On the helly defect of a graph

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 F_i of some set. A core of is any subset of F_i. 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(K^i–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 K^j(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 = K^2p(G). By Es-calante's result, K^2p(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 p— 1 vertices of C, and hence v Î V(H). Consequently, V(G[]_c) = V (H). Furthermore, both H and G[]_care subgraphs of G, but H is induced. Thus E(G[]_c) Í 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_qis that there exists a bi-jection between the subfamily of 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_q(G). If 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[]_ccontains a universal vertex x, then every clique of contains the q-complete of G that corresponds to x, that is, G[]_ccontains 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[]_ccontains 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 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[']_ccontains 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[]_ccontains a universal vertex. Moreover, by Lemma 2, H[]_cis a spanning subgraph of 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 = {C₁, . . . , 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_ihas a q⁺-core Q_i, for i = 1,...,k. It is clear that Q_i

C_i. Moreover, every two distinct Q_i, Q_j are contained in a same clique, since k > 3. Hence the sets Q₁, Q₁,..., 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_q() = {F_q(C₁),...,F_q(C_k)}. Since every C_iÎ contains at least p sets from Q₁, Q₁,..., Q_p+1, it is clear that the clique F_q(C_i) of if contains at least p vertices of C. Therefore, F_q(C_i) Î , for i = 1,..., k.

Let T be the (p + 1)-expansion of H relative to C. By Lemma 2, H[]_cis a spanning subgraph of T. Therefore, V(Q) Í V(T), for every Q Î . In particular, V(F_q(C_i)) Í 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_i) contains x, otherwise F_q(C_i) would not be maximal. Thus, ' has a 1⁺-core and if []_c contains a universal vertex. By Lemma 3, G[]_ccontains 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 6Let 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 V₁ = , and add the following edges:

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

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 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 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 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 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 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 F_q(K(H)) contains three vertices v₁,v₂,v₃corresponding to them. Since there is a clique in K(H) containing , , , it follows that {v₁,v₂,v₃} is a triangle 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 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 F_q(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 F_q(K(H)). All the cliques of the families C₃ and C₄ contain the vertex meaning that v₃ v₄ is also an edge of F_q(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 F_q(K(H)). By the same argument, we conclude that v₅ is adjacent to the vertices v₁and v₂, but not to v₃; and v₆ is adjacent to v₂ and v₃, but not to v₁. Therefore, no vertex v_i, for 1 < i < 6, is universal in T¹.

Denote A₄ = È {}È {a_i : 1 < i < q} and A₅ = È {} È {b_i : 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 A₅is

. 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 F_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.

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Publication Dates

• Publication in this collection
  16 Dec 2003
• Date of issue
  2001