Algorithmic Complexity of Isolate Secure Domination in Graphs
aa r X i v : . [ c s . D M ] F e b Algorithmic Complexity of Isolate SecureDomination in Graphs
J. Pavan Kumar , ∗ P. Venkata Subba Reddy Department of Computer Science and EngineeringNational Institute of TechnologyWarangal, Telangana, India 506004.
Abstract
A dominating set S is an Isolate Dominating Set ( IDS ) if the in-duced subgraph G [ S ] has at least one isolated vertex. In this paper,we initiate the study of new domination parameter called, isolate se-cure domination . An isolate dominating set S ⊆ V is an isolate securedominating set ( ISDS ), if for each vertex u ∈ V \ S , there exists a neigh-boring vertex v of u in S such that ( S \ { v } ) ∪ { u } is an IDS of G . Theminimum cardinality of an ISDS of G is called as an isolate secure dom-ination number , and is denoted by γ s ( G ). Given a graph G = ( V, E )and a positive integer k, the ISDM problem is to check whether G has anisolate secure dominating set of size at most k. We prove that ISDM isNP-complete even when restricted to bipartite graphs and split graphs.We also show that ISDM can be solved in linear time for graphs ofbounded tree-width.
Keywords : Domination, NP-complete, Secure domination : 05C69, 68Q25.* - Corresponding author1 - [email protected] - [email protected]
In this paper, every graph G = ( V, E ) considered is finite, simple (i.e., withoutself-loops and multiple edges) and undirected with vertex set V and edge set E . For a vertex v ∈ V , the ( open ) neighborhood of v in G is N ( v ) = { u ∈ V :1 u, v ) ∈ E } , the closed neighborhood of v is defined as N [ v ] = N ( v ) ∪ { v } . If S ⊆ V , then the (open) neighborhood of S is the set N ( S ) = ∪ v ∈ S N ( v ). Theclosed neighborhood of S is N [ S ] = S ∪ N ( S ). The degree of a vertex v is thesize of the set N ( v ) and is denoted by d ( v ). If d ( v ) = 0, then v is called an isolated vertex of G . If d ( v ) = 1, then v is called a pendant vertex . For a graph G = ( V, E ) , and a set S ⊆ V, the subgraph of G induced by S is defined as G [ S ] = ( S, E S ), where E S = { ( x, y ) : x, y ∈ S and ( x, y ) ∈ E } . A spanningsubgraph is a subgraph that contains all the vertices of the graph. If G [ S ] isa complete subgraph of G , then it is called a clique of G . A set S ⊆ V is an independent set if G [ S ] has no edges. A split graph is a graph in which thevertices can be partitioned into a clique and an independent set.In a graph G = ( V, E ), a set S ⊆ V is a Dominating Set ( DS ) in G if forevery u ∈ V \ S , there exists v ∈ S such that ( u, v ) ∈ E , i.e., N [ S ] = V . Theminimum size of a dominating set in G is called the domination number of G and is denoted by γ ( G ). Given a graph G = ( V, E ) and a positive integer k, thedomination decision (DOM) problem is to check whether G has a dominatingset of size at most k. The DOM problem is known to be NP-complete [5].The literature on various domination parameters in graphs has been surveyedin [7, 8]. An important domination parameter called secure domination hasbeen introduced by E.J. Cockayne in [2]. A dominating set S ⊆ V is a SecureDominating Set ( SDS ) of G , if for each vertex u ∈ V \ S , there exists aneighboring vertex v of u in S such that ( X \ { v } ) ∪ { u } is a dominating setof G (in which case v is said to defend u ). The minimum size of a SDS in G is called the secure domination number of G and is denoted by γ s ( G ). Let S ⊆ V . Then a vertex w ∈ V is called a private neighbor of v with respectto S if N [ w ] ∩ S = { v } . If further w ∈ V \ S , then w is called an externalprivate neighbor (epn) of v . The secure domination decision problem (SDOM)is known to be NP-complete for general graphs [4] and remains NP-completeeven when restricted to bipartite graphs and split graphs [9].Another domination parameter called isolate domination has been intro-duced by Hamid and Balamurugan in [6]. A dominating set S is an IsolateDominating Set ( IDS ) if the induced subgraph G [ S ] has at least one isolatedvertex. The isolate domination number γ ( G ) is the minimum size of an IDS of G . In [11], N.J. Rad has proved that the isolate domination decision problemis NP-complete, even when restricted to bipartite graphs. A dominating set S is said to be a connected dominating set (CDS), if the induced subgraph G [ S ] is connected. A CDS S is said to be a secure connected dominating set (SCDS) in G if for each u ∈ V \ S , there exists v ∈ S such that uv ∈ E and( S \ { v } ) ∪ { u } is a CDS in G . Algorithmic complexity of secure connecteddomination problem has been studied in [10]. In this paper, we initiate thestudy of a variant of domination called isolate secure domination . An isolatedominating set S ⊆ V is an Isolate Secure Dominating Set ( ISDS ), if for eachvertex u ∈ V \ S , there exists a neighboring vertex v of u in S such that( S \ { v } ) ∪ { u } is an IDS of G . The minimum size of an ISDS of G is called as2n isolate secure domination number , and is denoted by γ s ( G ). Given a graph G = ( V, E ) and a positive integer k, the ISDM problem is to check whether G has an isolate secure dominating set of size at most k. Motivation
A communication network is modeled as a graph G = ( V, E )where each node represents a communicating device and each edge representsa communication link between two devices. All the nodes in the network needto communicate and exchange information to perform a task. However, in thenetworks where the reliability of the nodes is not guaranteed, every node v can be a potential malicious node. A virtual protection device at node v can(i) detect the malicious activity at any node u in its closed neighborhood (ii)migrate to the malicious node u and repair it. One is interested to deployminimum number of virtual protection devices such that every node shouldhave at least one virtual protection device within one hop distance even aftervirtual protection device is migrated to malicious node. This problem can besolved by finding a minimum secure dominating set of the graph G . Further,if two virtual protection devices are deployed adjacently then there is a chanceof one virtual protection device getting damaged or corrupted if other oneis corrupted. In some scenarios it is desirable to have at least one virtualprotection device isolated from all other devices before and after each potentialmigration to act as a backup device. This problem can be solved by finding aminimum isolate secure dominating set of the graph G . In this section, some precise values and bounds for new domination param-eter called isolate secure domination in frequently encountered graph classesare presented. The graphs for which isolate secure dominating set exist thefollowing observation holds.
Observation 1.
For a graph G , γ s ( G ) ≥ γ s ( G ) . Proposition 1.
For the complete graph K n with n vertices, γ s ( K n ) = 1 . Theorem 1.
Let G be a graph with n vertices. Then γ s ( G ) = 1 if and onlyif G = K n .Proof. Suppose γ os ( G ) = 1 and let S = { v } be a minimum size ISDS of G .Suppose G = K n , then there exists x, y ∈ V ( G ) such that d ( x, y ) ≥
2. Then( S \ { v } ) ∪ { x } = { x } , which is not a dominating set of G , since ( x, y ) / ∈ E .Therefore, G = K n . The converse is true by Proposition 1. Remark 1.
There is no isolate secure dominating set possible for the completebipartite graph.
Proposition 2.
Let P n be a path graph with n ( ≥ vertices. Then γ s ( P n ) = ⌈ n ⌉ . roof. From [2], we know that γ s ( P n ) = ⌈ n ⌉ . By this and Proposition 1, wehave γ s ( P n ) is at least ⌈ n ⌉ . Hence in order to complete the proof, we need toexhibit an ISDS of P n of size ⌈ n ⌉ . • a • b • c • d • e • f • gFigure 1: Path graph P Suppose P n = ( v , v , . . . , v n ) and n = 7 m + r, where m ≥ ≤ r ≤ . Define two sets X = m − [ i =0 { v i +2 , v i +4 , v i +6 } and Y = ∅ , if r = 0 { v m +1 } , if r = 1 or 2 { v m +1 , v m +3 } , if r = 3 or 4 { v m +1 , v m +3 , v m +5 } , if r = 5 or 6Clearly | X ∪ Y | = ⌈ n ⌉ . It can be noted that the set X ∪ Y forms an independentset. It can also be verified that X ∪ Y forms an ISDS of P n . Hence theresult. Observation 2. If G ′ is a spanning subgraph of graph G , then γ s ( G ′ ) ≥ γ s ( G ) . Proposition 3.
Let C n be a cycle graph with n vertices. Then γ s ( C n ) = ⌈ n ⌉ .Proof. From Observation 2, γ s ( P n ) ≥ γ s ( C n ). From Proposition 2, it can benoted that γ s ( C n ) ≤ ⌈ n ⌉ . From Observation 1, we know that γ s ( G ) ≤ γ s ( G ).It is known that γ s ( C n ) = ⌈ n ⌉ [2]. Hence the result. In this section, the complexity of a new domination parameter called isolate se-cure domination is investigated for bipartite graphs, split graphs and boundedtree-width graphs.The decision version of isolate secure domination problem is defined as follows.
ISOLATE SECURE DOMINATION DECISION problem ( ISDM ) Instance:
A simple, undirected graph G = ( V, E ) and a positive integer p . Question:
Does there exist a ISDS of size at most p in G ?The following proposition has been proved by Cockayne et al [2].4 • x • x • x • x • x • y • y • y • y • y • a • a • b • c • d • e • f • g • b • c • d • e • f • g Figure 2: Example construction of a graph G ′ Proposition 4. ([2]) X is a SDS of G if and only if for each u ∈ V \ X , thereexists v ∈ X such that G [ epn ( v, X ) ∪ { u, v } ] is complete. The following corollary follows from Proposition 4.
Corollary 1.
For any bipartite graph G with SDS S , | epn ( v, S ) | ≤ , ∀ v ∈ V . The DOM problem in bipartite graphs (DSDPB) has been proved as NP-complete by A.A. Bertossi [1].
Theorem 2.
ISDM is NP-complete for bipartite graphs.Proof.
It can be shown that ISDM is in NP, since if a set S ⊆ V , such that | S | ≤ k is given as a witness to a yes instance then it can be verified inpolynomial time that S is an ISDS of G . We reduce the DSDPB probleminstance to ISDM problem instance for bipartite graphs as follows. Given abipartite graph G = ( X, Y, E ), we construct a graph G ′ = ( X ′ , Y ′ , E ′ ) where X ′ = X ∪ { b , d , f , g , a , c , e } , Y ′ = Y ∪ { a , c , e , b , d , f , g } and E ′ = E ∪ { ( a , v ) : v ∈ X } ∪ { ( a , v ) : v ∈ Y } ∪ { ( a i , b i ) , ( a i , f i ) , ( a i , g i ) , ( b i , c i ),( c i , d i ) , ( d i , e i ), ( e i , f i ) : 1 ≤ i ≤ } . Clearly, G ′ is a bipartite graph and can beconstructed from G in polynomial time. An example construction of a graph G ′ from a graph G is depicted in figure 2.Next, we prove that G has a dominating set of size at most k if and only if G ′ has an ISDS of size at most p = k + 6. Let D be a dominating set of sizeat most k in G and D ∗ = D ∪ { a , c , e , a , c , e } . Clearly D ∗ is an ISDS ofsize at most k + 6 in G ′ .Conversely, suppose that G ′ has an ISDS D ∗ of size at most p = k +6. It canbe observed that | D ∗ ∩{ a , b , c , d , e , f }| ≥ | D ∗ ∩{ a , b , c , d , e , f }| ≥ | D ∗ ∩ ( X ∪ Y ) | ≤ k . Since g and g are pendant vertices, | D ∗ ∩{ a , g }| ≥ | D ∗ ∩ { a , g }| ≥
1. Let D = D ∗ ∩ { a , g } and D = D ∗ ∩ { a , g } . Nowwe consider the following four possible cases. case ( i ). | D | = 1 and | D | = 1, then all vertices in X \ D ∗ can be dominatedby Y ∩ D ∗ and all vertices in Y \ D ∗ can be dominated by X ∩ D ∗ hence D ∗ ∩ ( X ∪ Y ) is a dominating set of size at most k in G . case ( ii ). | D | = 1 and | D | = 2, then all vertices in X \ D ∗ can be dominatedby Y ∩ D ∗ and from Corollary 1, | epn ( a , D ∗ ) | ≤ , therefore D ′ ∪ epn ( a , D ∗ )is a dominating set of size at most k in G . case ( iii ). | D | = 2 and | D | = 1, this case is analogous to case (ii). case ( iv ). | D | = 2 and | D | = 2, then if D ∗ ∩ ( X ∪ Y ) is a dominating setof G , we are done. Otherwise, let D ′ = D ∗ ∩ ( X ∪ Y ). From Corollary 1, | epn ( a i , D ∗ ) | ≤ , for 1 ≤ i ≤ | D ′ | ≤ k −
2. In such a case D = D ′ ∪ { epn ( a i , D ∗ ) : 1 ≤ i ≤ } is a dominating set of size at most k in G .Hence the proof.The decision version of secure domination problem is defined as follows. SECURE DOMINATION DECISION problem (SDOM)Instance:
A simple, undirected graph G = ( V, E ) and a positive integer k . Question:
Does there exist a SDS of size at most k in G ?It has been proved that the SDOM is NP-complete for split graphs by H.B.Merouane et al. [9] Theorem 3.
ISDM is NP-complete for split graphs.Proof.
It is known that ISDM is in NP. We reduce SDOM problem instancefor split graphs to ISDM problem instance for split graphs as follows. Givena split graph G = ( C, I, E ), construct a graph G ′ = ( C ′ , I ′ , E ′ ), where C ′ = C ∪ { x , y } , I ′ = I ∪ { x , y } and E ( G ′ ) = E ∪ { ( x , v ) , ( y , v ) : v ∈ C } ∪{ ( x , y ) , ( x , x ) , ( y , y ) } . Note that G ′ is a split graph and can be constructedfrom G in polynomial time. Figure 3 illustrates an example construction. • x • y • x • y • b • c • a • d • e • r • q • pG Figure 3: Example construction of a split graph G ′ G has a SDS of size at most k if and only if G ′ has an ISDS of size at most p = k + 2. Let D be a SDS of size at most k in G and D ∗ = D ∪ { x , y } . It can be easily verified that D ∗ is an ISDS of size atmost k + 2 in G ∗ .Conversely, suppose that G ′ has an ISDS D ∗ of size at most p = k + 2. Let X ′ = D ∗ ∩ { x , x } , Y ′ = D ∗ ∩ { y , y } and D ′ = D ∗ \ ( X ′ ∪ Y ′ ). We nowhave the following cases. (i) | X ′ | = 1 and | Y ′ | = 1, then D ′ is a SDS of size atmost k in G . (ii) | X ′ | = 1 and | Y ′ | = 2, then for any v ∈ C \ D ∗ , D ′ ∪ { v } isa SDS of size at most k in G . (iii) | X ′ | = 2 and | Y ′ | = 1, this case is similarto the previous case. (iv) | X ′ | = 2 and | Y ′ | = 2, then for any two vertices u, v ∈ C \ D ∗ and u = v , D ′ ∪ { u, v } is a SDS of size at most k in G .Since the Domination problem is w[2]-complete for bipartite graphs and splitgraphs [12] and the reductions in Theorems 2 and 3 are in the function of theparameter k, the following two corollaries are immediate. Corollary 2.
ISDM is w[2]-hard for bipartite graphs.
Corollary 3.
ISDM is w[2]-hard for split graphs.
Let G be a graph, T be a tree and v be a family of vertex sets V t ⊆ V ( G )indexed by the vertices t of T . The pair ( T, v ) is called a tree-decompositionof G if it satisfies the following three conditions: (i) V ( G ) = S t ∈ V ( T ) V t , (ii) forevery edge e ∈ E ( G ) there exists a t ∈ V ( T ) such that both ends of e lie in V t ,(iii) V t ∩ V t ⊆ V t whenever t , t , t ∈ V ( T ) and t is on the path in T from t to t . The width of ( T, v ) is the number max {| V t | − t ∈ T } , and thetree-width tw ( G ) of G is the minimum width of any tree-decomposition of G .By Courcelle’s Thoerem, it is well known that every graph problem that canbe described by counting monadic second-order logic (CMSOL) can be solvedin linear-time in graphs of bounded tree-width, given a tree decomposition asinput [3]. We show that ISDM problem can be expressed in CMSOL. Theorem 4 ( Courcelle’s Theorem ) . ([3]) Let P be a graph property expressiblein CMSOL and let k be a constant. Then, for any graph G of tree-width atmost k , it can be checked in linear-time whether G has property P . Theorem 5.
Given a graph G and a positive integer k , ISDM can be expressedin CMSOL.Proof. First, we present the CMSOL formula which expresses that the graph G has a dominating set of size at most k. Dominating ( S ) = ( | S | ≤ k ) ∧ ( ∀ p )(( ∃ q )( q ∈ S ∧ adj ( p, q ))) ∨ ( p ∈ S ) where adj ( p, q ) is the binary adjacency relation which holds if and only if, p, q are two adjacent vertices of G. Dominating ( S ) ensures that for every vertex7 ∈ V , either p ∈ S or p is adjacent to a vertex in S and the cardinality of S is at most k. Isolate − Dom ( S ) = Dominating ( S ) ∧ ( ∃ p )( p ∈ S ∧ (( ∄ q )( q ∈ ( S \ { p } ) ∧ adj ( p, q ))) Now, by using the above two CMSOL formulas we can express ISDM inCMSOL formula as follows.
ISDM ( S ) = Isolate − Dom ( S ) ∧ ( ∀ x )(( x ∈ S ) ∨ (( ∃ y )( y ∈ S ∧ Isolate − Dom (( S \ { y } ) ∪ { x } )))) Therefore, ISDM can be expressed in CMSOL.Now, the following result is immediate from Theorems 4 and 5.
Theorem 6.
ISDM can be solvable in linear time for bounded tree-widthgraphs.
In this paper, it is shown that ISDM problem is NP-complete even whenrestricted to bipartite graphs, or split graphs. Since split graphs form aproper subclass of chordal graphs, this problem is also NP-complete for chordalgraphs. We have shown that ISDM problem is linear time solvable for boundedtree-width graphs. It will be interesting to investigate the algorithmic com-plexity of ISDM problem for subclasses of chordal and bipartite graphs.
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