1 Introduction
As usual, the underlying topology of an interconnection network is modeled by a connected graph , where is the set of processors and is the set of communication links between processors. A subgraph obtained from by removing a set of vertices is denoted by . A separating set (or vertexcut) of a connected graph is a set of vertices whose removal renders to become disconnected. If is not a complete graph, the connectivity is the cardinality of a minimum separating set of . By convention, the connectivity of a complete graph with vertices is defined to be . A graph is connected if .
The connectivity is an important topic in graph theory. In particular, it plays a key role in applications related to the modern interconnection networks, e.g., can be used to assess the vulnerability of the corresponding network, and is an important measurement for reliability and fault tolerance of the network [28]. However, to further analyze the detailed situation of the disconnected network caused by a separating set, it is natural to generalize the classical connectivity by introducing some conditions or restrictions on the separating set and/or the components of [14]. The most basic consideration is the number of components associated with the disconnected network. To figure out what kind of separating sets and/or how many sizes of a separating set can result in a disconnected network with a certain number of components, Chartrand et al. [5] proposed a generalization of connectivity with respect to separating set for making a more thorough study. In this paper, we follow this direction to investigate such kind of generalized connectivity on a class of interconnection networks called alternating group networks (defined later in Section 2).
For an integer , the generalized connectivity of a graph , denoted by , is the minimum number of vertices whose removal from results in a disconnected graph with at least components or a graph with fewer than vertices. A graph is connected if . A synonym for such a generalization was also called the general connectivity by Sampathkumar [26] or component connectivity (connectivity for short) by Hsu et al. [18], Cheng et al. [7, 8, 9] and Zhao et al. [29]. Hereafter, we follow the use of the terminology of Hsu et al. Obviously, . Similarly, for an integer , the generalized edgeconnectivity (edgeconnectivity for short) , which was introduced by Boesch and Chen [3], is defined to be the smallest number of edges whose removal leaves a graph with at least components if , and if . In addition, many problems related to networks on faulty edges haven been considered in [15, 16, 17, 25].
The notion of connectivity is concerned with the relevance of the cardinality of a minimum vertexcut and the number of components caused by the vertexcut, which is a good measure of robustness of interconnection networks. Accordingly, this generalization is called the cutversion definition of generalized connectivity. We note that there are other diverse generalizations of connectivity in the literature, e.g., Hager [12] gave the socalled pathversion definition of generalized connectivity, which is defined from the view point of Menger’s Theorem. Recently, Sun and Li [27] gave sharp bounds of the difference between the two versions of generalized connectivities.
For research results on connectivity of graphs, the reader can refer to [5, 10, 11, 7, 8, 9, 18, 23, 24, 26, 29]. At the early stage, the main work focused on establishing sufficient conditions for graphs to be connected, (e.g., see [5, 26, 23]). Also, several sharp bounds of connectivity related to other graph parameters can be found in [26, 11]. In addition, for a graph and an integer , a function called connectivity function is defined to be the minimum edgeconnectivity among all subgraphs of obtained by removing vertices from , and several properties of this function was investigated in [10, 24]. By contrast, finding connectivity for certain interconnection networks is a new trend of research at present. So far, the exact values of connectivity are known only for a few classes of networks, in particular, only for small ’s. For example, is determined on the dimensional hypercube for (see [18]) and (see [29]), the dimensional hierarchical cubic network (see [7]), the dimensional complete cubic network (see [8]), and the generalized exchanged hypercube for and (see [9]). However, determining connectivity is still unsolved for most interconnection networks. As a matter of fact, it has been pointed out in [18] that, unlike the hypercube, the results of the wellknown interconnection networks such as the star graphs [1] and the alternating group graphs [20] are still unknown.
Recently, we studied two types of generalized 3connectivities (i.e., the cutversion and the pathversion of the generalized connectivities as mentioned before) in the dimensional alternating group network , which was introduced by Ji [19] to serve as an interconnection network topology for computing systems. In [4], we already determined the 3component connectivity for . In this sequel, we continue the work and show the following result.
Theorem 1.
For , .
2 Background of alternating group networks
Let and denote the set of all even permutations over . For , the dimensional alternating group network, denoted by , is a graph with the vertex set of even permutations (i.e., ), and two vertices and are adjacent if and only if one of the following three conditions holds [19]:

(i) , , , and for .

(ii) , , , and for .

(iii) There exists an such that , , , , and for .
The basic properties of are known as follows. contains vertices and edges, which is a vertexsymmetric and regular graph with diameter and connectivity . For and , let be the subnetwork of induced by vertices with the rightmost symbol in its permutation. It is clear that is isomorphic to . In fact, has a recursive structure, which can be constructed from disjoint copies for such that, for any two subnetworks and , and , there exist edges between them. Fig. 1 depicts , where each part of shadows indicates a subnetwork isomorphic to .
A path (resp., cycle) of length is called a path (resp., cycle). For notational convenience, if a vertex belongs to a subnetwork , we simply write instead of . The disjoint union of two subnetworks and is denoted by . The subgraph obtained from by removing a set of vertices is denoted by . An edge with two end vertices and for is called an external edges between and . In this case, and are called outneighbors to each other. By contrast, edges joining vertices in the same subnetwork are called internal edges, and the two adjacent vertices are called inneighbors to each other. By definition, it is easy to check that every vertex of has inneighbors and exactly one outneighbor. Hereafter, for a vertex , we use to denote the set of inneighbors of , and the unique outneighbor of . Moreover, if is a subgraph of , we define as the inneighborhood of , i.e., the set composed of all inneighbors of those vertices in except for those belong to .
In what follow, we shall present some properties of , which will be used later. For more properties on alternating group networks, we refer to [6, 13, 19, 30, 31].
Lemma 1.
Lemma 2.
For and , let be a connected induced subgraph of . Then, the following properties hold:

(1) If , then is a cycle or a path. Moreover, if is a cycle (resp., a path), then (resp., ).

(2) If , then .
Proof. The two properties can easily be proved by induction on . Now, we only verify the subgraph in Fig. 1 for the basis case . Recall that every vertex has inneighbors in . For (1), the result of 3cycle is clear. If is a 2path, at most two adjacent vertices in can share a common inneighbor, it follows the . For (2), the condition means that the number of vertices in cannot exceed a half of those in . In particular, if , then is either a claw (i.e., ), a paw (i.e., plus an edge), or a path. Moreover, if is a paw, a claw or a 3path, then no two adjacent vertices, at most one pair of adjacent vertices, or at most two pair of adjacent vertices in can share a common inneighbor, respectively. This shows that when is a paw, when is a claw, and when is a 3path. Also, if , it is clear that .
For designing a reliable probabilistic network, Bauer et al. [2] first introduced the notion of super connectedness. A regular graph is (loosely) superconnected if its only minimum vertexcuts are those induced by the neighbors of a vertex, i.e., a minimum vertexcut is the set of neighbors of a single vertex. If, in addition, the deletion of a minimum vertexcut results in a graph with two components and one of which is a singleton, then the graph is tightly superconnected. More accurately, a graph is tightly superconnected provided it is tightly superconnected and the cardinality of a minimum vertexcut is equal to . Zhou and Xiao [30] pointed out that and are not superconnected, and showed that for is tightly superconnected. Moreover, to evaluate the size of the connected components of with a set of faulty vertices, Zhou and Xiao gave the following properties.
Lemma 3.
(see [30]) For , if is a vertexcut of with , then one of the following conditions holds:

(1) has two components, one of which is a trivial component (i.e., a singleton).

(2) has two components, one of which is an edge, say . In particular, if , is composed of all neighbors of and , excluding and .
Lemma 4.
(see [30]) For , if is a vertexcut of with , then one of the following conditions holds:

(1) has two components, one of which is either a singleton or an edge.

(2) has three components, two of which are singletons.
Through a more detailed analysis, Chang et al. [4] recently obtained a slight extension of the result of Lemma 3 as follows.
Lemma 5.
(see [4]) Let is a vertexcut of with . Then, the following conditions hold:

(1) If , then has two components, one of which is a singleton, an edge, a cycle, a path, or a paw.

(2) If , then has two components, one of which is a singleton, an edge, or a cycle.

(3) If , then has two components, one of which is either a singleton or an edge.
3 The 4component connectivity of
Since is a 3cycle, by definition, it is clear that . Also, in the process of the drawing of Fig. 1, we found by a bruteforce checking that the removal of no more than five vertices in (resp., eight vertices in ) results in a graph that is either connected or contains at most three components. Thus, the following lemma establishes the lower bound of for .
Lemma 6.
and .
Lemma 7.
For , .
Proof. Let be any vertexcut in such that . For convenience, vertices in (resp., not in ) are called faulty vertices (resp., faultfree vertices). By Lemma 4, if , then contains at most three components. To complete the proof, we need to show that the same result holds for . Let and for each . We claim that there exists some subnetwork, say , such that it contains faulty vertices. Since , if it is so, then there are at most two such subnetworks. Suppose not, i.e., every subnetwork for has faulty vertices. Since is connected, remains connected for each . Recall the property (3) of Lemma 1 that there are independent edges between and for each pair with . Since for , it guarantees that the two subgraphs and are connected by an external edge in . Thus, is connected, and this contradicts to the fact that is a vertexcut in . Moreover, for such subnetworks, it is sure that some of must be a vertexcut of . Otherwise, is connected, a contradiction. We now consider the following two cases:
Case 1: There is exactly one such subnetwork, say , such that it contains faulty vertices. In this case, we have for all and is a vertexcut of . Let be the subgraph of induced by the faultfree vertices outside , i.e., . Since every subnetwork in has faulty vertices, from the previous argument it is sure that is connected. We denote by the component of that contains as its subgraph, and let be the number of faulty vertices outside . Since , we have . Consider the following scenarios:
Case 1.1: . In this case, there are no faulty vertices outside . That is, . Indeed, this case is impossible because if it is the case, then every vertex of has the faultfree outneighbor in . Thus, belongs to , and it follows that is connected, a contradiction.
Case 1.2: . Let be the unique faulty vertex outside . That is, . Since is a vertexcut of , we assume that is divided into disjoint connected components, say . For each , if , then there is at least one vertex of with its outneighbor in , and thus belongs to . We now consider a component that is a singleton, say . If , then must be contained in , and thus belongs to . Clearly, there exists at most one component such that . In this case, has exactly two components and .
Case 1.3: . Let be the two faulty vertices outside . That is, . Since is a vertexcut of , we assume that is divided into disjoint connected components, say . For each , if , then there is at least one vertex of with its outneighbor in , and thus belongs to . We now consider a component with , i.e., is an edge, say . By the property (2) of Lemma 1, we have . If , then at least one of and must be contained in , and thus belongs to . Since and has inneighbors (not including and ) in , we have for . Thus, there exists at most one such component such that . If it is the case of existence, then has exactly two components and . Finally, we consider a component that is a singleton. Since and every vertex has degree in , we have for . Thus, at most two such components exist in , say and where . If , then both and must be contained in , and thus and belong to . Also, if either or , then has exactly two components, one of which is a singleton or . Finally, if , then has exactly three components, two of which are singletons and .
Case 1.4: . Let be the three faulty vertices outside . That is, . Since is a vertexcut of , we assume that is divided into disjoint connected components, say . For each , if , then there is at least one vertex of with its outneighbor in , and thus belongs to . We now consider a component with , i.e., is either a 3cycle or a 2path. Assume that . If there is a vertex for , then must be contained in , and thus belongs to . Since and, by Lemma 2, we have , it follows that there exists at most one such component such that . If it is the case of existence, then has exactly two components, one of which is either a 3cycle or a 2path. Next, we consider a component with , i.e., is an edge, say . From an argument similar to Case 1.3 for analyzing the membership of and in the set , we can show that has exactly two components and . Finally, we consider a component that is a singleton. Then, an argument similar to Case 1.3 for analyzing singleton components shows that at most two such components exist in . Thus, has either two components (where one of which is a singleton) or three components (where two of which are singletons).
Case 1.5: . Let be the four faulty vertices outside . That is, . Since is a vertexcut of , we assume that is divided into disjoint connected components, say . For each , if , then there is at least one vertex of with its outneighbor in , and thus belongs to . If , by Lemma 2, we have . Since , it follows that for . Thus, none of component with exists in . Next, we consider a component with and assume . By Lemma 2, we have . Since is no more than , at most one such component exists in . Furthermore, if such exists, then it is either a 3cycle or a 2path. Thus, an argument similar to Case 1.4 for analyzing the membership of , and in the set , we can show that has exactly two components, one of which is a 3cycle or a 2path. Finally, if we consider a component with , an argument similar to the previous cases shows that has either two components (where one of which is a singleton or an edge) or three components (where two of which are singletons).
Case 1.6: . Let be the five faulty vertices outside . That is, . Since is a vertexcut of , we assume that is divided into disjoint connected components, say . For each , if , then there is at least one vertex of with its outneighbor in , and thus belongs to . If or , by Lemma 2, we have . Since , it follows that for . Thus, none of component with or exists in . We now consider a component with . Since , by Lemma 2, if such exists, then it must be a 3cycle, and thus an argument similar to the previous cases shows that has exactly two components, one of which is a 3cycle. Finally, if we consider a component with , an argument similar to the previous cases shows that has either two components (where one of which is a singleton or an edge) or three components (where two of which are singletons).
Case 1.7: . In this case, we have . Since is isomorphic to and is a vertexcut of with no more than vertices, by Lemma 4, has at most three components as follows:
Case 1.7.1: has two components, one of which is either a singleton or an edge. Let and be such two components for which . More precisely, for . Clearly, the above inequality indicates that there exist some vertices of such that their outneighbors are contained in , even if all outneighbors of vertices in are contained in . Thus, belongs to . Also, if there is a vertex with its outneighbor in , then belongs to . Otherwise, has exactly two components, one of which is either a singleton or an edge.
Case 1.7.2: has three components, two of which are singletons. Let and be such three components for which and . Since for , there exist some vertices of such that their outneighbors are contained in . This shows that belongs to . Since has three components, the outneighbor of a vertex or cannot be contained in . Thus, has exactly three components, two of which are singletons.
Case 2: There exist exactly two subnetworks, say and , such that . Since is a vertexcut of , at least one of the subgraphs and must be disconnected. Let be the subgraph of induced by the faultfree vertices outside , i.e., . Since , we have for all . The bound of implies that is connected, and it follows that is also connected. We denote by the component of that contains as its subgraph. Since , we consider the following scenarios:
Case 2.1: . Clearly, . Since we have assumed , it follows that and there exist no faulty vertices outside . That is, . Indeed, this case is impossible because if it is the case, then there exist a vertex of such that its outneighbor is contained in . Thus, belongs to , and it follows that is connected, a contradiction.
Case 2.2: . Since , it implies . Since is isomorphic to and , by Lemma 5, if is disconnected, then it has exactly two component, one of which is either a singleton or an edge. Suppose , where and are disjoint connected components such that . More precisely, for , where the last term is the number of faulty vertices outside . Clearly, the above inequality indicates that there exist some vertices of such that their outneighbors are contained in , even if all outneighbors of vertices in are contained in . Thus, belongs to . Also, if there is a vertex of with its outneighbor in , then belongs to . By contrast, we can show that belongs to by a similar way if it is connected. Thus, contains at most one component (which is either a singleton or an edge) such that this component is a subgraph of . Similarly, since , contains at most one component (which is either a singleton or an edge) such that this component is a subgraph of . Thus, there are at most three components in . We claim that cannot simultaneously contain both an edge and a singleton as components. Suppose not and, without loss of generality, assume and . Then, at least two outneighbors of and are not contained in . Otherwise, produces a 4cycle or 5cycle, which contradicts to the property (1) of Lemma 1. Thus, the number of faulty vertices of requires at least , a contradiction. Similarly, we claim that cannot simultaneously contain two disjoint edges and as components. Suppose not. By an argument similar above, we can show that either has faulty vertices for or it contains a 4cycle or 5cycle. However, both the cases are not impossible. Consequently, if contains three component, then two of which are singletons, one is a vertex of and the other is of .
Case 2.3: . Clearly, . Since is isomorphic to and , it is tightly superconnected. Also, since , if is a vertexcut of , then it must be a minimum vertexcut. Particularly, there are two components in , one of which is a singleton, say . That is, all inneighbors of are faulty vertices (i.e., ). Otherwise, is connected and thus belongs to . On the other hand, we consider all situations of as follows. Clearly, if is connected, then it belongs to , and this further implies that must be disconnected. In this case, contains exactly two components, one of which is a singleton . We now consider the case that is not connected and claim that it has at most two disjoint connected components. Suppose not. Since is isomorphic to , by Lemma 5, the number of faulty vertices in is at least . Since , it follows that . Thus, this situation is a symmetry of Case 2.1 by considering the exchange of and , which leads to a contradiction. Suppose , where and are disjoint connected components such that . Since for , where the last term is the number of faulty vertices outside . Clearly, the above inequality indicates that there exist some vertices of such that their outneighbors are contained in , even if all outneighbors of vertices in are contained in . Thus, belongs to . Also, if there is a vertex of with its outneighbor in , then belongs to . Otherwise, is a component of . By Lemma 2, since when , we have . Moreover, since when , if , then must be a 3cycle. If , then is either a singleton or an edge. Note that if is a 3cycle or an edge, then cannot contain the the singleton as its component. Otherwise, an argument similar to Case 2.2 shows that either has more than faulty vertices or produces a 4cycle or 5cycle, a contradiction.
Corollary 8.
For , if is a vertexcut of with , then one of the following conditions holds:

(1) has two components, one of which is either a singleton, an edge, a cycle, or a path.

(2)
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