## Print version ISSN 0104-6500On-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 Ki(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 êSê= k, a k+-set when êSê > k, and a k-set when êSê < k. This same notation will also apply for families of sets.

Let be a family of subsets Fi of some set. A core of is any subset of Fi. Let p,q be integers, p > 1 and q > 0. Say that is (p,q)-intersecting when every p-subfamily of it has a q+-core. In addition, is (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 Ki(G), is defined as follows: K0(G) = G, while Ki(G) = K(Ki–1(G)), i > 0.

The Helly defect of a graph G is the smallest i such that Ki(G) is clique-Helly. If Ki(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 Ki(G) = G, for some i. In addition, the smallest i satisfying Ki(G) = G is the period of a periodic graph G.

Escalante [9] proved that if Ki(G) is clique-Helly, then Kj(G) is clique-Helly for any j > i, and that if G is clique-Helly, then K2(G) is an induced subgraph of G.

Let G be a periodic clique-Helly graph, and let p > 1 be its period. Clearly, G = K2p(G). By Es-calante's result, K2p(G) is an induced subgraph of K2(G), therefore G = K2(G). This implies that p < 2. This argument shows that if H is a periodic graph with period strictly greater than 2, then Ki(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 1 vertices of C, and hence v Î V(H). Consequently, V(G[]c) = V (H). Furthermore, both H and G[]c are subgraphs of G, but H is induced. Thus E(G[]c) Í E(H).

Let q be a positive integer. The graph Fq(G) is defined in the following way: the vertices of Fq(G) correspond to the q-completes of G, two vertices being adjacent in Fg(G) if the corresponding q-completes in G are contained in a common clique. We remark that Fq is precisely the operator Fq,2q, studied in [10]. An interesting property of Fq is that there exists a bi-jection between the subfamily of q+-cliques of G and the family of cliques of Fq(G) [5]. The following definitions are possible due to this property: If C is a q+-clique of G, denote by Fq(C) the clique that corresponds to C in Fq(G). If C' is a clique of Fq(G), denote by (C')the q+-clique that corresponds to C' in G. If = is a subfamily of q+-cliques of G, define Fq() = {Fq(C) |C Î }. If is a subfamily of cliques of Fq(G), define = {(C) |C Î }.

Lemma 3 [5] Let G be a graph, a subfamily of q+ -cliques of it, = Fq (), and H = Fq(G). Then H[]c contains a universal vertex if and only if G[]c contains q universal vertices.

Proof. If H[]ccontains a universal vertex x, then every clique of contains the q-complete of G that corresponds to x, that is, G[]c contains 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[]c contains a universal vertex.

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 Fq(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 Fq(G), such that T contains no universal vertex.

Let be the subfamily of cliques of H = Fq(G) that contain at least p vertices of C. Let = (). Consider a p-subfamily ' Í . Let = Fq ('). By Lemma 4, ' has a 1+-core. That is, H[']c contains a universal vertex. This implies, by Lemma 3, that 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[]c contains a universal vertex. Moreover, by Lemma 2, H[]c is a spanning subgraph of T. However, T contains no universal vertex. This is a contradiction. Therefore, every (p + 1)-expansion of H = Fq(G) contains a universal vertex.

Conversely, assume by contradiction that G is not (p, q)-clique-Helly. Let = {C1, . . . , Ck} 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 \Ci has a q+-core Qi, for i = 1,...,k. It is clear that Qi Ci. Moreover, every two distinct Qi, Qj are contained in a same clique, since k > 3. Hence the sets Q1, Q1,..., Qp+1 correspond to a (p + 1)-complete C in Fq(G).

Let be the subfamily of cliques of H = Fq(G) that contain at least p vertices of C. Let ' = Fq() = {Fq(C1),...,Fq(Ck)}. Since every Ci Î contains at least p sets from Q1, Q1,..., Qp+1, it is clear that the clique Fq(Ci) of if contains at least p vertices of C. Therefore, Fq(Ci) Î , for 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(Fq(Ci)) Í V(T), for i = 1,..., k. By hypothesis, T contains a universal vertex x. Then x is adjacent to all the vertices of Fq(Ci)\{x}, for i = 1,..., k. This implies that Fq(Ci) contains x, otherwise Fq(Ci) would not be maximal. Thus, ' has a 1+-core and if []c contains a universal vertex. By Lemma 3, G[]c contains 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 Ga,Gb,Gc. Add six vertices forming the set V1 = , and add the following edges:

Add now q vertices for each copy of G, forming the set V2 = {a1,a2,...,aq, b1,b2,...,bq, c1,c2, ...,cq}, and add the edges:

For each vertex w Î Q, consider its three copies in Ga,Gb,Gc, 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 V3 =

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 Ga,Gb,Gc, 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 V2È V3.

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 Ci, 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 C1, C2, and C3, 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 Fq (K(H)) contains three vertices v1,v2,v3 corresponding to them. Since there is a clique in K(H) containing , , , it follows that {v1,v2,v3} is a triangle T of Fq (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 Ci, 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 C4,C5, and C6, respectively. Since any clique in C4 contains , it follows that is a complete of K(H). Analogously, and are completes of K(H). Consider the cliques È { a1}, È { b1}, and È {, c1} that belong to the families C4, C5, and C6, 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 v1,v2,v3 of Fq (K(H)), corresponding to , , and , respectively, form an independent set.

The following argument shows that v4,v5,v6 belong to T'. All the cliques of the families C1and C4 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 v1 v4 is an edge of Fq (K(H)). All the cliques of the families C3 and C4 contain the vertex meaning that v3 v4 is also an edge of Fq (K(H)). Select a clique of the family C4 and another of the family C2. 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 v2 v4 is not an edge of Fq (K(H)). By the same argument, we conclude that v5 is adjacent to the vertices v1 and v2, but not to v3; and v6 is adjacent to v2 and v3, but not to v1. Therefore, no vertex vi, for 1 < i < 6, is universal in T1.

Denote A4 = È {}È {ai : 1 < i < q} and A5 = È {} È {bi : 1 < i < q }. Note that all the cliques of 64 and 65 are subsets of A4 and A5, respectively. Note also that A4È A5 = Æ, and the only edge joining vertices of A4and A5 is . Therefore, any clique sharing vertices simultaneoulsy with the cliques of C4 and C5 must contain this edge. The vertices that are adjacent simultaneously to and form the set { : 1 < i < q} È Q. Consequently, each clique that contains and contains solely vertices of this latter set. Since Q is a complete and { :1 < i < q} is an independent set, there are q cliques of H that contain the edge , and they correspond exactly to the cliques of C1. Therefore the only cliques of H that share vertices with all the cliques of C4 and C5 are the cliques of C1 . 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 v1 is the only one vertex simultaneously adjacent to v4 and v5 in Fq (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

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*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