On d -distance m -tuple ( ℓ,r )-domination in graphs
GGeneralized Liar’s Dominating Set in Graphs
Sangram K. Jena Ramesh K. Jallu Gautam K. Das ∗ Indian Institute of Technology Guwahati, India 781039 { sangram,j.ramesh,gkd } @iitg.ac.in Abstract
In this article, we study generalized liar’s dominating set problem in graphs. Let G = ( V, E )be a simple undirected graph. The generalized liar’s dominating set, called as the distance- d ( m, (cid:96) )-liar’s dominating set, is a subset L ⊆ V such that (i) each vertex in V is distance- d dominated by at least m vertices in L , and (ii) each pair of distinct vertices in V is distance- d dominated by at least (cid:96) vertices in L , where m, (cid:96), d are positive integers and m < (cid:96) . Here, avertex v is distance- d dominated by another vertex u means the shortest path distance between u and v is at most d in G .We first consider distance-1 ( m, (cid:96) )-liar’s dominating set problem and prove that it is NP-complete. Next, we consider distance- d ( m, (cid:96) )-liar’s dominating set problem and show thatit is also NP-complete. These liar’s dominating set problems are generalized version of liar’sdominating set problem as researcher studied only distance-1 (2 , m, (cid:96) )-liar’s dominating set problem cannot beapproximated within a factor of ( − ε ) ln | V | for any ε >
0, unless NP ⊆ DTIME( | V | O (log log | V | ) ),and (ii) distance- d ( m, (cid:96) )-liar’s dominating set problem cannot be approximated within a factorof ( − ε ) ln | V | for any ε >
0, unless NP ⊆ DTIME( | V | O (log log | V | ) ). Let G = ( V, E ) be a simple connected and undirected graph. For a vertex v ∈ V , the closedneighborhood of v in G is denoted by N G [ v ] and is defined as N G [ v ] = { u ∈ V | ( u, v ) ∈ E } ∪ { v } .A dominating set of G is a subset D of V such that every vertex in V is in either D or adjacentto at least one vertex in D . In other words, | N G [ v ] ∩ D | ≥ v ∈ V . The vertices in D are called as dominators and the rest are called as dominatees. A dominator dominates all itsneighbors and itself.A k -tuple dominating set of G is a dominating set with the restriction that every vertex in V mustbe dominated by at least k ≥ | N G [ v ] ∩ D | ≥ k for each v ∈ V .The goal of the k -tuple dominating set problem is to find a k -tuple dominating set of minimumsize. A liar’s dominating set D of G is a subset of V having the following two properties: (i) D isa 2-tuple dominating set, and (ii) | ( N G [ u ] ∪ N G [ v ]) ∩ D | ≥
3, for every pair of distinct vertices u and v in V . The objective of the liar’s dominating set problem is to find a liar’s dominating set ofminimum size in a given graph G .Given a simple connected undirected graph G = ( V, E ), δ G ( v i , v j ) denotes the length of a shortestpath between the vertices v i and v j in G . For an integer d >
0, the distance- d neighborhood ∗ Corresponding author a r X i v : . [ c s . CC ] J u l f a vertex v i ∈ V is denoted by N dG [ v i ] and defined as N dG [ v i ] = { v j ∈ V | δ G ( v i , v j ) ≤ d } . A distance- d ( m, (cid:96) ) -liar’s dominating set (distance- d ( m, (cid:96) )-LDS) of G is a subset L ⊆ V such that(i) for every v i ∈ V , | N dG [ v i ] ∩ L | ≥ m , and (ii) for every pair of distinct vertices v i , v j ∈ V , | ( N dG [ v i ] ∪ N dG [ v j ]) ∩ L | ≥ (cid:96) , where m, (cid:96), d are positive integers and m < (cid:96) . The objective of thedistance- d ( m, (cid:96) )-LDS problem is to find a minimum size distance- d ( m, (cid:96) )-LDS in a given graph G , and we call this problem as the minimum distance- d ( m, (cid:96) ) -liar’s dominating set problem. InFigure 1, the set of vertices { e, f, i } form a distance-3 (2 , d ( m, (cid:96) )-liar’s dominating set with d = 1 , m = 2, and (cid:96) = 3. a bc d gh j k l m ef i Figure 1: The set { e, f, i } is the distance-3 (2 , G = ( V, E ) is a possible location for an intruder such as a thief, a saboteur, a fire orsome possible fault. Assume also that there is exactly min( (cid:96) − m, (cid:100) (cid:96)/ (cid:101) −
1) intruders in the systemrepresented by G . A protection device placed at a vertex v is assumed to be able to (i) detectthe intruder at any vertex in its closed distance- d neighborhood N dG [ v ], and (ii) report the vertex u ∈ N dG [ v ] at which the intruder is located. We are interested in deploying protection devicesat a minimum number of vertices so that the intruder can be detected and identified correctly.This can be solved by finding a minimum cardinality m -tuple dominating set, say D , of G anddeploying protection devices at all the vertices of D . If any one protection device can fail to detectthe intruder, then to correctly detect and identify the intruder one needs to place the protectiondevices at all the vertices of a minimum cardinality 2 m -tuple dominating set of G . Now it may sohappen that all the protection devices detect the intruder location correctly but while reportingsome of these protection devices can misreport or lie (either deliberately or through a transmissionerror) about the intruder location. Assume that at most min( (cid:96) − m, (cid:100) (cid:96)/ (cid:101) −
1) protection devicesin the closed distance- d neighborhood of an intruder location can lie. Under these circumstances,to protect the network we have to install the protection devices at all the vertices of a minimumdistance- d ( m, (cid:96) )-liar’s dominating set. In 2009, Slater [11] first introduced minimum distance-1 (2 , , n ( ≥
3) vertices lies between ( n +1)and n . In the same paper, Slater observed for a subclass of trees for which there exist only onedistance-1 (2 , G = ( V, E )having n vertices and m edges γ LR ( G ) ≥ (2 n − m ), where γ LR ( G ) is the cardinality of a liar’sdominating set of minimum size in G , and γ LR ( G ) ≥ (6 / (3∆ + 2)) n , where ∆ is the maximumdegree of a vertex in G . For a tree T having n vertices, γ LR ( T ) = n if and only if each v ∈ V ( T )2s an endpoint or at least one component of T − v has cardinality at most two. Later, Rodenand Slater [10] characterized distance-1 (2 , ( n + 1). They showed that even for bipartite graphs the minimum distance-1 (2 , K a,b with 1 ≤ a ≤ b as follows: (i) γ LR ( K ,n − ) = n , (ii) γ LR ( K ,b − ) = b + 1, and (iii) γ LR ( K a,b ) = min { a + 1 , } for3 ≤ a ≤ b .For different graph classes like split graphs and chordal graphs Panda and Paul [6] proved its NP-hardness and also proposed a linear time algorithm to compute a minimum distance-1 (2 , O (ln ∆)-factor approximation algorithm, where ∆ is the maximum degree of the given graph. For properinterval graphs also Panda and Paul [7] considered the problem and proposed a linear time algo-rithm. They also studied the minimum distance-1 (2 , p -claw free graphs [9]. Sterling [12] presented bounds on liar’s domination number by consid-ering the problem on two-dimensional grid graphs.Alimadadi et al. [1] provided the characterization of graphs and trees for which the distance-1(2 , | V | and | V | −
1, respectively. The authors observed that a connectedgraph G with number of vertices n ≥ , n if and only if everyvertex v in G satisfies at least one of the following conditions (i) deg ( v ) = 1, (ii) at least onecomponent of G \ { v } has at most two vertices, (iii) v belongs to an end-block (a block havingat most one cut vertex of G ) having 3 vertices. For connected graphs with girth (the length ofa shortest cycle) at least five, they obtained an upper bound for the ratio between the distance-1(2 , , , , connected distance- , -LDS is a distance-1 (2 , total distance- , -LDS is a dominating set L with the following two properties(i) for every v ∈ V , | ( N G [ v ] \ { v } ) ∩ L | ≥
2, and (ii) for every distinct pair of vertices u and v , | ( N G [ u ] \ { u } ) ∪ ( N G [ v ] \ { v } )) ∩ L | ≥
3. The objective of both problems is to find connecteddistance-1 (2 , , O (ln ∆)-factor approximation algorithms. They alsoproved that the problems are APX-complete for graphs with maximum degree 4. Jallu and Das [3]studied the geometric version of the minimum distance-1 (2 , , In this article we have considered generalized version of the liar’s dominating set problem, namelydistance- d ( m, (cid:96) )-liar’s dominating set (distance- d ( m, (cid:96) )-LDS). We prove that the distance- d ( m, (cid:96) )-LDS problem is NP-complete by showing the following decision version of the distance- d ( m, (cid:96) )-LDSproblem is NP-complete. Decision version of distance- d ( m, (cid:96) ) -LDS:Input. A simple connected undirected graph G = ( V, E ) with at least (cid:96) vertices and three positiveintegers m , d ( ≤ | V | − k ( ≤ | V | ), where m < (cid:96) .3 uestion. Does G have a distance- d ( m, (cid:96) )-LDS of size at most k ?We also prove that the distance- d ( m, (cid:96) )-LDS problem cannot be approximated within a factor of( − ε ) ln | V | for any ε >
0, unless NP ⊆ DTIME( | V | O (log log | V | ) ). m, (cid:96) ) -LDS problem In this section, we show that the distance-1 ( m, (cid:96) )-LDS problem in graphs is NP-complete byreducing the dominating set (DS) problem to it, which is known to be NP-complete [2].The decision versions of the distance-1 ( m, (cid:96) )-LDS problem and DS problem are defined below.
Decision Version of the distance- m, (cid:96) ) -LDS problem:Instance: An undirected graph G = ( V, E ), m, (cid:96) , and a positive integer k . Question:
Does there exist a distance-1 ( m, (cid:96) )-LDS L of G such that | L | ≤ k ? Decision Version of the DS problem:Instance:
An undirected graph G = ( V, E ) and a positive integer k . Question:
Does there exist a dominating set D of G such that | D | ≤ k ? Theorem 3.1.
The decision version of the distance- m, (cid:96) ) -LDS problem is NP-complete.Proof. For any given set L ⊆ V and a positive integer k , we can verify whether L is a distance-1( m, (cid:96) )-LDS of size at most k or not in polynomial time by checking both the conditions of distance-1( m, (cid:96) )-LDS. Therefore, distance-1 ( m, (cid:96) )-LDS is in NP.Now, we prove the hardness of the distance-1 ( m, (cid:96) ) -LDS problem by reducing the decision versionof DS problem, which is known to be NP-complete [2], to it. Let < G = ( V, E ) , k > be an instanceof dominating set (DS) problem, where G = ( V, E ) is an undirected graph and k is an integer. Alsoassume V = { v , v , . . . , v n } . Now, we construct an instance < G (cid:48) = ( V (cid:48) , E (cid:48) ) , m, (cid:96) > of the decisionversion of distance-1 ( m, (cid:96) )-LDS problem as follows: V (cid:48) = V ∪ V ∪ V V = { v , v , . . . , v n } ,V = { v , v , . . . v (cid:96) − } ,V = { v , v } E (cid:48) = E ∪ E ∪ E ∪ E E = { ( v i , v j ) | ( v i , v j ) ∈ E } ,E = { ( v i , v j ) | ≤ i < j ≤ (cid:96) − } ,E = { ( v i , v j ) | ≤ i ≤ n, ≤ j ≤ (cid:96) − } ,E = { ( v i , v ) , ( v i , v ) | ≤ i ≤ (cid:96) − } Observe that, G (cid:48) = ( V (cid:48) , E (cid:48) ) can be constructed in polynomial time and | V (cid:48) | = n + (cid:96) + 1, where n = | V | and (cid:96) < n . The construction of G (cid:48) from G is shown in Figure 2(a). Claim 1: G has a dominating set of size at most k if and only if G (cid:48) has a distance- m, (cid:96) ) -LDSof size at most k + (cid:96) . Proof:
Let D be a dominating set of G and | D | = k . Let L = { v i | v i ∈ D } ∪ V ∪ { v } . Now, wewill show that L is a distance-1 ( m, (cid:96) )-LDS in G (cid:48) . (i) Observe that for each v ∈ V (cid:48) , | N G (cid:48) [ v ] ∩ L | ≥ m as m < (cid:96) and each v ∈ V (cid:48) is dominated by (cid:96) − V . 4 v v v v (cid:48) v (cid:48) v (cid:48) v (cid:48) G = ( V, E ) G’=(V’,E’) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) v (cid:48) ( a ) ( b ) v v v · · · v (cid:96) − v v · · · v n v v v Figure 2: (a) A graph G (cid:48) = ( V (cid:48) , E (cid:48) ) constructed for the instance of distance-1 ( m, (cid:96) )-LDS problem.(b) A graph G (cid:48) = ( V (cid:48) , E (cid:48) ) constructed for the instance of distance- d ( m, (cid:96) )-LDS problem. (ii) Let u and v be any two distinct vertices V (cid:48) . Case 1: { u, v } ∈ V ∪ V , then | ( N G (cid:48) [ u ] ∪ N G (cid:48) [ v ]) ∩ L | = { ( V ∪ V \ { v } ) } ≥ (cid:96) . Case 2: { u, v } ∈ V , then | ( N G (cid:48) [ u ] ∪ N G (cid:48) [ v ]) ∩ L | ≥ (cid:96) , because every vertex v i ∈ V is adjacentwith (cid:96) − V and D is a dominating set in G . Case 3: u ∈ V and v ∈ V ∪ V , then | ( N G (cid:48) [ u ] ∪ N G (cid:48) [ v ]) ∩ L | = |{ c } ∪ V | ≥ (cid:96) , where c ∈ { v i | v i ∈ D } ∩ N G (cid:48) [ v ].Thus L is a distance-1 ( m, (cid:96) )-LDS in G (cid:48) and | L | ≤ k + (cid:96) .Conversely, let L is a distance-1 ( m, (cid:96) )-LDS of G (cid:48) of size at most k + (cid:96) . Since | ( N G (cid:48) [ v ] ∪ N G (cid:48) [ v ]) ∩ L | ≥ (cid:96) , there must be (cid:96) vertices from V ∪ V in L (see Figure 2(a)). Let D (cid:48) = L \ ( V ∪ V )and D = { v i ∈ V | v i ∈ V (cid:48) ∩ D (cid:48) } . It remains to prove that D is a dominating set of the graph G . Suppose that D is not a dominating set of G . Then there exist only one vertex v ∈ V suchthat D ∩ N G [ v ] = φ . If more than one vertex exists say, { u, v } ∈ V such that D ∩ N G [ u ] = φ and D ∩ N G [ v ] = φ , then | ( N G (cid:48) [ u ] ∪ N G (cid:48) [ v ]) ∩ L | ≤ |{ v , v , . . . , v (cid:96) − }| < (cid:96) , which is a contradiction tothe fact that L is a distance-1 ( m, (cid:96) )-LDS of G (cid:48) . Therefore, if D is not a dominating set of G thenthere can be at most one vertex v ∈ V such that D ∩ N G [ v ] = φ . Observe that V ∪ V ⊆ L . Oncontrary assume that V ∪ V (cid:42) L i.e., | ( V ∪ V ) ∩ L | ≤ (cid:96) . Then there exist a vertex u ∈ V ∪ V and u / ∈ L . Case 1: u ∈ V and u / ∈ L . Now, | ( N G (cid:48) [ v ] ∪ N G (cid:48) [ u ]) ∩ L | = |{ v , v , . . . , v (cid:96) − }| = (cid:96) −
1, which is acontradiction to the fact that L is a distance-1 ( m, (cid:96) )-LDS of graph G (cid:48) . Case 2: u ∈ V and u / ∈ L . Now, | ( N G (cid:48) [ v ] ∪ N G (cid:48) [ v ]) ∩ L | ≤ ( | V | −
1) + |{ v }| = (cid:96) −
1, which is acontradiction to the fact that L is a distance-1 ( m, (cid:96) )-LDS of graph G (cid:48) .Thus, V ∪ V ⊆ L . Delete v from L and introduce v , i.e., L = ( L \ { v } ) ∪ { v } . So D is adominating set of G and | D | ≤ k .Therefore, we conclude, distance-1 ( m, (cid:96) )-LDS problem is NP-complete.5 Hardness of the distance- d ( m, (cid:96) ) -LDS problem In this section, we show that distance- d ( m, (cid:96) )-LDS problem is NP-complete. Here, we prove thatthe decision version of the distance- d ( m, (cid:96) )-LDS problem is NP-complete, which leads to the claimof this section. For fixed constant d ≥
2, the decision version of the distance- d ( m, (cid:96) )-LDS problemis defined as follows. Instance:
An undirected connected graph G = ( V, E ) with | V | ≥ (cid:96) and two positive integers k ≤ | V | , and d ≤ | V | − Question:
Does G have a distance- d ( m, (cid:96) )-LDS of size at most k ?We prove that decision version of distance- d ( m, (cid:96) )-LDS problem ( d ≥
2) is NP-complete by reducingthe decision version of the distance-1 ( m, (cid:96) )-LDS problem to it in polynomial time. Note thatdistance-1 ( m, (cid:96) )-LDS problem is NP-complete (see Section 3). Recall, the decision version ofdistance-1 ( m, (cid:96) )-LDS problem:
Instance:
An undirected connected graph G = ( V, E ) with | V | ≥ (cid:96) and a positive integer k ≤ | V | . Question:
Does G have a distance-1 ( m, (cid:96) )-LDS of size at most k ? Theorem 4.1.
The decision version of the distance- d ( m, (cid:96) ) -LDS problem is NP-complete.Proof. The decision version of distance- d ( m, (cid:96) )-LDS problem is in NP as for a given certificate (asubset of V ) we can verify whether it is satisfying both the conditions of distance- d ( m, (cid:96) )-LDS ornot in polynomial time.We now describe a polynomial time reduction from an arbitrary instance of the decision version ofdistance-1 ( m, (cid:96) )-LDS to an instance of the decision version of distance- d ( m, (cid:96) )-LDS.Let G = ( V = { v , v , . . . , v n } , E ) be an arbitrary instance of the decision version of distance-1( m, (cid:96) )-LDS problem. We construct an instance, a graph G (cid:48) = ( V (cid:48) , E (cid:48) ), of the decision version ofdistance- d ( m, (cid:96) )-LDS problem as follows: V (cid:48) = { v (cid:48) i | v i ∈ V } ∪ (cid:91) v i ∈ V { v (cid:48) i , v (cid:48) i , . . . , v (cid:48) id − } ( see F igure b ) f or an example ) E (cid:48) = { ( v (cid:48) i , v (cid:48) j ) | ( v i , v j ) ∈ E } ∪ (cid:91) v i ∈ V { ( v (cid:48) i , v (cid:48) i ) , ( v (cid:48) i , v (cid:48) i ) , . . . , ( v (cid:48) id − , v (cid:48) id − ) } Claim 2: G has a distance- m, (cid:96) ) -LDS of cardinality at most k if and only if G (cid:48) has a distance- d ( m, (cid:96) ) -LDS of cardinality at most k . Necessity:
Let L be a distance-1 ( m, (cid:96) )-LDS of G such that | L | ≤ k . Let L (cid:48) = { v (cid:48) i ∈ V (cid:48) | v i ∈ L } .We can argue that L (cid:48) is a distance- d ( m, (cid:96) )-LDS in G (cid:48) and | L (cid:48) | ≤ k . Since | L (cid:48) | = | L | and | L | ≤ k ,so | L (cid:48) | ≤ k . As each vertex v ∈ V satisfies distance-1 ( m, (cid:96) )-LDS properties and each vertex in G (cid:48) is at most d − L (cid:48) , L (cid:48) suffices to ensure distance- d ( m, (cid:96) )-LDS ingraph G (cid:48) for d ≥ Sufficiency:
Let L (cid:48) be a distance- d ( m, (cid:96) )-LDS in G (cid:48) such that | L (cid:48) | ≤ k . We shall show that, by up-dating (i.e., removing or replacing) some of the vertices in L (cid:48) , at most k vertices from { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } can be chosen such that the set of corresponding vertices in V is a distance-1 ( m, (cid:96) )-LDS in G . Let L (cid:48)(cid:48) = L (cid:48) . For each vertex v (cid:48) ij ∈ V (cid:48) , (1 ≤ j ≤ d − ≤ i ≤ n ) we do the following: if v (cid:48) ij ∈ L (cid:48)(cid:48) ,6hen replace it with its associated vertex v (cid:48) i if v (cid:48) i is not already in L (cid:48)(cid:48) , otherwise, replace it with anyvertex in N G (cid:48) [ v (cid:48) i ] ∩ { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } which is not in L (cid:48)(cid:48) . If all the vertices of N G (cid:48) [ v (cid:48) i ] ∩ { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } are in L (cid:48)(cid:48) (i.e., ( N G (cid:48) [ v (cid:48) i ] ∩ { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } ) ⊆ L (cid:48)(cid:48) ), then remove v (cid:48) ij from L (cid:48)(cid:48) . Therefore, | L (cid:48)(cid:48) | ≤ k . Let L = { v i ∈ V | v (cid:48) i ∈ L (cid:48)(cid:48) } . Now, we prove that L is a distance-1 ( m, (cid:96) )-LDS in G such that | L | ≤ k .Since | L (cid:48)(cid:48) | ≤ k , then | L | ≤ k . We first prove the first condition (i.e., for every v ∈ V , | N G [ v ] ∩ L | ≥ m )of distance-1 ( m, (cid:96) )-LDS.Consider a vertex v (cid:48) i ∈ V (cid:48) , for some 1 ≤ i ≤ n , let s be the number of vertices in L (cid:48) from the set { v (cid:48) i , v (cid:48) i , . . . , v (cid:48) id − } . Case 1. s = 0 . Since L (cid:48) is distance- d ( m, (cid:96) )-LDS, there must exist at least m vertices, say { v (cid:48)(cid:48) , v (cid:48)(cid:48) , . . . , v (cid:48)(cid:48) m } in { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } ∩ L (cid:48) such that { v (cid:48)(cid:48) , v (cid:48)(cid:48) , . . . , v (cid:48)(cid:48) m } ⊆ N dG (cid:48) [ v (cid:48) i,d − ], otherwise, L (cid:48) isnot a feasible solution as v (cid:48) id − does not have m distance- d ( m, (cid:96) )-dominators. Therefore, | N G (cid:48) [ v (cid:48) i ] ∩ ( { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } ∩ L (cid:48)(cid:48) ) | ≥ m . Case 2. s ≥ . Let v (cid:48) ij , v (cid:48) ij , . . . , v (cid:48) ij t ∈ L (cid:48) , for some 1 ≤ j , j , . . . , j t ≤ d −
1. By our constructionof L (cid:48)(cid:48) each vertex in { v (cid:48) ij , v (cid:48) ij , . . . , v (cid:48) ij t } is replaced by one of the vertices in N G (cid:48) [ v (cid:48) i ] ∩{ v (cid:48) , v (cid:48) , . . . , v (cid:48) n } .Therefore, in this case also | N G (cid:48) [ v (cid:48) i ] ∩ ( { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } ∩ L (cid:48)(cid:48) ) | ≥ m .Thus, by our construction of L from L (cid:48)(cid:48) , | N G [ v i ] ∩ L | ≥ m is true.Now we prove the second condition of distance-1 ( m, (cid:96) )-LDS (i.e., for every pair of distinct vertices u, v ∈ V , | ( N G [ u ] ∪ N G [ v ]) ∩ L | ≥ (cid:96) ).Let v i and v j be two distinct vertices in G . Consider the vertices v (cid:48) id − and v (cid:48) jd − in G (cid:48) . As L (cid:48) is adistance- d ( m, (cid:96) )-LDS of G (cid:48) , it satisfies the second property of distance- d ( m, (cid:96) )-LDS in G (cid:48) . Thusthere exist at least (cid:96) dominators dominating v (cid:48) id − and v (cid:48) jd − in L (cid:48) , i.e., | ( N dG (cid:48) [ v (cid:48) id − ] ∪ N dG (cid:48) [ v (cid:48) jd − ]) ∩ L (cid:48) | ≥ (cid:96) . These dominators are either from N G [ v (cid:48) i ] ∪ N G [ v (cid:48) j ] or from { v (cid:48) i , v (cid:48) i , . . . , v (cid:48) id − } and/or from { v (cid:48) j , v (cid:48) j , . . . , v (cid:48) jd − } . As per our construction of L (cid:48)(cid:48) from L (cid:48) , we are replacing each dominator in { v (cid:48) i , v (cid:48) i , . . . , v (cid:48) id − } ∪ { v (cid:48) j , v (cid:48) j , . . . , v (cid:48) jd − } (if any) by a vertex in ( N G (cid:48) [ v (cid:48) i ] ∪ N G (cid:48) [ v (cid:48) j ]) ∩ { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } .Since G is connected and | V | ≥ (cid:96) , so is G (cid:48) . Therefore, L (cid:48)(cid:48) contains at least (cid:96) vertices from ( N G (cid:48) [ v (cid:48) i ] ∪ N G (cid:48) [ v (cid:48) j ]) ∩ { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } , i.e., | ( N G (cid:48) [ v (cid:48) i ] ∪ N G (cid:48) [ v (cid:48) j ]) ∩ { v (cid:48) , v (cid:48) , . . . , v (cid:48) n } ∩ L (cid:48)(cid:48) | ≥ (cid:96) . Therefore, accordingto the construction of L from L (cid:48)(cid:48) , | ( N G [ v i ] ∪ N G [ v j ]) ∩ L | ≥ (cid:96) . Thus, L is a distance-1 ( m, (cid:96) )-LDSof the graph G having cardinality at most k .Therefore, the decision version of distance- d ( m, (cid:96) )-LDS problem is NP-complete. m, (cid:96) ) -LDS In this section, we shall prove that the distance-1 ( m, (cid:96) )-LDS problem cannot be approximatedwithin a factor of ( − ε ) ln( | V | ) for any ε >
0, unless NP ⊆ DTIME( | V | log log | V | ). We argue theclaim by showing that if distance-1 ( m, (cid:96) )-LDS can be approximated within a factor of ( − ε ) ln( | V | )for any ε > G (cid:48) , then the dominating set problem can be approximated within a factorof (1 − ε ) ln( | V | ) for any ε > Theorem 5.1. [2] Minimum dominating set problem cannot be approximated within a factor of (1 − ε ) ln( | V | ) for any ε > , unless NP ⊆ DTIME ( | V | log log | V | ) . Theorem 5.2.
Minimum distance- m, (cid:96) ) -LDS problem cannot be approximated within a factorof ( − ε ) ln( | V | ) for any ε > , unless NP ⊆ DTIME ( | V | log log | V | ) . roof. Let G be a simple graph. Consider the construction of the graph G (cid:48) for any given graph G as discussed in Section 3. As per our construction, we proved that dominating set problem can bepolynomially reducible to distance-1 ( m, (cid:96) )-LDS problem.Let D ∗ and L ∗ be the optimal DS and distance-1 ( m, (cid:96) )-LDS in G and G (cid:48) , with cardinalities γ ds ( G )and γ LR ( G (cid:48) ), respectively. Now we can argue the following claim: γ LR ( G (cid:48) ) = γ ds ( G ) + (cid:96) . Theinequality γ LR ( G (cid:48) ) ≤ γ ds ( G ) + (cid:96) is trivial as per our construction in Section 3. On the otherhand, γ LR ( G (cid:48) ) ≥ γ ds ( G ) + (cid:96) follows from the sufficiency proof of Claim 1 in Section 3. So given adominating set D of G , one can find a distance-1 ( m, (cid:96) )-LDS L of G (cid:48) such that | L | = | D | + (cid:96) . Now, | L || L ∗ | = | D | + (cid:96) | D ∗ | + (cid:96) ≥ | D || D ∗ | .Suppose there exists a polynomial time algorithm that approximates distance-1 ( m, (cid:96) )-LDS problemwithin a factor of ( − ε ) ln N for graphs with N vertices. As per our construction of the graph G (cid:48) from G (see Figure 2(a)), G (cid:48) contains, N = n + (cid:96) + 1 ≤ n for n ≥ n is the totalnumber of vertices in G and (cid:96) < n . Therefore, | D || D ∗ | ≤ (1 − ε ) ln N ≤ (1 − ε ) ln n (1 + ln 2ln n ) . For sufficiently large n , the term (1 + ln 2ln n ) can be bounded by 1 + ε , where ε ≥ n . Now we have(1 − ε ) ln n (1 + ln 2ln n ) ≤ (1 − ε (cid:48) ) ln n, where ε (cid:48) < ε + ε . Therefore, for an arbitrary graph, we can approximate the dominating setproblem by a factor of (1 − ε (cid:48) ) ln n , which leads to a contradiction to Theorem 5.1. Thus, theminimum distance-1 ( m, (cid:96) )-LDS problem cannot be approximated within a factor of ( − ε ) ln( | V | )for any ε >
0, unless NP ⊆ DTIME( | V | log log | V | ). d ( m, (cid:96) ) -LDS problem In this section, we give a lower bound on the approximation ratio of any approximation algorithmfor the distance- d ( m, (cid:96) )-LDS problem by providing an approximation preserving reduction fromthe distance-1 ( m, (cid:96) )-LDS problem. Theorem 5.3.
For a given undirected graph G = ( V, E ) , the distance- d ( m, (cid:96) ) -LDS problem cannotbe approximated within a factor of ( − ε ) ln | V | , for any fixed constant d ≥ and ε > , unless NP ⊆ DTIME ( | V | O (log log | V | ) ) .Proof. Let G = ( V, E ) be an arbitrary instance of the distance-1 ( m, (cid:96) )-LDS problem with n vertices. Given G = ( V, E ), we construct a graph G (cid:48) = ( V (cid:48) , E (cid:48) ), an instance of the distance- d ( m, (cid:96) )-LDS problem as described in Section 4. Let L ∗ and L ∗ d be the optimal distance-1 ( m, (cid:96) )-LDS and distance- d ( m, (cid:96) )-LDS in G and G (cid:48) , with cardinalities γ LR ( G ) and γ dLR ( G (cid:48) ), respectively.Now we can argue the following claim: γ dLR ( G (cid:48) ) = γ LR ( G ). The inequality γ dLR ( G (cid:48) ) ≤ γ LR ( G ) istrivial as every distance-1 ( m, (cid:96) )-LDS of G is a distance- d ( m, (cid:96) )-LDS in G (cid:48) . On the other hand, γ dLR ( G (cid:48) ) = | L ∗ d | ≥ | L | follows from the sufficiency proof of Claim 2 in Section 4.Given any distance-1 ( m, (cid:96) )-LDS L of G , one can find a distance- d ( m, (cid:96) )-LDS L d of G (cid:48) with | L d | = | L | . Suppose there exist a polynomial time algorithm to approximate distance- d ( m, (cid:96) )-LDS8roblem within a factor of ( − ε ) ln | V (cid:48) | , where | V (cid:48) | = n + n ( d − ≤ n (see Section 4). Now | L || L ∗ | = | L d || L ∗ d | ≤ ( − ε ) ln n = ( − ε ) ln n ≤ ( − ε (cid:48) ) ln n , where ε (cid:48) ≤ ε . Therefore, the result followsfrom Theorem 5.2. In this paper, We have considered generalized version of the liar’s dominating set problem availablein literature. We showed that the distance-1 ( m, (cid:96) )-liar’s dominating set (distance-1 ( m, (cid:96) )-LDS)problem is NP-complete and proved that it cannot be approximated with in a factor of ( − ε ) ln | V | ,unless NP ⊆ DTIME( | V | O (log log | V | ) ), where V is the vertex set of input graph. We also provedthat distance- d ( m, (cid:96) )-liar’s dominating set (distance- d ( m, (cid:96) )-LDS) problem is NP-complete andproved that the problem cannot be approximated within a factor of ( − ε ) ln | V | , unless NP ⊆ DTIME( | V | O (log log | V | ) ), where V is the vertex set of input graph. References [1] Abdollah Alimadadi, Mustapha Chellali, and Doost Ali Mojdeh. Liar’s dominating sets ingraphs.
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