Structure connectivity and substructure connectivity of twisted hypercubes

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1 arxiv: v1 [math.co] Mar 018 Structure connectivity and substructure connectivity of twisted hypercubes Dong Li, Xiaolan Hu, Huiqing Liu Abstract Let G be a graph and T a certain connected subgraph of G. The T -structure connectivity κ(g; T ) (or resp., T -substructure connectivity κ s (G; T )) of G is the minimum number of a set of subgraphs F = {T 1, T,..., T m } (or resp., F = {T 1, T,..., T m}) such that T i is isomorphic to T (or resp., T i is a connected subgraph of T ) for every 1 i m, and F s removal will disconnect G. The twisted hypercube H n is a new variant of hypercubes with asymptotically optimal diameter introduced by X.D. Zhu. In this paper, we will determine both κ(h n ; T ) and κ s (H n ; T ) for T {K 1,r, P k }, respectively, where 3 r 4 and 1 k n. Keywords: Twisted hypercube; T -structure connectivity; T -substructure connectivity 1 Introduction Interconnection networks play an important role in parallel and distributed systems. An interconnection network can be represented by an undirected graph G = (V, E), where each vertex in V corresponds to a processor, and every edge in E corresponds to a communication link. The neighborhood N G (v) of a vertex v in a graph G = (V, E) is the set of vertices adjacent to v. For S V (G), the neighborhood N G (S) of S in G is defined as N G (S) = ( v S N G (v)) S. We use P k = v 1, v,..., v k and C k = v 1, v,..., v k, v 1 to denote a path and a cycle of order k, respectively. For S V (G), we use G[S] to denote the subgraph of G induced by S. For a subgraph H of a graph G, we use G H to denote the subgraph of Hubei Key Laboratory of Applied Mathematics, Faculty of Mathematics and Statistic, Hubei University, Wuhan 43006, PR China School of Mathematics and Statistics & Hubei Key Laboratory of Mathematical Sciences, Central China Normal University, Wuhan , PR China 1

2 G induced by V (G) V (H). For a set F = {T 1, T,..., T m }, where each T i is isomorphic to a connected subgraph of G, we use G F to denote the subgraph of G induced by V (G) V (T 1 ) V (T ) V (T m ). For graph definition and notation not mentioned here we follow [1]. The connectivity κ(g) of a graph G is the minimum number of vertices whose removal leaves the remaining graph disconnected or trivial. The connectivity is one of the most important parameters to measure the reliability and fault tolerance of an interconnection network [3], [4], [6], [7], [16]. The larger the connectivity is, more reliable the interconnection network is. However, this parameter has a deficiency. That is, it tacitly assumes that all vertices adjacent to the same vertex of G could fail at the same time, which is highly unlikely for large-scale systems. To compensate for this shortcoming, Fábrega [5] proposed the concept of g-extra connectivity. The g-extra connectivity of a graph G, denoted by κ g (G), is the minimum number of vertices of G whose deletion disconnects G and every remaining component has more than g vertices. Some recent results on the g-extra connectivity of graphs see [], [8], [17], [18]. Instead of considering the effect of vertices becoming faulty, Lin et al. [9] considered the effect caused by some structures becoming faulty, and introduced the concept of structure connectivity and substructure connectivity of graphs. Let T be a connected subgraph of a graph G, and F a set of subgraphs of G such that every element in F is isomorphic to T. Then F is called a T -structure-cut if G F is disconnected. The T -structure connectivity κ(g; T ) of G is defined as the cardinality of a minimum T -structure-cut of G. Similarly, let F be a set of subgraphs of G such that every element in F is isomorphic to a connected subgraph of T. Then F is called a T -substructure-cut if G F is disconnected. The T -substructure connectivity κ s (G; T ) of G is defined as the cardinality of a minimum T -substructure-cut of G. By definition, κ s (G; T ) κ(g; T ). Note that K 1 -structure connectivity and K 1 -substructure connectivity are exactly the classical vertex connectivity. Lin et al. [9] determined κ(q n ; T ) and κ s (Q n ; T ) for the hypercube Q n and T {K 1,1, K 1,, K 1,3, C 4 }, respectively. Sabir and Meng [14] generalized their results and established κ(q n ; T ) and κ s (Q n ; T ) for T {P k, C k, K 1,4 }, where 3 k n, they also determined κ(f Q n ; T ) and κ s (F Q n ; T ) for the folded hypercube F Q n and T {P k, C k, K 1,3 }, where n 7 and k n. Moreover, Mane [1] determined κ(q n ; Q m ) with m n and established the upper bound of κ(q n ; C k ) with k n 1. Recently, Lv et al. [11] explored κ(q k n; T ) and κ s (Q k n; T ) for the k-ary n-cube Q k n and T {K 1, K 1,1, K 1,, K 1,3 }. The interconnection network considered in this study is the twisted hypercube H n, which is a hypercube-like network with asymptotically optimal diameter introduced by Zhu [19].

3 H n has many attractive properties, such as low vertex degree, strong connectivity and super connectivity. Qi and Zhu [13] considered the fault-diameter and wide-diameter of the twisted hypercubes. Liu et al. [10] determined the R g -vertex-connectivity and established the g-good neighbor conditional diagnosability of the twisted hypercubes under the PMC model and MM model, respectively. In this paper, we establish both κ(h n ; T ) and κ s (H n ; T ) for T {K 1,r, P k }, where 3 r 4 and 1 k n. The rest of the paper is organized as follows. In Section, we introduce the definition of the n-dimensional twisted hypercube H n and provide preliminaries for our results. In Section 3, we determine κ(h n ; K 1,r ) and κ s (H n ; K 1,r ) for 3 r 4. In Section 4, we determine κ(h n ; P k ) and κ s (H n ; P k ) for 1 k n. Our conclusions are given in Section 5. Preliminaries In this section, we first introduce the definition of the n-dimensional twisted hypercube H n, then present some properties of H n. Denote by Z n the set of binary strings of length n. For x, y Z n, x y denotes the sum of x and y in the group Z n, i.e., (x y) i = x i + y i (mod ) (for x Z n, x i denotes the ith bit of x). If x is a binary string of length n 1 and y is a binary string of length n, then xy is the concatenation of x and y, which is a binary string of length n 1 + n. If Z is a set of binary strings, then let xz = {xy : y Z}. For x Z n, and for 1 i < j n, denote by x[i, j] the binary string x i x i+1... x j. We first present an integer function κ(n). Definition.1 Let κ be the integer function defined as the following: { 0, if n = 1, κ(n) = max{1, log n log log n }, otherwise. Next, a permutation φ of binary strings is given as follows. Definition. Assume x Z n. Then φ(x) Z n is the binary string such that φ(x)[1, κ(n)] = x[1, κ(n)] x[n κ(n) + 1, n], φ(x)[κ(n) + 1, n] = x[κ(n) + 1, n]. Note that the restriction of φ to Z n is indeed a permutation of Z n, with φ (x) = x. Now we give a recursive definition of the twisted hypercubes. 3

4 Definition.3 Set H 1 := K, with vertices 0 and 1. For n, H n is obtained from two copies of H n 1, 0H n 1 and 1H n 1, by adding edges connecting 0x and 1φ(x) for all x H n 1. The vertex set of H n is Z n. It follows from the definition that H 1 = K, H = C 4, H 3 and H 4 are depicted in Figure 1. H3 H4 Figure 1. H3 and H4 By the definition of Hn, the following lemma is straightforward. Lemma.4 Hn is triangle-free. It is seen that Hn is a kind of n-dimensional hypercube-like networks. Then Hn is a n-regular graph with connectivity n. Furthermore, Hn has the following properties. Lemma.5 [10] For any u, v V (Hn), u and v have at most two common neighbors. Lemma.6 [15] κ1(hn) = n, where n 3. Lemma.7 [] κ(hn) = 3n 5, where n 5. Let a {0, 1} be an integer and a the complement of a, i.e., a = 1 a. Let u = u1u un V (Hn). By the definition of the twisted hypercubes, u has only one neighbor u1φ(uu3 un) in u1hn 1, and n 1 neighbors in u1hn 1. For every neighbor w = w1w wn of u in u1hn 1, there exists some q (1 q n 1) such that wi = ui for 1 i q. We use u i to denote the neighbor of u with the same first i 1 bits as that of u, and u i is called the (i)-neighbor of u, where 1 i n. That is, u i = u1 ui 1uiφ(ui+1 un) = u1 ui 1ui(ui+1 un κ+1)(ui+ un κ+) (ui+κ un)ui+κ+1 un, where κ = κ(n i). 4

5 Note that there are n i neighbors of u i in u 1 u i 1 u i H n i, we use u i,j to denote the neighbor of u i with the same first i+j 1 bits as that of u i, and u i,j is called the (i, j)-neighbor of u, where 1 j n i. Denote κ i = κ(n i). Then u i [1, i](u i+1 u n κi +1) (u i+j 1 u n κi +j 1)(u i+j u n κi +j)φ(u i [i + j + 1, n]), u i,j if 1 j κ i, = u i [1, i](u i+1 u n κi +1) (u i+κi u n )u i+κi +1 u i+j 1 u i+j φ(u i+j+1 u n ), if κ i + 1 j n i. Specially, u i,1 = u 1 u i 1 u i (u i+1 u n )(u i+ u n )u i+3 u n, if κ i = κ i+1 = 1, u 1 u i 1 u i (u i+1 u n κ+1 )(u i+ u n κ+ u n κ+1 ) (u i+κ u n u n 1 )(u i+κ+1 u n )u i+κ+ u n, if κ i = κ i+1 = κ, u 1 u i 1 u i (u i+1 u n κ+1 )u i+ u n, if κ i = κ i = κ. Similarly, there are n i j neighbors of u i,j in u 1 u i 1 u i (u i+1 u n κi +1) (u i+j 1 u n κi +j 1)(u i+j u n κi +j)h n i j if 1 j κ i, in u 1 u i 1 u i (u i+1 u n κi +1) (u i+κi u n )u i+κi +1 u i+j 1 u i+j H n i j if κ i + 1 j n i. We use u i,j,k to denote the neighbor of u i,j with the same first i+j +k 1 bits as that of u i,j, and u i,j,k is called the (i, j, k)-neighbor of u, where 1 k n i j. Specially, u 1 u i 1 u i (u i+1 u n )(u i+ u n )φ(u i+3 u n ), if κ i = κ i+1 = 1, u 1 u i 1 u i (u i+1 u n 1 )(u i+ u n u n 1 )φ((u i+3 u n )u i+4 u n ), u i,1,1 if κ i = κ i+1 =, = u 1 u i 1 u i (u i+1 u n κ+1 )(u i+ u n κ+ u n κ+1 )φ(u i,1 [i + 3, n]), if κ = κ i = κ i+1 3, u 1 u i 1 u i (u i+1 u n κ+1 )u i+ φ(u i+3 u n ), if κ i = κ i = κ. At the same time, u 1 u i 1 u i (u i+1 u n )(u i+ u n )u i+3 φ(u i+4 u n ), if κ i = κ i+1 = 1, u 1 u i 1 u i (u i+1 u n 1 )(u i+ u n u n 1 )(u i+3 u n )φ(u i+4 u n ), if κ i = κ i+1 =, u 1 u i 1 u i (u i+1 u n )(u i+ u n 1 u n )(u i+3 u n u n 1 )φ(u i,1 [i + 4, n]), u i,1, = if κ i = κ i+1 = 3, u 1 u i 1 u i (u i+1 u n κ+1 )(u i+ u n κ+ u n κ+1 )(u i+3 u n κ+3 u n κ+ ) φ((u i+4 u n κ+4 u n κ+3 ) (u i+κ u n u n 1 )(u i+κ+1 u n )u i+κ+ u n ), if κ = κ i = κ i+1 4, u 1 u i 1 u i (u i+1 u n κ+1 )u i+ u i+3 φ(u i+4 u n ), if κ i = κ i = κ. 5

6 Let u h( ) be a vertex of H n. If the first i bits of u h( ) is u 1 u u i, then u h( ) has exactly one neighbor in u 1 u u i 1 u i H n i, and we denote this neighbor u h( ),i. For example, let u = 0010 H 4, then u 1 = 1010, u = 0110, u 3 = 0000, u 4 = 0011, u 1,1 = u,1 = 1110, u,1 = 0100, u 3,1 = 0001, u 3,1 = 1000, u 4,1 = 1111, u,1,1 = 1100, u 3,1,1 = Figure illustrates the neighbors of u. Figure. Some neighbors related to a vertex u of H n Note that u i+1 = u 1 u u i u i+1 φ(u i+ u i+3 u n ). Then u 1 u i 1 u i (u i+1 u n )(u i+ u n )u i+3 u n, if κ i = κ i+1 = 1, u u i+1,i 1 u i 1 u i (u i+1 u n κ+1 )(u i+ u n κ+1 u n κ+ ) = (u i+κ u n 1 u n )(u i+κ+1 u n )u i+κ+ u n, if κ i = κ i+1 = κ, u 1 u i 1 u i (u i+1 u n κ+1 )u i+ u n, if κ i = κ i = κ. It is seen that u i+1,i = u i,1. Then u i,1 is a common neighbor of u i and u i+1. Note that u is also a common neighbor of u i and u i+1. By Lemma.5, any two vertices have at most two common neighbors. So we have the following proposition. Proposition.8 Let u = u 1 u u n be any vertex of H n and N Hn (u) = {u 1, u,..., u n }. Then N Hn (u i ) N Hn (u i+1 ) = {u, u i,1 }, where 1 i n 1. 6

7 Recall that for u = u 1 u n V (H n ), u n = u 1 u n 1 u n, u n 1 = u 1 u n u n 1 u n and u n 1,1 = u 1 u u n u n 1 u n, we have u n,i = u 1 u i 1 u i φ(u i+1 u n 1 u n ), u n 1,1,1 = u 1 φ(u u n u n 1 u n ), u n 1,1, = u 1 u φ(u 3 u n u n 1 u n ). The following two propositions are straightforward. Proposition.9 For any u = u 1 u... u n V (H n ), N Hn (u) = {u 1, u,..., u n }, Then (a) u i,1 u j,1 for 1 i j n 1. (b) u n 1,1,1 u 1,1, u n 1,1,1 u 1,1,1 and u n 1,1,1 u 1,1,. (c) u n 1,1, u,1, u n 1,1, u,1,1 and u n 1,1, u,1,. (d) u n,i u i,1, u n,i u i,1,1 and u n,i u i,1,, where 1 i n 3. Proposition.10 For any u = u 1 u... u n V (H n ), N Hn (u) = {u 1, u,..., u n }, then u n,(n ) u n,1, u n,(n ),(n 3) u n 3,1,. u n,(n ),(n 3),..., u,1, u n,(n ),(n 3),...,,1 u 1,1. By Lemma.4 and Proposition.8, we have following proposition. Proposition.11 For any u = u 1 u... u n V (H n ), N Hn (u) = {u 1, u,..., u n }, we have (a) H n [{u i, u i,1, u i+1, u i,1,1 }] is a K 1,3 with center u i,1, where 1 i n. (b) H n [{u n, u n 1,1, u n,1, u n, }] is a K 1,3 with center u n. (c) H n [{u n 1, u n 1,1, u n, u n 1,1,1 }] is a K 1,3 with center u n 1,1. (d) H n [{u i, u i,1, u i+1, u i,1,1, u i,1, }] is a K 1,4 with center u i,1, where 1 i n 3. (e) H n [{u n, u n 1,1, u n,1, u n,, u n,3 }] is a K 1,4 with center u n. (f) H n [{u n 1, u n 1,1, u n, u n 1,1,1, u n 1,1, }] is a K 1,4 with center u n 1,1. (g) H n [{u n, u n,1, u n 1, u n,1,1, u n,1,1 }] is a K 1,4 with center u n,1. 7

8 3 κ(h n ; K 1,r ) and κ s (H n ; K 1,r ) for r {3, 4} In this section, we first show some properties of H n related to stars, then explore the T - structure connectivity and T -substructure connectivity of H n for T = K 1,3 and T = K 1,4, respectively. Lemma 3.1 Let K 1,r be a star in H n. If u is a vertex of H n K 1,r, then N Hn (u) V (K 1,r ). Proof. Let x be the center of the star K 1,r. If (x, u) E(H n ), then N Hn (u) V (K 1,r ) = {x} by Lemma.4. That is, N Hn (u) V (K 1,r ) = 1. If (x, u) / E(H n ), then N Hn (u) V (K 1,r ) as u and x have at most two common neighbors by Lemma.5. Lemma 3. Let K 1,3 be a star in H n. If u and v are two adjacent vertices of H n K 1,3, then N Hn ({u, v}) V (K 1,3 ) 3. Proof. Let V (K 1,3 ) = {x, x 1, x, x 3 } and E(K 1,3 ) = {(x, x 1 ), (x, x ), (x, x 3 )}. If (x, u) E(H n ), then N Hn (u) V (K 1,3 ) = {x} by Lemma.4. That is, N Hn (u) V (K 1,3 ) = 1. By Lemma 3.1, N Hn (v) V (K 1,3 ). Then N Hn ({u, v}) V (K 1,3 ) 3. Therefore (x, u) / E(H n ). By the symmetry of u and v, (x, v) / E(H n ). Then N Hn ({u, v}) V (K 1,3 ) {x 1, x, x 3 } and thus N Hn ({u, v}) V (K 1,3 ) 3. Lemma 3.3 Let K 1,r be a star in H n with center x. If u and v are two adjacent vertices of H n K 1,r, then N Hn ({u, v}) V (K 1,r ) 4. Moreover, if N Hn ({u, v}) V (K 1,r ) = 4, then (u, x), (v, x) / E(H n ) and u, v, x V (ah n 1 ) for some a {0, 1}. Proof. By Lemma 3.1, N Hn (u) V (K 1,r ) and N Hn (v) V (K 1,r ). Then N Hn ({u, v}) V (K 1,r ) 4. Suppose that N Hn ({u, v}) V (K 1,r ) = 4, in the following, we show that u, v, x V (ah n 1 ) for some a {0, 1}. Note that N Hn (u) V (K 1,r ) = and N Hn (v) V (K 1,r ) =, then (u, x), (v, x) / E(H n ). If u V (0H n 1 ) and v V (1H n 1 ), then x and v have at most one common neighbor if x V (0H n 1 ), and x and u have at most one common neighbor if x V (1H n 1 ), a contradiction. Therefore u, v V (ah n 1 ) for some a {0, 1}. If x V (ah n 1 ), then N Hn ({u, v}) N Hn (x) {u 1, v 1, x 1 }, i.e, N Hn ({u, v}) V (K 1,r ) 3, a contradiction. Therefore x V (ah n 1 ). Lemma 3.4 Let F 1, F,..., F n/ 1 be n 1 stars in H n. If u and v are two adjacent vertices of H n n/ 1 i=1 V (F i ), then N Hn ({u, v}) ( n/ 1 i=1 V (F i )) n 3. 8

9 Proof. Note that N Hn ({u, v}) = n, suppose to the contrary that N Hn ({u, v}) ( n/ 1 i=1 V (F i )) = n. By Lemma 3.3, N Hn ({u, v}) V (K 1,r ) 4, then n = N Hn ({u, v}) ( n/ 1 i=1 V (F i )) 4( n 1). Thus n is odd, N H n ({u, v}) V (F i ) = 4 for 1 i n/ 1 and N Hn ({u, v}) ( n/ i=1 V (F i)) = N Hn ({u, v}). Let x i be the center of F i, then (u, x i ), (v, x i ) / E(H n ) and u, v, x i V (ah n 1 ) for some a {0, 1} by Lemma 3.3. Assume, without loss of generality, that u 1 F 1. Then u 1 x 1 as (u, x 1 ) / E(H n ). On the other hand, (u 1, x 1 ) E(H n ), which implies x 1 V (ah n 1 ), a contradiction. Lemma 3.5 κ s (H n ; K 1,r ) n for 3 r 4 and n 4. Proof. Let F = {P 1,..., P }{{} 1, P,..., P, P }{{} 3,..., P 3, K }{{} 1,3,..., K 1,3 } if r = 3, and F = }{{} x 1 x x 3 x 4 {P 1,..., P }{{} 1, P,..., P, P }{{} 3,..., P 3, K }{{} 1,3,..., K 1,3, K }{{} 1,4,..., K 1,4 } if r = 4, where x }{{} i 0 and x 1 x x 3 x 4 x 5 1 i r + 1. Then F = r+1 i=1 x i. Suppose to the contrary that F n 1 and H n F is disconnected, then H n F has at least two components. Let C be the smallest component of H n F. We consider the following three cases. Case 1. V (C) = 1. In this case, C is an isolated vertex w. Note that N Hn (w) = n. By Lemmas.4 and 3.1, every element in F contains at most two neighbors of w, then F N Hn (w). Thus, ( n 1) n, a contradiction. Case. V (C) =. In this case, C is an edge (u, v). Note that N Hn ({u, v}) = n. If r = 3, every element in F contains at most three neighbors of {u, v} by Lemmas.4 and 3., then 3 F N Hn ({u, v}). Thus 3( n 1) n, a contradiction. Therefore r = 4. By Lemma 3.4, all elements in F contains at most n 3 neighbors of {u, v}, a contradiction. Case 3. V (C) 3. Note that F n 1 and every element in F contains at most five vertices, then V (F) 5( n 1). If n = 4, then V (F) 5. By Lemma.6, κ 1(H n ) = n = 6, which implies that we have to delete at least 6 vertices to separate C from H 4, a contradiction. Therefore n 5. By Lemma.7, κ (H n ) = 3n 5, which implies that we have to delete at least 3n 5 > 5( n 1) vertices to separate C from H n, a contradiction. Thus, κ s (H n ; K 1,3 ) n. 9

10 3.1 κ(h n ; K 1,3 ) and κ s (H n ; K 1,3 ) Lemma 3.6 κ(h n ; K 1,3 ) n for n 4. Proof. For any u = u 1 u n V (H n ), N Hn (u) = {u 1,..., u n }. Let T i be the subgraph induced by {u i, u i,1, u i+1, u i,1,1 } for 1 i n. If n is odd, we let T n be the subgraph induced by {u n, u n 1,1, u n,1, u n, }; and if n is even, let T n 1 the subgraph induced by {u n 1, u n 1,1, u n, u n 1,1,1 }. Then T i is a K 1,3 by Proposition.11(a-c) for 1 i n, and V (T i ) V (T j ) = by Proposition.9 for 1 i j n. Set S = {T 1, T 3,..., T n, T n } if n is odd; and S = {T 1, T 3,..., T n 3, T n 1 } if n is even. Then, in either case, H n S is disconnected, one component is {u} and S = n. Thus, κ(h n ; K 1,3 ) n. Note that κ s (H n ; K 1,3 ) κ(h n ; K 1,3 ), and thus, by Lemmas 3.5 and 3.6, we have the following result. Theorem 3.7 κ(h n ; K 1,3 ) = κ s (H n ; K 1,3 ) = n for n κ(h n ; K 1,4 ) and κ s (H n ; K 1,4 ) Lemma 3.8 κ(h n ; K 1,4 ) n for n 4. Proof. For any u V (H n ), N Hn (u) = {u 1,..., u n }. Let T i be the subgraph induced by {u i, u i,1, u i+1, u i,1,1, u i,1, } for 1 i n 3. If n is odd, we let T n be the subgraph induced by {u n, u n,1, u n 1, u n,1,1, u n,1,1 } and T n the subgraph induced by {u n, u n 1,1, u n,1, u n,, u n,3 }; and if n is even, let T n 1 be the subgraph induced by {u n 1, u n 1,1, u n, u n 1,1,1, u n 1,1, }. Then T i is a K 1,4 by Proposition.11(d-g) for 1 i n, and V (T i ) V (T j ) = by Proposition.9 for 1 i j n. Set S = {T 1, T 3,..., T n 4, T n, T n } if n is odd; and S = {T 1, T 3,..., T n 3, T n 1 } if n is even. Then, in either case, H n S is disconnected, one component is {u} and S = n. Thus, κ(h n ; K 1,4 ) n and κs (H n ; K 1,4 ) n. Note that κ s (H n ; K 1,4 ) κ(h n ; K 1,4 ), and hence, by Lemmas 3.5 and 3.8, we have the following result. Theorem 3.9 κ(h n ; K 1,4 ) = κ s (H n ; K 1,4 ) = n for n 4. 4 κ(h n ; P k ) and κ s (H n ; P k ) Recall that κ(h n ; P 1 ) = κ s (H n ; P 1 ) = κ(h n ) = n for n 3. So we assume k. 10

11 4.1 κ(h n ; P ) and κ s (H n ; P ) Lemma 4.1 If κ(n 1), then H n does not consists of n 1 quadrilaterals sharing a common edge. Proof. For any u = u 1 u... u n V (H n ), we show that for any 1 i n, u i and u n have no other common neighbors other than u, and u n 1 and u 1 have no other common neighbors other than u. Recall that u i = u 1 u i 1 u i (u i+1 u n κi +1) (u i+κi u n )u i+κi +1 u n 1 u n, where κ i = κ(n i). First we assume i n. We only need to show that (u i ) j (u n ) k for 1 j i n and 1 k n 1. If 1 j i n 1, then (u i ) j [n] = u n u n = (u n ) k [n], thus (u i ) j (u n ) k. Therefore j = n. If i k, then (u i ) n [i] (u n ) k [i]; and if i = k, then (u i ) n [i + κ i ] = u i+κi u n u i+κi u n = (u n ) i [i + κ i ]. Thus (u i ) n (u n ) k. Now we assume i = n 1. We only need to show that (u 1 ) p (u n 1 ) q for p n and 1 q n 1 n. If p n, then (u 1 ) p [n 1, n] = u n 1 u n u n 1 u n = (u n 1 ) q [n 1, n] for 1 q n and (u 1 ) p [n 1, n] = u n 1 u n u n 1 u n = (u n 1 ) n [n 1, n], and thus (u 1 ) p (u n 1 ) q. If p = n, then (u 1 ) n [n 1, n] = u n 1 u n u n 1 u n = (u n 1 ) q [n 1, n] for 1 q n and (u 1 ) n [n 1, n] = u n 1 u n u n 1 u n = (u n 1 ) n [n 1, n], and thus (u 1 ) n (u n 1 ) q. Therefore p = n 1. If q 1, then (u 1 ) n 1 [1] (u n 1 ) q [1]; and if q = 1, then (u n 1 ) 1 = u 1 (u u n κ1 +1) (u κ1 1 u n )(u κ1 u n 1 )(u κ1 +1 u n )u κ1 + u n u n 1 u n as κ 1 = κ(n 1) and thus (u 1 ) n 1 [κ 1 ] = (u κ1 u n 1 ) (u κ1 u n 1 ) = (u n 1 ) 1 [κ 1 ]. Hence (u 1 ) n 1 (u n 1 ) q. Lemma 4. κ(h n ; P ) n 1 if κ(n 1) = 1 and κ(h n ; P ) n if κ(n 1). Proof. we distinguish cases pertaining to the value of κ(n) in the following. Case 1. κ(n 1) = 1. In this case, we set u = Then u i = 0 0u i 0 0, where u i = 1, 1 i n. Let v = u n 1 = 0 010, then v i = 0 0v i 0 010, where v i = 1, 1 i n ; v n = Note that if κ(n 1) = 1, then (u i, v i ) E(H n ); and hence C 4 = u, u i, v i, v, u is a cycle of length 4. Set S = {u i v i : 1 i n and i n 1}. Then, H n S is disconnected, one component is {u, v} and S = n 1. Thus, κ(h n ; P ) n 1. Case. κ(n 1). 11

12 In this case, for any u = u 1 u... u n V (H n ), N Hn (u) = {u 1, u,..., u n }. By Lemma 4.1, H n does not consists of n 1 quadrilaterals sharing a common edge. So we can set that T i be the subgraph induced by {u i, u i,1 } for 1 i n 1, then T i is isomorphic to P. Set S = {T 1, T,..., T n 1, T n }, where T n is the subgraph induced by {u n, u n,1 }. Then, H n S is disconnected, one component is {u} and S = n. Thus, κ(h n ; P ) n. Lemma 4.3 κ s (H n ; P ) n 1 if κ(n 1) = 1 and κ s (H n ; P ) n if κ(n 1). Proof. we distinguish cases pertaining to the value of κ(n) in the following. Case 1. κ(n 1) = 1. Let F = {P 1,..., P }{{} 1, P,..., P } and F = x }{{} 1 + x n for x 1, x 0. Suppose to x 1 x the contrary that H n F is disconnected, then H n F has at least two components. Let C be the smallest component of H n F. We consider following two cases. Subcase 1.1. V (C) = 1. In this subcase, we set V (C) = {w}. Note that N Hn (w) = n. By Lemma.4, every element in F contains at most one neighbor of w. Thus, we have to delete at least n elements of F to isolate C. But it is impossible since F n < n. Subcase 1.. V (C). In this subcase, by Lemma.6, κ 1 (H n ) = n. This implies that we have to delete at least n vertices to isolate C. Since F n, we have V (F) (n ) = n 4 < n, a contradiction. Thus, κ s (H n ; P ) n 1 and κ(h n ; P ) n 1. Case. κ(n 1). In this case, n > 5. Let F = {P 1,..., P }{{} 1, P,..., P } and F = x }{{} 1 + x n 1 for x 1 x x 1, x 0. Suppose to the contrary that H n F is disconnected, then H n F has at least two components. Let C be the smallest component of H n F. We consider following three cases. Subcase.1. V (C) = 1. In this subcase, we set V (C) = {w}. Note that N Hn (w) = n. By Lemma.4, every element in F contains at most one neighbor of w. Thus, we have to delete at least n elements of F to isolate C. But it is impossible since F n 1 < n. Subcase.. V (C) =. 1

13 In this subcase, C is an edge of H n and N Hn (C) = n. By Lemma.6 and Lemma 4.1, we have to delete at least n +1 = n elements of F to separate C. However, F n 1 < n, a contradiction. Subcase.3. V (C) 3. In this subcase, by Lemma.7, κ (H n ) = 3n 5. This implies that we have to delete at least 3n 5 vertices to isolate C. Since F n 1, we have V (F) (n 1) = n < 3n 5, a contradiction with n > 5. Thus, κ s (H n ; P ) n and κ(h n ; P ) n. Note that κ s (H n ; P ) κ(h n ; P ), and hence, by Lemmas 4. and 4.3, we have the following result. Theorem 4.4 κ(h n ; P ) = κ s (H n ; P ) = n 1 if κ(n 1) = 1; κ(h n ; P ) = κ s (H n ; P ) = n if κ(n 1). 4. κ(h n ; P k ) and κ s (H n ; P k ) with k 3 Lemma 4.5 Let P k be a path of order k in H n with 1 k n. If v is a vertex of H n P k, then N Hn (v) V (P k ) k. Proof. By Lemma.4, v can be adjacent to at most one vertex of any two consecutive vertices on P k. Thus, the lemma follows. Lemma 4.6 Let P k be a path of order k in H n with 3 k n. If u and v are two adjacent vertices of H n P k, then N Hn ({u, v}) V (P k ) k + (k(mod 3)). Moreover, 3 N Hn ({u, v}) V (P k ) k 1. Proof. Let P k = v 1, v,..., v k. We first show that N Hn ({u, v}) {v i 1, v i, v i+1 } for i k 1. ( ) Otherwise, we suppose that N Hn ({u, v}) {v i 1, v i, v i+1 } 3 for some i ( i k 1). Without loss of generality, we assume that N Hn (u) {v i 1, v i, v i+1 }. By Lemma.4, we have uv i 1, uv i+1 E(H n ), and uv i / E(H n ). Moreover, N Hn (u) {v i 1, v i, v i+1 } =. Then N Hn (v) {v i 1, v i, v i+1 } 1 and vv i 1, vv i+1 / E(H n ) by Lemma.4. So vv i E(H n ), and thus {v, v i 1, v i+1 } N(u) N(v i ), a contradiction with Lemma.5. Now we prove this lemma by induction on k. For k = 3, by the above proof, N Hn ({u, v}) V (P 3 ). In the induction step, assume that the lemma is true for 3 k l. When k = l + 1, N Hn (v l+1 ) {u, v} 1 by Lemma.4, and then N Hn ({u, v}) V (P l+1 ) N Hn ({u, v}) V (P l ) + 1 l + (l(mod 3)) + 1 by the inductive hypothesis. 3 13

14 Note that if (l + 1)(mod 3) = 1 or, then (l + 1)(mod 3) = l(mod 3) + 1 and l = l+1, 3 3 and hence l l+1 + (l(mod 3)) + 1 = + ((l + 1)(mod 3)). So, in the following, we assume 3 3 that l 5 and (l+1)(mod 3) = 0. Then (l )(mod 3) = 0. By ( ), N Hn ({u, v}) V (P l+1 ) N Hn ({u, v}) V (P l ) + l l+1 + =. 3 3 Therefore the proof of the lemma is complete. Lemma 4.7 Let 3 k n. Then κ(h n ; P k ) n k+1 if k is odd and κ(h n; P k ) n k if k is even. Proof. For any u = u 1 u... u n V (H n ), N Hn (u) = {u 1, u,..., u n }, we distinguish cases pertaining to the parity of k in the following. Case 1. k is odd. In this case, we denote t := k+1. Let n = qt + r for some nonnegative integers q and r with 0 r k 1. Since n k, we have q 1. Subcase 1.1. r = 0. In this subcase, we set P 1 = u 1, u 1,1, u,..., u t 1, u t 1,1, u t, P = u t+1, u t+1,1, u t+,..., u k, u k,1, u k+1,. P q = u n t+1, u n t+1,1, u n t+,..., u n 1, u n 1,1, u n. Then V (P i ) V (P j ) = by Proposition.10 for 1 i j q, and then F = {P 1, P,..., P q } forms a P k -structure-cut of H n, one component of H n F is {u} and F = n k+1. Subcase r k 1. In this subcase, we set P 1 = u 1, u 1,1, u,..., u t 1, u t 1,1, u t, P = u t+1, u t+1,1, u t+,..., u k, u k,1, u k+1,. P q = u n r t+1, u n r t+1,1, u n r t+,..., u n r 1, u n r 1,1, u n r, P (q+1) = u n r+1, u n r+1,1, u n r+,..., u n, u n,(n ),..., u n,(n ),...,(n k+r ). Then V (P i ) V (P j ) = by Proposition.10 for 1 i j q + 1, and then F = {P 1, P,..., P q, P (q+1) } forms a P k -structure-cut of H n, one component of H n F is {u} and F = n k+1. Thus, in either subcase, κ(h n ; P k ) n k+1. Case. k is even. 14

15 In this case, we denote a := k. Let n = qa + r for some nonnegative integers q and r with 0 r k. Since n k, we have q 1. Subcase.1. r = 0. In this subcase, we set P 1 = u 1, u 1,1, u,..., u a, u a,1, P = u a+1, u a+1,1, u a+,..., u k, u k,1,. P q = u n a+1, u n a+1,1, u n a+,..., u n, u n,1. Then V (P i ) V (P j ) = by Proposition.10 for 1 i j q, and then F = {P 1, P,..., P q } forms a P k -structure-cut of H n, one component of H n F is {u} and F = n k. Subcase.. 1 r k. In this subcase, we set P 1 = u 1, u 1,1, u,..., u a, u a,1, P = u a+1, u a+1,1, u a+,..., u k, u k,1,. P q = u n r a+1, u n r a+1,1, u n r a+,..., u n r, u n r,1, P (q+1) = u n r+1, u n r+1,1, u n r+,..., u n, u n,(n ),..., u n,(n ),...,(n k+r ). Then V (P i ) V (P j ) = by Proposition.10 for 1 i j q + 1, and then F = {P 1, P,..., P q, P (q+1) } forms a P k -structure-cut of H n, one component of H n F is {u} and F = n k. Thus, κ(h n ; P k ) n k. The proof of the lemma is now complete. Lemma 4.8 Let 3 k n. Then κ s (H n ; P k ) n k+1 if k is odd and κs (H n ; P k ) n k if k is even. Proof. If no confusion should arise, we use F = {P 1,..., P }{{} 1, P,..., P }{{} x 1 x,..., P k,..., P }{{} k } to x k denote a set of connected subgraphs of P k with F = k i=1 x i for x i 0. If 3 n 4, then k = 3 or k = 4 as 3 k n. So in either case, it suffices to show that H n F is connected whenever F 1. If x k = 0, then H n P i is connected for 1 i k 1 since κ(h n ) = n k. Hence, we may assume that x k = 1, and consequently x 1 = = x k 1 = 0. That is, F = {P k }. Assume that H n P k is disconnected, then each component of H n P k contains at least two vertices as H n is n-regular and triangle-free. On the other hand, by Lemma.6, κ 1 (H n ) = n > k = V (P k ), a contradiction. 15

16 So, in the following, we may assume that n 5. We proceed this by contradiction. Case 1. k is odd. Suppose to the contrary that F n k+1 1 and H n F is disconnected, then H n F has at least two components. Without loss of generality, let C be the smallest component of H n F. We consider the following two cases. Case 1.1. V (C) = 1. In this subcase, C is an isolated vertex. Let V (C) = {w}, then N Hn (w) = n. By Lemma 4.5, every element in F contains at most k+1 neighbors of w. Thus, k+1 F n, i.e. k+1 n ( 1) n, a contradiction. k+1 Case 1.. V (C). In this subcase, C contains at least one edge. By Lemma.6, κ 1 (H n ) = n. This implies that we have to delete at least n vertices to separate C from H n. However, from the assumption F n k k+1 k+1 (n ) < n, a contradiction. Case. k is even. n n+k 1 1, we infer that V (F) k( 1) k( 1) = k+1 k+1 Suppose to the contrary that F n k 1 with H n F being disconnected, and let C be the smallest component of H n F. Consider the following three cases. Case.1. V (C) = 1. In this subcase, C is an isolated vertex. Let V (C) = {w}, then N Hn (w) = n. By Lemma 4.5, every element in F contains at most k k ( n 1) n, a contradiction. k Case.. V (C) =. neighbors of w. Thus, k F n, i.e. In this subcase, C is an edge of H n. Suppose that V (C) = {u, v}, then N Hn ({u, v}) = n. By Lemma.6, every element in F contains at most k 1 neighbors of {u, v}. It means that we have to delete at least n vertices of F to separate C. However, k 1 F n 1 n+k 1 = n < n, a contradiction. k k k k 1 Case.3. V (C) 3. In this subcase, V (F) k( n 1) from the assumption F n 1. On the other k k hand, κ (H n ) = 3n 5 by Lemma.7, which implied that we have to delete at least 3n 5 vertices to separate C from H n. However, it is easily to check that V (F) k( n k 1) < 3n 5 for n 5, a contradiction. Therefore we complete the proof of Lemma 4.8. Recall that κ s (H n ; P k ) κ(h n ; P k ), and hence, by Lemmas 4.7 and 4.8, we have the 16

17 following result. Theorem 4.9 Let 3 k n. Then κ(h n ; P k ) = κ s (H n ; P k ) = n if k is odd; κ(h k+1 n; P k ) = κ s (H n ; P k ) = n if k is even. k 5 Conclusions In this paper, we consider the T -structure connectivity and T -substructure connectivity of the twisted hypercube H n for T {K 1,r, P k } and show that κ(h n ; K 1,r ) = κ s (H n ; K 1,r ) = n for n 4 and 3 r 4; n, if k = 1, n 1, if k = and κ(n 1) = 1, κ(h n ; P k ) = κ s (H n ; P k ) = n, if k = and κ(n 1), n, k+1 if 3 k n and k is odd, n, k if 4 k n and k is even. However, determining the K 1,r -structure connectivity and K 1,r -substructure connectivity of H n with r 5 remain open. To explore κ(h n ; C k ) and κ s (H n ; C k ), the approach used in this paper is invalid as H n is a nonbipartite graph. But one may explore the structure connectivity and substructure connectivity of other interconnection networks by the approach used in this paper. Acknowledgments Xiaolan Hu is partially supported by NNSFC under grant number and NSF of Hubei Province under grant number 016CFB146. Huiqing Liu is partially supported by NNSFC under grant numbers and References [1] J.A. Bondy, U.S.R. Murty, Graph Theory, Springer, 008. [] N.W. Chang, S.Y. Hsieh, {, 3}-extraconnectivities of hypercube-like networks, J. Comput. System Sci. 79(5)(013) [3] Y.C. Chen, J.J.M. Tan, L.H. Hsu, Super connectivity and super-edge-connectivity for some interconnection networks, Appl. Math. Comput. 140()(003)

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