Cycle Embedding on the Edge Fault Star Graphs

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(1)Cycle Embedding on the Edge Fault Star Graphs Tseng-Kuei Li. ∗. Abstract The star graph is one of the famous interconnection networks. Edge fault tolerance is an important issue for a network. And the cycle embedding problem is widely discussed in many researches. In this paper, we show that the n-dimensional star graph can be embedding cycles of even length from 6 to n! when the number of edge fault does not exceed n − 3. Since the graph is bipartite and (n − 1) regular, our result is optimal.. Keywords: star graph, cycle embedding, fault tolerant, pancyclic.. 1. Introduction. Network topology is a crucial factor for a network since it determines the performance of the network. For convenience of discussing their properties, networks are usually represented by graphs. In this paper, a network topology is represented by a simple undirected graph, which is loopless and without multiple edges. For the graph definition and notation we follow [2]. G = (V, E) is a graph if V is a finite set and E is a subset of {(a, b) | a = b ∈ V }, where (a, b) denotes an unordered pair. We call V the vertex set and E the edge set. We say that vertices a and b are adjacent if and only if (a, b) ∈ E. A path is a sequence of adjacent vertices, denoted by v0 , v1 , · · · , vk , in which v0 , v1 , · · · , vk are distinct except that possibly v0 = vk . The length of the path is k. We say that the path is a cycle if v0 = vk . A path (or a cycle) is hamiltonian with respect to a graph G if it crosses all vertices of G. ∗. Correspondence to: Assistant Professor T.K. Li, Department of Computer Science and Information Engineering, Ching Yun Institute of Technology, JungLi, Taiwan 320, R.O.C. e-mail: tealee@ms8.hinet.net.. 1.

(2) To find a cycle of given length in a graph is a cycle embedding problem. Particularly, to find cycles of all lengthes in a graph, i.e., of length from 3 to the number of vertices in the graph, is the pancyclic problem. In this paper, we focus on such a problem on the star graphs proposed by Akers et. al [1]. Since the star graphs are bipartite graphs, they contain no odd length cycles. Moreover, the minimum length of cycles of the star graphs is 6. Jwo et. al [5] showed that an n-dimensional star graph contains cycles of even length from 6 to n!. Thus, we may say that the star graphs are weak even pancyclic. However, there is no result about such a property on the faulty star graphs. Since components in a network would fail sometimes, to study the graphs with faults in the researches is more practice. In this study, we discuss the even cycles embedding problem on the edge fault star graphs. We show that an n-dimensional star graph contains cycles of even length from 6 to n! when the number of edge faults does not exceed n − 3. In the next section, we introduce the definition of star graphs. In Section 3, we show the main result in the finally.. 2. Definition and Basic Properties. The following is the definition of the star graphs. For convenience, we always use v1 v2 · · · vn as the digital representation of vertex v in Sn in this paper. Definition 1 The n-dimensional star graph, denoted by Sn , is the graph (V, E), where V = {v | v is a permutation of 1, 2, · · · , n} and E = {(u, v) | v = ui u2 · · · ui−1 u1 ui+1 · · · un }. By the definition, Sn contains n! vertices and each vertex is of degree (n − 1). For example, 1234 is a vertex in S4 and connects to 2134, 3214, and 4231. We use N (u) to denote the neighborhood of u, e.g., N (1234) = {2134, 3214, 4231}. S1 , S2 , and S3 are a vertex, an edge, and a cycle of length 6, respectively. We show S4 in Figure 1. It is easy to observe that there are four vertex-disjoint S3 ’s embedded in S4 . The following lemma states this property. Lemma 1 There are n vertex-disjoint Sn−1 ’s embedded in Sn for n ≥ 2. 2.

(3) 1234. 4231. 3214. 2134. 3241. 2431. 2314. 3124. 2341. 3421. 1324. 4321. 3412. 2413. 4312. 1432. 4213. 1423. 1342. 4132. 1243. 4123. 3142. 2143. Figure 1: 4-dimensional star graph. Proof. Let i be some integer between 2 and n. Let H i:j = (V i:j , E i:j ) for V i:j = {u ∈ V (Sn ) | ui = j} and E i:j = {(u, v) ∈ E(Sn ) | u, v ∈ V i:j } for 1 ≤ j ≤ n. Clearly, V i:1 , V i:2 , · · · , V i:n is a partition of V (Sn ). It is also not difficult to see that H i:j is isomorphic to Sn−1 . Thus, the lemma follows.. 2. We use Sni:j to denoted the subgraph induced by the vertex set {u | ui = j} for 2 ≤ i ≤ n. Specifically, we use Snj as the abbreviation of Snn:j and call it as the jth (n − 1)-dimensional subgraph of Sn . In our proof, we will use an important property of the edge fault star graphs, called k-edge-fault hamiltonian laceable. A graph G is hamiltonian laceable if and only if for any two vertices in G, there is a hamiltonian path of G between them. Furthermore, A graph G is k-edge-fault hamiltonian laceable if and only if for any two vertices in G, there is a hamiltonian path of G between them, where G contains at most k edge faults. The following lemma is proved by [4] Lemma 2 Sn is (n − 3)-edge-fault hamiltonian laceable. By the lemma, we have the following result: 3.

(4) x. u. n! -2 vertices. v. y. Figure 2: Path of length n! − 3 between u and v. Lemma 3 Let n ≥ 4 and u, v ∈ V (Sn ) with u and v in different partite sets. Then there are two paths of length n! − 3 and n! − 1 between u and v. Proof. Since Sn is hamiltonian laceable, there is a path of length n! − 1 between u and v. Then consider the path of length n! − 3. Let x ∈ N (v) with x = u and y ∈ N (x) with y = v. Let F = {(x, z) | z ∈ N (x) − {v, y}}. Then |F | = n − 3, u and y are in different partite sets, and x only connects to v and y in Sn − F . (See Figure 2.) Since Sn is (n − 3)-edge fault tolerant hamiltonian laceable, there is a hamiltonian path between u and y in Sn − F . Obviously, the vertex sequence v, x, y is in the path. Thus, the path should be of the form u, · · · , v, x, y. Then we have a path of length n! − 3 in Sn between 2. u and v.. Note that the result is only applied to the fault-free star graphs. We will use this result in our main proof frequently.. 3. Main Result. In this section, we propose our main result. Our proof is by induction. For the base case, we enumerate all required cycles in the following lemma. Lemma 4 There is a cycle of each even length from 6 to n! in Sn − F , where F is a set of edge faults with |F | ≤ n − 3 for n = 3 and 4. Proof. For n = 3, S3 is C6 . Since |F | = 0, the lemma follows. 4.

(5) For n = 4, |F | = 1. Since S4 is edge-symmetric, we may assume that F = {(1243, 3241)}. Then we construct cycles of length 6, 8, 10, · · · , 24 in the following: C6 : 1234, 4231, 2431, 1432, 4132, 2134, 1234. C8 : 1234, 4231, 2431, 1432, 3412, 2413, 4213, 3214, 1234. C10 : Replace (4231, 2431) in C6 by 4231, 3241, 2341, 4321, 3421, 2431. C12 : Replace (4231, 2431) in C8 by 4231, 3241, 2341, 4321, 3421, 2431. C14 : Replace (1432, 4132) in C10 by 1432, 3412, 4312, 1342, 3142, 4132. C16 : Replace (1432, 3412) in C12 by 1432, 4132, 3142, 1342, 4312, 3412. C18 : Replace (2134, 1234) in C14 by 2134, 3124, 1324, 2314, 3214, 1234. C20 : Replace (2413, 4213) in C16 by 2413, 1423, 4123, 2143, 1243, 4213. C24 : Replace (3214, 1234) in C20 by 3214, 2314, 1324, 3124, 2134, 1234. C22 : Replace the subpath 4231, · · · , 3412 in C24 by 4231, 2431, 3421, 4321, 2341, 1342, 3142, 4132, 1432, 3412. 2. Hence, the lemma follows.. In fact, the cycles in the above proof are constructed by the method similar to that in Lemma 8 except for C22 and we find C22 by programs however. For the inductive step, we need to establish (n − 2) disjoint paths crossing the given number of subgraphs to guarantee that there is still at least one of these paths crossing no faulty edge when the number of edge faults in Sn does not exceed (n − 3). Moreover, the endpoints of these paths must be in the same subgraph. The following three lemmas discuss how to establish the disjoint paths. Then we use the paths to complete our proof in the latest lemma. For ease of description, we use ui , uji , and (uj )ki to denote the ith digits of vertices u, uj , and (uj )k , respectively. We use P or v0 , P, vk to denote the same 5.

(6) path where the later points the two endpoints of path P . For two paths P = x, P, y and Q = u, Q, v with y and u being adjacent, we use x, P, y, u, Q, v to denote the path concatenating P with Q. Lemma 5 Let x and y be two vertices in Sn for n ≥ 3 with x1 = y1 . Then d(x, y) ≥ 3. Proof. By definition, d(x, y) = 1. Suppose that d(x, y) = 2. Then there is a vertex z adjacent to x and y. Let x = zi z2 · · · zi−1 z1 zi+1 · · · zn and y = zj z2 · · · zj−1 z1 zj+1 · · · zn . Since x = y, i = j and thus, x1 = zi = zj = y1 . We get a contradiction. So d(x, y) ≥ 3. 2 Lemma 6 Let x1 , x2 , · · · xn−2 ∈ V (Sni ) with x11 = x21 = · · · = x1n−2 for n ≥ 3. If uj ∈ , then the 2(n − 2) vertices are distinct. N (xj ) for 1 ≤ j ≤ n − 2 with u11 = u21 = · · · = un−2 1 Proof. Consider xi1 and xi2 for 1 ≤ i1 < i2 ≤ n − 2. Since xi11 = xi12 , by Lemma 5, d(xi1 , xi2 ) ≥ 3. It is clearly that xi1 , xi2 , ui1 , and ui2 are distinct.. 2. Lemma 7 There are (n−2) disjoint paths of length 2m−1 crossing m (n−1)-dimensional subgraphs of Sn such that the endpoints of these paths are in Sni for any 1 ≤ i ≤ n and 3 ≤ m ≤ n for n ≥ 3. n Proof. Without loss of generality, assume that i = m. Let a1 , a2 , · · · , an−2 ∈ V (Sm ). be distinct (n − 2) vertices with aj1 = 1 for all 1 ≤ j ≤ n − 2. Consider the (n − 2) paths: aj = (aj )0 , (aj )1 , (aj )2 , · · · , (aj )2m−1 for all 1 ≤ j ≤ n − 2, where 1. ((aj )2k−2 , (aj )2k−1 ): (aj )2k−1 = (aj )2k−2 and (aj )2k−1 = (aj )2k−2 for 1 ≤ k ≤ m, i.e., 1 1 n n exchanging the first digit and the last digit; and j 2k−1 j 2k−1 and (aj )2k such that (aj )2k−1 = k+1 2. ((aj )2k−1 , (aj )2k ): (aj )2k 1 = (a )l l = (a )1 l. for 1 ≤ k ≤ m − 1, i.e., exchanging the first digit and the lth digit which is equal to k + 1. (See example below the proof.) It is not difficult to check that each path goes though Snm , Sn1 , Sn2 , · · · , Snm−1 and then returns to Snm . Clearly, all the paths are of length 2m − 1 and the endpoints of each path 6.

(7) are in Snm . Since there are exactly two vertices of each path in one (n − 1)-dimensional subgraph, by Lemma 6, the 2(n − 2) vertices of all paths in anyone subgraph are distinct. Hence, all the vertices in these (n − 2) paths are distinct.. 2. For example, n = 6, m = 3, and a1 = 124563, a2 = 125463, a3 = 142563, a4 = 145263. Then we have four disjoint paths as following: 124563, 324561, 234561, 134562, 314562, 214563, 125463, 325461, 235461, 135462, 315462, 215463, 142563, 342561, 243561, 143562, 341562, 241563, 145263, 345261, 245361, 145362, 345162, 245163.. So we may easily find a path of length 2m − 1 crossing m (n − 1)-dimensional subgraphs in Sn − F , where F is a set of edge faults with |F | ≤ n − 3. And such a path uses exactly one edge in each (n − 1)-dimensional subgraph. Note that the two endpoints of the path are not adjacent except for m = 3, 4. Now we show the inductive step: Lemma 8 There are cycles of all even length from 6 to n! in Sn − F , where F is a set of edge faults with |F | ≤ n − 3 for n ≥ 5. Proof. Assume that the statement is true for all 3 ≤ k ≤ n − 1. Since Sn is edgesymmetric, we may assume that there is at least one faulty edge between two (n − 1)dimensional subgraphs, i.e., not in any (n−1)-dimensional subgraph. Thus, each subgraph contains at most (n − 4) faulty edges and is still hamiltonian laceable. Moreover, since |F | ≤ n − 3, there are at least four (n − 1)-dimensional subgraphs containing no faulty edge. Without loss of generality, assume that Sn1 and Snn contain no faulty edge. By the hypothesis, we have cycles of each even length from 6 to (n − 1)! in Sn1 − F , i.e., we have cycles of each even length from 6 to (n − 1)! in Sn − F .. 7.

(8) Now we construct cycles of even length from (n − 1)! + 2 to (n − 1)! + 2(n − 2). Suppose that l = (n − 1)! − 4 + 2m for some 3 ≤ m ≤ n. By Lemma 7, there is a path P of length 2m − 1 crossing m (n − 1)-dimensional subgraphs of Sn − F with two endpoints in Sn1 . Assume that the two endpoints are u and v. By Lemma 3, there is a path Q of length (n − 1)! − 3 in Sn1 between u and v. Then u, P, v, Q, u forms a cycle of length (n − 1)! − 4 + 2m in Sn − F . Then we construct cycles of even length from (n − 1)! + 2(n − 1) to n!. Let l = k((n − 1)! − 2) + 2n + 2h for some 1 ≤ k ≤ n − 1 and 0 ≤ h ≤. (n−1)!−2 2. − 1 such that. (n − 1)! + 2(n − 1) ≤ l ≤ n!. Then l may be any even integer between (n − 1)! + 2(n − 1) and n!. In the following, we construct cycles of length l in Sn − F . By Lemma 7, there is a path P = u1 , v 2 , u2 , v 3 , · · · , v n , un , v 1 in Sn − F such that ui , v i ∈ V (Sni ). Consider two cases: Case 1: h = 1. If k ≥ 2, replace each edge (v i , ui ) in P by a hamiltonian path of Sni −F for 2 ≤ i ≤ k. If h ≥ 2, by the symmetric property of Sn−1 and hypothesis, we may find a cycle of length 2h+2 crossing the edge (v n , un ). So we may replace the edge (v n , un ) in P by a path of length 2h+1. Now the length of P is (k −1)((n−1)!−2)+2n−1+2h = l − ((n − 1)! − 1). By Lemma 3, there is a path Q of length (n − 1)! − 1 in Sn1 between v 1 and u1 . Thus, u1 , P, v 1 , Q, u1 forms a cycle of length l in Sn − F . Case 2: h = 1. If k ≥ 2, replace each edge (v i , ui ) in P by a hamiltonian path of Sni − F for 2 ≤ i ≤ k. Since there is no fault in Snn , there is a cycle of length 6 crossing the edge (v n , un ). So we may replace the edge (v n , un ) in P by a path of length 5. Now the length of P is (k − 1)((n − 1)! − 2) + 2n − 1 + 2h = l − ((n − 1)! − 3). By Lemma 3, there is a path Q of length (n − 1)! − 3 in Sn1 between v 1 and u1 . Thus, u1 , P, v 1 , Q, u1 forms a cycle of length l in Sn − F . 2. Hence, the lemma follows. So we have the following result: 8.

(9) Theorem 1 There are cycles of all even length from 6 to n! in Sn − F , where F is a set of edge faults with |F | ≤ n − 3 for n ≥ 3.. References [1] S.B. Akers and B. Krishnamurthy, “A group-theoretic model for symmetric interconnection networks,” IEEE Trans. Computers, vol. 38, 1989, pp. 555-566. [2] J.A. Bondy and U.S.R. Murty, Graph Theory with Applications, North-Holland, New York, 1980. [3] F. Harary and M. Lewinter, “Hypercubes and other revursively defined hamilton laceable graphs,” Congressus Numerantium 60, 1987, pp. 81-84. [4] S.Y. Hsieh, G.H. Chen, and C.W. Ho, “Longest fault-free paths in star graphs with edge faults,” IEEE Trans. Computers, vol. 50, no. 9, pp. 960–971, Sep. 2001. [5] J.S. Jwo, S. Lakshmivarahan, and S.K. Dhall, “Embedding of cycles and grids in star graphs,” J. Circuits Syst. Comput., vol 1, 1991, pp. 43-74. [6] S. Latifi, S. Zheng, and N. Bagherzadeh, “Optimal ring embedding in hyperucbes with faulty links,” Fault-Tolerant Computing Symp., 1992, pp. 178-184. [7] Y.C. Tseng, S.H. Chang, and J.P. Sheu, “Fault-tolerant ring embedding in star graph with both link and node failures,” IEEE Trans. Parallel and Distributed Systems, vol. 8, 1997, pp. 1185-1195.. 9.

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