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## Journal of the Brazilian Computer Society

##
*Print version* ISSN 0104-6500*On-line version* ISSN 1678-4804

### J. Braz. Comp. Soc. vol.7 no.3 Campinas 2001

#### http://dx.doi.org/10.1590/S0104-65002001000200007

**ARTICLES**

**On the helly defect of a graph**

**Mitre C. Dourado ^{*}; Fábio Protti^{}; Jayme L. Szwarcfiter^{}**

**ABSTRACT**

The *Helly defect *of a graph *G *is the smallest integer *i *such that the iterated clique graph *K ^{i}*(

*G*) is clique-Kelly. We prove that it is NP-hard to decide whether the Helly defect of

*G*is at most 1.

**Keywords: **Clique graphs, clique-Helly graphs, Helly defect

**1 Introduction**

In this work, we consider the following question, on iterated clique graphs. Given a graph *G *and an integer *i >* 0, is the

*i*-iterated clique graph of G a clique-Helly graph? Since clique-Helly graphs can be recognized in polynomial time [11], for

*i =*0 the answer of this question can be given in polynomial time. In this work, we prove that the above problem is NP-hard for

*i =*1. In fact, the NP-hardness holds for a more general problem stated in Theorem 6.

In general, write that a set *S *is a *k*-set when ê*S*ê= *k, *a *k*^{+}-set when ê*S*ê* > k, *and a

*k*

^{–}-set when ê

*S*ê

*This same notation will also apply for families of sets.*

__<__k. Let be a family of subsets *F _{i}* of some set. A

*core*of is any subset of

*F*Let

_{i}.*p,q*be integers,

*p*1 and

__>__*q*0. Say that is (

__>__*p,q*)

*-intersecting*when every p

^{–}-subfamily of it has a

*q*In addition, is (

^{+}-core.*p, q*)

*-Helly*when every (

*p,q*)-intersecting subfamily of it has a

*q*

^{+}-core. Consequently, the classical

*p-Helly*families of sets [2, 3] correspond to the case (

*p,*1)-Helly, while the Helly families correspond to the notation (2, 1)-Helly.

Let *G* be a graph. A *complete *of *G *is a subset of pairwise adjacent vertices of it. A *clique *is a maximal complete. A *clique-transversal *of *G* is a subset of vertices intersecting all cliques. The problem of deciding whether a given subset is a clique-transversal of *G* has been proved to be Co-NP-Complete by Erdös, Gallai and Tuza [8]. Say that G is (*p,q*)*-clique-Helly *when the family of the cliques of G is (*p,q*)-Helly. When G is (2,1)-clique-Helly, write simply *clique-Helly.*

The *clique graph K*(*G*) of G is the intersection graph of the cliques of *G*. The *i-th iterated clique graph *of *G*, denoted *K ^{i}*(

*G*)

*,*is defined as follows:

*K*

^{0}(

*G*)

*= G,*while

*K*(

^{i}*G*)

*= K*(

*K*

^{i}^{–1}(

*G*))

*, i >*0.

The *Helly defect *of a graph G is the smallest *i *such that *K ^{i}*(

*G*) is clique-Helly. If

*K*(

^{i}*G*) is not clique-Helly, for any finite

*i,*say that its Helly defect is infinite. Trivially, the Helly defect of a clique-Helly graph is 0.

A graph *G* is *periodic *when *K ^{i}*(

*G*)

*= G*, for some

*i.*In addition, the smallest

*i*satisfying

*K*(

^{i}*G*) =

*G*is the

*period*of a periodic graph

*G*.

Escalante [9] proved that if *K ^{i}*(

*G*) is clique-Helly, then

*K*(

^{j}*G*) is clique-Helly for any

*j*and that if

__>__i,*G*is clique-Helly, then

*K*(

^{2}*G*) is an induced subgraph of

*G*.

Let *G* be a periodic clique-Helly graph, and let *p >* 1 be its period. Clearly,

*G*=

*K*(

^{2p}*G*)

*.*By Es-calante's result,

*K*(

^{2p}*G*) is an induced subgraph of

*K*(

^{2}*G*)

*,*therefore

*G*=

*K*(

^{2}*G*)

*.*This implies that

*p*2. This argument shows that if

__<__*H*is a periodic graph with period strictly greater than 2, then

*K*(

^{i}*H*) cannot be clique-Helly for any

*i*0, and thus its Helly defect is infinite.

__>__ Bandelt and Prisner [1] proved that the Helly defect of a chordal graph is at most 1. In [4] there are examples of graphs with Helly defect *i, *for any finite *i.*

In this paper, we prove that it is NP-hard to decide whether the clique graph of *G* is (2,q)-Helly, for any fixed *q >* 1. Consequently, it is NP-hard to decide whether

*G*has Helly defect at most 1.

**2 The proof**

In this section, we prove that it is NP-hard to recognize if the Helly defect of a graph is at most 1. We need first some definitions and results.

Let *G *be a graph and *T *a triangle of it. The *extended triangle *of *G, *relative to *T, *is the subgraph of *G *induced by subset of vertices of *G *which form a triangle with (at least) two vertices of *T.*

**Theorem 1 **[6, 11] *G is a clique-Helly graph if and only if every of its extended triangles contains a universal vertex.*

Extended triangles can be generalized as follows: let C be a *p*-complete of a graph *G, p > *3. The

*p-expansion of G relative to C*is the subgraph of

*G*induced by the vertices forming a

*p*-complete with

*p –*1 vertices of

*C.*

Let be a subfamily of cliques of *G. *The *clique subgraph induced by* * in G, *denoted by *G*[]* _{c }, *is the subgraph of

*G*formed exactly by the vertices and edges belonging to the cliques of

*.*

**Lemma 2 **[5] *Let G be a graph, C a p-complete of it, H the p-expansion of G relative to C, and ** the subfamily of cliques of G that contain at least p *1 *vertices of C. Then G*[]_{c} is a spanning subgraph of H.

**Proof. **We have to show that *V*(*G*[]* _{c}*)

*= V*(

*H*)

*.*Let

*v*Î

*V*(

*H*)

*.*Then

*v*is adjacent to at least

*p -*1 vertices of

*C.*Hence,

*v*together with those

*p *1 vertices form a

*p*-complete, which is contained in a clique that contains at least

*p *1 vertices of

*C.*Therefore,

*v*Î

*V*(

*G*[]

*)*

_{c}*.*Now, consider

*v*Î

*V*(

*G*[]

*)*

_{c}*.*Then

*v*belongs to some clique containing

*p *1 vertices of

*C.*That is,

*v*is adjacent to at least

*p*1 vertices of

*C*, and hence

*v*Î

*V*(

*H*)

*.*Consequently,

*V*(

*G*[]

*)*

_{c}*= V*(

*H*)

*.*Furthermore, both

*H*and

*G*[]

*are subgraphs of*

_{c}*G,*but

*H*is induced. Thus

*E*(

*G*[]

*) Í*

_{c}*E*(

*H*)

*.*

Let *q *be a positive integer. The graph F* _{q}*(

*G*) is defined in the following way: the vertices of F

*(*

_{q}*G*) correspond to the

*q*-completes of

*G*, two vertices being adjacent in F

*(*

_{g}*G*) if the corresponding

*q*-completes in

*G*are contained in a common clique. We remark that F

_{q}is precisely the operator F

_{q,2q}, studied in [10]. An interesting property of F

*is that there exists a bi-jection between the subfamily of*

_{q}*q*

^{+}-cliques of

*G*and the family of cliques of F

*(*

_{q}*G*) [5]. The following definitions are possible due to this property: If

*C*is a

*q*

^{+}-clique of

*G,*denote by F

*(*

_{q}*C*) the clique that corresponds to

*C*in F

*(G). If*

_{q}*C'*is a clique of F

*(*

_{q}*G*), denote by (

*C'*)the

*q*

^{+}-clique that corresponds to

*C'*in

*G*. If = is a subfamily of

*q*

^{+}-cliques of

*G*, define F

*()*

_{q}*=*{F

*) |*

_{q}(C*C*Î }. If is a subfamily of cliques of F

*(*

_{q}*G*), define = {(

*C*) |

*C*Î }.

**Lemma 3 **[5] *Let G be a graph, ** a subfamily of q ^{+} -cliques of it, *

*=*F

*()*

_{q}*, and H =*F

*(*

_{q}*G*).

*Then H*[]

*[]*

_{c}contains a universal vertex if and only if G

_{c }contains q universal vertices. **Proof. **If* H*[]* _{c}*contains a universal vertex

*x,*then every clique of contains the

*q*-complete of

*G*that corresponds to

*x,*that is,

*G*[]

*contains*

_{c}*q*universal vertices. Conversely, if

*G*[]

_{c}contains

*q*universal vertices forming a

*q*-complete

*Q*of

*G*, then every clique of contains the vertex of

*H*that corresponds to

*Q,*that is,

*H*[]

*contains a universal vertex.*

_{c}

The proof of the next lemma is easy, and thus we will omit it:

**Lemma 4 **[5] *Let C be a *(*p+ *1)*-complete of a graph G, and let ** be a p ^{–}-subfamily of cliques of G such that every clique of *

*contains at least p vertices of C. Then*

*has a*1

^{+}

*-core.*

The following theorem is a characterization of (*p, q*)*-*clique-Helly graphs:

**Theorem 5 **[5] *Let p > *1

*be an integer. A graph G is*(

*p, q*)

*- clique- Helly if and only if every*(p+1)

*-expansion of*F

*(*

_{q}*G*)

*contains a universal vertex.*

**Proof. **Suppose that *G* is a (*p,*1)-clique-Helly graph and there exists a (*p + *1)-expansion *T, *relative to a (*p + *1)-complete *C* of F* _{q}*(

*G*), such that

*T*contains no universal vertex.

Let be the subfamily of cliques of *H = *F* _{q}*(

*G*) that contain at least

*p*vertices of

*C*. Let = (). Consider a

*p*

^{–}-subfamily

*'*Í

*.*Let

*=*F

*(*

_{q}*'*)

*.*By Lemma 4, ' has a 1

^{+}-core. That is,

*H*[']

*contains a universal vertex. This implies, by Lemma 3, that*

_{c}*G*[

*'*]

_{C}contains

*q*universal vertices. Thus,

*'*has a

*q*

^{+}-core, that is, is (

*p, q*)-intersecting. Since

*G*is (

*p, q*)-clique-Helly, we conclude that has a

*q*core and

^{+}-*G*[]

_{C}contains

*q*universal vertices. By using Lemma 3 again,

*H*[]

*contains a universal vertex. Moreover, by Lemma 2,*

_{c}*H*[]

*is a spanning subgraph of*

_{c}*T.*However,

*T*contains no universal vertex. This is a contradiction. Therefore, every (

*p +*1)-expansion of

*H =*F

*(*

_{q}*G*) contains a universal vertex.

Conversely, assume by contradiction that *G* is not (*p, q*)-clique-Helly. Let * = *{*C*_{1}, . . . , *C _{k}*} be a minimal (

*p, q*)-intersecting subfamily of cliques of

*G*which does not have a

*q*-core. Clearly,

*k*>

*p.*

By the minimality of , the subfamily \*C _{i} *has a

*q*core

^{+}-*Q*, for

_{i}*i =*1,...,

*k*. It is clear that

*Q*

_{i}*C*Moreover, every two distinct

_{i}.*Q*,

_{i}*Q*are contained in a same clique, since

_{j }*k*3. Hence the sets

__>__*Q*

_{1},

*Q*

_{1}

*,*...,

*Q*

_{p+1}correspond to a (

*p*+ 1)-complete

*C*in F

*(*

_{q}*G*).

Let be the subfamily of cliques *of H = *F* _{q}*(

*G*) that contain at least

*p*vertices of

*C.*Let ' = F

*() = {F*

_{q}*(*

_{q}*C*

_{1}),...,F

*(*

_{q}*C*)}. Since every

_{k}*C*Î contains at least

_{i}*p*sets from

*Q*

_{1},

*Q*

_{1}

*,...*,

*Q*

_{p+1}

*,*it is clear that the clique F

*(*

_{q}*C*) of if contains at least

_{i}*p*vertices of

*C.*Therefore, F

*(*

_{q}*C*) Î , for

_{i}*i*= 1,...,

*k.*

Let *T *be the (*p* + 1)-expansion of *H *relative to *C. *By Lemma 2, *H*[]* _{c} *is a spanning subgraph of

*T.*Therefore,

*V*(

*Q*) Í

*V*(

*T*)

*,*for every

*Q*Î . In particular,

*V*(F

*(*

_{q}*C*)) Í

_{i}*V*(

*T*), for

*i*= 1,...,

*k.*By hypothesis,

*T*contains a universal vertex

*x.*Then

*x*is adjacent to all the vertices of F

*(*

_{q}*C*)\{

_{i}*x*}, for

*i =*1,...,

*k.*This implies that F

*(*

_{q}*C*) contains

_{i}*x,*otherwise F

*(*

_{q}*C*) would not be maximal. Thus, ' has a 1

_{i}^{+}-core and if []

_{c}contains a universal vertex. By Lemma 3,

*G*[]

*contains*

_{c}*q*universal vertices, that is, has a

*q*core

^{+}-*.*This contradicts the assumption for

*.*Hence,

*G*is a (

*p,q*)-clique-Helly graph.

Next, we state the main result of this paper.

**Theorem 6** *Let q >* 1

*be a fixed integer. Given a graph G, it is NP-hard to decide whether K*(

*G*)

*is*(2,

*q*)

*-clique-Helly.*

**Proof.**

The proof is a transformation from the following problem. Given a graph *G *and a clique *Q *of it, does *Q *intersect all the cliques of G? (In other words, is *Q *a *clique-transversal *of *G?*) This problem was shown to be Co-NP-complete in [7]*.*

Given a graph *G *and a clique *Q *of it, we have to construct a graph *H *such that *Q* is a clique-transversal of *G *if and only if *K*(*H*) is (2,g)-clique-Helly. The construction of if is as follows: consider first the graph formed by three disjoint copies of *G, *denoted by G_{a},G_{b},G_{c}. Add six vertices forming the set *V*_{1}* = *, and add the following edges:

Add now *q* vertices for each copy of *G*, forming the set *V _{2} = *{

*a*

_{1},

*a*

_{2},...,

*a*,

_{q}*b*

_{1},

*b*

_{2},...,

*b*

_{q},

*c*

_{1},

*c*

_{2}, ...,

*c*}, and add the edges:

_{q} For each vertex *w* Î *Q*, consider its three copies in *G _{a},G_{b},G_{c}, *and collapse them into a single vertex, preserving all its adjacencies. For simplicity, the vertex of

*H*corresponding to a vertex

*w*Î

*Q*will also be denoted by

*w,*and we will refer to

*Q*as a complete of

*H.*

Finally, add more *3q *vertices forming the set *V*_{3}* =*

The construction of if is completed (see Figure 1). Clearly, the clique *Q* of *G* corresponds to the complete *Q of H, *while any other clique *Q' *¹* Q *of *G *corresponds to three complete sets *, *located at *G _{a},G_{b},G_{c}, *respectively.

By construction, observe that every clique of if is formed by a copy of a clique of G together with three new vertices: two from *V *and one from *V*_{2}È *V*_{3}*.*

We proceed with the proof of the theorem.

Suppose that Q is a clique-transversal of *G. *This implies that *Q *intersects any clique *Q' *of *G. *Since each clique of *H *contains a copy of a clique of *G *by construction, *Q *intersects any clique of *H. *Therefore any clique of *H *that contains *Q* is a clique-transversal of *H. *Consider the following cliques of *H:*

Since these *q *cliques contain *Q, *they are clique-transversals of *H. *Clearly, a clique which is a clique-transversal of a graph *G' *is a universal vertex of *K*(*G'*)*. *So the clique graph of *H *contains *q *universal vertices, hence *K*(*H*) is (2,*q*)-clique-Helly.

Conversely, suppose that *Q *is not a clique-transversal of *G. *Consider the following three families of cliques of *H:*

It is clear that every clique in *C _{i}, i = *1,2,3, is a vertex of

*K*(

*H*)

*.*Let , , be the subsets of vertices of

*K*(

*H*) corresponding to the cliques in the families

*C*

_{1},

*C*

_{2}, and

*C*

_{3}, respectively. Since all these cliques contain

*Q*, all the vertices of Ç Ç are pairwise adjacent. Therefore , and are completes of

*K*(

*H*)

*.*Moreover, it implies the existence of a clique containing these 3 completes. Since each one of these completes contains

*q*vertices, the graph F

*(*

_{q}*K*(

*H*)) contains three vertices

*v*corresponding to them. Since there is a clique in

_{1},v_{2},v_{3}*K*(

*H*) containing , ,

*,*it follows that {

*v*} is a triangle

_{1},v_{2},v_{3}*T*of F

*(*

_{q}*K*(

*H*))

*.*Let

*T'*be the extended triangle relative to

*T.*Consider the following additional families of cliques of

*H:*

where , , *, *are the three copies of a clique *Q ^{' }*of

*G*that does not intersects

*Q.*Clearly,

*Q'*exists because

*Q*is not a clique-transversal of

*G.*Clearly, every clique in

*C*,

_{i}*i =*4, 5, 6, is a vertex of

*K*(

*H*)

*.*Let , , be the subsets of vertices of

*K*(

*H*) corresponding to the cliques in the families

*C*

_{4},

*C*

_{5}, and

*C*

_{6}

*,*respectively. Since any clique in

*C*

_{4}contains

*,*it follows that is a complete of

*K*(

*H*)

*.*Analogously, and are completes of

*K*(

*H*)

*.*Consider the cliques È {

*a*

_{1}}, È {

*b*

_{1}}

*,*and È {

*, c*

_{1}} that belong to the families

*C*

_{4},

*C*5, and

*C*

_{6}, respectively. Since they are disjoint, their corresponding vertices in

*K*(

*H*) are not adjacent. Consequently, there exists no clique in

*H*containing a pair of completes from , ,

*.*Therefore the vertices

*v*F

_{1},v_{2},v_{3}of*(*

_{q}*K*(

*H*))

*,*corresponding to

*,*

*,*and

*,*respectively, form an independent set.

The following argument shows that *v _{4},v_{5},v_{6}* belong to

*T'.*All the cliques of the families

*C*

_{1}and

*C*

_{4}contain the vertex

*.*Hence the corresponding vertices in

*K*(

*H*) are pairwise adjacent. Therefore there is a clique in

*K*(

*H*) containing the completes and

*.*Consequently

*v*is an edge of F

_{1}v_{4}*(*

_{q}*K*(

*H*))

*.*All the cliques of the families

*C*

_{3}and

*C*

_{4}contain the vertex meaning that

*v*is also an edge of F

_{3}v_{4}*(*

_{q}*K*(

*H*))

*.*Select a clique of the family

*C*

_{4}and another of the family

*C*

_{2}. Since they do not intersect, the corresponding vertices in

*K*(

*H*) are not adjacent. Hence there is no clique in

*H*containing the completes and

*.*Consequently

*v*is not an edge of F

_{2}v_{4}*(*

_{q}*K*(

*H*))

*.*By the same argument, we conclude that

*v*is adjacent to the vertices

_{5}*v*and

_{1}*v*but not to

_{2},*v*; and

_{3}*v*is adjacent to

_{6 }*v*and

_{2}*v*, but not to

_{3}*v*Therefore, no vertex

_{1}.*v*, for 1

_{i}*6, is universal in*

__<__i__<__*T*

^{1}. Denote *A*_{4} = È {}È {*a _{i}* : 1

*} and*

__<__i__<__q*A*È {} È {

_{5}=*b*: 1

_{i}*}*

__<__i__<__q*.*Note that all the cliques of 64 and 65 are subsets of A

_{4}and

*A*, respectively. Note also that

_{5}*A*

_{4}È

*A*Æ, and the only edge joining vertices of

_{5}=*A*

_{4}and

*A*is

_{5}*.*Therefore, any clique sharing vertices simultaneoulsy with the cliques of

*C*

_{4}and

*C*

_{5}must contain this edge. The vertices that are adjacent simultaneously to and form the set { : 1

*} È*

__<__i__<__q*Q.*Consequently, each clique that contains and contains solely vertices of this latter set. Since

*Q*is a complete and { :1

*} is an independent set, there are*

__<__i__<__q*q*cliques of

*H*that contain the edge

*,*and they correspond exactly to the cliques of

*C*

_{1}

*.*Therefore the only cliques of

*H*that share vertices with all the cliques of

*C*

_{4}and

*C*

_{5}are the cliques of

*C*

_{1}

*.*Since their cardinalities are all equal to

*q,*is the only

*q*-complete of

*K*(

*H*) which is contained in a clique that contains or

*.*Hence

*v*is the only one vertex simultaneously adjacent to

_{1}*v*and

_{4}*v*in F

_{5}*(*

_{q}*K*(

*H*))

*.*Consequently,

*T*does not have a universal vertex. By Theorem 5, it follows that

^{'}*K*(

*H*) is not (2,

*q*)-clique-Helly.

**Corollary 7** *It is NP-hard to verify whether the Helly defect of a graph is at most one.*

**3 Conclusions**

We have proved that the problem of recognizing whether a given graph has Helly defect at most one is NP-hard.

A related problem is to recognize whether the Helly defect of a given graph is finite. It is not known whether this problem is decidable.

**References**

[1] H.-J. Bandelt and E. Prisner. Clique graphs and Helly graphs. *Journal of Combinatorial Theory B, *51:34-45, 1991. [ Links ]

[2] C. Berge. *Hypergraphes. *Gauthier-Villars, Paris, 1987. [ Links ]

[3] C. Berge and P. Duchet. A generalization of Gilmore's theorem. In M. Fiedler, editor, *Recent Advances in Graph Theory, *pages 49-55. Acad. Praha, Prague, 1975. [ Links ]

[4] C. F. Bornstein and J. L. Szwarcfiter. On clique convergent graphs. *Graphs and Combinatorics, *11:213-220, 1995. [ Links ]

[5] M. C. Dourado, F. Protti, and J. L. Szwarcfiter. A generalization of the Helly property applied to the cliques of a graph. Technical Report 01-02, NCE/UFRJ, Rio de Janeiro, RJ, Brazil, april 2002 (Submitted). [ Links ]

[6] F. F. Dragan. *Centers of Graphs and the Helly Property. *PhD thesis, Moldava State University, Chisinau, Moldava, 1989. In Russian. [ Links ]

[7] G. Durán, M. C. Lin, and J. L. Szwarcfiter. On clique transversals and clique independent sets. *Annals of Operations Research, *2002. To appear. [ Links ]

[8] P. Erdös, T. Gallai, and Zs. Tuza. Covering the cliques of a graph with vertices. *Discrete Mathematics, *108:279-289, 1992. [ Links ]

[9] F. Escalante. Über iterierte Clique-Graphen. *Abhandlungender Mathematischen Seminar der Universität Hamburg, *39:59-68, 1973. [ Links ]

[10] E. Prisner. *Graph Dynamics. *Pitman Research Notes in Mathematics 338, Longman, 1995. [ Links ]

[11] J. L. Szwarcfiter. Recognizing clique-Helly graphs. *Ars Combinatoria, *45:29-32, 1997. [ Links ]

*Universidade Federal do Rio de Janeiro, COPPE, Caixa Postal 68511, 21945-970, Rio de Janeiro, RJ, Brasil. Partially supported by the Conselho Nacional de Desenvolvimento Científico e Tecnológico - CNPq, Brasil. E-mail: rnitre@cos.ufrj.br

Universidade Federal do Rio de Janeiro, NCE, Caixa Postal 2324, 20001-970, Rio de Janeiro, RJ, Brasil. Partially supported by the Conselho Nacional de Desenvolvimento Científico e Tecnológico - CNPq, Brasil. E-mail: fabiop@nce.ufrj.br

Universidade Federal do Rio de Janeiro, Instituto de Matemática, NCE and COPPE, Caixa Postal 2324, 20001-970, Rio de Janeiro, RJ, Brasil. Partially supported by the Conselho Nacional de Desenvolvimento Científico e Tecnológico -CNPq, and Fundação de Amparo à Pesquisa do Estado do Rio de Janeiro - FAPERJ, Brasil. E-mail: jayme@nce.ufrj.br