Research Article | Open Access

Volume 2020 |Article ID 7607856 | https://doi.org/10.1155/2020/7607856

Lidan Pei, Xiangfeng Pan, "The Bondage Number of Generalized Petersen Graphs ", Discrete Dynamics in Nature and Society, vol. 2020, Article ID 7607856, 11 pages, 2020. https://doi.org/10.1155/2020/7607856

# The Bondage Number of Generalized Petersen Graphs

Revised10 Mar 2020
Accepted14 Mar 2020
Published09 May 2020

#### Abstract

The domination number of a nonempty graph is the minimum cardinality among all subsets such that . The bondage number of a graph is the smallest number of edges whose removal from results in a graph with larger domination number. The exact value of for and the bounds of for are determined.

#### 1. Introduction

For terminology and notation on graph theory not given here, we refer the reader to [1]. Let be a finite, undirected graph without loops, multiple edges, and isolated vertices, where is the vertex set and is the edge set. A graph G is nonempty if . For a vertex , let be the open neighborhood and be the closed neighborhood. For a set , the open neighborhood is defined to be and the closed neighborhood is . We use to denote the subgraph of induced by .

A subset is a dominating set of if . The domination number is the minimum cardinality of a dominating set of . A dominating set is called a -set if its cardinality is . Studies on domination number have long attracted graph theorists for their applications and theoretical interest. The two outstanding domination books by Haynes et al. [2, 3] have a comprehensive research of domination.

The bondage number of a nonempty graph is the smallest number of edges whose removal from results in a graph with domination number greater than the domination number of . That is,

The bondage number of a nonempty graph is well defined since the domination number of every spanning subgraph of a nonempty graph is at least as great as . The bondage number is proposed by Fink et al. [4] in 1990. However, it is quite difficult to compute the exact value of the bondage number for general graphs. Hu and Xu [5] showed that the problem of determining bondage number for arbitrarily given graph is NP-hard. Therefore, it is important to determine bondage numbers for some graphs or networks with simple structure. The exact value of the bondage number has been worked out for only a few classes of graphs, such as complete graphs, paths, cycles, complete t-partite graphs [4], trees [68], complete -partite digraphs [9], de Bruijn and Kautz digraphs [10], the Cartesian product of two cycles [1113], the Cartesian product of two paths [14], the strong product of two paths [15], and Mycielski Graphs [16]. For a further study on the advances of bondage number, the readers are suggested to refer to [17, 18].

For positive integer numbers and , where , the generalized Petersen graph is a graph on vertices with

Although the generalized Petersen graph has a special construction, the complexity and difficulty are ubiquitous throughout the study on it for each . Even so, many researchers still insist on and obtain many graph parameter values other than the bondage number of [1925]. Yet, the studies on the domination number and its variations of are attractive. Recall that the problems of determining the domination number and bondage number for arbitrarily given graph are NP-complete [26] and NP-hard, respectively. Therefore, researchers tend to study the domination problem of generalized Petersen graph with a given small value for or a given relationship between and [2733]. It is worth mentioning that Fu et al. [28] showed the exact domination number of generalized Petersen graph . Hence, it is meaningful to study the bondage number of generalized Petersen graph in this paper. The rest of the paper is organized as follows. Section 2 shows the lower bound of . In Section 3, we present the upper bounds of for and , respectively. In particular, the upper bound of for is just the exact value.

#### 2. Lower Bound on Bondage Number of

Let , that is, and . Fu et al. [28] proved that and showed a minimum dominating set of graph , where

In order to determine the lower bound on bondage number of , we need to first prove the following lemma by the definition of bondage number.

Lemma 1. , where and are any two edges of .

Proof. By the definition of graph , we need to consider the following six cases: , ; , ; , ; , ; , ; , , . Actually, we only show the proof for , and , , since the proofs for the other four cases are similar (for reference, see Appendix).
Assume that is the minimum dominating set of graph constructed in [28]. We prove for in the following. Without loss of generality, we assume that and , where . If , then . In view of and , we only need to prove for . LetWe claim that for each , there exists an such that where . We consider , respectively.
For , if (mod 5), let ; then, (mod 5) such that . If (mod 5), let ; then, (mod 5) such that .
For , if (mod 5), let . Then, and (mod 5) such that . If (mod 5), let . Then, (mod 5) and (mod 5) such that .
For , let when (mod 5); then, (mod 5) and such that . Let when (mod 5); then, (mod 5) and such that .
For , let when (mod 5), when (mod 5), and when (mod 5). Thus, .
For , let if (mod 5) and if (mod 5). Then, .
It is easy to verify that is a dominating set of graph for . Hence, is a dominating set of graph . In consideration of , we get that for any , and . The result follows.
Next, we prove for . Let , . We claim that is a dominating set of for , whereFor , letFor and , is a -set of , and it is a dominating set of . Thus, . For and , letFor, letFor , letFor , letThe integers and are obtained such that for , respectively. Thus, is a dominating set of graph . As a consequence, .
It is well known that for any two edges . Combining Lemma 1, we obtain that , which implies the following theorem.

Theorem 1. .

#### 3. Upper Bound on Bondage Number of

Let , where and . The main results we get in this section are for and for . We prove the above results mainly by mathematical induction on in Lemma 3. As a recursive condition in mathematical induction, the inequation shown in Lemma 2 is important and necessary. In addition, due to the complexity inherent in the study of domination number and generalized Petersen graph, the proof for Lemma 2 is inevitably tedious and difficult. Let ; we first present an algorithm which constructs from a smaller graph .

Figure 1 gives an illustration for Algorithm 1.

 Input: the graph with , , . Output: the graph . Step 1. Delete the subset of vertices , along with their 18 incident edges. Denote the resulting graph by . Step 2. Define the graph to have vertex set and edge set . Return .

Observation 1. For each , the graph returned by Algorithm 1 is isomorphic to .

Lemma 2. For each , .

Proof. We assume that , and show . Let be a -set of and and be the graphs returned by Algorithm 1. Then, . We will identify with and . Then, , where . Let . Then, we get that and the set can dominate all vertices in , except possibly vertices in . We consider the following cases.

Case 1. .

Since dominates the vertices in and dominates the vertices in , we have that forms a dominating set of . Hence, .

Case 2. . Then, .
The vertices in can dominate at most three vertices in ; otherwise, all the four vertices in belong to . Thus, cannot be dominated by since . Therefore, .

Subcase 1. . Since , we get that . Thus, is a dominating set of , a contradiction with the minimality of . Hence, this case does not exist.

Subcase 2. . In this case, there is only one vertex dominated by ; then, is a dominating set of . Thus, .

Subcase 3. .
Suppose that are dominated by ; thus, and may not dominated by . If for some vertex , then is a dominating set of and the result follows. If , then there exist the following three possible cases (see Figure 2). To make the following proof easier to understand, we attach the corresponding graph for each case. Furthermore, in each figure a vertex colored dark indicates a vertex in , a vertex indicates a possible vertex in , and a vertex indicates a vertex not dominated by and . Since , the vertices should be dominated by the two vertices in .(1) (see Figure 2(a)), that is, are dominated by .It is obvious that or is a dominating set of in this case. For dominating vertex in graph , at least one vertex in belongs to . Hence,or(2) (see Figure 2(b)), that is, are dominated by .In this case, can be a dominating set of . If , then for dominating vertices and in graph . It implies that vertex cannot be dominated by , a contradiction. Therefore, . It follows that(3) (see Figure 2(c)), that is, are dominated by .Since and , at least one vertex of and belongs to . Since , is a dominating set of if one vertex of belongs to . Since , is a dominating set of if one vertex of belongs to . In the following, we just need to consider . Then, is a dominating set of . Hence,

Subcase 4. ; then, and there exist three vertices in dominated by . Suppose that are dominated by . If where , then is a dominating set of , which means that

Otherwise, we have the following seven possible cases by the symmetry (see Figure 3).(1) (see Figure 3(a)); then, are dominated by and could be a dominating set of .It is obvious that at least one vertex of belongs to to dominate . If or , then . If , then is a dominating set of (in graph , is dominated by ). Thus, .(2) (see Figure 3(b)); then, are dominated by , and or can be a dominating set of in this case.For dominating vertex in graph , at least one vertex in belongs to . Hence, .(3) (see Figure 3(c)), that is, are dominated by and are not dominated by .Since the remaining one vertex in belongs to , or to dominate . If , then , and thus the vertex cannot be dominated by , a contradiction. If , then is a dominating set of . So, .(4) (see Figure 3(d)); then, are dominated by and are not dominated by .Because the only one vertex in cannot dominate and simultaneously, or . Actually, is a dominating set of , which follows that .(5) (see Figure 3(e)); then, are dominated by .At least one vertex of belongs to to dominate . If or , then is a dominating set of . If , then is a dominating set of . It follows that .(6) (see Figure 3(f)); then, are dominated by and are not dominated by .The only one vertex in cannot dominate and simultaneously; hence, or . For , we get that is a dominating set of . In view of , . For , is a -set of if , and is a dominating set of if or . Therefore, .(7) (see Figure 3(g)), that is, are dominated by and are not dominated by .

Clearly, or is a dominating set of in this case. For dominating vertex in graph , we have that . Hence, .

Case 3. .
The vertices in can dominate at most two vertices in ; otherwise, (S ∩ T) ⊆ (S ∩ Q) , which means that cannot be dominated by . Therefore, .

Subcase 5. . We have nothing to do, and is a dominating set of and .

Subcase 6. . By symmetry, or not, we have the following four cases (see Figure 4).(1) (see Figure 4(a)); then, is dominated by .The two vertices in cannot dominate and at the same time; otherwise cannot be dominated by , a contradiction. Hence, at least one vertex of belongs to . Therefore, is a dominating set of and .(2) (see Figure 4(b)); then, is dominated by .The two vertices in cannot dominate and at the same time, that is, . Otherwise, cannot be dominated by , a contradiction.   Hence, at least one vertex of belongs to . In addition, for dominating in graph . If or , then can be a dominating set of . If , in view of at least one vertex of belonging to , then is a dominating set of . Therefore, .(3) (see Figure 4(c)); then, is dominated by .The two vertices in cannot dominate and at the same time; otherwise, cannot be dominated by , a contradiction. Hence, at least one vertex of belongs to . It implies that is a dominating set of and .(4) (see Figure 4(d)).If , then to dominate . And is a dominating set of .
If , then ; otherwise, . Thus, to dominate . Since and is a dominating set of , at least one vertex of belongs to . If , then is a dominating set of . If , then is a dominating set of .

Subcase 7. ; then, and the vertices should be dominated by the only one vertex in . We consider eleven possible cases.(1) (see Figure 5(a)); then, is dominated by .We claim that ; otherwise, cannot be dominated by . Hence, for dominating . In addition, to dominate . If or , then is a dominating set of . If , then is a dominating set of . Thus, .(2) (see Figure 5(b)); then, is dominated by .If or , then cannot be dominated by , which is a contradiction to the definition of . If , then is a dominating set of . Hence, we assume that