otherwise , and also equal to the degree of the selected vertex and both and to .
Updating state:
In the post order traversal of , for each non leaf vertex like , we define the following sets:

,

,

.
Next, we consider the following cases:

: In this case we have two situation to consider:

If is selected, then, all of the children of are dominated. So, in calculating the domination number we do not consider the leaf children. Thus,
(8) To calculate the maximum cover number, we first add the degree of to and then, for each child , according to the values of

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