Liar's Domination in Unit Disk Graphs
LLiar’s Domination in Unit Disk Graphs ∗ Ramesh K. Jallu †‡ , Sangram K. Jena § , and Gautam K. Das ¶ The Czech Academy of Sciences, Institute of Computer Science Department of Mathematics, Indian Institute of Technology GuwahatiMay 29, 2020
Abstract
In this article, we study a variant of the minimum dominating set problemknown as the minimum liar’s dominating set (MLDS) problem. We prove that theMLDS problem is NP-hard in unit disk graphs. Next, we show that the recentsub-quadratic time -factor approximation algorithm [2] for the MLDS problem iserroneous 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-timeapproximation scheme. keywords: Dominating set, Liar’s dominating set, Unit Disk Graph, Approxima-tion scheme
Given a simple undirected graph G = ( V, E ), the open and closed neighborhoods ofa vertex v i ∈ V are defined by N G ( v i ) = { v j ∈ V | ( v i , v j ) ∈ E and v i (cid:54) = v j } and N G [ v i ] = N G ( v i ) ∪ { v i } , respectively. A dominating set D of G is a subset of V such thatevery vertex in V \ D is adjacent to at least one vertex in D . That is, each vertex v i ∈ V is either in D or there exists a vertex v j ∈ D such that ( v i , v j ) ∈ E . Observe that for ∗ Preliminary version of this paper appeared in COCOON, 2018 † [email protected] ‡ Ramesh K. Jallu was supported by the Czech Science Foundation, grant number GJ19-06792Y, andby institutional support RVO:67985807 § [email protected] ¶ [email protected] a r X i v : . [ c s . CC ] M a y ny dominating set D ⊆ V , | N G [ v i ] ∩ D | ≥ v i ∈ V . We say that a vertex v i is dominated by v j in G , if v j ∈ D and ( v i , v j ) ∈ E . The dominating set problem asksto find a dominating set of minimum size in a given graph. A set D ⊆ V is a k -tupledominating set in G , if each vertex v i ∈ V is dominated by at least k vertices in D . Inother words, | N G [ v i ] ∩ D | ≥ k for each v i ∈ V . The minimum cardinality of a k -tupledominating set of a graph G is called the k -tuple domination number of G .A liar’s dominating set (LDS) in a simple undirected graph G = ( V, E ), is a dominat-ing set D having the following two properties: (i) for every v i ∈ V , | N G [ v i ] ∩ D | ≥
2, and(ii) for every pair of distinct vertices v i and v j in V , | ( N G [ v i ] ∪ N G [ v j ]) ∩ D | ≥
3. For agiven graph G , the problem of finding an LDS in G of minimum cardinality is known asthe minimum liar’s dominating set (MLDS) problem. The cardinality of an MLDS in agraph G is known as the liar’s domination number of G . Every 3-tuple dominating set isa liar’s dominating set as it satisfies both conditions, so the liar’s domination number liesbetween 2-tuple and 3-tuple domination numbers.Our interest in the LDS problem arises from the following scenario. Consider a graphin 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 protectiondevice such as a camera or a sensor placed at a node can not only detect (and report) theintruder’s presence at it, but also at its neighbors. Our objective is to place a minimumnumber of protection devices such that the intrusion of the intruder at any vertex isdetected and reported. In this situation, one must place the devices at the vertices ofa minimum dominating set of the graph to achieve the goal. The protection devicesare 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 totransmission error. Assume that at most one protection device in the closed neighborhoodof the intruder can lie (misreport). In this context, one must place the protection devicesat the vertices of an MLDS of the graph to achieve the objective. The first property inthe definition of LDS deals with single device fault-tolerance, while the second propertydeals with the case in which two distinct locations about the intruder are reported. 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 incase of trees by proving that the size of any liar’s dominating set of a tree of order n is between ( n + 1) and n . Later, Roden and Slater [14] characterized tree classes withliar’s domination number equal to ( n + 1). In the same paper, they also showed that the2LDS problem is NP-hard even for bipartite graphs. Panda and Paul [10] proved thatthe problem is NP-hard for split graphs and chordal graphs. They also proposed a lineartime algorithm for computing an MLDS in case of trees.Panda et al. [13] studied the approximability of the problem and presented an O (ln ∆)-factor approximation algorithm, where ∆ is the degree of the graph. Panda and Paul [11]considered the problem for proper interval graphs and proposed a linear time algorithm forcomputing a minimum cardinality liar’s dominating set. The problem is also studied forbounded degree graphs, and p -claw free graphs [13]. Sterling [17] considered the problemon 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 theliar’s domination number is | V | and | V | −
1, respectively. Panda and Paul [9, 12] studiedvariants of liar’s domination, namely, connected liar’s domination and total liar’s domi-nation. A connected liar’s dominating set (CLDS) is an LDS whose induced subgraph isconnected. A total liar’s dominating set (TLDS) is a dominating set D with the followingtwo properties: (i) for every v ∈ V , | N G ( v ) ∩ D | ≥
2, and (ii) for every distinct pairof vertices u and v , | ( N G ( u ) ∪ N G ( v )) ∩ D | ≥
3, where N G ( · ) is the open neighborhoodof a vertex. The objective of both problems is to find CLDS and TLDS of minimumsize, respectively. The authors also proved that both problems are NP-hard and proposed O (ln ∆)-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 geometricversion of the MLDS problem, and presented constant factor approximation algorithmswith high running time. Recently, Banerjee and Bhore [2] proposed a -factor approx-imation algorithm in sub-quadratic time. However, unfortunately, their approximationanalysis is erroneous and the approximation factor is at least 11 (refer Section 4). We study the MLDS problem on a geometric intersection graph model, particularly inUDGs. A unit disk graph (UDG) is an intersection graph of equal radii disks in the plane.Given a set { d , d , . . . , d n } of n circular disks in the plane, each having radius 1, thecorresponding UDG G = ( V, E ) is defined as follows: each vertex v i ∈ V corresponds toa disk d i , and there is an edge between two vertices v i and v j if and only if the Euclideandistance between the corresponding disk centers d i and d j 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 algorithmand a PTAS in Section 4 and Section 5, respectively. Finally, we conclude the paper inSection 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 G = ( V, E ) and a positive integer k . Question:
Does there exist a liar’s dominating set D in G such that | D | ≤ k ?. The vertex cover problem in planar graphs ( Vc-Pla ) Instance:
A simple planar graph G with maximum degree 3 and a positive integer k . Question:
Does there exist a vertex cover C of G such that | C | ≤ k ?. Lemma 3.1 ([18]) . A planar graph G = ( V, E ) with maximum degree 4 can be embeddedin the plane using O ( | V | ) area in such a way that its vertices are at integer coordinatesand its edges are drawn so that they are made up of line segments of the form x = i or y = j , for integers i and j . 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 graphwith the property that the number of bends along each edge is at most 2 (see Figure 1).
Corollary 3.2.
A planar graph G = ( V, E ) with maximum degree 3 and | E | ≥ can beembedded in the plane such that its vertices are at (4 i, j ) and its edges are drawn as asequence of consecutive line segments on the lines x = 4 i or y = 4 j , for integers i and j . Lemma 3.3.
Let G = ( V, E ) be an instance of Vc-Pla with | E | ≥ . An instance G (cid:48) = ( V (cid:48) , E (cid:48) ) of Lds-Udg can be constructed from G in polynomial-time.Proof. We construct G (cid:48) in four phases. Phase 1: Embedding of G into a grid of size × G in the plane as discussed previously using one of the algorithmsin [5, 6]. An edge in the embedding is a sequence of connected line segment(s) of lengthfour units each. If the total number of line segments used in the embedding is (cid:96) , then thesum of the lengths of the line segments is 4 (cid:96) as each line segment has length 4 units. Wename the points in the embedding correspond to the vertices of G by node points (seeFigure 1(b)). 4 v v v v v (a) p p p p p p (b) p p p p p p (c) Figure 1: (a) A planar graph G with maximum degree 3, (b) its embedding on a grid,and (c) construction of an UDG from the embedding. Phase 2: Adding extra points to the embedding
Divide the set of line segments in the embedding into two categories, namely, proper andimproper. We call a line segment proper if none of its end points correspond to a vertexin G . A line segment is improper if it is not a proper segment. For each edge ( p i , p j ) oflength 4 units we add two points at distances 1 and 1.5 units of p i and p j , respectively(thus adding four points in total, see the edge ( p , p ) in Figure 1(c)). For each edge oflength greater than 4 units, we also add points as follows: for each improper line segmentwe add four points at distances 1, 1.5, 2.5, and 3.5 units from the endpoint correspondingto a vertex in G , and for each proper line segment we add four points at distances 0.5 and1.5 units from its endpoints (see Figure 1(c)). We name the points added in this phase joint points . 5 hase 3: Adding extra line segments and points Add a line segment of length 1.4 units (on the lines x = 4 i or y = 4 j for some integers i or j ) for every point p i , which corresponds to a vertex v i in G , without coinciding with theline segments that had already been drawn. Observe that adding this line segment on thelines x = 4 i or y = 4 j is possible without losing the planarity as the maximum degree of G is 3. Now, add three points (say x i , y i , and z i ) on these line segments at distances 0.2,1.2, and 1.4 units, respectively, from p i . We name the points added in this phase supportpoints . Phase 4: Construction of UDG
For convenience, let us denote the set of node points, joint points, and support pointsby N , J , and S , respectively. Let N = { p i | v i ∈ V } , J = { q , q , . . . , q (cid:96) } , and S = { x i , y i , z i | v i ∈ V } . We construct a UDG G (cid:48) = ( V (cid:48) , E (cid:48) ), where V (cid:48) = N ∪ J ∪ S andthere is an edge between two points in V (cid:48) if and only if the Euclidean distance betweenthe points is at most 1 (see Figure 1(c)). Observe that, | N | = | V | (= n ), | J | = 4 (cid:96) , where (cid:96) is the total number of line segments in the embedding, and | S | = 3 | V | (= 3 n ). Hence, | V (cid:48) | = 4( n + (cid:96) ) and (cid:96) is bounded by a polynomial of n . Therefore G (cid:48) can be constructedin polynomial-time. Theorem 3.4.
Lds-Udg is NP-complete.Proof.
Lds-Udg ∈ N P , since for any given set D ⊆ V and a positive integer k , we canverify whether D is a liar’s dominating set of size at most k or not in polynomial-time.We prove the hardness of Lds-Udg by reducing
Vc-Pla to it. Let G = ( V, E ) bean instance of
Vc-Pla . Construct an instance G (cid:48) = ( V (cid:48) , E (cid:48) ) of Lds-Udg as discussedin Lemma 3.3. We now prove the following claim: G has a vertex cover of size at most k if and only if G (cid:48) has a liar’s dominating set of size at most k + 3 (cid:96) + 3 n . Necessity:
Let C ⊆ V be a vertex cover of G such that | C | ≤ k . Let N (cid:48) = { p i ∈ N | v i ∈ C } , i.e., N (cid:48) is the set of vertices (or node points) in G (cid:48) that correspond to the verticesin C . From each segment in the embedding we choose 3 vertices (joint points). Theset of chosen vertices, say J (cid:48) ( ⊆ J ), together with N (cid:48) and S will form an LDS of desiredcardinality in G (cid:48) . We now discuss the process of obtaining the set J (cid:48) . Initially J (cid:48) = ∅ . As C is a vertex cover, every edge in G has at least one of its end vertices in C . Let ( v i , v j )be an edge in G and v i ∈ C (the tie can be broken arbitrarily if both v i and v j are in C ).Note that the edge ( v i , v j ) is represented as a sequence of line segments in the embedding.Start traversing the segments (of ( v i , v j )) from p i , where p i corresponds to v i , and add allthe vertices to J (cid:48) except the first one from each segment encountered in the traversal (see( p , p ) in Figure 2 (b). The bold vertices are part of J (cid:48) while traversing from p ).6 v v v v v (a) p p p p p p (b) Figure 2: (a) A vertex cover { v , v , v } in G , and (b) the construction of J (cid:48) in G (cid:48) (thetie between v and v , and v and v is broken by choosing v )Apply the above process to each edge in G . Observe that the cardinality of J (cid:48) is 3 (cid:96) as we have chosen 3 vertices from each segment in the embedding. Let D = N (cid:48) ∪ J (cid:48) ∪ S .Now, we argue that D is a liar’s dominating set in G (cid:48) .1. Each p i ∈ N is dominated by x i in S . If p i ∈ N (cid:48) (i.e., the corresponding vertex v i ∈ C in G ), then | N G (cid:48) [ p i ] ∩ D | ≥ |{ p i , x i }| = 2. If p i / ∈ N (cid:48) , then there must existat least one vertex q j in J (cid:48) dominating p i . The existence of q j is guaranteed by theway we constructed J (cid:48) . Hence, | N G (cid:48) [ p i ] ∩ D | ≥ |{ q j , x i }| = 2. In either case everyvertex in N is dominated by at least two vertices in D . It is needless to say thatvertex in J is dominated by at least two vertices in N (cid:48) ∪ J (cid:48) . Similarly, every vertexin S is dominated by itself, by its neighbor(s) in S , and, perhaps, by one vertex in N (cid:48) . Therefore, every vertex in V (cid:48) is double dominated by vertices in D .2. Consider a pair of distinct vertices in V (cid:48) . Of course, every pair of distinct verticesin S satisfy the liar’s second condition. We prove that remaining pairs of distinctvertices also do satisfy the liar’s second condition by considering all possible cases. Case a. p i , p j ∈ N : If at least one of p i , p j belongs to N (cid:48) (without loss of generalitysay p i ∈ N (cid:48) ), then | ( N G (cid:48) [ p i ] ∪ N G (cid:48) [ p j ]) ∩ D | ≥ |{ x i , x j , p i }| = 3. If none of p i , p j belongs to N (cid:48) , then there must exists some q i , q j ∈ J (cid:48) such that q i , q j dominate p i , p j , respectively. Hence, | ( N G (cid:48) [ p i ] ∪ N G (cid:48) [ p j ]) ∩ D | ≥ |{ x i , x j , q i , q j }| = 4. Case b. q i , q j ∈ J : If both q i , q j ∈ J (cid:48) , then it is trivial that | ( N G (cid:48) [ q i ] ∪ N G (cid:48) [ q j ]) ∩ D | ≥
3. Suppose one of q i , q j belongs to J (cid:48) (without loss of generality let us assume q i ∈ J (cid:48) ). As every vertex in G (cid:48) is double dominated, q j must be dominated by twovertices in J (cid:48) or by either some q k in J (cid:48) and some p l in N (cid:48) . In either case we get | ( N G (cid:48) [ q i ] ∪ N G (cid:48) [ q j ]) ∩ D | ≥
3. A similar argument works even if none of q i , q j belongto J (cid:48) . 7 ase c. p i ∈ N and q j ∈ J : If none of p i and q j belong to D , then the argument istrivial as each one is dominated by at least two vertices in D . If both belong to D ,then | ( N G (cid:48) [ p i ] ∪ N G (cid:48) [ q j ]) ∩ D | ≥ |{ p i , x i , q j }| = 3. If p i ∈ D and q j / ∈ D (the othercase is similar), then | ( N G (cid:48) [ p i ] ∪ N G (cid:48) [ q j ]) ∩ D | ≥ q j is double dominated.Likewise, we can argue for other pair combinations too. Therefore, every pair ofdistinct vertices in V (cid:48) is dominated by at least 3 vertices in D .Therefore D is an LDS in G (cid:48) and | D | = | N (cid:48) | + | J (cid:48) | + | S | ≤ k + 3 (cid:96) + 3 n . Sufficiency:
Let D ⊆ V (cid:48) be an LDS of size at most k + 3 (cid:96) + 3 n . We prove that G has avertex cover of size at most k with the aid of the following claims.(i) S ⊂ D .(ii) Every segment in the embedding must contribute at least 3 vertices to D and hence | J ∩ D | ≥ (cid:96) , where (cid:96) is the total number of segments in the embedding.(iii) If p i and p j correspond to end vertices of an edge ( v i , v j ) in G , and if both p i , p j are not in D , then there must be at least 3 (cid:96) (cid:48) + 1 vertices in D form the segment(s)representing the edge ( v i , v j ), where (cid:96) (cid:48) is the number of segments representing theedge ( v i , v j ) in the embedding.Claim (i) directly follows from the definition of liar’s dominating set. Observe thatwe added points x i , y i , z i such that p i is adjacent to x i , x i is adjacent to y i , and y i isadjacent to z i in G (cid:48) , i.e., { ( p i , x i ) , ( x i , y i ) , ( y i , z i ) } ⊂ E (cid:48) , for each i . Hence, z i and y i must be in D due to the first condition of the liar’s domination. Also, every connectedcomponent of D in G must contain at least three vertices due to the second condition ofliar’s domination. Hence, x i ∈ D . Therefore, any liar’s dominating set of G (cid:48) must contain { x i , y i , z i } , ≤ i ≤ n , i.e., S ⊂ D .Claim (ii) follows from the fact that only consecutive points are adjacent (in G (cid:48) ) onany segment in the embedding. Let η be a segment in the embedding having vertices q i , q i +1 , q i +2 , and q i +3 . On contrary, assume that η has only two of its vertices in D . Notethat both q i +1 and q i +2 can not be in D simultaneously. If both are present in D , thenthey do not satisfy the second condition as q i and q i +3 are not in D , i.e., | ( N G (cid:48) [ q i +1 ] ∪ N G (cid:48) [ q i +2 ]) ∩ D | = |{ q i +1 , q i +2 }| = 2; contradiction to D is an LDS. Without loss ofgenerality we assume that q i +2 / ∈ D (the similar argument works even if q i +1 / ∈ D ). If q i and q i +1 are in D , then | ( N G (cid:48) [ q i +1 ] ∪ N G (cid:48) [ q i +2 ]) ∩ D | = |{ q i , q i +1 }| = 2. If q i and q i +3 are in D , then | ( N G (cid:48) [ q i +1 ] ∪ N G (cid:48) [ q i +2 ]) ∩ D | = |{ q i , q i +3 }| = 2. If q i +1 and q i +3 are in D , then | ( N G (cid:48) [ q i +1 ] ∪ N G (cid:48) [ q i +2 ]) ∩ D | = |{ q i +1 , q i +3 }| = 2. In either case we arrived at acontradiction. 8laim (iii) follows from Claim (ii). Let ( v i , v j ) be an edge in G such that p i and p j are not in D . By Claim (ii) every segment must contribute at least three vertices to D .Hence, the number of vertices in D from the segments representing the edge ( v i , v j ) is atleast 3 (cid:96) (cid:48) . We argue that if both p i and p j are not in D , then the number of vertices in D from the segments representing the edge ( v i , v j ) is at least 3 (cid:96) (cid:48) + 1. Suppose that there areexactly 3 (cid:96) (cid:48) vertices in D from the segments. That is, no segment representing the edge( v i , v j ) contains more than three vertices in D . Let p i , q , q , . . . , q (cid:96) (cid:48) , p j be the verticesencountered while traversing the segments from p i . If (cid:96) (cid:48) = 1, the argument can be provenas in the proof of Claim (ii). Assume (cid:96) (cid:48) > p i p j p i p j q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q ( a ) ( b ) Figure 3: Illustration of Claim (iii). The vertices marked red must be in D . Case a. (cid:96) (cid:48) is even:
Since p i and p j are not in D and due to the second condition of theliar’s domination, the vertices q , q , q from the first segment and q (cid:96) (cid:48) − , q (cid:96) (cid:48) − , q (cid:96) (cid:48) from the last segment must be in D . The vertices q and q (cid:96) (cid:48) − can not be in D aswe assumed that each segment contains exactly three vertices in D . If we continuein the same manner for the rest of the segments from both sides, we end up innot choosing the vertices q (cid:96) (cid:48) and q (cid:96) (cid:48) +1 from the (cid:96) (cid:48) -th and ( (cid:96) (cid:48) + 1)-th segments,respectively. Note that q (cid:96) (cid:48) is the last vertex on (cid:96) (cid:48) -th segment and q (cid:96) (cid:48) +1 is the firstvertex on ( (cid:96) (cid:48) +1)-th segment and ( q (cid:96) (cid:48) , q (cid:96) (cid:48) +1 ) is an edge in G (cid:48) (see Figure 3(a)). Also,note that | ( N G (cid:48) [ q (cid:96) (cid:48) ] ∪ N G (cid:48) [ q (cid:96) (cid:48) +1 ]) ∩ D | = |{ q (cid:96) (cid:48) − , q (cid:96) (cid:48) +2 }| = 2. Implies, the vertices q (cid:96) (cid:48) and q (cid:96) (cid:48) +1 are not satisfying the second condition, which is a contradiction toour assumption that D is an LDS of G (cid:48) . Case b. (cid:96) (cid:48) is odd:
If we proceed as in Case a, we can observe that D must contain all the four vertices on ( (cid:96) (cid:48) + 1)-th segment, i.e., the middle segment, (see Figure 3(b)).9hich is a contradiction to our assumption that no segment, representing the edge( v i , v j ), contains more than three vertices in D .We now shall show that, by removing and/or replacing some vertices in D , a set of k vertices from N can be chosen such that the corresponding vertices in G is a vertex cover.The vertices in S account for 3 n vertices in D (due to Claim (i)). Let D = D \ S and C = { v i ∈ V | p i ∈ D ∩ N } . If any edge ( v i , v j ) in G has none of its end vertices in C ,then we do the following: consider the sequence of segments representing the edge ( v i , v j )in the embedding. Since, both p i and p j are not in D , there must exist a segment havingall its vertices in D (due to Claim (iii)). Consider the segment having its four vertices in D . Delete any one of the vertices on the segment and introduce p i (or p j ). Update C andrepeat the process till every edge has at least one of its end vertices in C . Due to Claim(ii), C is a vertex cover in G with | C | ≤ k . Therefore, Lds-Udg is NP-complete.
Banerjee and Bhore [2] in their recent paper proposed an approximation algorithm andclaim that their algorithm achieves a 5.5-factor approximation ratio for the MLDS prob-lem in UDGs. However, their approximation analysis is erroneous. We first provide acounterexample defying their claim and then propose a simple 7.31-factor approximationalgorithm for the said problem.For completeness here we give the idea of the algorithm proposed in [2] briefly. Asa first step, the point set P (i.e., the set of disk centers) is sorted according to their x -coordinates. Now consider the left most point, say p i , and consider p i in the solution.Next, compute the set of points of P that are inside the circle centered at p i and of radius , 1, and . Let these sets be Cov ( C ( p i )), Cov ( C ( p i )), and Cov ( C ( p i )), respectively.The points which lie outside the set Cov ( C ( p i )), their corresponding disks of radius 1do not contain any point from Cov ( C ( p i )). So, it suffice to consider the points inside Cov ( C ( p i )) to ensure liar’s domination for Cov ( C ( p i )). Since p i is the left most point in P , the set Q = Cov ( C ( p i )) \ Cov ( C ( p i )) can contain at most five mutually independentpoints (i.e., the mutual distance between those five points is greater than one. In otherwords, the unit radius disks centered at those points do not contain the centers of otherdisks). In the next step (call it Case 1), for each point q i ∈ Q , the algorithm choosesat most two points from the set S ( q i ) = Cov ( C ( p i )) ∩ Cov ( C ( q i )) in the solution, ifavailable, where Cov ( C ( q i )) is the set of points lying in the unit disk centered at q i .After selection of these points, Q is updated to Q \ Cov ( C ( q i )) and proceed to next pointin Q . Thus, the algorithm picks at most 5 × = (cid:83) q i ∈ Q S ( q i ). However, S could be an empty set due to either Q = ∅ or S ( q i ) = ∅ foreach q i ∈ Q (call it Case 2). If S = ∅ or | S | <
2, then the algorithm chooses at most4 points (including p i ) from Cov ( C ( p i )) depending on the cardinality of Cov ( C ( p i )).Thus, in this case the algorithm picks fewer than 11 points from Cov ( C ( p i )). The pointschosen so far ensures the liar’s domination for the points in Cov ( C ( p i )). Now, P isupdated to P \ Cov ( C ( p i )), and the process is repeated (with the next leftmost point,say p j ) until P is empty.For each point p i ∈ P , any optimal solution should contain at least two points from Cov ( C ( p i )) due to the first condition of liar’s domination, and the algorithm choosesat most 11 points. Thus, the authors claim that the proposed algorithm is a -factorapproximation by the charging argument 11 points in the solution returned by algorithmcan be charged to two points in the optimal solution. But, the same two points in theoptimal solution could be charged multiple times.Suppose p i and p j 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 of11 points in the solution to dominate Cov ( C ( p j )) for which the same optimal solutionfor Cov ( C ( p i )) is enough to ensure liar’s domination for Cov ( C ( p j )). We elaborate ourclaim in detail with an example.Consider the set of points in Figure 4(a) as an instance to the algorithm. The pointsare sorted according to their x -coordinates. Let the leftmost point be p i (see Figure 4(b)).The points { q , q , q , q , q } ∈ Q and are five mutually independent points chosen by thealgorithm such that Cov ( C ( p i )) ∩ Cov ( C ( q j )) = 2, for j = 1 , , . . . ,
5. Along with p i ,the total number of points chosen in this iteration is 11. Update P = P \ Cov ( C ( p i )).In the next iteration, p j is the leftmost point (see Figure 4(c)) and { q , q , q , q , q } ∈ Q are five mutually independent points chosen by the algorithm so that Cov ( C ( p j )) ∩ Cov ( C ( q j )) = 2, for i = 6 , , . . . ,
10. The algorithm chooses 11 points (including p j ) inthe 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)) aresufficient to ensure the liar’s domination first (resp. second) condition for the point sets Cov ( C ( p i )) and Cov ( C ( p j )). After a few iterations p k will be chosen as the next leftmost point and the algorithm chooses 11 points (by Case 1) in the solution (see Figure4(d)). However, the same three red points ensures liar’s domination for Cov ( C ( p i )), Cov ( C ( p j )), and Cov ( C ( p k )). So the approximation factor of the algorithm proposedin [2] is at least 11. 11 a) p i q q q q q p j q q p k q (b) p i q q q q q p j q q q q q p k q q (c) p i q q q q q p j q q q q q p k q q q q q (d) Figure 4: (a) A point set: an instance, (b) 11 points chosen from
Cov ( C ( p i )) out ofwhich 3 red points are in optimal solution, and (c) and (d) the selected red points for Cov ( C ( p i )) will ensure liar’s domination for Cov ( C ( p j )) and Cov ( C ( p k )).12 .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 thealgorithm is: sequentially compute three maximal independent sets in the given UDGand add extra vertices, if necessary, to ensure liar’s domination. In [15] Shang et. al.established a relation between maximal independent set and minimum k -dominating set in UDGs. By using their result, we can have a 10-factor approximation algorithm forliar’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 andminimum liar’s dominating set to obtain a 7.31-factor approximation algorithm for theMLDS problem in UDGs. Lemma 4.1.
Let G = ( V, E ) be a UDG. If I and D opt denote a maximal independent setand an MLDS of G , respectively, then | I | ≤ (cid:113) | D opt | .Proof. Let I (cid:48) = I ∩ D opt , X = I \ I (cid:48) , Y = D opt \ I (cid:48) . For u, v ∈ X , let c u,v denotethe number of vertices in Y which lie in the closed neighborhoods of u and v in G , i.e., c u,v = | ( N [ u ] ∪ N [ v ]) ∩ Y | . As D opt is a liar’s dominating set of G , c u,v ≥ u, v ∈ X , and we get (cid:80) u,v ∈ X c u,v ≥ · | X | ( | X |− . For u (cid:48) , v (cid:48) ∈ Y , analogues to c u,v ,let d u (cid:48) ,v (cid:48) = | ( N [ u (cid:48) ] ∪ N [ v (cid:48) ]) ∩ X | . As G is a UDG, for each vertex in Y there can beat most 5 independent vertices in its neighborhood, and thus d u (cid:48) ,v (cid:48) ≤
10 for each u (cid:48) and v (cid:48) in Y . Hence, we get 10 · | Y | ( | Y |− ≥ (cid:80) u (cid:48) ,v (cid:48) ∈ Y d u (cid:48) ,v (cid:48) . Note that the number ofedges in E induced between X and Y is (cid:80) u,v ∈ X c u,v (= (cid:80) u (cid:48) ,v (cid:48) ∈ Y d u (cid:48) ,v (cid:48) ). Thus, we have3 · | X | ( | X |− ≤ · | Y | ( | Y |− , which implies | X | ≤ (cid:113) | Y | . Therefore, | I | = | X | + | I (cid:48) | ≤ (cid:113) | Y | + | I (cid:48) | ≤ (cid:113) | D opt | . Lemma 4.2.
The set D returned by Algorithm 1 is an LDS of G .Proof. Algorithm 1 sub-sequentially computes three maximal independent sets I , I , and I in G (see line numbers 2-7). Let I = I ∪ I ∪ I . Note that any vertex not in I hasa neighbor (dominator) in each I , I , and I . Thus, each vertex (resp. every pair ofdistinct vertices) not in I ∪ I ∪ I satisfies the first (resp. second) condition of liar’sdomination. Also, every vertex in I is adjacent to a vertex in I and I . Thus, thevertices in I satisfy both the conditions of the lair’s domination. Similarly, every vertexin I is adjacent to a vertex in I (otherwise, I cannot be a maximal independent set)and, hence, the vertices in I satisfy both the conditions of the lair’s domination. For any A minimum k -dominating set D of G is a minimum dominating set of G with the property that everyvertex not in D should have at least k dominators in D . lgorithm 1 Liar’s dominating set in UDG
Require:
An UDG G = ( V, E ) Ensure:
A liar’s dominating set D of G = ( V, E ) i ← I i ← ∅ , and D ← ∅ for ( i = 1 to do if ( V (cid:54) = ∅ ) then I i ← M IS ( V ) (cid:46) M IS ( · ) returns a maximal independent set D ← D ∪ I i ; V ← V \ I i end if end for for every u ∈ I do if N G ( u ) ∩ ( I ∪ I ) = ∅ then let v ∈ N G ( u ) D = D ∪ { v } else if | N G ( u ) ∩ ( I ∪ I ) | = 1 then let w be a neighbor of v ∈ N G ( u ) ∩ ( I ∪ I ) such that w (cid:54) = u D = D ∪ { w } end if end for return D vertex u ∈ I , if N G ( u ) ∩ ( I ∪ I ) = ∅ , then the algorithm adds an arbitrary neighbor v of u to D (see line number 11). If N G ( u ) ∩ ( I ∪ I ) (cid:54) = ∅ , then u has neighbor in I ∪ I .In either case the vertices in I satisfy the two conditions of the liar’s domination.For any pair of distinct vertices u ∈ I and v ∈ V \ I , the second condition is alreadysatisfied as v has a neighbor in each I , I , and I . Similarly, for any pair of distinctvertices u ∈ I , and v ∈ I , if v has multiple neighbors in I or u has multiple neighborsin I , then the second condition is satisfied. If u is the only neighbor of v in I and viceversa, then the algorithm adds an arbitrary neighbor w of v to D , see line number 14,and thus the second condition is ensured for u and v . If v ∈ I , the second condition istrivially holds as v has a neighbor in each I and I . Therefore, D is an LDS in G . Theorem 4.3.
For a given UDG G = ( V, E ) , Algorithm 1 achieves approximation ratio7.31 for the MLDS problem in O ( | V | + | E | ) time.Proof. Let D ∗ be an MLDS of G . Algorithm 1 sequentially computes three maximalindependent sets I , I , and I in G and I = I ∪ I ∪ I is not necessarily be an LDSof G as there might be some vertices with one of the following cases: (i) a vertex u ∈ I not satisfying the first condition, or (ii) a pair of distinct vertices u ∈ I and v ∈ I notsatisfying the second condition of the liar’s domination. In the former case we add anarbitrary neighbor of u , and in the latter case we add an neighbor w of v . Note that ineither case such a neighbor is guaranteed to exist in G , and | D | ≤ | I | + | I | . Without loss14f generality we can assume that | I | ≤ | I | ≤ | I | . Therefore, | D | ≤ | I | ≤ · (cid:113) | D ∗ | ≤ . | D ∗ | (by Lemma 4.1). The running time follows as Algorithm 1 uses the subroutine M IS ( · ) (in line number 4) to compute a maximal independent set. In this section, we propose a PTAS for the MLDS problem in UDGs, i.e., for a givenUDG G = ( V, E ) and a parameter (cid:15) >
0, we propose an algorithm which produces a liar’sdominating set of size no more than (1 + (cid:15) ) times the size of a minimum liar’s dominatingset in G. We use δ G ( u, v ) to denote the number of edges on a shortest path between u and v in G . For A, B ⊆ V , δ G ( A, B ) denotes the distance between A and B and is defined as δ G ( A, B ) = min u ∈ A,v ∈ B { δ G ( u, v ) } . For A ⊆ V , D ( A ) and D opt ( A ) denote an LDS and anoptimal (minimum size) LDS of A in G , respectively. We define the closed neighborhoodof a set A ⊆ V as N G [ A ] = (cid:83) v ∈ A N G [ v ]. S S S S Figure 5: A 4-separated collection S = { S , S , S , S } The proposed PTAS is based on the concept of m -separated collection of subsets of V ( m ≥ G = ( V, E ) be a UDG. A collection S = { S , S , . . . , S k } such that S i ⊆ V for i = 1 , , . . . , k , is said to be an m -separated collection, if δ G ( S i , S j ) > m , for1 ≤ i ≤ k and 1 ≤ j ≤ k (see Figure 5 for a 4-separated collection). Nieberg and Hurink[8] introduced 2-separated collection to propose a PTAS for the minimum dominating setproblem in UDGs and our PTAS follows form it. However, the algorithm in [8] cannotbe directly applied as in intermediate steps of the algorithm we need to add extra nodes(see line numbers 12-21 in Algorithm 2) to ensure the liar’s domination. We argue thatthe extra nodes added are small enough and do not effect the approximation factor.15 emma 5.1. Let S = { S , S , . . . , S k } be an m -separated collection. If | S i | ≥ for ≤ i ≤ k , then (cid:80) ki =1 | D opt ( S i ) | ≤ | D opt ( V ) | .Proof. Observe that N G [ S i ] ∩ N G [ S j ] = ∅ for i (cid:54) = j and 1 ≤ i, j ≤ k . Also, D opt ( S i ) ∩ D opt ( S j ) = ∅ as S i and S j are m -separated. Let S (cid:48) i = { u ∈ V | v ∈ S i and δ G ( u, v ) ≤ } for i = 1 , , . . . , k . Observe that S i ⊆ S (cid:48) i and S (cid:48) i ∩ D opt ( V ) is a liar’s dominating setof S i for i = 1 , , . . . , k . Since, δ G ( S i , S j ) > m ( ≥
4) for i (cid:54) = j , implies S (cid:48) i ∩ S (cid:48) j = ∅ .Therefore, ( S (cid:48) i ∩ D opt ( V )) ∩ ( S (cid:48) j ∩ D opt ( V )) = ∅ and (cid:83) ki =1 ( S (cid:48) i ∩ D opt ( V )) ⊆ D opt ( V ). Also, | D opt ( S i ) | ≤ | S (cid:48) i ∩ D opt ( V ) | for i = 1 , , . . . , k as S (cid:48) i ∩ D opt ( V ) is a liar’s dominating setof S i , and D opt ( S i ) is a minimum size liar’s dominating set. Thus, (cid:80) ki =1 | D opt ( S i ) | ≤ (cid:80) ki =1 | S (cid:48) i ∩ D opt ( V ) | ≤ | D opt ( V ) | . Lemma 5.2.
Let S = { S , S , . . . , S k } be an m -separated collection, and N , N , . . . , N k be subsets of V with S i ⊆ N i for all i = 1 , , . . . , k . If there exists ρ ≥ such that | D opt ( N i ) | ≤ ρ | D opt ( S i ) | holds for all i = 1 , , . . . , k , and if (cid:83) ki =1 D opt ( N i ) is a liar’sdominating set in G , then the value of (cid:80) ki =1 | D opt ( N i ) | is at most ρ times the size of aminimum liar’s dominating set in G .Proof. (cid:80) ki =1 | D opt ( N i ) | ≤ ρ (cid:80) ki =1 | D opt ( S i ) | ≤ ρ | D opt ( V ) | . The latter inequality followsfrom Lemma 5.1. In this section, we discuss the construction of a 4-separated collection S = { S , S , . . . , S k } and subsets N , N , . . . , N k of V such that S i ⊆ N i for all i = 1 , , . . . , k . The algorithmproceeds in an iterative manner. Initially V = V . In the i -th iteration the algorithmcomputes S i and N i . For a given (cid:15) >
0, the i -th iteration of the algorithm starts with anarbitrary vertex v ∈ V i and increases the value of r (= 2 , , . . . ) as long as | D ( N r +4 G [ v ]) | >ρ | D ( N rG [ v ]) | holds. Here, D ( N r +4 G [ v ]) and D ( N rG [ v ]) are liar’s dominating sets of N r +4 G [ v ]and N rG [ v ], respectively, and ρ = 1 + (cid:15) . The smallest r violating the above condition, sayˆ r , is obtained. Set S i = N ˆ rG [ v ] and U i = N ˆ r +4 G [ v ]. Now, the removal of U i from V i maylead to some isolated (i) vertex u ∈ V i , and/or (ii) connected component with two vertices u, w ∈ V i . In case (i), for each such vertex u find x, y ∈ U i such that { u, x, y } forms aconnected component and update U i as follows: U i = U i \ { x, y } . In case (ii), for eachsuch pair of vertices u, w find x ∈ U i such that { u, w, x } forms a connected componentand update U i as follows: U i = U i \ { x } . Set N i = U i and V i +1 = V i \ N i . The processstops if V i +1 = ∅ and returns the sets S i s and N i s. The collection of the sets S i s is a4-separated collection. The pseudo code is given in Algorithm 2.16 lgorithm 2 Liar’s dominating set
Require:
A unit disk graph G = ( V, E ) with | V | ≥ (cid:15) > Ensure:
A liar’s dominating set D of V i ← V i +1 ← V D ← ∅ and ρ ← (cid:15) while ( V i +1 (cid:54) = ∅ ) do pick an arbitrary v ∈ V i +1 N [ v ] ← v and r ← while | ( D ( N r +4 G [ v ]) | > ρ | D ( N rG [ v ]) | do (cid:46) call Algorithm 1 r ← r + 1 end while ˆ r ← r (cid:46) the smallest r violating while condition in step 6 i ← i + 1 (cid:46) the index i keeps track of the number of iterations S i ← N ˆ rG [ v ] and U i ← N ˆ r +4 G [ v ] if ( V i +1 \ N ˆ r +4 G [ v ] contains isolated components of size 1 and/or 2) then for (each component, { u } , of size 1) do find x, y ∈ U i such that { u, x, y } is a connected component U i ← U i \ { x, y } end for for (each component, { u, w } , of size 2) do find x ∈ U i such that { u, w, x } is a connected component U i ← U i \ { x } end for end if N i ← U i D ← D ∪ D ( N i ) (cid:46) call Algorithm 1 V i +1 ← V i \ N i end while return D The liar’s dominating set of a r -th neighborhood of a vertex v , D ( N rG [ v ]), can becomputed with respect to G as described in Algorithm 1. Algorithm 1 successively findsmaximal independent sets I , I , and I . Now, I ∪ I ∪ I is a liar’s dominating set for N rG [ v ] \ I as every vertex not in I either belongs to I ∪ I or is adjacent to at least onevertex in each I , I , and I . To ensure the liar’s domination for the vertices in I , for eachvertex u in I we add at most a vertex (see line numbers 8-16 in Algorithm 1). In summary,Algorithm 2 deals with obtaining an m -separated collection S = { S , S , . . . , S k } andcollection N = { N , N , . . . , N k } such that S i ⊆ N i ⊆ V and using Algorithm 1 (thatdeals with obtaining a liar’s dominating set of the r -th neighborhood of a vertex) as asub-routine, Algorithm 2 computes a liar’s dominating set for G . Lemma 5.3. D ( N rG [ v ]) is an LDS of N rG [ v ] in G and | D ( N rG [ v ]) | ≤ O ( r ) .Proof. Algorithm 1 computes D ( N rG [ v ]) by first computing maximal independent sets17 , I , and I subsequently and then it adds at most one vertex for each vertex in I toensure that D ( N rG [ v ]) is a feasible solution. We can show D ( N rG [ v ]) is an LDS of N rG [ v ] in G as in the proof of Lemma 4.2. Hence, | D ( N rG [ v ]) | ≤ · | I | ≤ · π ( r +1) π (1) = O ( r ). Thelatter inequality follows from the standard area argument, the number of non-intersectingunit disks can be packed in a larger disk of radius r + 1 centered at v . Lemma 5.4.
In each iteration of Algorithm 2, there exists an r violating the condition | ( D ( N r +4 G [ v ]) | > ρ | D ( N rG [ v ]) | , where ρ = 1 + (cid:15) .Proof. We prove the lemma by contradiction. Suppose there exists v ∈ V such that | ( D ( N r +4 G [ v ]) | > ρ | D ( N rG [ v ]) | for r = 2 , , . . . . Observe that | D ( N G [ v ]) | ≥ G .If r = 4 k , 4( r + 1) ≥ | ( D ( N rG [ v ]) | > ρ | D ( N r − G [ v ]) | > · · · > ρ r | D ( N G [ v ]) | ≥ ρ r , andif r = 4 k + s for 1 ≤ s ≤ r + 1) ≥ | ( D ( N rG [ v ]) | > ρ | D ( N r − G [ v ]) | > · · · > ρ r − | D ( N G [ v ]) | ≥ ρ r − .In both the cases the first inequality follows from Lemma 5.3. Hence,4( r + 1) > ρ r , if r is 4 kρ r − , if r is 4 k + s (1)The right hand part in inequality (1) is an exponential function in r and the left handpart is a polynomial in r , for arbitrarily large r none of the inequalities can be true. Hencewe arrived at contradiction. Thus there exists an r violating the condition.The following lemma suggests that the smallest r violating inequality (1) is boundedby a constant that depends only on (cid:15) . Lemma 5.5.
The smallest r violating the inequality (1) is bounded by O ( (cid:15) log (cid:15) ) .Proof. Let ˆ r be the smallest r violating the inequalities in (1). Using the inequalities (i)log(1 + (cid:15) ) > (cid:15) for 0 < (cid:15) <
1, (ii) log x < x for x >
1, and (iii) log (cid:15) ≥ (cid:15) ≤ , we showˆ r ≤ O ( (cid:15) log (cid:15) ). Let x = c(cid:15) log (cid:15) . Consider the inequality 4( x + 1) ≤ x ) ≤ (1 + (cid:15) ) x .The former inequality trivially holds for the choice of x and for any (cid:15) >
0, and takingthe logarithm on both sides of the latter inequality, we get log 4+8 log 2 xx ≤ log(1 + (cid:15) ). Byinequality (i), now, it suffice to show that log 4+8 log 2 xx ≤ (cid:15) . Using the inequalities (ii) and(iii), the choice of x satisfies the inequality for any constant c satisfying log 2 c +2 < c . Lemma 5.6.
For a given v ∈ V , liar’s dominating set D opt ( N i ) of N i can be computedin polynomial time. roof. Note that N i ⊆ N r +4 G [ v ]. The size of a liar’s dominating set D ( N i ) of N i is boundedby O ( r ) (by Lemma 5.3). Again, r = O ( (cid:15) log (cid:15) ) by Lemma 5.5. Therefore, the size ofthe minimum size liar’s dominating set D opt ( N i ) of N i is bounded by a constant. Theprocess of checking whether a given set is a liar’s dominating set or not can be donein polynomial-time. Therefore, we can consider every subset of N i as a possible liar’sdominating set and check whether it is a liar’s dominating set or not in polynomial-time.Finally, the minimum size liar’s dominating set is reported. Lemma 5.7.
For the collection of neighborhoods { N , N , . . . , N k } created by Algorithm2, the union D = (cid:83) ki =1 D ( N i ) is a liar’s dominating set in G .Proof. We first prove that for every v ∈ V, | N G [ v ] ∩ D| ≥
2. Observe that (cid:83) ki =1 N i = V as V i +1 = V i \ N i and N i ⊆ V i . Thus, every vertex v ∈ N i for some 1 ≤ i ≤ k . By Lemma5.3, | N G [ v ] ∩ D| ≥ u, v ∈ V be any two arbitrary vertices. Thefollowing cases may arise. Case 1. u, v ∈ N i for some 1 ≤ i ≤ k Since D ( N i ) is the liar’s dominating set of N i in G , we have, | ( N G [ u ] ∪ N G [ v ]) ∩ D ( N i ) | ≥ u, v ∈ N i . Hence, | ( N G [ u ] ∪ N G [ v ]) ∩ D| ≥ u, v ∈ V . Case 2. u ∈ N i and v ∈ N j for some i (cid:54) = j and 1 ≤ i, j ≤ k If u and v are not adjacent in G , the proof is trivial. Hence, we assume that ( u, v ) ∈ E i.e., u and v are adjacent in G . Now the following sub-cases may arise.(a) u ∈ D ( N i ) and v ∈ D ( N j )Observe that | N G [ u ] ∩ D ( N i ) | ≥ | N G [ v ] ∩ D ( N j ) | ≥ D ( N i ) and D ( N j ) are liar’sdominating sets of N i and N j , respectively. Hence, u has a neighbor, say w , in D ( N i ),similarly v has also a neighbor, say x , in D ( N j ). However, maybe w = x or maybe not.In either case | ( N G [ u ] ∪ N G [ v ]) ∩ D| ≥ u / ∈ D ( N i ) and v ∈ D ( N j ) (the other case proof is similar)Since D ( N i ) is a liar’s dominating set of N i , we have | N G [ u ] ∩ D ( N i ) | ≥
2. Hence, | ( N G [ u ] ∪ N G [ v ]) ∩ D| ≥ v is part of the solution.(c) u / ∈ D ( N i ) and v / ∈ D ( N j ) (The proof is similar to the previous cases). Corollary 5.8.
For the collection N = { N , N , . . . , N k } created by Algorithm 2, theunion D ∗ = (cid:83) ki =1 D opt ( N i ) is a liar’s dominating set. Theorem 5.9.
For a given UDG, G = ( V, E ) , and an (cid:15) > , we can design a (1 + (cid:15) ) -factor approximation algorithm to find an LDS in G with running time n O ( c ) , where c = O ( (cid:15) log (cid:15) ) . roof. Note that Algorithm 2 generates the collection of sets S = { S , S , . . . , S k } and N = { N , N , . . . , N k } such that S is a 4-separated collection of V with S i ⊆ N i for each i ∈ { , , . . . , k } and (cid:83) ki =1 N i = V with N i ∩ N j = ∅ for i (cid:54) = j . Corollary 5.8 suggests that D ∗ = (cid:83) ki =1 D opt ( N i ) is a liar’s dominating set of G . The approximation bound followsfrom Lemma 5.1, and Lemma 5.2. Let | N i | = n i for 1 ≤ i ≤ k . By Lemma 5.6, an optimalliar’s dominating set D opt ( N i ) of N i can be computed in n O ( (cid:15) log (cid:15) ) i time. Therefore, thetotal running time to compute D ∗ is (cid:80) ki =1 n O ( (cid:15) log (cid:15) ) i ≤ n O ( (cid:15) log (cid:15) ) . In this article, we studied the minimum liar’s dominating set problem (MLDS) in UDGs.We proved that the decision version of the MLDS problem is NP-complete. We proposeda simple 7.31-factor approximation algorithm and a PTAS for the problem. We believethat it is possible to get much better approximation ratios by exploring inherent geomet-ric properties of UDGs. As a future direction, we work on it and hope to design suchalgorithms for the problem.
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