The super connectivity of shuffle-cubes

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1 Information Processing Letters 96 (2005) The super connectivity of shuffle-cubes Jun-Ming Xu a,,minxu b, Qiang Zhu c a Department of Mathematics, University of Science and Technology of China, Hefei, Anhui , China b Institute of Applied Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing , China c Department of Mathematics, XiDian University, Xi an, Shanxi , China Received 21 January 2005; received in revised form 23 July 2005; accepted 28 July 2005 Available online 25 August 2005 Communicated by F.Y.L. Chin Abstract The shuffle-cube SQ n,wheren 2 (mod ), a new variation of hypercubes proposed by Li et al. [T.-K. Li, J.J.M. Tan, L.-H. Hsu, T.-Y. Sung, The shuffle-cubes and their generalization, Inform. Process. Lett. 77 (2001) 35 1], is an n-regular n-connected graph. This paper determines that the super connectivity of SQ n is 2n and the super edge-connectivity is 2n 2forn Elsevier B.V. All rights reserved. Keywords: Combinatorial problems; Shuffle-cubes; Super connectivity; Super edge-connectivity; Hypercubes 1. Introduction The shuffle-cube, denoted by SQ n, where n 2 (mod ), as an interconnection network topology proposed by Li et al. [], is a new variation of hypercubes Q n obtained by changing some links. For an n-bit binary string u = u n 1 u n 2...u 1 u 0 V(SQ n ),letp j (u) = u n 1 u n 2...u n j and s i (u) = u i 1 u i 2...u 1 u 0. The n-dimensional shuffle-cube SQ n, n 2 (mod ), is recursively defined as fol- The work was supported by NNSF of China (Nos and ). * Corresponding author. address: xujm@ustc.edu.cn (J.-M. Xu). lows: SQ 2 is Q 2.Forn 3, SQ n consists of 16 subcube SQ i 1i 2 i 3 i n s, where i j {0, 1} for 1 j and p (u) = i 1 i 2 i 3 i for all vertices u in SQ i 1i 2 i 3 i n.the vertices u = u n 1 u n 2...u 1 u 0 and v = v n 1 v n 2... v 1 v 0 in different (n )-dimensional subcubes are linked by an edge in SQ n if and only if s n (u) = s n (v) and p (u) p (v) V s2 (u), where the symbol denotes the addition with modulo 2 and V 00 ={1111, 0001, 0010, 0011}, V 01 ={0100, 0101, 0110, 0111}, V 10 ={1000, 1001, 1010, 1011}, V 11 ={1100, 1101, 1110, 1111} /$ see front matter 2005 Elsevier B.V. All rights reserved. doi: /j.ipl

2 12 J.-M. Xu et al. / Information Processing Letters 96 (2005) simple connected graph. For x V(G),letN G (x) be the set of neighbors of x and d G (x) = N G (x), the degree of x. Forxy E(G), letn G (xy) = N G (x) N G (y) \{x,y}, and let ζ G (xy) = NG (xy), ζ(g)= min { ζ G (xy): xy E(G) } ; ξ G (xy) = d G (x) + d G (y) 2, ξ(g) = min { ξ G (xy): xy E(G) }. Fig. 1. A shuffle-cube SQ 6. We illustrate SQ 6 in Fig. 1 showing only edges incident at vertices in SQ and omitting others. It is convenient to let n = k + 2 and u = u n 1 u n 2...u 1 u 0 = u k uk 1...u 1 u0, where u0 = u 1u 0 and u j = u j+1u j u j 1 u j 2 for 1 j k. Then two vertices u and v in SQ n are linked by an edge if and only if one of the following conditions holds: (1) u j vj V u 0 for exactly one j satisfying 1 j k and u j = vj for all 0 j j k. (2) u 0 v0 {01, 10} and uj = vj for all 1 j k. It has been shown that SQ n is n-regular n-connected in []. In this paper, we further discuss its super connectivity, a more refined parameter than the connectivity for measuring the reliability and the fault tolerance of a network [2,3]. Let G = (V, E) be a graph. A subset S V (respectively F E)iscalledasuper vertex-cut (respectively super edge-cut)ifg S (respectively G F )is not connected and every component contains at least two vertices. The super connectivity κ (G) (respectively super edge-connectivity λ (G)) is the minimum cardinality over all super vertex-cuts (respectively super edge-cuts) in G if they exist. In [2], Esfahanian proved that κ (Q n ) = λ (Q n ) = 2n 2 for n 3. In this paper, we prove that κ (SQ n ) = 2n and λ (SQ n ) = 2n 2, where n 2 (mod ) and n Some lemmas We follow [1] for graph-theoretical terminology and notation not defined here. Let G = (V, E) be a Lemma 1 [3]. λ (G) ξ(g) for any graph G with order at least four and not a star. Lemma 2. ζ(sq n ) = 2n, and the edge uv which attains this value is only if u = u k...u1 u0 and v = u k...ui+1 (u i e)ui 1...u 0, where e {0001, 0010, 0011} V 00 and u 0 = v0 = 00. Proof. Let uv be an edge in SQ n with u = u k uk 1... u 1 u0 and v = vk vk 1...v 1v0, where u0 = u 1u 0. Then u k...ui+1 (u i e 1)u i 1...u 1 u0 for i 0, e 1 V v = u 0, or u k uk 1...u 1 (u0 e 2) for i = 0,e 2 {01, 10}. For convenience, we denote v = u k...ui+1 (u i e 3 )u i 1...u 1 u0 for the possible two cases. If u and v have no neighbors in common, then ζ SQn (uv) = 2n 2 > 2n. Suppose now that u and v have a neighbor w in common. Since w is a neighbor of u, then u k...uj+1 (u j e 1 )uj 1...u 1 u0 for j 0,e 1 w = V u 0, or u k uk 1...u 1 (u0 e 2 ) for j = 0,e 2 {01, 10}. For convenience, we denote w = u k...uj+1 (u j e )u j 1...u 1 u0 for the possible two cases. Since w is a neighbor of v, then v k...vs+1 (v s e 1 )vs 1...v 1v0 for s 0,e 1 w = V v 0, or v kvk 1...v 1(v0 e 2 ) for s = 0,e 2 {01, 10}.

3 J.-M. Xu et al. / Information Processing Letters 96 (2005) We denote w = v k...vs+1 (v s e )v s 1...v 1v0 for the possible two cases. Then i = j = s, u i e = (u i e 3) e = u i (e 3 e ). If i = 0, then e = e 3 e does not holds for {e 3,e,e } {01, 10}. If i 0, then e = e 3 e holds only for u 0 = 00 and {e 3,e,e }={0001, 0010, 0011}. The two vertices u k...ui+1 (u i e )u i 1...u 1 u0 and uk...ui+1 (u i e )u i 1...u 1 u0 are all neighbors of both u and v. So ζ SQn (uv) = 2n and the lemma follows by the arbitrary choice of the edge uv. Lemma 3. κ (SQ n ) 2n for n 6. Proof. By Lemma 2, let uv be an edge of SQ n such that ζ SQn (uv) = 2n, where u = u k...u1 u0 and v = u k...ui+1 (u i e)ui 1...u 0, e V 00 and u 0 = v 0 = 00. We prove that N SQ n (uv) is a super vertex-cut, which means κ (SQ n ) ζ SQn (uv) = 2n. To the end, we need to prove that SQ n (N SQn (uv) {u, v}) has no isolated vertices. Suppose that w = w k...w1 w0 is a vertex in SQ n (N SQn (uv) {u, v}). We now prove N SQn (w) N SQn (uv), where w 0 {00, 01, 10, 11}. If w 0 = 00, since w/ N SQ n (uv) {u, v}, then the vertex w = w k...w1 (w0 01) is a neighbor of w and w / N SQn (uv). If w 0 {01, 10}, the vertex w = w k...w1 11 is a neighbor of w and w / N SQn (uv). If w 0 = 11, the vertex w = w k...wi+1 (w i e )w i 1...w 1w0 is a neighbor of w and w / N SQn (uv) where e V 11. Owning to the above discussion, we have N SQn (w) N SQn (uv) for n 6. Lemma. Let n = k + 2, k 1. Then the number of non-adjacent edges between any two distinct (k 2)- subcubes is at least 2 k. Proof. Let SQ i 1i 2 i 3 i k 2 and SQ j 1j 2 j 3 j k 2 be two distinct (k 2)-subcubes in SQ k+2. Obviously, the edges between SQ i 1i 2 i 3 i k 2 and SQ j 1j 2 j 3 j k 2 are non-adjacent since i 1 i 2 i 3 i j 1 j 2 j 3 j. By the definition of SQ k+2, every edge uv between SQ i 1i 2 i 3 i k 2 and SQ j 1j 2 j 3 j k 2 satisfies s k 2 (u) = s k 2 (v), and p (u) p (v) = i 1 i 2 i 3 i j 1 j 2 j 3 j V s2 (u). Therefore, s 2 (u) is determined. Then the number of non-adjacent edges between SQ i 1i 2 i 3 i k 2 and SQ j 1j 2 j 3 j k 2 is at least 2 k+2 2 = 2 k. 3. Main results Theorem 1. κ (SQ n ) = 2n, where n = k + 2 and k 1. Proof. By Lemma 3, we only need to prove κ (SQ n ) 2n. To the end, let F be an arbitrary set of vertices in SQ n such that F 2n 5 and SQ n F has no isolated vertices. We prove that SQ n F is connected. By definition, SQ n consists of 16 subcube SQ n s. We partition 16 subcube SQ n s of SQ n into two subsets S 1 and S 2, where S 1 ={SQ n SQ n contain at least n vertices in F }, S 2 = {SQ n SQ n contain at most n 5 vertices in F }. Then S 1 consists of at most three subcube SQ n s since (n )>2n 5forn 6, and so S 2. We prove that SQ n F is connected from the following claims. Claim 1. S 2 F is connected. Proof. Let n = k + 2, k 1. Since every (k 2)- subcube in S 2 is (k 2)-connected and contains at most n 5 (= k 3) vertices in F, it is connected in S 2 F. If k = 1, then n = 6, F 7 and every subcube SQ 2 in S 2 contains at most one vertex in F. We decompose S 2 into two subgraphs H 1 and H 2, where H 1 = {SQ 2 V(SQ 2 ) F }and H 2 ={SQ 2 V(SQ 2 ) F = }. Then H 2 contains at least 9 subcube SQ 2 s. It is easy to observe that H 2 is connected. If H 1 =, then the claim follows. Assume H 1 below. Let G = SQ 2 be a subcube in H 1. Since G contains at most one vertex in F, G F is connected and G has at least 11 neighbors in other subcube SQ 2 s, at least one of them is in H 2 since there are at most 7 subcubes in H 1 S 1. Since H 2 is connected, each subcube in H 1 connects with H 2 in SQ 6 F, and so S 2 F is connected. If k 2, let G 1 and G 2 be two arbitrary distinct (k 2)-subcubes in S 2. We only need to prove that G 1 connects with G 2 in S 2 F.LetE 12 be the set of non-adjacent edges between G 1 and G 2. Then E 12 2 k by Lemma. Since 2 k > 2(k 3) = 2(n

4 126 J.-M. Xu et al. / Information Processing Letters 96 (2005) ) for k 2, G 1 connects with G 2 in S 2 F for k 2. If S 1 =, then there is nothing to do by Claim 1. Assume S 1 below. Claim 2. Let G 1 be a subcube SQ n in S 1 and T a connected component with t vertices in G 1 F. Then T connects with S 2 F. Proof. If t = 1, since SQ n F has no isolated vertices, the vertex u 1 in T connects with a vertex u 2 in (SQ n G 1 ) F. If t 2, for any edge xy in T,wehave V(T) N G1 (xy) t 2, and N G1 (xy) 2(n ) by Lemma 2. Since every vertex in G 1 has the same kth -bit, different vertices in G 1 have different neighbors in SQ n G 1.SoT has t neighbors in SQ n G 1. Thus, T has at least 2(n ) (t 2) + t neighbors in SQ n T. Since 2(n ) (t 2) + t = 2n + 3t 10 > 2n 5 F and T is a component of G 1 F, there exist a vertex u 1 in T and a vertex u 2 in (SQ n G 1 ) F such that u 1 u 2 E(SQ n F). Let G 2 be a subcube SQ n that contains u 2.If G 2 S 2, then the claim follows. Assume G 2 S 1 below. Since each of G 1 and G 2 contains at least n vertices in F and (2n 5) 2(n ) = 3, F contains at most three vertices of SQ n G 1 G 2.By Lemma 2, u 1 and u 2 have at most two neighbors in common. Note that each vertex in a subcube SQ n has four neighbors in other subcubes. Thus, {u 1,u 2 } has at least four neighbors in SQ n G 1 G 2, at least one of them, say u 3, is not in F and V(G 3 ) F 1, where G 3 is the (n )-subcube containing u 3. Since V(G 3 ) F 1 n 5, G 3 S 2, and so the claim follows. By the above discussion, we prove that SQ n F is connected, which means κ (SQ n ) 2n forn 6. The theorem follows. Theorem 2. λ (SQ n ) = 2n 2, where n = k + 2 and k 1. Proof. By Lemma 1, we only need to prove λ (SQ n ) 2n 2forn 6. Let F be an arbitrary set of edges in SQ n such that F 2n 3 and SQ n F has no isolated vertices. We prove that SQ n F is connected. We partition 16 subcube SQ n s of SQ n into two subsets S 1 and S 2, where S 1 ={SQ n SQ n contain at least n edges in F }, S 2 ={SQ n SQ n contain at most n 5 edges in F }. Then S 1 consists of at most four subcube SQ n s since 5(n ) >2n 3 for n 6, and so S 2. We complete the proof by the following claims. Claim 1. S 2 F is connected. Proof. Let n = k + 2, k 1. Since SQ n is n-regular n-connected in [], we conclude that SQ n is n-edgeconnected. Then every (k 2)-subcube in S 2 is also connected in S 2 F.LetG 1 and G 2 be two arbitrary distinct (k 2)-subcubes in S 2. We only need to prove G 1 connects with G 2 in S 2 F.LetB 12 beaset of edges between G 1 and G 2. Then B 12 2 k.if B 12 F, then there is noting to do. Assume B 12 F below. Since S 1 consists of at most four (k 2)-subcubes, S For each of other 10 subcubes in S 2 F,if at most one of subcube in {G 1,G 2 } connects with it, then F (10 + 1) B k > F, a contradiction. Thus, there exists a (k 2)-subcube, say G 3, such that G 3 connects each of G 1 and G 2 in S 2 F. This implies G 1 and G 2 are connected in S 2 F. If S 1 =, then there is nothing to do by Claim 1. Assume S 1 below. Claim 2. Let G 1 be a subcube SQ n in S 1 and T connected component with t vertices in G 1 F. Then T connects with S 2 F. Proof. If t = 1, since there is no isolated vertex in SQ n F, the vertex u 1 in T connects with a vertex u 2 in (SQ n G 1 ) F. If t = 2, there exist two vertices u 1 in T and u 2 in SQ n G 1 such that u 1 u 2 / F for ξ(g) = 2n 2 and F 2n 3. If t 3, let xy be an edge in T and A the set of edges that are incident with x or y in G 1. Then A =2(n ) 2 and E(T ) A t. Since G 1 is a subcube SQ n and every vertex in G 1 has the same kth -bit, different vertices in G 1 have different

5 J.-M. Xu et al. / Information Processing Letters 96 (2005) neighbors in SQ n G 1. So there are t other edges between T and SQ n G 1. Thus, there are at least 2(n ) 2 t + t edges between T and SQ n T. Since 2(n ) 2 t + t = 2n + 3t 10 > 2n 3 F and T is a component of G 1 F, there exist two vertices u 1 in T and u 2 in SQ n G 1 such that u 1 u 2 / F. Let G 2 be a subcube SQ n that contains u 2.If G 2 S 2, the claim follows. Assume G 2 S 1 below. Since each of G 1 and G 2 contains at least n edges in F and (2n 3) 2(n ) = 5, F contains at most five edges of E(SQ n G 1 G 2 ).By Lemma 2, u 1 and u 2 have at most two neighbors in common. Note that each vertex of 16 subcube SQ n s has only four neighbors in other subcubes (see Fig. 1). Thus, {u 1,u 2 } connects with at least four neighbors in SQ n G 1 G 2 by six edges, at least one of them, say u 3, satisfies u 1 u 3 / F or u 2 u 3 / F and E(G 3 ) F 1, where G 3 is the (n )-subcube containing u 3. Since E(G 3 ) F 1 n 5, G 3 S 2, and so the claim follows. By the above discussion, we prove that SQ n F is connected, which means λ (SQ n ) 2n 2forn 6, and so the theorem follows. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan, London, [2] A.H. Esfahanian, Generalized measures of fault tolerance with application to n-cube networks, IEEE Trans. Comput. 38 (11) (1989) [3] A.H. Esfahanian, S.L. Hakimi, On computing a conditional edge-connectivity of a graph, Inform. Process. Lett. 27 (1988) [] T.-K. Li, J.J.M. Tan, L.-H. Hsu, T.-Y. Sung, The shuffle-cubes and their generalization, Inform. Process. Lett. 77 (2001) 35 1.