Some notes on the isolate domination in graphs

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AKCE International Journal of Graphs and Combinatorics 14 (2017) 112–117

www.elsevier.com/locate/akcej

Some notes on the isolate domination in graphs

Nader Jafari Rad

Department of Mathematics, Shahrood University of Technology, Shahrood, Iran

Received 13 July 2015; received in revised form 21 January 2017; accepted 23 January 2017 Available online 6 February 2017

Abstract

A subsetSof vertices of a graphGis adominating setofGif every vertex inV(G)−Shas a neighbor inS. Thedomination numberγ (G)is the minimum cardinality of a dominating set ofG. A dominating setSis anisolate dominating setif the induced subgraphG[S]has at least one isolated vertex. Theisolate domination numberγ₀(G)is the minimum cardinality of an isolate dominating set ofG. In this paper we study the complexity of the isolate domination in graphs, and obtain several bounds and characterizations on the isolate domination number, thus answering some open problems.

⃝c 2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

Keywords:Domination; Isolate domination; Total domination; Complexity

1. Introduction

For notation and graph theory terminology, we in general follow [1,2]. Specifically, letGbe a graph with vertex setV(G) = V of order|V| = n and size|E(G)| = m, and letv be a vertex in V. Theopen neighborhoodof v is NG(v) = {u ∈ V | uv ∈ E(G)}and theclosed neighborhood of vis NG[v] = {v} ∪N(v). The degree ofv is deg_G(v)= |N_G(v)|. If the graphGis clear from the context, we simply writeN(v)and deg(v)rather than NG(v) and deg_G(v), respectively. For a set S ⊆ V, itsopen neighborhood is the set N(S) = ∪_v∈SN(v), and itsclosed neighborhoodis the setN[S] = N(S)∪S. A vertex of degree one is called aleaf and its unique neighbor asupport vertex. Apendant edgeis an edge which one of its end-points is a leaf. Astarof ordern≥3 is a tree that has precisely one vertex that is not a leaf. Aclaw-freegraph is a graph with no induced subgraph isomorphic to a star of order 4.

Adouble-staris a tree that has precisely two vertices that are not leaves. We referS(a,b)as a double-star which its central vertices have degreeaandb, respectively. For a subsetSof vertices ofGwe denote byG[S]the subgraph of G inducedbyS. For two subsets of verticesX andY ofV(G), we denote byG[X,Y]the subgraph ofGinduced by X ∪Y. Thediameterof a graphG, denoted bydi am(G), is the maximum distance between pairs of vertices ofG.

ThegirthofG, denoted byg(G), is the length of a shortest cycle contained inG.

Peer review under responsibility of Kalasalingam University.

E-mail address:n.jafarirad@gmail.com.

http://dx.doi.org/10.1016/j.akcej.2017.01.003

0972-8600/ c⃝2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

A subsetS of vertices of a graph Gis a dominating setof Gif every vertex inV(G)−S has a neighbor inS.

Thedomination numberγ (G)is the minimum cardinality of a dominating set ofG. We refer a dominating set of cardinalityγ (G)as aγ (G)-set. A dominating set S in a graph with no isolated vertex is called atotal dominating setofG ifG[S]has no isolated vertex. Thetotal domination number γt(G)is the minimum cardinality of a total dominating set ofG. We refer a total dominating set of cardinalityγt(G)as aγt(G)-set. A total dominating setSin Gis called anefficient total dominating setif the open neighborhoods of the vertices ofSform a partition forV(G).

For a subsetSof vertices ofG, and a vertexx ∈ S, we say that a vertexy ̸∈ Sis anexternal private neighbor of x with respect to SifN(y)∩S= {x}. We denote byepn(x,S)the set of all external private neighbors ofxrespect toS.

Hamid and Balamurugan [3] initiated the study ofisolate dominationin graphs. A dominating setS is anisolate dominating setif the induced subgraphG[S]has at least one isolated vertex. Theisolate domination numberγ₀(G) is the minimum cardinality of an isolate dominating set ofG. The concept of isolate domination was further studied, for example in [4–9]. Hamid et al. [3] showed that for a cubic graphG,γ (G)≤γ₀(G)≤γ (G)+1. They presented several bounds, and properties for the isolate domination number, and proposed the following problem(s).

Problems: (1) Characterize cubic graphsGwithγ₀(G)=γ (G)+1.

(2) Characterize graphsGwithγ₀(G)=γ (G), orγ₀(G)=ⁿ₂. (3) Find bounds forγ0(G).

In this paper we first study the complexity of the isolate domination number in graph by showing that the decision problem for this variant is NP-complete, even when restricted to bipartite graphs. We then answer all of the above problems. We present several bounds, and characterizations for the isolate domination number in a graph.

In the following we state some known results that we need for the next. Thecorona graphof a graphG, denoted byGoK1, is the graph obtained fromGby adding a pendant edge to every vertex ofG.

Theorem 1 ([1]).For a graph G of order n with no isolated vertex,γ (G) ≤ ⁿ₂, with equality if and only if each component of G is a C4or the corona H oK1for any connected graph H .

Lemma 2 ([3]).For paths and cycles of order n,γ0(Pn)=γ0(Cn)= ⌈ⁿ₃⌉.

Theorem 3 ([2]).If G is a graph of order n and with no isolated vertex thenγ_t(G)≥ _∆(G)ⁿ .

2. Complexity

In this section we show that the decision problem for the isolate domination is NP-complete, even when restricted to bipartite graphs. We use a transformation from the 3-SAT problem. Atruth assignment for a setU of Boolean variables is a mappingt :U → {T,F}. A variableuis said to betrue(orfalse) undert ift(u)=T (ort(u)=F). If uis a variable inU, thenuanduareliteralsoverU. The literalu is true undert if and only if the variableu is true undert, and the literaluis true if and only if the variableuis false. AclauseoverUis a set of literals overU, and it is satisfiedby a truth assignment if and only if at least one of its members is true under that assignment. A collectionC of clauses overUis satisfiable if and only if there exists some truth assignment forUthat simultaneously satisfies all the clauses inC. Such a truth assignment is called a satisfying truth assignment forC. The 3-SAT problem is specified as follows.

3-SAT problem:

Instance:A collection C = {C1,C2, . . . ,Cm}of clauses over a finite set U of variables such that|Cj| = 3 for j =1,2, . . . ,m.

Question:Is there a truth assignment forUthat satisfies all the clauses inC?

Note that the 3-SAT problem was proven to be NP-complete in [10]. Consider the following decision problem.

Isolate dominating set (IDS):

Instance:A graphG=(V,E)and a positive integerk.

Question:DoesGhave an isolate dominating set of size at mostk?

Theorem 4. The isolate domination problem is NP-complete for bipartite graphs.

Proof. It is clear that the isolate domination problem belongs to NP, since it is easy to verify a “yes” instance of the isolate domination problem in polynomial time. We show the NP-hardness of the isolate domination problem by transforming the 3-SAT to it in polynomial time. Let U = {u₁,u₂, . . . ,u_n}, and C = {C₁,C₂, . . . ,C_m} be an arbitrary instance of the 3-SAT problem. We construct a graph G and an integer k such that C is satisfiable if and only if γ₀(G) ≤ k. The graph G is constructed as follows. For i = 1,2, . . . ,n, let H_i be a 6-cycle u_iv_iu_id_ib_ia_iu_i being consecutive vertices. Fori = 1,2, . . . ,n, corresponding to each variableu_i ∈ U, associate the graph H_i. Corresponding to each clauseC_j = {x_j,y_j,z_j} ∈C, associate a single vertexc_j and add the edge-set E_j = {c_jx_j,c_jy_j,c_jz_j}for j =1,2, . . . ,m. LetGbe the resulted graph. ClearlyGis a bipartite graph.

We show thatC is satisfiable if and only ifγ₀(G) ≤ 2n. Assume that C is satisfiable. Lett:U → {T,F}be a satisfying truth assignment forC. We construct a subset Dof vertices of G as follows. Ift(u_i) = T, then we put the verticesu_i andd_i inD; ift(u_i)=F, then put the verticesu_i anda_i inD. Then it can be easily checked thatD is an isolate dominating set forGof cardinality 2n. Conversely, assume thatS is an isolate dominating set forGof cardinality at most 2n. Clearly|V(H_i)∩S| ≥2 fori =1,2, . . . ,n. Since|S| ≤2n, we obtain|V(H_i)∩S| =2 for i =1,2, . . . ,n, andS∩ {c₁, . . . ,c_m} = ∅. If{u_i,u_i} ⊆ S for somei ∈ {1,2, . . . ,n}thenb_i is not dominated by S, a contradiction. Thus|{u_i,u_i} ∩S| ≤ 1 for eachi =1,2, . . . ,n. If{u_i,u_i} ∩S = ∅for somei ∈ {1,2, . . . ,n}

then{b_i, v_i} ⊆ S, and then we replace{b_i, v_i}by{u_i,d_i}. Thus we may assume that|{u_i,u_i} ∩S| = 1 for each i =1,2, . . . ,n. Clearly for eachcj, (j =1,2, . . . ,m) there isi ∈ {1,2, . . . ,n}such thatcj is dominated by some vertex inS∩V(Hi). Lett:U → {T,F}be a mapping defined byt(ui)=T ifui ∈S, andt(ui)=F ifui ∈S. For each j ∈ {1,2, . . . ,m}, there is an integeri ∈ {1,2, . . . ,n}such thatcj is dominated byS∩ {ui,ui}. Assume that ui ∈Sandcj is dominated byui. By the construction ofG, the literalui is in the clauseCj. Thent(ui)=T, which implies that the clauseCj is satisfied byt. Next assume thatui ∈Sandcjis dominated byui. By the construction of G, the literalu_i is in the clauseC_j. Thent(u_i)=F. Thustassignsu_i the truth valueT, that is,tsatisfies the clause C_j. HenceCis satisfiable.

Since the construction of the p-reinforcement instance is straightforward from a 3-SAT instance, the size of the p-reinforcement instance is bounded above by a polynomial function of the size of 3-SAT instance. It follows that this is a polynomial transformation, as desired.

3. On graphsGwithγ₀(G)=γ (G)

In this section we derive some properties of graphsGwithγ₀(G)=γ (G). We begin with the following obvious observation.

Observation 5. For a graph G,γ₀(G)=γ (G)if and only if there is aγ (G)-set S such that G[S]has some isolated vertex.

The following follows immediately fromObservation 5.

Corollary 6. For a graph G,γ0(G)̸=γ (G)if and only if every minimum dominating set is a total dominating set.

We say thatγ (G)andγt(G)arestrongly equal, denoted byγ (G)≡γt(G)if everyγ (G)-set is also aγt(G)-set.

The strong equality between two domination parameters was first introduced by Haynes et al. [11]. FromCorollary 6 we obtain the following.

Theorem 7. For a graph G,γ0(G)=γ (G)if and only if γ (G)̸≡γt(G).

Note that in [11] all treesT withγ (T)≡γt(T)was constructively characterized. In the next proposition we state some properties of graphsGwithγ0(G)̸=γ (G). The proof is straightforward, and thus is omitted.

Proposition 8. If γ₀(G)̸=γ (G), thenγ (G)=γ_t(G)≤ ⁿ₃, and|epn(x,S)| ≥2for anyγ (G)-set S, and any vertex x∈S.

The following follows immediately fromProposition 8.

Proposition 9. If di am(G)=2thenγ₀(G)=γ (G).

Theorem 10. For any claw-free graph G,γ0(G)=γ (G).

Proof. LetGbe a claw-free graph. Suppose thatγ0(G)̸= γ (G). Let S be aγ (G)-set. ByProposition 8,γ (G)= γt(G), and|epn(u,S)| ≥ 2 for any vertexu ∈ S. Letx be a vertex ofSwith deg_G[S](x)= 1, andy ∈ N(x)∩S.

IfG[epn(x,S)]is a complete graph then(S− {x})∪ {w}is aγ (G)-set which is not a total dominating set, where w ∈ epn(x,S). This contradictsCorollary 6. ThusG[epn(x,S)]is not a complete graph. Sinceyis adjacent to no vertex ofepn(x,S), and|epn(x,S)| ≥2, we find thatGcontains a claw, a contradiction.

As noted, it is shown in [3] that for a cubic graph G,γ (G) ≤ γ0(G)≤ γ (G)+1. In the following we answer Problem (1).

Theorem 11. For any cubic graph G,γ0(G)=γ (G).

Proof. LetGbe a cubic graph of ordern. Suppose thatγ0(G)̸=γ (G). Thenγ0(G)=γ (G)+1. LetSbe aγ (G)-set.

ByTheorem 3andProposition 8,γ (G)=γ_t(G)= |S| = ⁿ₃. ByTheorem 7,S = ^{γ (G)}₂ K₂, andSis an efficient total dominating set. Thus|epn(x,S)| =2 for anyx∈ S. For eachx ∈ S, letepn(x,S)= {x¹,x²}. Clearlyx¹̸∈ N(x²) for anyx ∈S, since any minimum set is a total dominating set. Letx₁∈S, andx₂²∈ N(x₁²). Thenx₂²is adjacent to a unique vertexx₂∈S. LetS₁=(S−{x₁,x₂})∪{x₁¹,x₂²}. ClearlyG[S₁]has some isolated vertex. Assume thatS₁is not an isolate dominating set forG. Thenx₂¹is not dominated byS1. Letx₃¹∈ N(x₂¹). Thenx₃¹is dominated by a vertex x3∈ S1. LetS2=(S1− {x₃})∪ {x¹₃}. Assume that S2is not an isolate dominating set. Thenx₃²is not dominated by S₂. Letx²₄∈ N(x₃²). Thenx₄²is dominated by a vertexx₄∈S₂. LetS₃=(S₂− {x₄})∪ {x₄²}. Continuing this process, we obtain a setS_ksuch that|S_k| = |S|andS_kis an isolate dominating set. Consequently,γ₀(G)=γ (G).

4. Bounds

In this section we provide several bounds and characterization for the isolate domination number of a graph.

Theorem 12. For any graph G,γ₀(G) ≤ γ (G)−1+ p₀, where p₀ = min{|epn(x,D)| : D is aγ (G)-set,x ∈ D and epn(x,D)̸= ∅}.

Proof. LetDbe aγ (G)-set. Clearly p₀≥1. The result is obvious ifG[D]has an isolated vertex. Thus assume that G[D]has no isolated vertex. Letx₀∈ Dbe a vertex such that 0<|epn(x₀,D)| ≤ |epn(x,D)|for any vertexx∈ D withepn(x,D)̸= ∅. Let X be a minimum independent dominating set forepn(x₀,D). Then(D− {x₀})∪X is an isolate dominating set forG, as desired.

It is clear that in the above proof,|epn(x₀,D)| ≤ ⌊^{n−γ (G)}_{γ (G)} ⌋ = ⌊_{γ (G)}ⁿ ⌋ −1. Thus we obtain the following.

Corollary 13. For any graph G of order n,γ₀(G)≤max{γ (G), γ (G)−2+ ⌊_{γ (G)}ⁿ ⌋}.

Furthermore, since|epn(x₀,D)| ≤∆(G)−1, for any connected graphGof ordern≥3,γ₀(G)≤γ (G)+∆(G)−

2. We next answer the second part of Problem (2).

It can be easily seen that if each component ofGis aC₄or the coronaH oK₁for any connected graph H, then γ₀(G) = γ (G). LetE_n be the class of all bipartite graphsG of even ordern such thatG can be obtained from a double-starS(ⁿ₂,ⁿ₂)by adding edges betweenXandY such thatγ₀(G[X,Y])≥ ⁿ₂−1, whereXis the set of all leaves of a partite set ofS(ⁿ₂,ⁿ₂), andY is the set of all leaves of the other partite set ofS(ⁿ₂,ⁿ₂).

Theorem 14. If G is a graph of order n≥2and with no isolated vertex, thenγ0(G)≤ ⁿ₂. Equality holds if and only if G∈En or each component of G is a C4or the corona H oK1for any connected graph H .

Proof. LetGbe a graph of ordern. Ifγ0(G)=γ (G)then clearly byTheorem 1,γ0(G)=γ (G)≤ ⁿ₂with equality if and only if each component of G is a C₄ or the corona H oK₁ for any connected graph H. Thus assume that γ₀(G) > γ (G). Clearlyγ (G)≥2.

First assume that γ (G) > 2. If γ₀(G) > ⁿ₂ then by Corollary 13, ⁿ₂ < γ₀(G) ≤ γ (G)−2+ _{γ (G)}ⁿ , and a simple calculation implies thatγ (G) > ⁿ₂, a contradiction. Thusγ₀(G)≤ ⁿ₂. Assume that equality holds. LetS be aγ (G)-set. ByProposition 8,|S| =γ (G)=γ_t(G)≤ ⁿ₃. ByProposition 8,|epn(x,S)| ≥ 2. Assume that there is

a vertexx ∈ Ssuch that|epn(x,S)| =2. Then(S− {x})∪epn(x,S)is an isolate dominating set forG, implying that ⁿ₂ = γ0(G) ≤ |S| +1 ≤ ⁿ₃ +1. This implies thatn ≤ 6. Since by Proposition 8,|epn(x,S)| ≥ 2 for any vertexx ∈ S, we obtain|S| =γ (G)=2, a contradiction. Thus assume that|epn(x,S)| ≥ 3 for anyx ∈ S. Then

|S| =γ (G)=γt(G)≤ ⁿ₄. Ifp0≤ ⁿ₄then byTheorem 12,ⁿ₂ =γ0(G)≤γ (G)−1+p0≤ ⁿ₂−1, a contradiction. Thus p₀≥ ⁿ₄+1. Thus|epn(x,S)| ≥ ⁿ₄+1 for anyx∈ S. This implies that|S| ≤3, and thus|S| =3. LetS= {x,y,z}.

Without loss of generality assume that|epn(x,S)| ≤ |epn(y,S)| ≤ |epn(z,S)|. LetXbe an independent dominating set forepn(x,S). Then|X| ≤ |epn(x,S)| ≤ ⁿ⁻³₃ . NowX ∪ {y,z}is an isolate dominating set forG, implying that

n

2 ≤2+ⁿ⁻³₃ . This inequality implies thatn≤6. This is a contradiction, since|epn(u,S)| ≥3 for anyu ∈S.

Next assume thatγ (G)= 2. ByProposition 8,γt(G)=2. LetS = {x,y}be aγ (G)-set. Let X be a minimum independent dominating set in G[epn(x,S)], andY be a minimum independent dominating set in G[epn(y,S)].

Assume that|X| ≤ |Y|. Then{y} ∪Xis an isolate dominating set of cardinality at mostⁿ₂, and soγ0(G)≤ ⁿ₂. Assume thatγ₀(G)= ⁿ₂. Then|X| = |Y| = ⁿ⁻²₂ ,X =N(x)− {y}andY =N(y)− {x}. Assume thatγ₀(G[X,Y]) < ⁿ₂−1.

LetDbe aγ₀(G[X,Y])-set. IfD∩X ̸= ∅andD∩Y ̸= ∅thenDis an isolate dominating set forG, a contradiction.

Thus assume that D ∩Y = ∅. Then(D ∩ X)∪ {y} is an isolate dominating set for G, a contradiction. Thus γ₀(G[X,Y])≥ ⁿ₂−1. ConsequentlyG∈E_n. Conversely it is straightforward to see thatγ₀(G)= ⁿ₂ifG∈E_n.

Theorem 15. If G is a graph of order n withδ(G) ≥ 2 and g(G) ≥ 5thenγ₀(G) ≤

n−⌊^g(G)₃ ⌋ 2



. This bound is sharp.

Proof. Let H be a graph obtained fromG by removing the vertices and edges of a shortest cycleC. We show that any vertex of V(H)has at most one neighbors inC. Letv ∈ V(H). Suppose thata,b ∈ N_G(v)∩V(C). Clearly d_G(a,b)≥3 andd_C(a,b)] ≥3, sinceg(G)≥5. Letaa₁a₂. . .a_k−1bbe the path onCfromatob. Then replacing this path withavbproduce a cycleC^′inGof length less thang(G), a contradiction. Thus any vertex ofV(H)has at most one neighbors inC. This implies thatHhas no isolated vertex, sinceδ(G)≥2. ByTheorem 14,γ0(H)≤ ⌈^n−g(G)₂ ⌉.

LetD1be aγ0(H)-set. IfD1is an isolate dominating set forG, thenγ0(G)≤ ⌈^n−g(G)₂ ⌉ ≤ ⌈^n−⌊

g(G) 3 ⌋

2 ⌉, as desired. Thus assume that D₁is not an isolate dominating set forG. LetD₂be aγ0(C)-set. Clearly byLemma 2,γ0(C)= ⌈^g(G)₃ ⌉.

Furthermore, for anyγ0(C)-setS, and any vertexx∈ S, there is a vertexy∈Ssuch thatd(x,y)≤3. IfD₁∪D₂is an isolate dominating set forGthen

γ₀(G)≤ |D₁| + |D₂| ≤

n−g(G) 2

 +

g(G) 3



≤

n− ⌊^g(G)₃ ⌋ 2

 .

Thus assume that D1∪D2is not an isolate dominating set forG. Then any vertex ofD2is adjacent to a vertex of D1inG. Let D3be obtained from D2by rotatingD2onC, clock-wise, such that any vertex is replaced by its next successive vertex onC. IfD₁∪D₃is an isolate dominating set forG then, as before,γ₀(G) ≤ ⌈^n−⌊

g(G) 3 ⌋ 2 ⌉, since

|D₂| = |D₃|. Thus assume that D₁∪D₃is not an isolate dominating set forG. Then any vertex ofD₃is adjacent to a vertex of D1inG. Now letD4 be obtained from D3 by rotating D3onC, clock-wise, such that any vertex is replaced by its next successive vertex onC. ThenD1∪D4is an isolate dominating set forG, sinceD1is not an isolate dominating set forG. This completes the proof of the bound. To see the sharpness consider cycles of length at least

five.

We finish by providing aNordhaus–Gaddum typeinequality for the isolate domination number.

Theorem 16. For any graph G of order n,γ₀(G)+γ₀(G)≤n+1, with equality if and only if G =K_nor G=K_n. Proof. If G has an isolated vertex then γ0(G) = 1, and so γ0(G)+γ0(G) ≤ n +1, sinceγ0(G) ≤ n. Thus assume that G has no isolated vertex, and similarly G has no isolated vertex. Now Theorem 14 implies that γ0(G)+γ0(G) ≤ ⁿ₂ + ⁿ₂ = n. Assume that equality holds. Then by the above proof,G or Ghave some isolated vertices. Assume thatGhas some isolated vertex. Then clearlyγ₀(G)=1. This means thatγ₀(G)=n. We conclude thatG=K_n. The converse is obvious.

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