On the computational complexity of defining sets

Discrete Applied Mathematics 149 (2005) 101–110

www.elsevier.com/locate/dam

On the computational complexity of defining sets�

Hamed Hatamia, Hossein Maserratb

aDepartment of Computer Science, University of Toronto, Toronto, Canada M5S 3G4bDepartment of Mathematical Sciences, Sharif University of Technology, Tehran, Iran

Received 21 September 2003; received in revised form 2 March 2005; accepted 16 March 2005Available online 10 May 2005

Abstract

Suppose we have a familyF of sets. For everyS ∈ F, a setD ⊆ S is adefining setfor (F, S) if Sis the only element ofF that containsD as a subset. This concept has been studied in numerous cases,such as vertex colorings, perfect matchings, dominating sets, block designs, geodetics, orientations,and Latin squares.

In this paper, first, we propose the concept of a defining set of a logical formula, and we prove thatthe computational complexity of such a problem is�2-complete.

We also show that the computational complexity of the following problem about the defining setof vertex colorings of graphs is�2-complete:

INSTANCE: A graphGwith a vertex coloringc and an integerk.QUESTION: If C(G) is the set of all�(G)-colorings ofG, then does(C(G), c) have a defining setof size at mostk?

Moreover, we study the computational complexity of some other variants of this problem.© 2005 Elsevier B.V. All rights reserved.

Keywords:Defining sets; Complexity; Graph coloring; Satisfiability

1. Introduction

In this paper, we consider a unification of the concepts already known as critical sets,forcing sets, and defining sets, where we formulate different natural problems in this regard.

� This research is done in Discrete Mathematics Laboratory of Department of Mathematics in Sharif Universityof Technology, Iran.

E-mail address:[email protected](H. Hatami).

0166-218X/$ - see front matter © 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.dam.2005.03.004

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102 H. Hatami, H. Maserrat / Discrete Applied Mathematics 149 (2005) 101–110

Specially, through considering such problems for 3SAT, by introducing suitable reductions,we prove that the decision problem related to the minimum defining set problem of graphcolorings1 is �2-complete.

Defining sets were studied for Latin squares[3,7], perfect matchings[1,2,10], orientations[5], geodetics[6], vertex colorings[13], designs[9], and dominating sets[4].

LetF be a family of sets. For everyS ∈ F, a setD ⊆ S is adefining setof (F, S), if Sis the only element inF which containsD as a subset. By abuse of language every definingset of(F, S) is also called a defining set ofF.

In what follows, we try to introduce a general formulation for the type of problems weare going to consider in the rest of this paper.

Suppose an inputI is given. The inputI might be a graph, a number, or any othermathematical object. Then letF(I ) be a family of sets which is defined according tothe setI. In this paper we are interested in the computational complexity of the followingthree general types of questions for specified inputs and definitions ofF.

1. Q1INSTANCE: I, a setS ∈ F(I ), and a setD ⊆ S.QUESTION: IsD a defining set of(F(I ), S)?

2. Q2INSTANCE: I, a setS ∈ F(I ), and an integerk.QUESTION: DoesShave a defining set of size at mostk?

3. Q3INSTANCE: I and an integerk.QUESTION: DoesF(I ) have a defining set of size at mostk?

The computational complexity of the problems related to defining sets was first studiedby Colbourn[7]. He studied Q1 whenF(n) is the set of Latin squares of ordern, and provedthat this question is CoNP-complete. Recently Adams et al.[1] studied Q2 whenF(G) isthe set of perfect matchings of a graphG, and proved that the question is NP-complete. In[2] it is shown that the question Q3 for this family is NP-complete. It is not hard to see thatthe question Q1 for this family is in Hatami and Tusserkani[12] studied Q2 and Q3 whenF(G) is the set of vertex colorings of a graphG, and proved that both of the questionsare NP-hard. In this paper we improve their result by showing that these problems are both�2-complete. In this regard we consider the family of all proper assignments to the variablesof a kCNF where akCNF is a Boolean expression in conjunctive normal form such thatevery clause has exactlyk variables. Let Q1-kSAT, Q2-kSAT, and Q3-kSAT stand for thethree questions Q1, Q2, and Q3 in this case, respectively. We show that Q1-3SAT is CoNP-complete, and Q2-3SAT and Q3-3SAT are both�2-complete. We also refer the reader tothe recent paper[11] for some other computational complexity results on the defining setsof vertex colorings.

We determine the computational complexity of Q1-3SAT, Q2-3SAT, and Q3-3SAT inSection 2. Section 3 is devoted to the study of the computational complexity of the questionsQ1, Q2, and Q3 for the set of vertex colorings of a graph.

1 Defined formally below.

H. Hatami, H. Maserrat / Discrete Applied Mathematics 149 (2005) 101–110 103

2. Defining sets and SAT

Let D andR be two sets, andf : D → R be a function. We can refer tof as the setf ={(x, f (x)) : x ∈ D}. This representation enables us to study the defining set of a familyof functions.

Let � be akCNF with variablesV = {v1, v2, . . . , vn}. For the sake of simplicity weuse the notation�(v) instead of�(v1, . . . , vn), where we would think ofv as a vector ofv1, v2, . . . , vn. Since any truth assignmentt : V → {true, false} is a function, we can studythe defining set of a family of assignments. Aproper assignmentof � is an assignmentwhich makes� true. LetS ⊆ V be a subset of the variables of�. A partial assignmentof� over the setS is a truth assignmentt : S → {true, false}. The setS is called the supportset oft, and this is denoted byS = supp(t). A partial assignment overS is calledproper ifevery clause of� contains at least one true literal from the variables inS.

For everykCNF�, letP(�) denote the family of proper assignments of�. We study thecomputational complexity of the general questions Q1, Q2, and Q3 for this special family.Let Q1-kSAT, Q2-kSAT, and Q3-kSAT stand for the three questions Q1, Q2, and Q3 in thiscase, respectively.

Duplicating a variable in a clause of a CNF does not change the family of its properassignments. Hence if all clauses of a CNF are of size at mostk (not necessarily equal to),then it can be converted to akCNF. Therefore without loss of generality, we may alwaysassume that all such expressions are inkCNF form.

In this section we show that Q1-3SAT is CoNP-complete, and Q2-3SAT and Q3-3SATare both�2-complete. From[14] we know that the following problem is�2-complete.

• ∃� 3SATINSTANCE: A 3CNF,�(x, y).QUESTION: Is ∃x�y �(x, y)?

Next, we define the∃∃!∗ kSAT problem, and prove that it is�2-complete. AkCNF �consisting of variablesV ={x1, . . . , xn, y1, . . . , ym} with a proper partial assignmentt overthe set{y1, y2, . . . , ym} is given. The question is:

“Is there a partial assignmentt ′ (not necessarily proper) over the set{x1, . . . , xn} suchthat� has a unique proper assignmentr which satisfiesr(xi) = t ′(xi) for every 1� i�n?”

Note that sincet is a proper partial assignment, if such at ′ exists, thenr(yj ) = t (yj ) for1�j �m.

• ∃∃!∗ kSATINSTANCE: A kCNF,�(x, y) and a proper partial assignmentt over the set of the variablesyj .QUESTION: Is ∃x∃!ty �(x, y)?

Theorem 1. The∃∃!∗ 4SATproblem is�2-complete.

Proof. The problem is in�2.To prove the completeness, we give a reduction from∃� 3SAT.Consider a 3CNF,�(x, y) and the problem∃x�y �(x, y). We construct an instance of


∃∃!∗ 4SAT, a 4CNF� with a proper partial assignmentt, as in the following. The expression� has all of the variables of� plus one more variablez. Let C1, C2, . . . , Cn be the clausesof �(x, y). Then

�(x, y, z) = (C1 ∨ z) ∧ (C2 ∨ z) ∧ · · · ∧ (Cn ∨ z) ∧ (z ∨ y1)

∧ (z ∨ y2) ∧ · · · ∧ (z ∨ ym). (1)

The partial assignmentt (z) = true and t (yj ) = true (1�j �m) is given. This partialassignment is proper because there exists a true literal in every clause amongz andyj ’s.For every proper assignmentuof �(x, y, z), if u(z)= true, thenu(yj )= true (1�j �m). Ifu(z) = false, then by ignoring the variablez in u, u is a proper assignment of�(x, y), andvice versa. So∃x�y �(x, y) if and only if ∃x∃!t (y, z) �(x, y, z). �

Next, we modify the proof of Theorem 1 so that we can conclude that∃∃!∗ 3SAT is�2-complete. Consider�(x, y, z), defined in (1). In�(x, y, z) every clause of the form(z∨ yj )

has two literals. But a clause of the form(Ci ∨z) has four literals. SupposeCi =a1∨a2∨a3,wherea1, a2, a3 are literals. We replace each clause(Ci ∨ z) in � by C′

i defined as follows:

C′i = (a1 ∨ a2 ∨ vi) ∧ (a3 ∨ z ∨ vi ) ∧ (a1 ∨ z ∨ vi)

∧ (a2 ∨ z ∨ vi) ∧ (a1 ∨ a3 ∨ vi) ∧ (a2 ∨ a3 ∨ vi),

wherevi ’s are new variables, and call the new expression as�′(x, y, v, z). Thus

�′(x, y, v, z) = C′1 ∧ · · · ∧ C′

n ∧ (z ∨ y1) ∧ (z ∨ y2) ∧ · · · ∧ (z ∨ ym).

Define the partial assignment asu(z) = true, u(yi) = true, u(vi) = true for 1� i�m.The following three observations imply that∃x∃!t (y, z) �(x, y, z) if and only if ∃x∃!u

(y, v, z) �′(x, y, v, z).

(a) u(z) = true, u(yj ) = true (1�j �m), andu(vi) = true (1� i�n) is a proper partialassignment of�′(x, y, v, z).

(b) Every truth assignment toa1, a2, a3, andzwhich assigns a true value to at least one ofthem is extended uniquely to a proper assignment ofC′

i .(c) Since every assignment which assigns a false value toa1, a2, a3, andzsimultaneously

is not a proper assignment ofC′i , every proper assignment of�′(x, y, v, z) leads to a

proper assignment of�(x, y, z) by ignoring the values ofvi ’s.

Note that any proper subset of the clauses ofC′i does not satisfy these properties. For

example consider the assignmentt (a1) = false, t (a2) = true, t (a3) = true, t (z) = false. Inthis case, regardless of what value is assigned tovi , the first five clauses are satisfied, andthe last clause is necessary to fix the value ofvi .

We conclude the following theorem from (a)–(c).

Theorem 2. The∃∃!∗ 3SATproblem is�2-complete.

In the proof of Theorem 2 the problem∃� 3SAT is reduced to∃∃!∗ 3SAT. In that proofby assuming that there are no variablesxi ’s in �(x, y) (i.e. the number of variables after the


first quantifier of∃� 3SAT is zero), we can obtain a reduction form� 3SAT to the problemwhich asks whether a given proper assignment of a 3CNF is its only proper assignment. Thisproblem is a restriction of Q1-3SAT in whichD, the set which is asked to be the definingset, is the empty set. Since� 3SAT is CoNP-complete, we have:

Theorem 3. Q1-3SATis CoNP-complete.

The next theorem determines the computational complexity of Q2-3SAT.

Theorem 4. Q2-3SATis�2-complete.

Proof. The problem is in�2. We reduce∃∃!∗ 3SAT to this problem. Let∃x∃!ty �(x, y)be an instance of∃∃!∗ 3SAT, where�(x, y) is a 3CNF with variablesx1, x2, . . . , xk andy1, y2, . . . , ym, and t is a proper partial assignment over variablesyj . We construct aninstance of Q2-3SAT, a 3CNF� with a proper assignmentt ′, such that(P(�), t ′) has adefining set of size at mostk, the number of the variablesxi , if and only if ∃x∃!ty �(x, y).In the following we describe how� is obtained from�(x, y).

For every 1� i�k, consider two new variablesvi andv′i , and replace everyxi in each

clause of�(x, y) with vi and everyxi with v′i .

For every 1�j �m, a literalaj is defined as follows. Ift (yj ) = false, thenaj is yj andotherwiseaj is yj . We add the following clauses to the expression in whichwi are newvariables.

(a1 ∨ a2 ∨ w1) ∧ (w1 ∨ a3 ∨ w2) ∧ · · · ∧ (wm−2 ∨ am ∨ wm−1). (2)

Note that by settingyj ’s according to the given assignmentt, wi ’s are forced to take thetruth valuetrue. The following clauses are also added to the expression.

(wm−1 ∨ v1 ∨ v′1) ∧ (wm−1 ∨ v1 ∨ v′

1) ∧ · · · ∧ (wm−1 ∨ vk ∨ v′k)

∧ (wm−1 ∨ vk ∨ v′k).

Call this new 3CNF,�(v, v′, y,w). Let t ′ be the assignmentt ′(vi)=t ′(v′i )=false(1� i�k),

t ′(wi)=true(1� i�m−1), andt ′(yj )=t (yj ) (1�j �m). Note thatt ′ is a proper assignmentof �.

We claim that(P(�), t ′) has a defining set of size at mostk, if and only if∃x∃!ty �(x, y).Suppose that∃x∃!ty �(x, y). This means that there is a partial assignmentu over x1,

x2, . . . , xk such that the only proper values foryj are the values that are assigned to themby the partial assignmentt. If u(xi) = true, we choose(v′

i , false), and ifu(xi) = false, wechoose(vi, false). Call this setS. We claim thatS is a defining set of(P(�), t ′).

Suppose thatSis not a defining set of(P(�), t ′). Then there is a proper assignmentt ′′ �= t ′which is an extension ofS. Sincet ′′(vi)=falseandt ′′(v′

j )=falsefor vi, v′j ∈ supp(S), it can

be easily seen that the assignmentr defined asr(xi) = u(xi) (1� i�k) andr(yj ) = t ′′(yj )

(1�j �m) is a proper assignment to the variables of�(x, y), which is a contradiction. Soall yj take the values that are assigned to them by the assignmentt. Hencewi ’s are true forall 1� i�m − 1. Since the two clauses(wm−1 ∨ vi ∨ v′

i ) and(wm−1 ∨ vi ∨ v′i ) are in�,

and exactly one ofvi or v′i is in supp(S), the value of the other one is also determined to be

false, and this is the assignmentt ′.


Next suppose that(P(�), t ′) has a defining setSof size at mostk. Then for every 1� i�k,at least one ofvi or v′

i is in supp(S). Otherwise we can change the values of bothvi andv′i to true, and still have a proper assignment. So a defining set of size at mostk includes

exactly one of(vi, false) or (v′i , false) for every 1� i�k. Let u be a partial assignment of

�(x, y) such thatu(xi) = true if vi ∈ supp(S), andu(xi) = falseif v′i ∈ supp(S).

We claim that�(x, y) has a unique proper assignmentr such thatr(xi) = u(xi) for every1� i�k. Suppose that there is a proper assignmentr for �(x, y) such thatr(xi)=u(xi) forall 1� i�k, but there exists at least one 1� i0�m such thatr(yi0) �= t (yi0).

Consider�(v, v′, y,w), and letr ′(yj ) = r(yj ) (1�j �k). Sincer(yi0) �= t (yi0), it ispossible to assign valuesr ′(wi) (1� i�m − 1) such thatr ′(wm−1) = falseand the clausesin (2) are true.

Note thatexactlyone ofvi or v′i is in supp(S). For every 1� i�k, if vi ∈ supp(S),

then definer ′(vi) = false, r ′(v′i ) = true; and if v′

i ∈ supp(S), then definer ′(vi) = true,r ′(v′

i ) = false.Sincet (wm−1)= false, the values assigned byr ′ do not make(wm−1∨vi ∨ v′

i )∧(wm−1∨vi ∨ v′

i ) false.Now all clauses are satisfied. So there exists another proper assignment containing the

defining set, which is a contradiction.�

Theorem 5. Q3-3SATis�2-complete.

Proof. The problem is in�2. We give a reduction from Q2-3SAT. Consider an instance ofQ2-3SAT, a 3CNF� with a proper assignmentt and an integerk. Let the variables of� bex1, x2, . . . , xn. We addn(k + 1) new variablesyij (1� i�n and 1�j �k + 1). For everyxi , if t (xi) = true, then we add the following clauses,

(xi ∨ yi1) ∧ (xi ∨ yi2) ∧ · · · ∧ (xi ∨ yi(k+1))

and if t (xi) = false, then we add the following clauses,

(xi ∨ yi1) ∧ (xi ∨ yi2) ∧ · · · ∧ (xi ∨ yi(k+1)).

The new 3CNF consists of� and thesen(k + 1) new clauses. Denote this 3CNF by�′.We claim thatP(�′) has a defining set of size at mostk, if and only if (P(�), t) has adefining set of size at mostk. Every defining set of(P(�), t) is also a defining set ofP(�′),because the assignmentt forces all of theyij to take a true value.

Next suppose that there is a defining set ofP(�′) which fixes a proper assignmentt ′. Forevery 1�x�n, if t ′(xi) �= t (xi), then since it is possible to assign every arbitrary values toyi1, yi2, . . . , yi(k+1), all thesek+1 variables are in the defining set. Hence in every definingset of size at mostk, all xi take the same values int ′ andt. Now, sincet ′(xi) = t (xi), byfixing the value ofxi , the values ofyij ’s are determined to be true, for 1�j �k + 1. Soif (yij , t

′(yij )) is in the defining set, then it is possible to replace it by(xi, t′(xi)). Thus

a defining set of size at mostk of P(�′) can be modified so that all its elements are in{(xi, t

′(xi)) : i = 1, . . . , n}, andt ′(xi) = t (xi). This is also a defining set of(P(�), t). �


3. Vertex coloring

For every graphG with vertex setV = {v1, . . . , vn}, every vertex coloringc of G is afunction which maps every vertexvi to a colorc(vi). For every partial coloringc of G,define supp(c) as the set of the vertices thatc assigns a color to them. Denote the family ofall �(G)-vertex colorings ofGbyC(G). In [8] it is shown that the uniqueness of colorabilityis CoNP-complete. This implies the following theorem.

Theorem 6. The problem Q1-VERTEX COLORING is CoNP-complete.

In this section we show that both of the problems Q2 and Q3 for this family are�2-complete.

• Q2-VERTEX COLORING

INSTANCE: A graphGwith a�(G)-vertex coloringc, and an integerk.QUESTION: Does(C(G), c) have a defining set of size at mostk?

• Q3-VERTEX COLORING

INSTANCE: A graphG, and an integerk.QUESTION: DoesC(G) have a defining set of size at mostk?

Theorem 7. Q2-VERTEX COLORING is�2-complete for graphs with� = 3.

Proof. The problem is in�2. To prove the completeness, we introduce a reduction fromQ2-3SAT. Consider an instance of Q2-3SAT: A proper assignmentt of �(x1, x2, . . . , xn)

and an integerk. We construct a graphG� with chromatic number 3 and a 3-vertex coloringct of G� such that(P(�), t) has a defining set of size at mostk if and only if (C(G�), ct )

has a defining set of size at mostk + 4.We begin by considering a cycle of size 3 with verticesw0, w1, andw2 which are

connected to four verticesw′1, w

′2, w

′3, andw′

4 as is shown inFig. 1(a). For every variablexi , add two verticesuxi

anduxiand edges{uxi

, uxi}, {uxi

, w2}, and{uxi, w2} to the graph.

This is illustrated inFig. 1(a).Consider a clauseCi = (a1 ∨ a2 ∨ a3) of �, whereaj (j = 1, 2, 3) is a literal. Sincet is

a proper assignment of�, without loss of generality we can assume thatt (a2) = true. Forevery such clause, we add a copy of the graph shown inFig. 1(b) to the graph, and connectits vertices to the other vertices as is shown inFig. 1(b). Note thatuaj

(j = 1, 2, 3) is oneof the verticesux1, ux2, . . . , uxn or ux1, ux2, . . . , uxn . Call this new graphG�.

One can easily check that assigning a 3-coloringct toua1,ua2, andua3 such thatct (w0)=0,ct (w1) = 1, andct (w2) = 2 and alsoct (ua2) = 1 determines the colors ofvi1, vi2, . . . , vi8uniquely. Letct be a 3-coloring ofG� defined as in the following:

• ct (w0) = 0, ct (w1) = 1, andct (w2) = 2.• ct (w

′1) = 1, ct (w

′2) = 2, ct (w

′3) = 0, andct (w

′4) = 2.

• For every 1� i�n if t (xi)=true, thenct (uxi)=1 andct (uxi

)=0, and otherwisect (uxi)=0

andct (uxi) = 1.

• Colors ofvij are determined uniquely by the colors of the vertices above.


w0 w2

w1 w1

w2

vi8

vi6 vi7

vi1 vi3

vi4 vi5

. . .

w′1 w′

3

w′2 w′

4

w1w0

w2

u x2u x2

u x1u x1

u xnu xn

u a1 u a2

u a2

u a3

v i2

(a) (b)

Fig. 1. (a) Verticesuxianduxi

are connected tow2. (b) For every clause we add a copy of this graph toG�.

The verticesw′1, w

′2, w

′3, andw′

4 are in every defining set (otherwise we can change theircolors). The colors of these four vertices determine the colors ofw0, w1, andw2 uniquely.

We claim that the size of the smallest defining set of(C(G�), ct ) is equal to the size of thesmallest defining set of(P(�), t) plus 4. Note that any partial coloring which only assigns0 or 1 toua1, ua2, ua3 and does not assign 0 to all of them can be extended to a propercoloring of the graph inFig. 1(b). Moreover, if all the verticesua1, ua2, ua3 are colored by0, then it can be easily seen thatvi8 is also forced to be colored by 0. Sincevi8 is connectedto w0 andw2, G� admits a 3-coloring, if and only if� has a proper assignment.

Suppose(C(�), t) has a defining set that consists ofk variablesxi1, xi2, . . . , xik . Thenassigning colors ofk + 4 verticesw′

1, w′2, w

′3, w

′4 and uxi1

, uxi2, . . . , uxik

constitutes adefining set of(C(G�), ct ).

Next suppose thatSis the smallest defining set of(C(G�), ct ). Thenw′1, w′

2, w′3, w′

4 arein supp(S). By assigning the colors of these vertices, the colors ofw1, w2, w3, and allvi8’sare determined uniquely. It can be verified easily that for every clauseCi =(a1, a2, a3) of �,sincect (ua2)=1, the colors ofvi1, vi2, . . . , vi7 are determined uniquely by fixing the colorof ua2, and the color ofuxi

determines the color ofuxi. Hence we can assume that supp(S)

containsw′1, w′

2, w′3, w′

4, and some ofuxi. Using the fact that any partial coloring which

only assigns 0 or 1 toua1, ua2, ua3 and does not assign 0 to all of them can be extended toa proper coloring of the graph inFig. 1(b), we conclude that the corresponding variables oftheseuxi

constitute a defining set of(C(�), t). �

Theorem 8. Q3-VERTEX COLORING is�2-complete for graphs with� = 3.

Proof. The problem is in�2. We give a reduction from Q2-VERTEX COLORINGwhen�=3.Consider an instance(C(G), c) of Q2-VERTEX COLORING, whereG is a graph andc is a3-vertex coloring ofG. Assume that the range ofc is the set{0, 1, 2}. An integerk is given,and it is asked that “Is there a defining set of size at mostk for (C(G), c)?” We construct anew graphH as follows:

1. First letH be the disjoint union ofG and a cyclew0w1w2 of size 3.


2. Then for every vertexui of G, let c1 andc2 be the two colors other thanc(ui). Add2k + 2 verticesvui ,cj ,1, vui ,cj ,2, . . . , vui ,cj ,k+1 (1�j �2) toH. For every 1� t �k + 1,connectvui ,cj ,t to bothui andwcj

. (Notice thatwcjis one ofw0, w1, or w2.)

3. Add four new verticesw′1, w′

2, w′3, andw′

4 toH, and connectw′1 andw′

2 to w0, and alsow′

3 andw′4 to w1.

Now we claim thatC(H) has a defining set of size at mostk + 4 if and only if(C(G), c)

has a defining set of size at mostk.First consider a defining set of size at mostk for (C(G), c), sayD. If we fix the colors

of the vertices inD and assign the colors 1 tow′1, 2 tow′

2, 0 tow′3, and 2 tow′

4, then thesek + 4 vertices constitute a defining set ofC(H).

Next suppose thatD is the smallest defining set ofC(H) which has at mostk+4 vertices.Without loss of generality assume that in the extension ofD to a 3-vertex coloringc′ of H,wi (0� i�2) is colored byi. Since the degrees ofw′

1, w′2, w′

3, andw′4 are equal to one, they

are in supp(D). Suppose that in the extension ofD to a 3-coloring ofH, a vertexui of G iscolored byc′(ui) which is not equal toc(ui). The verticesvui ,c

′(ui ),t (1� t �k +1) are onlyconnected toui andwc′(ui ). Since these two vertices are colored by the same colors, all thesek + 1 vertices are in the defining set, and with the four verticesw′

i , the size of the definingset is at leastk + 5. SinceD is of size at mostk + 4, for every vertexui , c′(ui) = c(ui).

We can suppose thatw′1 andw′

2 (and sow′3 andw′

4) are colored by different colors.Otherwise by changing the color ofw′

2 (and sow′4), D still remains a defining set ofC(H).

Sincew′1 andw′

2 and alsow′3 andw′

4 are colored by different colors, they determine thecolors ofw1, w2, andw3 uniquely. Therefore sinceD is the smallest defining set, none ofw1, w2, andw3 is in supp(D). Also if a vertexvui ,cj ,t (cj �= c(ui)) is in D, then we canreplace it withui . Now, if we remove the four verticesw′

i from the defining set, we obtaina defining set of(C(G), c). �

Acknowledgements

The authors wish to thank A. Daneshgar for many useful comments and suggestions inpreparing this paper, and E.S. Mahmoodian and P. Afshani for many valuable discussionstowards the results of this paper. We would also like to thank Oleg Verbitsky for pointingout some errors in the draft version of this paper.

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On the computational complexity of defining sets