# Liar's Domination in Unit Disk Graphs

In this article, we study a variant of the minimum dominating set problem known as the minimum liar's dominating set (MLDS) problem. We prove that the MLDS problem is NP-hard in unit disk graphs. Next, we show that the recent sub-quadratic time 11/2-factor approximation algorithm <cit.> for the MLDS problem is erroneous and propose a simple O(n + m) time 7.31-factor approximation algorithm, where n and m are the number of vertices and edges in the input unit disk graph, respectively. Finally, we prove that the MLDS problem admits a polynomial-time approximation scheme.

## Authors

• 5 publications
• 4 publications
• 7 publications
• ### Total Domination in Unit Disk Graphs

Let G=(V,E) be an undirected graph. We call D_t ⊆ V as a total dominatin...
07/23/2020 ∙ by Sangram K. Jena, et al. ∙ 0

• ### The Generalized Independent and Dominating Set Problems on Unit Disk Graphs

In this article, we study a generalized version of the maximum independe...
06/27/2020 ∙ by Sangram K. Jena, et al. ∙ 0

• ### On the Approximability and Hardness of the Minimum Connected Dominating Set with Routing Cost Constraint

In the problem of minimum connected dominating set with routing cost con...
11/29/2017 ∙ by Tung-Wei Kuo, et al. ∙ 0

• ### APX-Hardness and Approximation for the k-Burning Number Problem

Consider an information diffusion process on a graph G that starts with ...
06/25/2020 ∙ by Debajyoti Mondal, et al. ∙ 0

• ### Packing 2D disks into a 3D container

In this article, we consider the problem of finding in three dimensions ...
10/25/2021 ∙ by Helmut Alt, et al. ∙ 0

• ### Consensus under Network Interruption and Effective Resistance Interdiction

We study the problem of network robustness under consensus dynamics. We ...
09/11/2020 ∙ by S. Rasoul Etesami, et al. ∙ 0

• ### A 0.502·MaxCut Approximation using Quadratic Programming

We study the MaxCut problem for graphs G=(V,E). The problem is NP-hard, ...
04/29/2021 ∙ by Stefan Steinerberger, et al. ∙ 0

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## 1 Introduction

Given a simple undirected graph , the open and closed neighborhoods of a vertex are defined by and , respectively. A dominating set of is a subset of such that every vertex in is adjacent to at least one vertex in . That is, each vertex is either in or there exists a vertex such that . Observe that for any dominating set , for each . We say that a vertex is dominated by in , if and . The dominating set problem asks to find a dominating set of minimum size in a given graph. A set is a -tuple dominating set in , if each vertex is dominated by at least vertices in . In other words, for each . The minimum cardinality of a -tuple dominating set of a graph is called the -tuple domination number of .

A liar’s dominating set (LDS) in a simple undirected graph , is a dominating set having the following two properties: (i) for every , , and (ii) for every pair of distinct vertices and in , . For a given graph , the problem of finding an LDS in of minimum cardinality is known as the minimum liar’s dominating set (MLDS) problem. The cardinality of an MLDS in a graph is known as the liar’s domination number of . Every 3-tuple dominating set is a liar’s dominating set as it satisfies both conditions, so the liar’s domination number lies between 2-tuple and 3-tuple domination numbers.

Our interest in the LDS problem arises from the following scenario. Consider a graph in which each node is a possible location for an intruder such as a thief, or a saboteur. We would like to detect and report the intruder’s location in the graph. A protection device such as a camera or a sensor placed at a node can not only detect (and report) the intruder’s presence at it, but also at its neighbors. Our objective is to place a minimum number of protection devices such that the intrusion of the intruder at any vertex is detected and reported. In this situation, one must place the devices at the vertices of a minimum dominating set of the graph to achieve the goal. The protection devices are prone to failure and hence certain degree of redundancy is needed in the solution. Also, some times the devices may misreport the intruder’s location deliberately or due to transmission error. Assume that at most one protection device in the closed neighborhood of the intruder can lie (misreport). In this context, one must place the protection devices at the vertices of an MLDS of the graph to achieve the objective. The first property in the definition of LDS deals with single device fault-tolerance, while the second property deals with the case in which two distinct locations about the intruder are reported.

## 2 Related Work

The MLDS problem is introduced by Slater [16]. He showed that the problem is NP-hard for general graphs, and gave a lower bound on the liar’s domination number in case of trees by proving that the size of any liar’s dominating set of a tree of order is between and . Later, Roden and Slater [14] characterized tree classes with liar’s domination number equal to . In the same paper, they also showed that the MLDS problem is NP-hard even for bipartite graphs. Panda and Paul [12] proved that the problem is NP-hard for split graphs and chordal graphs. They also proposed a linear time algorithm for computing an MLDS in case of trees.

Panda et al. [9] studied the approximability of the problem and presented an -factor approximation algorithm, where is the degree of the graph. Panda and Paul [10] considered the problem for proper interval graphs and proposed a linear time algorithm for computing a minimum cardinality liar’s dominating set. The problem is also studied for bounded degree graphs, and -claw free graphs [9]. Sterling [17] considered the problem on two-dimensional grid graphs and presented bounds on the liar’s domination number.

Alimadadi et al. [1] provided the characterization of graphs and trees for which the liar’s domination number is and , respectively. Panda and Paul [11, 13] studied variants of liar’s domination, namely, connected liar’s domination and total liar’s domination. A connected liar’s dominating set (CLDS) is an LDS whose induced subgraph is connected. A total liar’s dominating set (TLDS) is a dominating set with the following two properties: (i) for every , , and (ii) for every distinct pair of vertices and , , where is the open neighborhood of a vertex. The objective of both problems is to find CLDS and TLDS of minimum size, respectively. The authors also proved that both problems are NP-hard and proposed -factor approximation algorithms. They also proved that the problems are APX-complete for graphs with maximum degree 4. Jallu and Das [7] first studied the geometric version of the MLDS problem, and presented constant factor approximation algorithms with high running time. Recently, Banerjee and Bhore [2] proposed a -factor approximation algorithm in sub-quadratic time. However, unfortunately, their approximation analysis is erroneous and the approximation factor is at least 11 (refer Section 4).

### 2.1 Our Contribution

We study the MLDS problem on a geometric intersection graph model, particularly in UDGs. A unit disk graph (UDG) is an intersection graph of equal radii disks in the plane. Given a set of circular disks in the plane, each having radius 1, the corresponding UDG is defined as follows: each vertex corresponds to a disk , and there is an edge between two vertices and if and only if the Euclidean distance between the corresponding disk centers and is at most 1.

We show that the decision version of the MLDS problem is NP-complete in UDGs (refer to Section 3). We propose a simple linear time 7.31-factor approximation algorithm and a PTAS in Section 4 and Section 5, respectively. Finally, we conclude the paper in Section 6.

## 3 Hardness of the MLDS Problem in UDGs

In this section, we show that the MLDS problem in UDGs is NP-complete by reducing the vertex cover problem defined in planar graphs to it, which is known to be NP-complete [4]. The decision versions of both the problems are formally defined below.

The MLDS problem in UDGs

(Lds-Udg)

Instance:

A unit disk graph and a positive integer .

Question:

Does there exist a liar’s dominating set in such that ?.

The vertex cover problem in planar graphs

(Vc-Pla)

Instance:

A simple planar graph with maximum degree 3 and a positive integer .

Question:

Does there exist a vertex cover of such that ?.

###### Lemma 3.1 ([18]).

A planar graph with maximum degree 4 can be embedded in the plane using area in such a way that its vertices are at integer coordinates and its edges are drawn so that they are made up of line segments of the form or , for integers and .

This kind of embedding is known as orthogonal drawing of a graph. Biedl and Kant [3] gave a linear time algorithm that produces an orthogonal drawing of a given graph with the property that the number of bends along each edge is at most 2 (see Figure 1).

###### Corollary 3.2.

A planar graph with maximum degree 3 and can be embedded in the plane such that its vertices are at and its edges are drawn as a sequence of consecutive line segments on the lines or , for integers and .

###### Lemma 3.3.

Let be an instance of Vc-Pla with . An instance of Lds-Udg can be constructed from in polynomial-time.

###### Proof.

We construct in four phases.
Phase 1: Embedding of into a grid of size
Embed the instance in the plane as discussed previously using one of the algorithms in [5, 6]. An edge in the embedding is a sequence of connected line segment(s) of length four units each. If the total number of line segments used in the embedding is , then the sum of the lengths of the line segments is as each line segment has length 4 units. We name the points in the embedding correspond to the vertices of by node points (see Figure 1(b)).

Phase 2: Adding extra points to the embedding
Divide the set of line segments in the embedding into two categories, namely, proper and improper. We call a line segment proper if none of its end points correspond to a vertex in . A line segment is improper if it is not a proper segment. For each edge of length 4 units we add two points at distances 1 and 1.5 units of and , respectively (thus adding four points in total, see the edge in Figure 1(c)). For each edge of length greater than 4 units, we also add points as follows: for each improper line segment we add four points at distances 1, 1.5, 2.5, and 3.5 units from the endpoint corresponding to a vertex in , and for each proper line segment we add four points at distances 0.5 and 1.5 units from its endpoints (see Figure 1(c)). We name the points added in this phase joint points.

Phase 3: Adding extra line segments and points
Add a line segment of length 1.4 units (on the lines or for some integers or ) for every point , which corresponds to a vertex in , without coinciding with the line segments that had already been drawn. Observe that adding this line segment on the lines or is possible without losing the planarity as the maximum degree of is 3. Now, add three points (say , , and ) on these line segments at distances 0.2, 1.2, and 1.4 units, respectively, from . We name the points added in this phase support points.
Phase 4: Construction of UDG
For convenience, let us denote the set of node points, joint points, and support points by , , and , respectively. Let , , and . We construct a UDG , where and there is an edge between two points in if and only if the Euclidean distance between the points is at most 1 (see Figure 1(c)). Observe that, , , where is the total number of line segments in the embedding, and . Hence, and is bounded by a polynomial of . Therefore can be constructed in polynomial-time. ∎

###### Theorem 3.4.

Lds-Udg is NP-complete.

###### Proof.

Lds-Udg, since for any given set and a positive integer , we can verify whether is a liar’s dominating set of size at most or not in polynomial-time.

We prove the hardness of Lds-Udg by reducing Vc-Pla to it. Let be an instance of Vc-Pla. Construct an instance of Lds-Udg as discussed in Lemma 3.3. We now prove the following claim: has a vertex cover of size at most if and only if has a liar’s dominating set of size at most .

Necessity: Let be a vertex cover of such that . Let , i.e., is the set of vertices (or node points) in that correspond to the vertices in . From each segment in the embedding we choose 3 vertices (joint points). The set of chosen vertices, say , together with and will form an LDS of desired cardinality in . We now discuss the process of obtaining the set . Initially . As is a vertex cover, every edge in has at least one of its end vertices in . Let be an edge in and (the tie can be broken arbitrarily if both and are in ). Note that the edge is represented as a sequence of line segments in the embedding. Start traversing the segments (of ) from , where corresponds to , and add all the vertices to except the first one from each segment encountered in the traversal (see in Figure 2 (b). The bold vertices are part of while traversing from ).

Apply the above process to each edge in . Observe that the cardinality of is as we have chosen 3 vertices from each segment in the embedding. Let . Now, we argue that is a liar’s dominating set in .

1. Each is dominated by in . If (i.e., the corresponding vertex in ), then . If , then there must exist at least one vertex in dominating . The existence of is guaranteed by the way we constructed . Hence, . In either case every vertex in is dominated by at least two vertices in . It is needless to say that vertex in is dominated by at least two vertices in . Similarly, every vertex in is dominated by itself, by its neighbor(s) in , and, perhaps, by one vertex in . Therefore, every vertex in is double dominated by vertices in .

2. Consider a pair of distinct vertices in . Of course, every pair of distinct vertices in satisfy the liar’s second condition. We prove that remaining pairs of distinct vertices also do satisfy the liar’s second condition by considering all possible cases.
Case a. : If at least one of belongs to (without loss of generality say ), then . If none of belongs to , then there must exists some such that dominate , respectively. Hence, .
Case b. : If both , then it is trivial that . Suppose one of belongs to (without loss of generality let us assume ). As every vertex in is double dominated, must be dominated by two vertices in or by either some in and some in . In either case we get . A similar argument works even if none of belong to .
Case c. and : If none of and belong to , then the argument is trivial as each one is dominated by at least two vertices in . If both belong to , then . If and (the other case is similar), then holds as is double dominated.

Likewise, we can argue for other pair combinations too. Therefore, every pair of distinct vertices in is dominated by at least 3 vertices in .

Therefore is an LDS in and .

Sufficiency: Let be an LDS of size at most . We prove that has a vertex cover of size at most with the aid of the following claims.

1. [label=()]

2. .

3. Every segment in the embedding must contribute at least 3 vertices to and hence , where is the total number of segments in the embedding.

4. If and correspond to end vertices of an edge in , and if both are not in , then there must be at least vertices in form the segment(s) representing the edge , where is the number of segments representing the edge in the embedding.

Claim (i) directly follows from the definition of liar’s dominating set. Observe that we added points such that is adjacent to , is adjacent to , and is adjacent to in , i.e., , for each . Hence, and must be in due to the first condition of the liar’s domination. Also, every connected component of in must contain at least three vertices due to the second condition of liar’s domination. Hence, . Therefore, any liar’s dominating set of must contain , i.e., .

Claim (ii) follows from the fact that only consecutive points are adjacent (in ) on any segment in the embedding. Let be a segment in the embedding having vertices , and . On contrary, assume that has only two of its vertices in . Note that both and can not be in simultaneously. If both are present in , then they do not satisfy the second condition as and are not in , i.e., ; contradiction to is an LDS. Without loss of generality we assume that (the similar argument works even if ). If and are in , then . If and are in , then . If and are in , then . In either case we arrived at a contradiction.

Claim (iii) follows from Claim (ii). Let be an edge in such that and are not in . By Claim (ii) every segment must contribute at least three vertices to . Hence, the number of vertices in from the segments representing the edge is at least . We argue that if both and are not in , then the number of vertices in from the segments representing the edge is at least . Suppose that there are exactly vertices in from the segments. That is, no segment representing the edge contains more than three vertices in . Let be the vertices encountered while traversing the segments from . If , the argument can be proven as in the proof of Claim (ii). Assume .

Case a. is even:

Since and are not in and due to the second condition of the liar’s domination, the vertices from the first segment and from the last segment must be in . The vertices and can not be in as we assumed that each segment contains exactly three vertices in . If we continue in the same manner for the rest of the segments from both sides, we end up in not choosing the vertices and from the -th and -th segments, respectively. Note that is the last vertex on -th segment and is the first vertex on -th segment and is an edge in (see Figure 3(a)). Also, note that . Implies, the vertices and are not satisfying the second condition, which is a contradiction to our assumption that is an LDS of .

Case b.

is odd:

If we proceed as in Case a, we can observe that must contain all the four vertices on -th segment, i.e., the middle segment, (see Figure 3(b)). Which is a contradiction to our assumption that no segment, representing the edge , contains more than three vertices in .

We now shall show that, by removing and/or replacing some vertices in , a set of vertices from can be chosen such that the corresponding vertices in is a vertex cover. The vertices in account for vertices in (due to Claim (i)). Let and . If any edge in has none of its end vertices in , then we do the following: consider the sequence of segments representing the edge in the embedding. Since, both and are not in , there must exist a segment having all its vertices in (due to Claim (iii)). Consider the segment having its four vertices in . Delete any one of the vertices on the segment and introduce (or ). Update and repeat the process till every edge has at least one of its end vertices in . Due to Claim (ii), is a vertex cover in with . Therefore, Lds-Udg is NP-complete. ∎

## 4 Approximation Algorithm

Banerjee and Bhore [2] in their recent paper proposed an approximation algorithm and claim that their algorithm achieves a 5.5-factor approximation ratio for the MLDS problem in UDGs. However, their approximation analysis is erroneous. We first provide a counterexample defying their claim and then propose a simple 7.31-factor approximation algorithm for the said problem.

For completeness here we give the idea of the algorithm proposed in [2] briefly. As a first step, the point set (i.e., the set of disk centers) is sorted according to their -coordinates. Now consider the left most point, say , and consider in the solution. Next, compute the set of points of that are inside the circle centered at and of radius , 1, and . Let these sets be , , and , respectively. The points which lie outside the set , their corresponding disks of radius 1 do not contain any point from . So, it suffice to consider the points inside to ensure liar’s domination for . Since is the left most point in , the set can contain at most five mutually independent points (i.e., the mutual distance between those five points is greater than one. In other words, the unit radius disks centered at those points do not contain the centers of other disks). In the next step (call it Case 1), for each point , the algorithm chooses at most two points from the set in the solution, if available, where is the set of points lying in the unit disk centered at . After selection of these points, is updated to and proceed to next point in . Thus, the algorithm picks at most points in this iteration. Let . However, could be an empty set due to either or for each (call it Case 2). If or , then the algorithm chooses at most 4 points (including ) from depending on the cardinality of . Thus, in this case the algorithm picks fewer than 11 points from . The points chosen so far ensures the liar’s domination for the points in . Now, is updated to , and the process is repeated (with the next leftmost point, say ) until is empty.

For each point , any optimal solution should contain at least two points from due to the first condition of liar’s domination, and the algorithm chooses at most 11 points. Thus, the authors claim that the proposed algorithm is a -factor approximation by the charging argument 11 points in the solution returned by algorithm can be charged to two points in the optimal solution. But, the same two points in the optimal solution could be charged multiple times.

Suppose and are the left most points considered in two successive iterations, respectively. There may be a case that the algorithm could end up by choosing a set of 11 points in the solution to dominate for which the same optimal solution for is enough to ensure liar’s domination for . We elaborate our claim in detail with an example.

Consider the set of points in Figure 4(a) as an instance to the algorithm. The points are sorted according to their -coordinates. Let the leftmost point be (see Figure 4(b)). The points and are five mutually independent points chosen by the algorithm such that , for . Along with , the total number of points chosen in this iteration is 11. Update . In the next iteration, is the leftmost point (see Figure 4(c)) and are five mutually independent points chosen by the algorithm so that , for . The algorithm chooses 11 points (including ) in the solution. Observe that in both the iterations the algorithm doesn’t enter Case 2 and, hence, chooses 22 points. In fact, any two (resp. three) red points (see Figure 4 (c)) are sufficient to ensure the liar’s domination first (resp. second) condition for the point sets and . After a few iterations will be chosen as the next left most point and the algorithm chooses 11 points (by Case 1) in the solution (see Figure 4(d)). However, the same three red points ensures liar’s domination for , , and . So the approximation factor of the algorithm proposed in [2] is at least 11.

### 4.1 A 7.31-factor approximation algorithm

In this Subsection, we propose a 7.31-factor approximation algorithm (see Algorithm 1) for minimum liar’s dominating set (MLDS) problem in UDGs. The basic idea of the algorithm is: sequentially compute three maximal independent sets in the given UDG and add extra vertices, if necessary, to ensure liar’s domination. In [15] Shang et. al. established a relation between maximal independent set and minimum -dominating set111A minimum -dominating set of is a minimum dominating set of with the property that every vertex not in should have at least dominators in . in UDGs. By using their result, we can have a 10-factor approximation algorithm for liar’s dominating set in UDGs. In the following lemma, the proof idea is similar to [15], we establish a relation between the cardinalities of maximal independent set and minimum liar’s dominating set to obtain a 7.31-factor approximation algorithm for the MLDS problem in UDGs.

###### Lemma 4.1.

Let be a UDG. If and denote a maximal independent set and an MLDS of , respectively, then .

###### Proof.

Let , , . For , let denote the number of vertices in which lie in the closed neighborhoods of and in , i.e., . As is a liar’s dominating set of , for each , and we get . For , analogues to , let . As is a UDG, for each vertex in there can be at most 5 independent vertices in its neighborhood, and thus for each and in . Hence, we get . Note that the number of edges in induced between and is . Thus, we have , which implies . Therefore, . ∎

###### Lemma 4.2.

The set returned by Algorithm 1 is an LDS of .

###### Proof.

Algorithm 1 sub-sequentially computes three maximal independent sets , , and in (see line numbers 4-9). Let . Note that any vertex not in has a neighbor (dominator) in each , and . Thus, each vertex (resp. every pair of distinct vertices) not in satisfies the first (resp. second) condition of liar’s domination. Also, every vertex in is adjacent to a vertex in and . Thus, the vertices in satisfy both the conditions of the lair’s domination. Similarly, every vertex in is adjacent to a vertex in (otherwise, cannot be a maximal independent set) and, hence, the vertices in satisfy both the conditions of the lair’s domination. For any vertex , if , then the algorithm adds an arbitrary neighbor of to (see line number 13). If , then has neighbor in . In either case the vertices in satisfy the two conditions of the liar’s domination.

For any pair of distinct vertices and , the second condition is already satisfied as has a neighbor in each , and . Similarly, for any pair of distinct vertices , and , if has multiple neighbors in or has multiple neighbors in , then the second condition is satisfied. If is the only neighbor of in and vice versa, then the algorithm adds an arbitrary neighbor of to , see line number 16, and thus the second condition is ensured for and . If , the second condition is trivially holds as has a neighbor in each and . Therefore, is an LDS in . ∎

###### Theorem 4.3.

For a given UDG , Algorithm 1 achieves approximation ratio 7.31 for the MLDS problem in time.

###### Proof.

Let be an MLDS of . Algorithm 1 sequentially computes three maximal independent sets , , and in and is not necessarily be an LDS of as there might be some vertices with one of the following cases: (i) a vertex not satisfying the first condition, or (ii) a pair of distinct vertices and not satisfying the second condition of the liar’s domination. In the former case we add an arbitrary neighbor of , and in the latter case we add an neighbor of . Note that in either case such a neighbor is guaranteed to exist in , and . Without loss of generality we can assume that . Therefore, (by Lemma 4.1). The running time follows as Algorithm 1 uses the subroutine (in line number 6) to compute a maximal independent set. ∎

## 5 Approximation Scheme

In this section, we propose a PTAS for the MLDS problem in UDGs, i.e., for a given UDG and a parameter , we propose an algorithm which produces a liar’s dominating set of size no more than times the size of a minimum liar’s dominating set in G. We use to denote the number of edges on a shortest path between and in . For , denotes the distance between and and is defined as . For , and denote an LDS and an optimal (minimum size) LDS of in , respectively. We define the closed neighborhood of a set as .

The proposed PTAS is based on the concept of -separated collection of subsets of (). Let be a UDG. A collection such that for , is said to be an -separated collection, if , for and (see Figure 5 for a 4-separated collection). Nieberg and Hurink [8] introduced 2-separated collection to propose a PTAS for the minimum dominating set problem in UDGs and our PTAS follows form it. However, the algorithm in [8] cannot be directly applied as in intermediate steps of the algorithm we need to add extra nodes (see line numbers 14-23 in Algorithm 2) to ensure the liar’s domination. We argue that the extra nodes added are small enough and do not effect the approximation factor.

###### Lemma 5.1.

Let be an -separated collection. If for , then .

###### Proof.

Observe that for and . Also, as and are -separated. Let for . Observe that and is a liar’s dominating set of for . Since, for , implies . Therefore, and . Also, for as is a liar’s dominating set of , and is a minimum size liar’s dominating set. Thus, . ∎

###### Lemma 5.2.

Let be an -separated collection, and be subsets of with for all . If there exists such that holds for all , and if is a liar’s dominating set in , then the value of is at most times the size of a minimum liar’s dominating set in .