## 1. Introduction

The graph we considered throughout this paper is simple and undirected. Let be a graph, where is the vertex-set of and is the edge-set of . The degree of a vertex is the number of incident edges, written or when the context is clear. The minimum degree of is and the maximum degree is .

For any subset , the closed neighborhood of is defined to be all neighbors of any vertex together with , denoted by , while the open neighborhood of is , denoted by . If , then we write and , respectively. The subgraph induced by is denoted by . A matching of is a set of independent edges of . For other standard graph notations not defined here please refer to [1].

Lin et al. [9] introduced structure and substructure connectivity to evaluate the fault tolerance of a network from the perspective of a single vertex, as well as some special structures of the network. Let be a set of pairwise disjoint connected subgraphs of and let . Then is a subgraph-cut of provided that is disconnected or trivial. Let be a connected subgraph of , then is an -structure-cut if is a subgraph-cut, and each element in is isomorphic to . The -structure connectivity of , written , is the minimum cardinality over all -structure-cuts of . Similarly, if is a subgraph-cut and each element of is isomorphic to a connected subgraph of , then is called an -substructure-cut. The -substructure connectivity of , written , is the minimum cardinality over all -substructure-cuts of .

Structure and substructure connectivity of some famous interconnection networks have been determined, such as hypercube [9], -ary -cube [11], folded hypercube [12], balanced hypercube [10], arrangement graph [7], alternating group graphs [8]. A natural question arise: what is computational complexity of structure and substructure connectivity in general graphs? In this paper, we study this problem.

## 2. NP-complete of structure connectivity

3-dimensional matching, 3DM for short, is one of the most standard NP-complete problems to prove NP-complete results. An instance of 3DM consists of three disjoint sets , and with equal cardinality , and a set of triples . For convenience, let . The question is to decide whether there is a subset covering , that is, and each element of occurs in exactly one triple of . This instance can be associated with a bipartite graph as follows. Each element of and each triple of is represented by a vertex of . There is an edge between an element and a triple if and only if the element is a member of the triple.

It has been proved in [3, 4] that 3DM is NP-complete when each element of appears in only two or three triples of , i.e., each vertex in the partite set of has degree two or three only. We shall show that the decision problem of structure connectivity is NP-complete by reducing from 3DM stated previously.

To this end, we state the following decision problem.

Problem: The -structure connectivity of an arbitrary graph.

Instance: Given a nonempty graph , a subgraph of and a positive integer .

Question: Is ?

Now we are ready to prove the following theorem.

###### Theorem 1

. The -structure connectivity is NP-complete when for any integer .

###### Proof.

Obviously, the structure connectivity problem is in NP, because we can check in polynomial time whether a set of disjoint s is a structure cut or not. It remains to show that the structure connectivity is NP-hard when for any integer . We prove this argument by reducing 3DM to it.

Let be an instance of 3DM defined previously. For convenience, let and . We make a further assumption that each vertex in the partite set of has degree two or three only.

Now we construct a graph from as follows (see Fig. 1).

Set

The vertex set of is . The subgraph of induced by is for each . Similarly, The subgraph induced by is vertex disjoint and the subgraph induced by is .

So the edge set of is

where , and .

We show that has a 3DM covering if and only if . First suppose that has a subset covering , that is, and each element of occurs in exactly one triple of . We show that . Clearly, for each vertex and hence the subgraph of induced by is isomorphic to . Thus, is a structure cut of with .

Next suppose that is a structure cut of with . We shall show that has a subset of covering . Recall that each vertex in the partite set of has degree two or three only, we may assume that the number of vertices with degree two (resp. three) in of is and (resp. ), and consequently, , which implies that .

Since each element of is a graph isomorphic to , we focus on the center vertex of . Let be the the set of center vertices of all , and let and . Since each vertex in has degree less than , any vertex in can not be center vertices of that is, . We claim that covers . Suppose on the contrary that does not cover , we shall show that is connected. Thus, two cases arise.

Case 1. . So . If there exists an edge such that and , then it is not hard to see that is connected, contradicting that is a structure cut of . Hence, all vertices in are covered by . Clearly, components in restricted on form a 3-dimensional matching of .

Case 2. . Note that for any vertex () and or 3 for any vertex (). By the structure of , each vertex in (as the center vertex of a ) can subvert at most one vertex in . Similarly, each vertex in can subvert precisely one vertex in together with three vertices in .

Since , we have . This implies that there exists an edge such that and . Obviously, after subverting vertices in , each clique either joins to or disappears, and similarly clique is decidedly joined to via one clique that joins. So is connected, a contradiction again.

This complete the proof. ∎

## 3. NP-complete of substructure connectivity

A vertex cover of is a subset such that for each edge , at least one of and belongs to . The decision version of the vertex cover problem is one of Karp’s 21 NP-complete problems [6] and is therefore a classical NP-complete problem.

We present the decision problem of the substructure connectivity as follows.

Problem: The substructure connectivity of an arbitrary graph.

Instance: Given a nonempty graph with , a subgraph of and a positive integer .

Question: Is ?

The following lemma will be used later.

###### Theorem 2

. The -substructure connectivity is NP-complete when .

###### Proof.

Obviously, the substructure connectivity problem is in NP, because we can check in polynomial time whether a set of disjoint subgraphs of is a substructure cut. It remains to show that the substructure connectivity is NP-hard when . We prove this argument by reducing vertex cover to this problem.

Given a graph with , we construct a graph from as follows (see Fig. 2).

Set

The vertex set of is . The subgraph of induced by is a complete graph for each .

So the edge set of is

where .

We show that has a vertex cover of size at most if and only if .

First suppose that has a vertex cover with . We show that there exists a substructure cut of with . For any vertex , let . Then consists of a spanning subgraph of with center vertex . Let . Clearly, and . Thus, is disconnected with independent cliques , indicating that is a -substructure-cut of size at most . So .

Next suppose that is a -substructure-cut of with . We show that has a vertex cover of size at most .

Since each element of is a subgraph of , we focus on the center vertex of for (since each center vertex of covers all its neighbors). Let be the set of center vertices of , , and let . Thus, two cases arise.

Case 1. . Then there are vertices of that are adjacent to the vertices of not covered by . Since each vertex of is adjacent exactly one vertex in , the number of vertices of not covered by is not greater than the number of vertices in . Therefore, there exists a vertex cover of of size at most .

Case 2. . Then . Since there are disjoint cliques of size in , each () with center vertices in could subvert at most vertices in each . Thus, the remaining subgraph of consists of at least two cliques of size at least and the vertices in separate cliques are not adjacent. It implies that is connected and not a complete graph, which is a contradiction.

This completes the proof. ∎

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