Algorithm and Hardness results on Liar's Dominating Set and k-tuple Dominating Set

02/27/2019 ∙ by Sandip Banerjee, et al. ∙ TU Wien 0

Given a graph G=(V,E), the dominating set problem asks for a minimum subset of vertices D⊆ V such that every vertex u∈ V∖ D is adjacent to at least one vertex v∈ D. That is, the set D satisfies the condition that |N[v]∩ D|≥ 1 for each v∈ V, where N[v] is the closed neighborhood of v. In this paper, we study two variants of the classical dominating set problem: k-tuple dominating set (k-DS) problem and Liar's dominating set (LDS) problem, and obtain several algorithmic and hardness results. On the algorithmic side, we present a constant factor (11/2)-approximation algorithm for the Liar's dominating set problem on unit disk graphs. Then, we obtain a PTAS for the k-tuple dominating set problem on unit disk graphs. On the hardness side, we show a Ω (n^2) bits lower bound for the space complexity of any (randomized) streaming algorithm for Liar's dominating set problem as well as for the k-tuple dominating set problem. Furthermore, we prove that the Liar's dominating set problem on bipartite graphs is W[2]-hard.

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1 Introduction

The dominating set problem is regarded as one of the fundamental problems in theoretical computer science which finds its applications in various fields of science and engineering [7, 2]. A dominating set of a graph is a subset of such that every vertex in is adjacent to at least one vertex in . The domination number, denoted as , is the minimum cardinality of a dominating set of . Garey and Johnson [5] showed that deciding whether a given graph has domination number at most some given integer is NP-complete. For a vertex , the open neighborhood of the vertex denoted as is defined as and the closed neighborhood of the vertex denoted as is defined as .

-tuple Dominating Set (-Ds):

Fink and Jacobson [4] generalized the concept of dominating sets as follows.

[colback=red!5!white] -tuple Dominating Set (-DS) Problem
Input:
A graph and a non-negative integer .
Goal: Choose a minimum cardinality subset of vertices such that, for every vertex , .

The -tuple domination number is the minimum cardinality of a -DS of . A good survey on -DS can be found in [3, 11]. Note that, -tuple dominating set is the usual dominating set, and -tuple and -tuple dominating set are known as double dominating set [6] and triple dominating set [15], respectively. Further note that, for a graph , , if there exists no -DS of . Klasing et al. [9] studied the -DS problem from hardness and approximation point of view. They gave a -approximation algorithm for the -tuple domination problem in general graphs, and showed that it cannot be approximated within a ratio of , for any .

Liar’s Dominating Set (LDS):

Slater [17] introduced a variant of the dominating set problem called Liar’s dominating set problem. Given a graph , in this problem the objective is to choose minimum number of vertices such that each vertex is double dominated and for every two vertices there are at least three vertices in from the union of their neighborhood set. The LDS problem is an important theoretical model for the following real-world problem. Consider a large computer network where a virus (generated elsewhere in the Internet) can attack any of the processors in the network. The network can be viewed as an unweighted graph. For each vertex , an anti-virus can: detect the virus at as well as in its closed neighborhood , and find and report the vertex at which the virus is located. Notice that, one can make network virus free by deploying the anti-virus at the vertices , where is the minimum size dominating set. However, in certain situations the anti-viruses may fail. Hence to make the system virus free it is likely to double-guard the nodes of the network, which is indeed the -tuple DS. But, if the anti-viruses detect the location of the viruses but fails to cure it properly due to some software error or corrupted circumstances. In this scenario, LDS serves the purpose. We define the problem formally below.

[colback=red!5!white] Liar’s Dominating Set (LDS) Problem
Input:
A graph and a non-negative integer .
Goal: Choose a subset of vertices of minimum cardinality such that

  • for every vertex , ,

  • for every pair of vertices of distinct vertices .

1.1 Our Results

In this paper, we obtain seveal algorithmic and hardness results for LDS and -DS problems on various graph families. On the algorithmic side in section 2, we present a constant factor ()-approximation algorithm for the Liar’s dominating set (LDS) problem on unit disk graphs. Then, we obtain a PTAS for the -tuple dominating set (-DS) problem on unit disk graphs. On the hardness side in section 3, we show a bits lower bound for the space complexity of any (randomized) streaming algorithm for Liar’s dominating set problem as well as for the -tuple dominating set problem. Furthermore, we prove that the Liar’s dominating set problem on bipartite graphs is W[2]-hard.

2 Algorithmic Results

2.1 Approximation Algorithm for LDS on Unit Disk Graphs

A unit disk graph (UDG) is an intersection graph of a family of unit radius disks in the plane. Formally, given a collection of unit disks in the plane, a UDG is defined as a graph , where each vertex corresponds to a disk and there is an edge between two vertices and if and only if their corresponding disks and contain and , respectively. Here, we consider the geometric variant of LDS known as Euclidean Liar’s dominating set problem, which is defined as follows:

[colback=red!5!white] Liar’s Dominating Set on UDG (LDS-UDG) Problem
Input:
A unit disk graph , where is a set of disk centers.
Output: A minimum size subset such that

  • for each point , .

  • for each pair of points , .

Jallu et al. [8] studied the LDS problem on unit disk graphs, and proved that this problem is NP-complete. Furthermore, given an unit disk graph and an , they have designed a -factor approximation algorithm to find an LDS in with running time , where . In this section, we design a -factor approximation algorithm that runs in sub-quadratic time.

For a point , let denote the disk centered at the point . For any two points , if , then we say that is a neighbor of (sometimes we say is covered by ) and vice versa. Since for any Liar’s dominating set , () holds, we assume that and for all the points , . For a point , let and denote the and coordinates of , respectively. Let , , denote the set of points inside the circle centered at and of radius , and unit, respectively.

The basic idea of our algorithm is as follows. Initially, we sort the points of based on their -coordinates. Now, consider the leftmost point (say ). We compute the sets , and . Next, we compute the set . Further, for each point , we compute the set . Finally, we compute the set . Moreover, our algorithm is divided into two cases.

Case 1 ():

For each point such that , we further distinguish between the following cases.

  1. If : we pick two arbitrary points from the set , and include them in the output set (see Figure 1(a)).

  2. If : in this case, we choose those two (or one) points (say ) in the output set (see Figure 1(b)).

Once these points are selected, we remove the remaining points from at this step from the set . Notice that the points that lie in are already -dominated. Later, we can pick those points if required. However, observe that we may choose only point from while we are in the case of . So we maintain a counter in Case 1. This counter keeps track of how many points we are picking from the set in total, for each point . If is at least , we simply add to the output set and do not enter into Case 2. Otherwise, we proceed to Case 2.

Case 2 ( or ):

here, we further distinguish between the following cases.

  1. If : then we choose points arbitrarily in the output set (see Figure 2(a)).

  2. If : let be these points. We include both of them in the output set . This settles the first condition of LDS for them. However, in order to fulfill the second condition of LDS for and , we must include at least one extra point here. First, we check the cardinality of . If , then we pick an arbitrary point from , and include in . Otherwise, we include two points and , and include them in (see Figure 2(b)). Note that, in this case, we know that and exist due to the input constraint of an LDS problem.

  3. If : then we check and include two points and arbitrarily from (see Figure 2(c)).

This fulfills the criteria of LDS of points in . Then we delete the remaining points of from . Next we select the left-most point from the remaining and repeat the same procedure until is empty. The pseudocode is given in Algorithm 1.

Figure 1: An illustration of Case 1; (a) , (b) .
Figure 2: An illustration of Case 2; (a) , (b) , (c) .

Input: A set of points in the plane.
Output: A subset of points , that is a liar’s dominating set of the unit disk graph defined on the points of .

1:  Sort the points of based on their -coordinates.Let be the sorted order.
2:  Let
3:   /* is a counter*/
4:  while  do
5:     
6:     Compute , ,
7:     Compute
8:     for  to  do
9:        Compute /**/
10:     Compute
11:     if  then
12:        for  to  do
13:           if  then
14:              if  then
15:                  /* chosen arbitrarily*/
16:                 
17:           else if  then
18:               /* are included in */
19:              
20:           else if  then
21:               /* are included in */
22:              
23:           
24:     if  then
25:         /*Since already points are present from in */
26:        Jump to line 
27:     else
28:         return
29:     
30:      /* is the next left-most point in */
31:  return  .
Procedure 1 Approximation Algorithm for LDS-UDG ()

Input: A point , the sets .
Output: A subset of points .

1:  Set
2:  Set
3:  if  then
4:      /* chosen arbitrarily.*/
5:  else if  then
6:      /* is the other point in .*/
7:     if  then
8:         /* is chosen arbitrarily.*/
9:     else
10:         /* and is chosen arbitarily.*/
11:  else
12:      /* chosen arbitrarily.*/
13:  return  
Procedure 2
Lemma 1

The set obtained from Algorithm 1, is a liar’s dominating set of the unit disk graph defined on the points of .

Proof

Algorithm 1 primarily relies on two cases. We begin with the leftmost point (say ). We first compute the sets , and . Then, we consider the points that lie in . If the disks centered at these points of radius  share points with , we choose points from their intersection to . This choice is based on the number of points available in the intersection. For every point , if (), we arbitrarily choose points in . Otherwise, we choose one or two points in , based on the cardinality of . Once these points are selected, we remove the remaining points of at this step from the set . Notice that the points lying in are already -dominated. Later, we can pick those points if required. Clearly, after this process any point that lies in and shares points with is at least -dominated or -dominated, Notice that for any point that lies outside , its corresponding disk does not share any region with . Later, for any points in if we have to choose points from to fulfill the criteria of LDS, we already took them in the output set at this stage. Meanwhile, we maintain a counter that keeps track of how many points we are picking from , for each point . If is at least , we know . We simply add to the output set and do not enter into Case 2. Otherwise, we proceed to Case 2. Besides, if there is no point , such that the disk centered at shares points with , we are in Case 2.

Now we show the correctness of Case 2. In this case, if , then we include three points arbitrarily including (say in ). Note that, the distance between any two points is at most . Since we are including points from in , it fulfills the criteria of LDS of the points contained inside the unit disk centered at . Next, consider the case if . Let be the other point in . We compute . Note that, due to the definitions of the problem, we know and . When we include an additional point arbitrarily (say ) in to fulfill the second condition of the LDS of the points . While , we include points and (both the points are chosen arbitrarily) from and , respectively. Notice that this makes the points and individually dominated. Now, for any pair of points from , at least points are chosen (that is, ). Finally, when , we include two points arbitrarily in to fulfill the criteria of LDS for . Note that the existence of two points and in the set follows from the problem constraint.

Thus, at every iteration, we made sure that the considered disk and the points inside that disk, fulfills the criteria of LDS. Furthermore, due to Case 1, we have taken sufficient points from the considered disk for the disks that intersect with it. This completes the proof.

Lemma 2

Algorithm 1 outputs a liar’s dominating set of the unit disk graph defined on the points of with approximation ratio .

Proof

For each point , we consider the set and choose points from this set. Consider the disk of radius . Notice that the points that lie outside the set , their corresponding disks of radius , do not share any points with the set . Thus, we only need to consider the points inside . The half-perimeter of the of the disk of radius is (since we are only considering the points which are to the right of ). Thus, we can pack at most disks of radius  inside such that they mutually do not contain the center of other disks (see Figure 3). We consider the points that lie in . If the disks centered at these points of radius  share points with , we choose points from their intersection to . Now, for each point , we choose at most two points from (see Procedure 1). Once these points are selected, we remove the remaining points at this step from the set (recall that ). The points that lie inside this disk, is already -dominated. Later, we can pick the point itself, if required. Besides we also pick . In this case, we pick at most .

Figure 3: An illustration of Lemma 2; at most disks of radius  can be packed such that they mutually do not contain the center of other disks.

Now, if there is no point in , whose corresponding disk that shares a region , or the number of points picked from is less than , we enter into Case 2 (see Procedure 2). Here, we include at most three points including the point . However, we enter into this case only when sufficient points have not been picked in Case 1. Thus, in this case we have picked fewer than points from (precisely, we pick points).

Now, any optimal solution contains at least two points from , for each point . Since we have to fulfill the first constraint of the LDS that each disk is individually dominated. Hence, we get the approximation factor .

Algorithm 1 clearly runs in polynomial time (to be precise in sub-quadratic time). Thus, from Lemma 1 and Lemma 2 we conclude the following theorem.

Theorem 2.1

Algorithm 1 computes a liar’s dominating set of the unit disk graph defined on the points of in sub-quadratic time with approximation ratio .

2.2 PTAS for -DS on Unit Disk Graphs

In this section we give a PTAS for the -tuple dominating set on unit disk graphs with a similar approach used by Nieberg and Hurink [12]. It might be possible to design a PTAS by using local search or shifting strategy for the -tuple dominating set problem on unit disk graphs. However, the complexity of these algorithms would be dependent on and . Thus we use the approach of Nieberg and Hurink [12], that gurantees a better running time.

Let be an unit disk graph in the plane. For a vertex , let and be the -th neighborhood and -th closed neighborhood of , respectively. For any two vertices , let be the distance between and in , that is the number of edges of a shortest path between and in . Let be the minimum -tuple dominating set of . For a subset , let be the minimum -tuple dominating set of the induced subgraph on . We prove the following theorem.

Theorem 2.2

There exists a PTAS for the -tuple dominating set problem on unit disk graphs.

Proof

The -separated collection of subsets is defined as follows: Given a graph , let be a collection of subsets of vertices , for , such that for any two vertices and with , . In the following lemma we prove that the sum of the cardinalities of the minimum -tuple dominating sets for the subsets of a -separated collection is a lower bound on the cardinality of .

Lemma 3

Given a graph , let be a 2-separated collection of subsets then, .

Proof

Consider a subset , and its neighborhood . We know from the properties of -separated collection of subsets that, any two subsets are disjoint for each . Any vertex outside has at least distance more than one to any vertex in . Hence, has to be a -tuple domination of the vertices of , where is the -tuple domination of . Otherwise the vertices of is not -tuple dominated, since no vertex outside can dominate a vertex of . On the other hand, we know . This implies . Hence, .

From Lemma 3, we get the lower bound of the minimum -tuple dominating set of . If we can enlarge each of the subset to a subset such that the -tuple dominating set of (that is ) is locally bounded to the -tuple dominating set of (that is ), then by taking the union of them we get an approximation of the -tuple dominating set of . For each subset , let there is a subset (where ), and let there exists a bound () such that . Then, if we take the union of the -tuple dominating sets of all , this is a -approximation of the -tuple dominating sets of the union of subsets (for ).

Now, we describe the algorithm. We begin with an arbitrary vertex , and compute the -tuple dominating sets of minimum cardinality for these neighborhoods until (for a constant ). Then, this process is used iteratively on the remaining graph induced by (where is the first point at th iteration when the condition is violated). Let be the total number of iterations, where . Let be the neighborhoods achieved from this process. The following lemma shows that a -tuple dominating set for the entire graph is given by the union of the sets .

Lemma 4

Let be the set of neighborhoods created by the above algorithm. The union forms a -tuple dominating set of .

Proof

Consider the set , and we know . Thus, . The algorithm stops while , which means . Besides, . Thus, if we compute the -tuple dominating set of each , their union clearly is the -tuple dominating set of the entire graph.

These subsets , for , created by the algorithm form a -separated collection in . Consider any two neighborhoods . We have computed on graph induced by (where ). So, for any two vertices and , the distance is at least . Thus we have the following corollary.

Corollary 1

The algorithm returns a -tuple dominating set of cardinality no more than the size of a dominating set .

It needs to be shown that this algorithm has a polynomial running time. The number of iterations is clearly bounded by . It is important to show that for each iteration we can compute the minimum -tuple dominating set in polynomial time for being constant or polynomially bounded.

Consider the -th neighborhood of a vertex , . Let be the maximal independent set of the graph induced by . From [12], we have . The cardinality of a minimum dominating set in is bounded from above by the cardinality of a maximal independent set in . Hence, . Now, we prove the following lemma.

Lemma 5

.

Proof

For a vertex , let be the -th closed neighborhood of . Let be the first maximal independent set of . We know .

Now, we take the next maximal independent set from , and take the union of them . Notice that every vertex has neighbors in , so they can be -tuple dominated by choosing vertices from . Also, every vertex can be -tuple dominated by choosing vertices from , since itself can be one and the other one can be picked from . Additionally, for each vertex in , we take a vertex from the neighborhood of in . Let be the union of these vertices. Now, every vertex can be -tuple dominated by choosing vertices from . So, . . Hence, . We continue this process times.

After steps, we get the union of the maximal independent sets . Additionally, we get the unions of . Notice that every vertex can be -tuple dominated by choosing vertices from . The cardinality of is at most , which is . We also know is upper bounded by . Thus, .

Nieberg and Hurink [12] showed that for a unit disk graph, there exists a bound on (the first value of that violates the property . This bound depends on the approximation not on the size of the of the unit disk graph given as input. Precisely, they have proved that there exists a constant such that , that is, the largest neighborhood to be considered during the iteration of the algorithm is bounded by a constant. Putting everything together, we conclude the proof.

3 Hardness Results

3.1 Streaming Lower Bound for LDS

In this section, we consider the streaming model: the edges arrive one-by-one in some order, and at each time-stamp we need to decide if we either store the edge or forget about it. We now show that any streaming algorithm that solves the LDS problem must essentially store all the edges.

Theorem 3.1

Any randomized111

By randomized algorithm we mean that the algorithm should succeed with probability

. streaming algorithm for LDS problem on -vertex graphs requires space.

Proof

We will reduce from the Index problem in communication complexity:

[colback=red!5!white] Index Problem
Input:
Alice has a string given by . Bob has an index .
Question: Bob wants to find , i.e., the bit of .

It is well-known that there is a lower bound of bits in the one-way randomized communication model for Bob to compute  [10]. We assume an instance of the Index problem where is a perfect square, and let . Fix any bijection from . Consequently we can interpret the bit string as an adjacency matrix for a bipartite graph with vertices on each side. Let the two sides of the bipartition be and

From the instance of Index, we construct an instance of the LDS. Assume that Alice has an algorithm that solves the -tuple dominating set problem using bits. First, we insert the edges corresponding to the edge interpretation of between nodes and : for each , Alice adds the edge if the corresponding entry in is 1. Alice then sends the memory contents of her algorithm to Bob, using bits.

Bob has the index , which he interprets as under the same bijection . He receives the memory contents of the algorithm, and proceeds to do the following:

  • Add two vertices and , and an edge

  • Add an edge from each vertex of to

  • Add an edge from each vertex of to

  • Add five vertices and edges and .

  • Add an edge from each vertex of to each vertex from

Let be a minimum LDS of . Note that has to be a double dominating set of . Since has only neighbors in , it follows that . Similarly . Note that also has only two neighbors in . Hence, we have that