# The complexity of the Bondage problem in planar graphs

A set S⊆ V(G) of a graph G is a dominating set if each vertex has a neighbor in S or belongs to S. Let γ(G) be the cardinality of a minimum dominating set in G. The bondage number b(G) of a graph G is the smallest number of edges A⊆ E(G), such that γ(G-A)=γ(G)+1. The problem of finding b(G) for a graph G is known to be NP-hard even for bipartite graphs. In this paper, we show that deciding if b(G)=1 is NP-hard, while deciding if b(G)=2 is coNP-hard, even when G is restricted to one of the following classes: planar 3-regular graphs, planar claw-free graphs with maximum degree 3, planar bipartite graphs of maximum degree 3 with girth k, for any fixed k≥ 3.

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## 1 Introduction

Given a graph , a set is called a dominating set if every vertex is an element of or is adjacent to an element of . When is a dominating set of we say that dominates . The minimum cardinality of a dominating set in is called the domination number and is denoted by . A dominating set with is called a minimum dominating set, for short a -set. For an overview of the topics in graph domination, we refer to the book of Haynes et al. [9]. The bondage number has been introduced by Fink et al. [5] has a parameter to measure the criticality of a graph, in respect to the domination number. The bondage number of a graph is the minimum number of edges whose removal from increases the domination number, that is, , where . We say that an edge of a graph is -critical if . Therefore a graph has a -critical edge if and only if . The Bondage problem is defined as follows:

.99 Bondage

[2pt]     Instance: a graph and a positive integer . Question: is ?

In a similar fashion the -Bondage is the problem of deciding if , for a graph and a fixed integer . The -Bondage problem has been shown to be NP-hard in [11]. Recently, it has been shown in [10] that the -Bondage problem remains NP-hard for bipartite graphs. To this date, the bondage number and related properties have been extensively studied. For more details we refer to the survey of Xu in [12].

Our aim in this paper is to determine the computational complexity of the -Bondage and -Bondage problems for class of subcubic planar graphs and for some of its subclasses.

## 2 Notations and preliminaries

The graphs considered in this paper are finite and simple, that is, without directed edges or loops or parallel edges. The reader is referred to [1] for definitions and notations in graph theory, and to [8] for definitions and terminology concerning complexity theory.

Let be a graph with vertex set and edge set . Let and . We say that and are the endpoints of the edge. Let and denote its minimum degree and its maximum degree, respectively. The degree of in is or simply when the referred graph is obvious. A -vertex is a vertex of degree . The graph is -regular whenever all its vertices have the same degree, that is, for each . We say that a graph is cubic if it is -regular, and it is subcubic if . We denote by the open neighborhood of a vertex in , and its closed neighborhood in . When it is clear from context, we note and . For a subset , let denote the subgraph of induced by , which has vertex set and edge set . We often refer to as an induced subgraph of when it is clear from the context. If a graph has no induced subgraph isomorphic to a fixed graph , we say that is -free. Let be an induced subgraph of . We refer to the outside neighbors of in as the set of neighbors that does not belong to vertices of , that is, . Let denote the girth of , that is the minimum length of a cycle. By convention, when is acyclic.

For a vertex , we write , where , and for a subset , we write , where . For an edge , we write , and for a subset , we write . For , the graph denotes the cordless path on vertices, that is, and . For a path , we say that is an even path if

is even, else it is an odd path. For

, the graph denotes the cordless cycle on vertices, that is, and . For , is called a hole. The triangle graph is . The claw is the 4-vertex star, that is, the graph with vertices , , , and edges , , . The paw is the graph with four vertices , and edges .

For two vertex disjoint induced subgraphs of : is complete to if is an edge for any and , and is anti-complete to if is not an edge for any and . A set is called a clique if any pairwise distinct vertices are adjacent. When is a clique then is a complete graph. A set is called a stable set or an independent set if any pairwise distinct vertices are non adjacent. A set is called a vertex cover if every edge of has an endpoint in . We denote by the size of a minimum vertex cover of and by the size of a maximum independent set of . A -set of is a minimum vertex cover of and an -set of is a maximum independent set of . We denote by -core and -core the set of vertices that belongs to every -set and every -set, respectively. The set of vertices that belongs to no -set and no -set is named -anticore, -anticore, respectively. We say that a graph has -cores or -anticores when -core or -anticore, where . Note that given an -set of , the set of vertices is a -set of , and vice-versa. Hence -core-anticore and -anticore-core.

## 3 Cores and anticores

Before giving hardness results for the Bondage problem in planar graphs, we need to introduce several intermediate problems. We show that these problems belong to the class of NP-hard problems.

Boros et al. studied -cores in [2]. They claimed that deciding if a given graph is such that -core is NP-complete. To this end, they use Cook Reducibility from the Minimum Independent Set Problem. It states that could be computed using only an oracle for the question -core. Yet we found some flaws in this proof. Unless P=NP, we can say that there does not exist a polynomial algorithm to decide if a graph is such that -core. But to show that a problem is NP-hard, one must give a polynomial reduction (Karp reduction) from an NP-hard problem. Moreover, unless P=NP, this problem is clearly not in NP since it seems impossible to verify in polynomial time that a set of vertices of size belongs to -cores. Hence we decided to propose a polynomial reduction to the two following problem:

.99 -core

[2pt]     Instance: a graph . Question: is there a vertex such that belongs to every maximum independent set of ?

.99 -anticore

[2pt]     Instance: a graph . Question: is there a vertex such that belongs to no minimum vertex cover of ?

Similarly, to the -core problem corresponds the -anticore problem, and to the -anticore problem corresponds the -core problem. We prove that the problems -core and -anticore are both NP-hard using a reduction from 3-SAT with exactly three variables per clause. Cook in [3] has shown that 3-SAT is NP-complete.

###### Theorem 3.1

The -core problem and the -anticore problem are NP-hard.

Proof:  Let be an instance of 3-SAT with variable variables and clauses with exactly three variables per clause. Let be the set of variables and be the set of clauses. From this instance we construct a graph such that is satisfiable if and only if -anticore and -core.

Each clause of is associated with a triangle , where each of the three vertices represent one literal of . Then we make a copy such that is complete to , and remove the edges and . Hence we have a partition of the triangles into such that and each clause is represented exactly once by a triangle in each set.

Let be a pair of opposite literals in two distinct clauses of , with be its pair of corresponding vertices in , that is, corresponds to and correspond to , such that belongs to a triangle of and belongs to a triangle of . Then we add an edge . We repeat this operation for every pair of opposite literals.

We create two independent sets and of order , and so that is complete to . Then we add edges to make complete to the triangles of , respectively. Last we add the vertices that form the path , and such that are complete to , respectively. At this point, our graph is fully constructed.

From our construction, we highlight the following two ways to construct an independent set of order :

• Select a vertex in each triangle of then pick all vertices of and either or ;

• Select a vertex in each triangle of then pick all vertices of and either or .

Note that from the given constructions, if then -core and -anticore.

We claim that . From our construction, there is no independent set of that contains vertices of and ; nor and ; and neither and . Moreover, if is an independent set that contains some vertices of either or then . Hence any independent set of that contains some vertices of either or is at most of order . So if there is an independent set of order , it is composed of one vertex of each triangle of , one vertex of each triangle of , and the vertices and . Hence if then -core and -anticore.

It remains to prove that if and only if is satisfiable. We construct an -set of order from a satisfying assignment of . For each clause , select one literal in that satisfies . Then pick the vertices and that correspond to the literal , and where belongs to a triangle of and belongs to a triangle of . Note that no vertices corresponding to the opposite literal are taken. Hence from our construction is an independent set of . Then is an -set of of order .

Conversely, we construct an assignment of , from a -set of order . From previous arguments, is composed of one vertex of , one vertex of each triangle of , and the vertices . Let be a pair of vertices such that and , where is a triangle of and is a triangle of . If correspond to the same variable , then they do not correspond to opposite literals. Hence we can construct a satisfying assignment as follows: for each vertex , where , add the associated literal of to . At this point, satisfies . This completes the proof.

Since for a graph , we have -core-anticore and -anticore-core, it follows:

###### Corollary 3.2

The problems -core and -anticore are both NP-hard.

We describe one operation that will be of use in next theorem’s proof. Given a graph and , we define the graph , where as follows: and .

###### Property 3.3

Let be a vertex of a graph , then -anticore iff .

Proof:  Let be a vertex of a graph . Any -set of contains either or to cover the edge . Therefore if -anticore then . Else there exists a -set of containing that is also a -set of .

We show that the -anticore problem remains NP-hard for planar graphs.

###### Theorem 3.4

The -anticore problem is NP-hard when restricted to planar graphs.

Proof:  Garey, Johnson and Stockmeyer in [7] proved that Vertex Cover (referenced as Node Cover) is NP-complete when restricted to planar graphs. To that end, they gave a polynomial reduction from the Vertex Cover with no restrictions on the graph. We claim that the graph constructed from has the following property: -anticore if and only if -anticore. First we give an overview of their proof.

Given a graph , we construct a graph . Embed in the plane, allowing edges to cross but so that no more than two edges meet at any point, and no edge cross a vertex other than its own endpoint. Then replace each crossing by a copy of as shown in Figure 1 to build . For each , where , let be the cardinality of a minimal vertex cover of such that and . Note that are the four corners of , see Figure 2. The following properties hold:

Hence and for every vertex cover of , . From a -set of , you can build a -set of by taking all corresponding vertices of in and exactly vertices for each copy of . So if then , where is the number of copy of (i.e. the number of crossings). This completes the overview of the proof of Garey et al.

We show that if -anticore then -anticore. Let be a vertex in -anticore. From Property 3.3 we know that . Since can be constructed from , we have and therefore -anticore. Hence if a vertex belongs to -anticore, then also belongs to -anticore.

Conversely, we show that if -anticore then -anticore. Suppose that -anticore. Let be two edges of that are crossing. We construct the graph , a copy of with the crossing point replaced by the graph as described above. W.l.o.g. we can assume that there exists two minimum vertex cover of such that and . Let be the set of black vertices represented in Figure 2. Note that and that is a minimum vertex cover of . Also is covering the edges and while is covering the edges and . Hence is a vertex cover of and since , it is minimum. Now consider the degree rotational symmetry of Figure 2. Let be the set of black vertices of this symmetry. Note that is covering the edges and while is covering the edges and . Hence is a vertex cover of and since , it is minimum. From these two -sets of , we have -anticore.

By iterating until there is no more edge crossing, we obtain such that -anticore. From previous arguments, it follows that if a vertex of belongs to some minimum vertex cover of , then belongs to some minimum vertex of . This completes the proof.

Since for a graph , -anticore-core, it follows:

###### Corollary 3.5

The -core problem is NP-hard for planar graphs.

## 4 Bondage in planar graphs

In this section, we show that the -Bondage and the -Bondage problem are NP-hard and coNP-hard, respectively, in planar graphs with maximum degree . The two following remarks will be extensively used in the proofs below.

###### Remark 4.1

Let be an edge of a graph . If there exists a minimum dominating set of such that or , then is not a -critical edge.

###### Remark 4.2

Let be a vertex of a graph . If there exists a minimum dominating set of such that and , then every edge incident with is not -critical.

Using Theorem 3.4, we show the NP-hardness of the -bondage problem in planar graphs.

###### Theorem 4.3

The -Bondage problem is NP-hard when restricted to planar graphs of maximum degree .

Proof:  We give a polynomial reduction from the -anticore problem which has been shown to be NP-hard for planar graphs in Theorem 3.4. From a connected planar graph , we build a connected planar graph with maximum degree . Let and .

For each vertex of we associate the connected component . The component is composed as follows:

• An induced cycle of length , where , that is,
;

• An induced path that is connected to by the edge ;

• To each pair of , where , is associated an induced paw where is a triangle, is an edge, and is connected to by the edges and .

For each edge of , we pick a copy of the graph depicted in Figure 3. Note that each component and is planar with degree maximum . Moreover each of , where , has degree . We show how the components connect to each other so that the graph is planar. Consider a planar representation of . Each component is assigned the same position on the plane as its associated vertex of . We do similarly with each component and its associated edge of so that no components overlap. Embed each induced cycle in the plane as a circle so that the path and the paws , where , are inside. Then, for each vertex of , iterate counterclockwise on the edges incident to from to . At each step, let be the considered edge incident to , add the edge where belongs to the component . Hence is planar with . This completes the construction.

In the first part we will prove that . In the last part we will prove that if and only if -anticore.

Let be the associated component of an edge of , and be the only outside neighbors of in . Recall that , and that and are two edges. Let be a minimum dominating set of . One can check the following:

• If then ;

• If and then . The case and is symmetric;

• Else .

From now, we say that is covered if .

Let be the associated component of a vertex of . We claim that any dominating set of is such that . Let . Let be a dominating set of . Since we intend to prove a lower bound, we assume that all outside neighbors of in are in , that is, . From we construct , a dominating set of . Let . Note that the set of vertices of is dominated by . Since is an induced path in where is the only vertex with neighbors outside , we add to . With similar arguments, for each paw , where , we add to . At this step , and it remains to dominate . Note that is the disjoint unions of induced paths of size . Let such that is dominating . From our construction, one can check the following: , and if then .

Let be a dominating set of . We denote by the set of vertices of , respectively. We would like to point out the following:

• If then from above, we have . In this case we say that is a non-dominating configuration of . An example of such a configuration is , where .

• If and , then we say that is a dominating configuration of . In this case covers each component , where . An example of a dominating configuration is , where .

• We say that is a covered configuration if or is covering and . Examples of covered configurations where or cover are displayed in Figure 3(a), 3(c) and 3(b).

Let be a dominating set of . We define as the set of that correspond to a dominating configuration in , and the set of that are not covered in . From previous arguments, we have the following lower bound:

 |S|≥4m+∑v∈V(G)(2dG(v)+1)+|SG|+|SH|≥8m+n+|SG|+|SH| (4.1)

Let and be a minimum vertex cover of . Let be a dominating set of . If can be obtained from as described below, we say that is the dominating set associated to . To each vertex corresponds a dominating configuration of . To each vertex corresponds a non-dominating configuration of . At this point, for each edge , the associated is covered by . Therefore we associate a covered configuration of each , depending on which of or is covering . Hence dominates with and . So we have .

Let . We show that . By contradiction, suppose there exists a dominating set of such that . From 4.1 we have . From we construct a vertex cover of . For each component of , we add its vertex to . Now it remains to cover the edges associated with the components of . For each component of , let be its associated edge in : we add one of its endpoints, say , to . Thus is a vertex cover of such that , a contradiction.

In the last part, we show that if -anticore then has no -critical edge.

First, we show that no edge with an endpoint in a component is -critical in . Let be a minimum vertex cover of and be an associated dominating set in . If then consider the vertices of highlighted in Figure 3(b). From Remark 4.1 and 4.2, no bold edge is -critical. From this symmetric covered configuration, no edge with an endpoint in is -critical in . Second, w.l.o.g. and . Then covers and therefore we consider the highlighted vertices of of Figure 3(a). From Remark 4.1 and 4.2, no bold edge is -critical. The case and is symmetric. Hence no edge with an endpoint in is -critical in .

Now we show that no edge of a component is -critical in . Let . First suppose that -core. Since -anticore, there exists and , two -sets of , such that and . Let be two associated dominating sets of , respectively. Since each is covered by in , we can assume that is as depicted in Figure 3(c). Note that corresponds to a non-dominating configuration while corresponds to a dominating configuration of . We give a non-dominating configuration for and a dominating configuration for :

1. where or ;

2. where or or .

Hence we can take any previous configurations of in , respectively, such that and are still -sets of . Then from Remark 4.1 and 4.2, one can check that no edge of is -critical in .

Now -core. Since -anticore, we can assume that for each edge of , there exists a minimum vertex cover of such that . Let be an associated dominating set in . Note that correspond to a dominating configuration of , respectively, and therefore is covered by both and . Hence we replace the vertices of by the dominating black vertices of depicted in Figure 3(c). Thus the vertex dominates the vertex of , where . We exposed some dominating sets of to replace in . Item . represents some dominating configuration of , while item are configurations that cover every component , where .

1. where or or ;

2. , if , where ;

3. , if .

Hence we can take any exposed configurations of in such that is still a -set of . Then from Remark 4.1 and 4.2, one can check the following: from . no edge incident to a vertex is -critical in ; from . no edge incident to is -critical in ; from . no edge incident to is -critical in . These configurations can be applied to any and therefore has no -critical edge in .

Last we claim that if -anticore then has a -critical edge. Let -anticore. We show that there is no -set of such that has a dominating configuration. By contradiction, let be a -set of such that is a dominating configuration of . From we construct a minimum vertex cover of . For each component of , the vertex is in to . For each component of , one of its endpoints, say , is in . From previous arguments, we have and thus . Yet is a minimum vertex cover of where a contradiction. Hence there is no -set of where has a dominating configuration. Let be a -set of . From previous arguments we have only two possible configurations of in , that are:

1. ;

2. .

Then in every -set of , the vertex is dominated by . Hence the edge is -critical in . So we can conclude that if and only if -anticore. This completes the proof.

###### Corollary 4.4

The -Bondage problem is coNP-hard when restricted to planar graphs of maximum degree .

Proof:  In the preceding proof, the graph has no -set such that . Since are isolated in , we have and therefore . Thus deciding if has not a bondage number two is equivalent to deciding if it has a bondage number one.

## 5 Bondage in restricted planar graphs

Extending the arguments of the proof of Theorem 4.3, we obtain the following.

###### Theorem 5.1

The -Bondage problem is NP-hard when restricted to planar claw-free graphs of maximum degree .

Proof:  The graph constructed in the proof of Theorem 4.3 contains the following claws: , , , for every vertex of , where and .

We describe the operation . Let be a vertex at the center of a claw in . We replace by a component as depicted in the right of Figure 5. Let be the graph obtained from this operation. Note that is planar and . The component satisfies the following properties. We have and is a -set. The graphs , ,