#### Abstract

Let be a simple graph. A set is a dominating set of , if every vertex in is adjacent to at least one vertex in . We denote the family of dominating sets of a graph with cardinality by . In this paper we introduce graphs with specific constructions, which are denoted by . We construct the dominating sets of by dominating sets of graphs , , and . As an example of , we consider . As a consequence, we obtain the recursive formula for the number of dominating sets of .

#### 1. Introduction

Let be a graph of order . For any vertex , the open neighborhood of is the set , and the closed neighborhood is the set . For a set , the open neighborhood is , and the closed neighborhood is . A set is a dominating set if , or equivalently, every vertex in is adjacent to at least one vertex in . The domination number is the minimum cardinality of a dominating set in , and the family of -sets is denoted by . For a detailed treatment of this parameter, the reader is referred to [1]. Let be the family of dominating sets of a graph with cardinality , and let . The domination polynomial of is defined as , where is the domination number of .

The domination polynomial of a graph has been introduced by Alikhani in his Ph.D. thesis [2]. More recently it has been investigated with respect to special graphs, zeros, and application in network reliability; see [2â€“9].

Obviously study of the dominating sets of graphs is a method for finding the coefficients of the domination polynomial of graphs. Authors studied the construction of dominating sets of some families of graphs to study their domination polynomials; see [10â€“13]. In this paper we would like to study some further results of this kind.

In the next section we introduce graphs with specific construction which is denoted by . As examples of these graphs, in Section 3 we study the dominating sets of paths and some other graphs. As a consequence, we give a recurrence relation for and .

As usual we use , for the smallest integer greater than or equal to and the largest integer less than or equal to , respectively. In this paper we denote the set simply by .

#### 2. Dominating Sets of Graphs

A path is a connected graph in which two vertices have degree 1, and the remaining vertices have degree 2. Let be the path with and ; see Figure 1(a).

In this section we introduce graphs with specific construction and study their dominating sets. Let be a path with vertices labeled by , for . Let be a graph obtained from by identifying a vertex of with an end vertex of . For example, if is a path , then is a path .

We need some lemmas and theorems to obtain main results in this section.

Lemma 1. if and only if or .

Proof. It follows from the definition of dominating set of graph.

We recall the following theorem.

Theorem 2 (see [14]). If , then .

The following lemma follows from Theorem 2.

Lemma 3. For any , .

Lemma 4 (see [15, page 371]). The domination number of path is .

A simple path is a path where all its internal vertices have degree two. In Figure 1(b), we have shown a graph which contains a simple path of length with vertices labeled . The following lemma follows from Lemma 4.

Lemma 5. If a graph contains a simple path of length , then every dominating set of must contain at least vertices of the path.

Lemma 6. If , and there exists , such that , then .

Proof. Suppose that . Since , contains at least one vertex labeled or . If , then , a contradiction. Hence, . But then in this case, , for any , also a contradiction.

Lemma 7. (i) If , then .
(ii) If then, .
(iii) If , then .

Proof. (i) Since , by Lemmas 1 and 3, or . In either case we have .
(ii) Suppose that , so by Lemma 1, we have or . If , then , and hence, , a contradiction. So we have , and hence, , also a contradiction.
(iii) Suppose that . Let , then at least one vertex labeled or is in . If , then by Lemma 5, at least one vertex labeled or is in . If or , then , a contradiction. If , then , a contradiction. Now suppose that , then by Lemma 5, at least one vertex labeled or is in . If or , then , a contradiction. If , then , a contradiction. Therefore .

Lemma 8. Suppose that , then(i) and if and only if ,(ii) and if and only if ,(iii) and if and only if ,(iv) and if and only if ,(v) and if and only if .

Proof. (i) () Since , by Lemmas 1 and 3, we have or . If , then , and by Lemma 1, , a contradiction. So we have , and since , together we have . So .
() If , then by Lemma 1, we have and .
(ii) () Since , by Lemmas 1 and 3, or . If , then , and hence , a contradiction. So we must have . Also since , we have . Therefore we have .
() If , then by Lemma 1, we have â€‰â€‰ and .
(iii) () Since , by Lemma 1, or . If , then , and by Lemma 1, , a contradiction. So we must have . But we also have because . Hence, we have .
() If , then by Lemma 1, , and .
(iv) Since , by Lemma 1, we have or . Since , by Lemma 1, we have . Therefore is not possible. Hence we must have . Thus or . But because . So we have .
If , then by Lemma 1, and .
(v) Since , and , then by applying Lemma 1, we have , and . So, by Lemma 3, , and hence .
If , then by Lemma 1 we have the result.

By Lemma 6, for the construction of , suffices to consider , and . By Lemma 7, we need only to consider the five cases in the following theorem.

Theorem 9. For every ,(i)if and , then â€‰â€‰; (ii)if and , then â€‰â€‰â€‰â€‰; (iii)if and , then â€‰â€‰â€‰â€‰, ; (iv)if and , then â€‰â€‰â€‰â€‰, ; (v)if , and are nonempty, then â€‰â€‰â€‰â€‰, â€‰â€‰â€‰â€‰â€‰â€‰, .

Proof. (i) Obviously . Now suppose that , then at least one of the vertices or is in . If then by Lemma 5 at least one of the vertices , or is in . If or , then , a contradiction. If , then , a contradiction. Now suppose that , then by Lemma 5 at least one of the vertices , or is in . If or , then , a contradiction. If , then . So .
(ii) By Lemma 8(ii), . If we suppose that has labeled with numbers in , then .
(iii) It is obvious that â€‰â€‰â€‰â€‰â€‰â€‰, â€‰â€‰â€‰â€‰. Now let . Then or is in . If , then by Lemma 5, at least one vertices , or is in . If or is in , then , a contradiction, because . Hence, , and . Therefore for some . Now suppose that and . By Lemma 5, at least one vertex labeled , or is in . By Lemma 8(iii), , and on the other hand since , then . So together we have . Therefore , and so all these sets in have minimum cardinality. Therefore, since and , then . Hence or is in , but . Thus for some . So â€‰â€‰, â€‰â€‰, .
(iv) By Lemma 8, . If we suppose that has labeled with numbers in , then â€‰â€‰â€‰â€‰â€‰â€‰, â€‰â€‰, .
(v) and . Let , so at least one vertex labeled or is in . If or , then . Now let , then or is in . If or , then . Now let , then or is in . If , then , for . If , then . Therefore we have , â€‰â€‰, â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰. Now, let , then or . If , then by Lemma 5, at least one vertex labeled , or is in . If or , then for some . If , and , then for some . Now suppose that and , then by Lemma 5, at least one vertex labeled , or is in . If or , then for some . If , and , then for some . So â€‰â€‰â€‰â€‰, â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰, â€‰â€‰â€‰â€‰â€‰â€‰.

Theorem 10. For any and ,

Proof. It follows from Theoremâ€‰9.

Theorem 11. For any ,

Proof. We have the result by definition of domination polynomial and Theorem 10.

#### 3. Dominating Sets of Paths and Some Other Graphs

First we investigate the dominating sets of paths. Since , we use the results for graph in the previous section to obtain properties of dominating sets of path. We suppose that is labeled as in Figure 1(a) and denote simply by .

Lemma 12. Suppose that , then one has(i) and if and only if and for some ,(ii) and if and only if ,(iii), and if and only if and for some ,(iv), and if and only if ,(v), and if and only if .

Proof. (i) () By Lemmas 4 and 8(i), we have , which give us and for some .
() It follows from Lemmas 4 and 8(i).
(ii)â€‰It follows from Lemma 8(ii).
(iii) () By Lemmas 4 and 8(iii), we have , which give us and for some .
() It follows from Lemmas 4 and 8(iii).
(iv)â€‰It follows from Lemma 8(iv).
(v) It follows from Lemmas 4 and 8(v).

The following theorem specify .

Theorem 13. For every and ,

Proof.
Case 1. and . By Lemma 12(i), and for some . Therefore .
Case 2. It follows from Theorem 9(ii).
Case 3. By Theorem 9(iii), . By Lemma 12(iii), and for some . Since , then . Therefore â€‰â€‰.
Cases 4 and 5 follow from Theorem 9(iv) and (v), respectively.

Example 14. We use Theorem 13 to construct for . Since and , by Theorem 13, . Since , and , we get . Since , , , , , , and , then by Theorem 13, , , , , , . And, for construction , , , , , , , , , , by Theorem 13, , , , , . Finally, since , , , , , , , , , , , , and , then , , , , , , , , , , , , , .

Theorem 15. If is the family of dominating set of with cardinality , then .

Proof. It follows from Theorem 10.

Remark 16. We have obtained [12, Lemma 12 and Theoremâ€‰â€‰5] with a different approach.

As other examples of graph , we study the dominating sets of some other graphs. By Theorem 10, the recursive formula in this theorem is true for every graphs which contain a gluing path with at least 3 vertices. Also, if we have the domination number of graphs, then we can study the properties of their dominating sets by Theorem 9 and Lemma 8. Here, we consider a special case of graph which is a tree. We denote this tree by . In other words is a tree, such that , and for , where is its vertex set, , and the edge set ; see Figure 2. For obtaining the domination number of , we need the following lemma which follows from observation.

Lemma 17. For every and , .

Theorem 18. For every and , .

Proof. By induction on . If , then . Now suppose that the theorem is true for all numbers less than or equal , and we prove it for . By applying Theorem 2 for , we have the following inequalities: Similarly by applying Theorem 2 for and induction hypothesis, we have the following inequalities; respectively, Now if for some , then by (4) and (5) we have â€‰â€‰â€‰â€‰â€‰+â€‰.
If for some , then by (4) and (6) we have .
Now if for some , we will consider the following cases.(i)â€‰One has for some . By applying Theorem 2 for , we have the following inequalities: â€‰Now, by (4) and (7), we have .(ii)â€‰One has for some . By Theorem 2 for we have â€‰By (4) and (8), we have .(iii)â€‰One has for some . We will prove that . We do it by induction on . If , then by Lemma 17, . So the result is true for . Now suppose that the result is true for all number less than , and we prove it for . By Lemma 17 and the induction hypothesis, we have the following equalities for : â€‰Ifâ€‰ , then for some , . Again by Lemma 17, â€‰Finally, for , we have â€‰â€‰â€‰â€‰â€‰â€‰â€‰+â€‰â€‰â€‰â€‰+â€‰â€‰+â€‰. Therefore in all cases .

Let or simply be the family of dominating set of with cardinality . By Lemma 8 and Theorem 9 we prove the following lemma, which is some properties of .

Lemma 19. Suppose that , then(i) and if and only if , , for some ,(ii), and if and only if ,(iii), and if and only if , , for some ,(iv), and , if and only if ,(v) and if and only if .

Proof. (i) By Lemma 8(i) and Theorem 18, , which give us , for some .
(ii) By Lemma 8(ii), .
(iii) By Lemma 8(iii) and Theorem 18, . Therefore and for some .
(iv) By Lemma 8(iv), .
(v) By Lemma 8(v) and Theorem 18, .

Here, we suppose that the tree labeled as shown in Figure 3. The following theorem constructs from , and .

Theorem 20. (i) If and , then
(ii) If and , then
(iii) If , and , then
(iv) If , and , then