#### 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).

**(a)**

**(b)**

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
*