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Fault hamiltonicity of augmented cubes
Parallel Computing 31 (2005) 131–145 www.elsevier.com/locate/parco
Fault hamiltonicity of augmented cubes Hong-Chun Hsu a
c
a,*
, Liang-Chih Chiang b, Jimmy J.M. Tan b, Lih-Hsing Hsu c
Department of Computer Science and Information Management, Providence University, Shalu, Taichung, Taiwan 43301, ROC b Department of Computer and Information Science, National Chiao Tung University, Hsinchu, Taiwan 300, ROC Department of Computer and Information Engineering, Ta Hwa Institute of Technology, Hsinchu, Taiwan 300, ROC Received 11 December 2003; revised 25 September 2004; accepted 9 October 2004 Available online 23 December 2004
Abstract In this paper, we consider the fault hamiltonicity and the fault hamiltonian connectivity of the augmented cubes AQn. Assume that F V(AQn) [ E(AQn) and n P 4. We prove that AQn F is hamiltonian if jFj 6 2n 3 and that AQn F is hamiltonian connected if jFj 6 2n 4. Moreover, these bounds are tight. Ó 2004 Elsevier B.V. All rights reserved. Keywords: Fault-tolerant; Hamiltonian; Hamiltonian connected; Augmented cubes
1. Introduction Network topology is a crucial factor for the interconnection networks since it determines the performance of the networks. Many interconnection network topologies have been proposed in the literature for the purpose of connecting
*
Corresponding author. E-mail address: [email protected] (H.-C. Hsu).
0167-8191/$ - see front matter Ó 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.parco.2004.10.002
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thousands of processing elements. Network topology is usually represented by a graph where vertex represent processors and edges represent links between processors. For the graph definition and notation, we follow [3]. G = (V, E) is a graph if V is a finite set and E is a subset of {(u, v)j(u, v) is an unordered pair of V}. We say that V is the vertex set and E is the edge set. For any vertex x of V, degG(x) denotes its degree in G. We use d(G) to denote min{degG(x)jx 2 V(G)}. Two vertices u and v are adjacent if (u, v) 2 E. A path, denoted by hv0, v1, v2, . . . , vki, is a sequence of distinct vertices where vi and vi+1 are adjacent for all i, 0 6 i 6 k 1. The length of a path P, l(P), is the number of edges in P. We also write the path hv0, v1, v2, . . . , vki as hv0, P1, vi, vi+1, . . . , vj, P2, vt, . . . , vki, where P1 is the subpath hv0, v1, . . . , vii and P2 is the subpath hvj, vj+1, . . . , vti. It is possible to write a path as hv0, v1, P, v1, v2, . . . , vki if l(P) is 0. Sometimes, a path can also be represented by hv0, v1, . . . , vi, e, vi+1, . . . , vni to emphasize that e is the edge (vi, vi+1). We use d(u, v) to denote the distance between u and v, i.e., the length of a shortest path joining u and v. A path is a hamiltonian path if its vertices are distinct and span V. A cycle is a path with at least three vertices such that the first vertex is the same as the last vertex. There are a lot of interconnection network topologies proposed in literature [7]. Among these topologies, the n-dimensional hypercube, denoted by Qn, is a popular one. Several variations of the hypercubes have been investigated to improve the efficiency of the hypercubes [1,2,4,5,10]. The augmented cube AQn, recently proposed by Choudum and Sunitha [6], is one of such variations. For any positive integer n, AQn is a vertex transitive, (2n 1)-regular, and (2n 1)-connected graph with 2n vertices. In this paper, we consider two important parameters of the augmented cubes, fault hamiltonicity and fault hamiltonian connectivity. These two parameters for interconnection networks are proposed by Huang et. al. [8]. A hamiltonian cycle of G is a cycle that traverses every vertex of G exactly once. A graph is hamiltonian if it has a hamiltonian cycle. A hamiltonian graph G is k-fault hamiltonian if G F remains hamiltonian for every F V(G) [ E(G) with jFj 6 k. The fault hamiltonicity, Hf(G), is defined to be the maximum integer k such that G is k-fault hamiltonian if G is hamiltonian and is undefined otherwise. Clearly, Hf (G) 6 d(G) 2 if Hf (G) is defined. In this paper, we prove that Hf (AQn) = 2n 3 if n 62 {1, 3}, Hf(AQ3) = 2, and Hf(AQ1) is undefined. Since d(AQn) = 2n 1, the fault hamiltonicity of the augmented cube AQn is exactly 2n 3 for n P 4. Therefore, we obtain the optimal result. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two vertices of G. All hamiltonian connected graphs except the complete graphs K1 and K2 are hamiltonian. A graph G is k-fault hamiltonian connected if G F remains hamiltonian connected for every F V(G) [ E(G) with jFj 6 k. The fault hamiltonian connectivity, H kf ðGÞ, is defined to be the maximum integer k such that G is k-fault hamiltonian connected if G is hamiltonian connected and is undefined otherwise. Clearly, H kf ðGÞ 6 dðGÞ 3 if H kf ðGÞ is defined and jV(G)j P 4. In this paper, we prove that H kf f ðAQn Þ ¼ 2n 4 if n 62 1,3, H kf ðAQ3 Þ ¼ 1, and H kf ðAQ1 Þ ¼ 0. Again, this result concerning the fault hamiltonian connectivity of the augmented cube AQn is optimal because d(AQn) = 2n 1.
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In the following section, we give the definition of the augmented cubes and discuss some properties of AQn. Then we present the fault hamiltonian properties of AQ3 and AQ4 in Section 3. In the final section, we discuss Hf (AQn) and H kf ðAQn Þ. 2. Augmented cubes Let n P 1 be an integer. The graph of the n-dimensional augmented cube [6], denoted by AQn has 2n vertices, each labeled by an n-bit binary string V(AQn) = {u1u2 . . . unjui 2 {0, 1}}. AQ1 is the graph K2 with vertex set {0, 1}. For n P 2, AQn can be recursively constructed by two copies of AQn1, denoted by AQ0n1 and AQ1n1 and by adding 2n edges between AQ0n1 and AQ1n1 as follows: Let V ðAQ0n1 Þ ¼ fð0u2 u3 . . . un Þ j ui ¼ 0 or 1 for 2 6 i 6 n} and V ðAQ1n1 Þ ¼ fð1v2 v3 . . . vn Þ j vi ¼ 0 or 1 for 2 6 i 6 n}. A vertex u = (0u2u3 . . . un) of AQ0nl is joined to a vertex v = (1v2v3 . . . vn) of AQ1n1 if and only if either (i) ui = vi for 2 6 i 6 n; in this case, (u, v) is called a hypercube edge and we set v = uh, or (ii) ui ¼ vi for 2 6 i 6 n; in this case, (u, v) is called a hypercube edge and we set v = uc . The augmented cubes AQ1, AQ2, and AQ3 are illustrated in Fig. 1. It is proved in [6] that AQn is a vertex transitive, (2n 1)-regular, and (2n 1)-connected graph with 2n vertices for any positive integer n. Let Ehn ¼ fðu,uh Þ j u 2 V ðAQ0n1 Þg and Ecn ¼ fðu,uc Þ j u 2 V ðAQ0n1 Þg. Obviously, Ehn and Ecn are two perfect matchings between the vertices of AQ0n1 and AQ1n1 . Hence, j Ehn j¼j Ecn j¼ 2n1 . Let C n ¼ fðu,vÞ j u ¼ 0u1 . . . un and v ¼ 0 u1 . . . un g. In other words, C n is the set of all complement 0 edges in AQn1 . The following lemmas are derived directly from the definition. Lemma 1. Let x and y be any two vertices in AQ0n1 with n P 4. Suppose that ðx,yÞ 62 C n . Then xh, xc, yh and yc are all distinct. Lemma 2. Assume that ðu,vÞ 2 AQ0n1 . Then ðuh ,vh Þ 2 AQ1n1 and ðuc ,vc Þ 2 AQ1n1 .
Fig. 1. The argumented cubes: (a) AQ1; (b) AQ2 and (c) AQ3.
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Lemma 3. Assume there is a vertex u in A0n1 . Let A ¼ fv j ðu,vÞ 2 EðAQ0n1 Þg and B = {vh, vcjv 2 A}. Then jAj = 2n 3 and jBj = 4n 8. To discuss the fault hamiltonicity and the fault hamiltonian connectivity of AQn, we need to introduce the following two terms for graphs. A graph G has property 2H if it satisfies the following conditions: Let {w, x} and {y, z} be two pairs of four distinct vertices of G. There exist two disjoint paths P1 and P2 of G such that (1) P1 joins w to x, (2) P2 joins y to z, and (3) P1 [ P2 span G. A hamiltonian connected graph G = (V, E) is said to be 1-vertex hamiltonian connected if G {u} is still hamiltonian connected after removing any vertex u 2 V. Thus, any k-fault hamiltonian connected graph is 1-vertex hamiltonian connected for k P 1. Furthermore, any 1-vertex hamiltonian connected graph is hamiltonian connected. Let F V(AQn) [ E(AQn) be any faulty set of AQn. An edge (u, v) is called F -fault free if (u, v) 62 F, u 62 F, and v 62 F; otherwise it is called F-fault. Let H = (V 0 , E 0 ) be a subgraph of AQn. We use F(H) to denote the set (V 0 [ E 0 ) \ F. We need the following lemmas. The proof of these lemmas are put in Appendix. Lemma 4. Hf(AQ3) = 2. Lemma 5. H kf ðAQ3 Þ ¼ 1. Lemma 6. Hf(AQ4) = 5. and H kf ðAQ4 Þ ¼ 4. Moreover, AQ4 has property 2H. Lemma 7. Let n P 4. Suppose that AQn1 is 1-vertex hamiltonian connected and has property 2H. Then AQn also has property 2H. Lemma 8. Let n P 5. Suppose that AQn1 is (2n 5)-fault hamiltonian, (2n 6)fault hamiltonian connected, and has property 2H. Then AQn is (2n 3)-fault hamiltonian. Lemma 9. Let n P 5. Suppose that AQn1 is (2n 5)-fault hamiltonian, (2n 6)fault hamiltonian connected, and has property 2H. Then AQn is (2n 4)-fault hamiltonian connected.
3. Fault hamiltonian and fault hamiltonian connectivity of the augmented cubes Theorem 10. Let n be a positive integers with n P 4. Then AQn is (2n 3)-fault hamiltonian, (2n 4)-fault hamiltonian connected, and has property 2H. Proof. We prove this theorem by induction on n. The induction base is n = 4. By Lemmas 7 and 6, the theorem holds for n = 4. With Lemmas 7–9, we prove the induction step. h
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With Lemmas 4 and 5, we have the following corollary. Corollary 11. Let n be a positive integer. (1) Hf(AQn) = 2n 3 and H kf ðAQn Þ ¼ 2n 4 if n 62 {1, 3}; (2) Hf(AQ1) is undefined and H kf ðAQ1 Þ ¼ 0; and (3) Hf(AQ3) = 2 and H kf ðAQ3 Þ ¼ 1.
Appendix A Proof of Lemma 4. For the proof, we have the following observation. Let C1 and C2 be any two cycles of K4. Then E(C1) \ E(C2) 5 ;. Now, we prove that Hf (AQ3) < 3. Let F = {001, 010, 100}. Then AQ3 F is the graph in Fig. 2(a). Obviously, AQ3 F is non-hamiltonian. Hence, Hf(AQ3) < 3. Then we prove that Hf (AQ3) P 2. Let F V(AQ3) [ E(AQ3) with jFj 6 2. Without loss of generality, we assume that j F ðAQ02 Þ jPj F ðAQ12 Þ j. Case 1: j F ðAQ02 Þ j¼ 2. Since AQ2 is a complete graph K4, there is a hamiltonian path P0 of AQ02 F joining some u and v. Obviously, there is a hamiltonian path P1 of AQ12 joining uh and vh. Hence, hu, uh, P1 vh, v, P0, ui forms a hamiltonian cycle of AQ3 F. Case 2: j F ðAQ02 Þ j¼ 1. Suppose that j F ðAQ12 Þ j¼ 1. Hence, there is a hamiltonian cycle C0 in AQ02 F ðAQ02 Þ there is a hamiltonian cycle C1 in AQ12 F ðAQ12 Þ. By Lemma 3, there exists an edge (u, v) in C0 such that both (u, uh) and (v, vh) are F-fault free and the edge (uh, vh) is in C1. Then we may write C0 as hu, P1, v, ui and write C1 as huh, P2, vh, uhi. Then hu, uh, P2, vh, v, P1, ui forms a hamiltonian cycle of AQ3 F. Suppose that j F ðAQ12 Þ j¼ 0. Hence, AQ12 is hamiltonian connected. Let C0 be the hamiltonian cycle of AQ02 F ðAQ02 Þ. Then for any edge (u, v) in C0, there exists a hamiltonian path P2 of AQ12 joining uh and vh. Then hu, uh, P2, vh, v, P1, ui forms a hamiltonian cycle of AQ3 F.
Fig. 2. The graphs of AQ3 F.
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Case 3: j F ðAQ02 Þ j¼ 0. Hence, both AQ02 and AQ12 are hamiltonian connected. Then we choose any two distinct F-fault free edges in Eh [ Ec say (u, v) and (w, x), which u,w 2 AQ02 and v,x 2 AQ12 . Let P1 be a hamiltonian path of AQ02 joining u and w, and P2 be a hamiltonian path of AQ12 joining v and x. Thus, hu, v, P2, x, w, P1, ui forms a hamiltonian cycle of AQ3 F. h Proof of Lemma 5. Now, we prove that H kf ðAQ3 Þ < 2. Let F = {001, 111}. Then AQ3 F is the graph in Fig. 2(b). Obviously, there is no hamiltonian path of AQ3 F joining 010 to 100. Hence, H kf ðAQ3 Þ < 2. Then we prove that H kf ðAQ3 Þ ¼ 1. Let F V(AQ3) [ E(AQ3) with jFj 6 1. By the symmetric property of AQ3, we may assume that F is the following three cases: (1)F = {000}, (2) F = {(000, 100)}, and (3) F = {(000, 111)}. The corresponding hamiltonian paths are listed below. Hence, AQ3 F is hamiltonian connected. h
F = {000}
F = {(000, 001)}
F = {(000, 111)}
h001, 010, 110, 100, 101, 111, 011i h001, 011, 111, 101, 100, 110, 010i h001, 010, 011, 111, 101, 110, 100i h001, 011, 010, 110, 100, 111, 101i h001, 010, 011, 111, 101, 100, 110i h001, 011, 010, 110, 100, 101, 111i h010, 001, 101, 100, 110, 111, 011i h010, 001, 011, 111, 101, 110, 100i h010, 011, 001, 110, 100, 111, 101i h010, 001, 011, 111, 101, 100, 110i h010, 011, 001, 110, 100, 101, 111i h011, 010, 001, 110, 101, 111, 100i h011, 010, 001, 110, 100, 111, 101i h011, 010, 001, 101, 111, 100, 110i h011, 010, 001, 110, 100, 101, 111i h100, 111, 110, 010, 011, 001, 101i h100, 111, 101, 001, 011, 010, 110i h100, 110, 101, 001, 010, 011, 111i h101, 100, 111, 011, 001, 010, 110i h101, 100, 110, 010, 001, 011, 111i h110, 100, 101, 001, 010, 011, 111i
h000, 010, 011, 100, 101, 111, 110, 001i h000, 011, 001, 101, 100, 111, 110, 010i h000, 010, 001, 101, 100, 111, 110, 011i h000, 010, 011, 001, 101, 111, 110, 100i h000, 010, 011, 001, 110, 111, 100, 101i h000, 010, 011, 001, 101, 100, 111, 110i h000, 010, 011, 001, 101, 100, 110, 111i h001, 010, 000, 100, 110, 101, 111, 011i h001, 011, 000, 111, 101, 100, 110, 010i h001, 010, 011, 000, 111, 101, 110, 100i h001, 011, 000, 010, 110, 100, 111, 101i h001, 010, 000, 011, 111, 101, 100, 110i h001, 011, 000, 010, 110, 100, 101, 111i h010, 001, 101, 000, 100, 110, 111, 011i h010, 001, 011, 000, 111, 101, 110, 100i h010, 000, 011, 001, 110, 100, 111, 101i h010, 001, 011, 000, 111, 101, 100, 110i h010, 000, 011, 001, 110, 100, 101, 111i h011, 000, 010, 001, 110, 101, 111, 100i h011, 000, 010, 001, 110, 100, 111, 101i h011, 000, 010, 001, 101, 111, 100, 110i h011, 000, 010, 001, 110, 100, 101, 111i h100, 111, 110, 010, 000, 011, 001, 101i h100, 111, 101, 001, 011, 000, 010, 110i h100, 110, 101, 001, 010, 000, 011, 111i h101, 100, 111, 000, 011, 001, 010, 110i h110, 100, 101, 001, 010, 000, 011, 111i
h000, 010, 011, 100, 101, 111, 110, 001i h000, 011, 001, 101, 100, 111, 110, 010i h000, 010, 001, 101, 100, 111, 110, 011i h000, 010, 011, 001, 101, 111, 110, 100i h000, 010, 011, 001, 110, 111, 100, 101i h000, 010, 011, 001, 101, 100, 111, 110i h000, 010, 011, 001, 101, 100, 110, 111i h001, 010, 000, 100, 110, 101, 111, 011i h001, 011, 000, 100, 101, 111, 110, 010i h001, 010, 000, 011, 111, 101, 110, 100i h001, 011, 000, 010, 110, 100, 111, 101i h001, 010, 000, 011, 111, 101, 100, 110i h001, 011, 000, 010, 110, 100, 101, 111i h010, 001, 101, 000, 100, 110, 111, 011i h010, 001, 000, 011, 111, 101, 110, 100i h010, 000, 011, 001, 110, 100, 111, 101i h010, 001, 000, 011, 111, 101, 100, 110i h010, 000, 011, 001, 110, 100, 101, 111i h011, 000, 010, 001, 110, 101, 111, 100i h011, 000, 010, 001, 110, 100, 111, 101i h011, 000, 010, 001, 101, 111, 100, 110i h011, 000, 010, 001, 110, 100, 101, 111i h100, 111, 110, 010, 000, 011, 001, 101i h100, 111, 101, 001, 011, 000, 010, 110i h100, 110, 101, 001, 010, 000, 011, 111i h101, 111, 100, 000, 011, 001, 010, 110i h110, 100, 101, 001, 010, 000, 011, 111i
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Proof of Lemma 6. To prove this lemma is very tedious. With long and detail discussion, we have completed theoretical proof for this lemma. Nevertheless, we do not present it in this paper for reducing complexity. However, we can also verify this lemma using computer. Our computer program was coded by depth first search method (a kind of brute force method). h Proof of Lemma 7. Let {w, x} and {y, z} be two pairs of four distinct vertices of AQn. We are going to construct two disjoint spanning paths, R1 and R1, of AQn such that R1 joins w and x and R2 joins y and z. We consider the following four cases. Case 1: {w, x, y, z} are all in AQ0n1 . By the assumption of this lemma, AQn1 has property 2H. There exist two disjoint paths hw, P1, xi and hy, P2, zi spanning AQ0n1 . Without loss of generality, we assume that l(P1) 6 l(P2). Suppose that n = 3. Then AQ02 is isomorphic to the complete graph K4. Obviously, these two paths of AQ02 are hw, xi and hy, zi. Since AQ12 is K4, there is a hamiltonian path hwh, P3, xhi of AQ12 . Thus, hw, wh, P3, xh, xi and hy, zi form the desired disjoint spanning paths of AQ3. Suppose that n P 4. Then l(P2) P 4. Hence, P2 can be written as hy, P4, u, v, zi. By Lemma 2, ðuh ,vh Þ 2 EðAQ1n1 Þ. By the assumption of this lemma, AQ1n1 is 1-vertex hamiltonian connected. Hence, AQ1n1 is hamiltonian connected. Thus, there exists a hamiltonian path huh, P3, vhiin AQ1n1 . Then R1 = hw, P1, xi and R2 = hy, P4, u, uh, P3, vh,v, zi form the desired disjoint spanning paths of AQn. See Fig. 3(a) for an illustration. Case 2: {w, x, y} are in AQ0n1 and z is in AQ1n1 . Let y* be a vertex in {yh, yc} such that y* 5 z. By the assumption of this lemma, there exists a hamiltonian path hy*, P2, zi in AQ1n1 and there exists a hamiltonian path hw*, P1, xi in AQ0n1 fyg. Then R1 = hw, P1, xi and R2 = hy, y*, P2, zi form the desired disjoint spanning paths of AQn. See Fig. 3(b) for an illustration.
Fig. 3. Illustrations for Lemma 7.
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Case 3: {w, x} are in AQ0n1 and {y, z} are in AQ1n1 . By the assumption of this lemma, there is a hamiltonian path R1 = hw, P1, xiin AQ0n1 and there is a hamiltonian R2 = hy, P2, zi in AQ1n1 . Obviously, R1 and R2 form the desired disjoint spanning paths of AQn. See Fig. 3(c) for an illustration. Case 4: {w, y} are in AQ0n1 and {x, z} are in AQ1n1 . Subcase 4.1: {(w, x),(y, z)} \ E(AQn) 5 ;. Without loss of generality, we assume that (w, x) 2 E(AQn). Since n P 3, there exists a vertex u in AQ0n1 such that u 5 y and uh 5 z. By the assumption of this lemma, there exists a hamiltonian path hy, P1, ui of AQ0n1 fwg there exists a hamiltonian path huh, P2, zi of AQ1n1 fxg. Obviously, R1 = hw, xi and R2 = hy, P1, u,uh, P2, zi form the desired disjoint spanning paths of AQn. See Fig. 3(d) for an illustration. Subcase 4.2: {(w, x),(y, z)} \ E(AQn) = ;. Then one of wh and wc is not z, say w*, and one of zh and zc is not w, say z*. By the assumption of this lemma, there exists a hamiltonian path hw*, P1, xi in AQ1n1 fzg and there exists a hamiltonian path hy, P2, z*i in AQ0n1 fwg. Thus, R1 = hw, w*, P1, xi and R2 = hy, P2, z*, zi form the desired disjoint spanning paths of AQn. See Fig. 3(e) for an illustration. Hence, the lemma is proved. h Proof of Lemma 8. Let F be any subset of V(AQn) [ E(AQn) with jFj 6 2n 3. Without loss of generality, we may assume that j F ðAQ0n1 Þ jPj F ðAQ1n1 Þ j. We are going to construct a hamiltonian cycle of AQn F depending on the following cases. Case 1: j F ðAQ0n1 Þ j6 2n 6. In this case, AQ0n1 F ðAQ0n1 Þ and AQ1n1 F ðAQ1n1 Þ are hamiltonian connected. Since j Ehu [ Ecn j j F jP 2n ð2n 3Þ P 25, there are two distinct F-fault free edges (u, v) and (x, y) such that u and x are in AQ0n1 F ðAQ0n1 Þ, and v and y are in AQ1n1 F ðAQ1n1 Þ. By the assumption of this lemma, there exist hamiltonian paths hu, P1, xi in AQ0n1 F ðAQ0n1 Þ and hv, P2, yi in AQ1n1 F ðAQ1n1 Þ. Then hu, v, P2, y, x, P1, ui forms a hamiltonian cycle of AQn F. See Fig. 4(a) for an illustration. Case 2: j F ðAQ0n1 Þ j¼ 2n 5. Hence, j F F ðAQ0n1 Þ j6 2. Since AQn1 is (2n5)fault hamiltonian, there is a hamiltonian cycle C in AQ0n1 F ðAQ0n1 Þ. Since n1 lðCÞ=2 P 2 2nþ5 > 5, there exists an edge (u, v) in C such that (u, uh) and 2 h (v, v ) are F-fault free. We can write C as hu, P1, v, ui. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path huh, P2, vhi of AQ1n1 F ðAQ1n1 Þ. Then hu, uh, P2, vh, v, P1, ui is a hamiltonian cycle of AQn F. See Fig. 4(b) for an illustration. Case 3: j F ðAQ0n1 Þ j¼ 2n 4. Suppose that there exists a hamiltonian cycle C of AQ0n1 F ðAQ0n1 Þ. As in Case 2, there exists an edge (u, v) in C such that (u, uh) and (v, vh) are F-fault free. We can write C as hu, P1, v, ui. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path P2 of
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Fig. 4. Illustrations for Lemma 8.
AQ1n1 F ðAQ1n1 Þ. joining uh and vh. Thus, hu, uh, P2, vh, v, P1, ui forms a hamiltonian cycle of AQn F. Suppose that there is no hamiltonian cycle of AQ0n1 F ðAQ0n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle Ci of AQ0n1 ðF ðAQ0n1 Þ ffi gÞ for every fi 2 F ðAQ0n1 Þ, We can write Ci as hui, Pi, vi, fi, uii. Since j F ðAQ0n1 Þ jP 6 and j F F ðAQ0n1 Þ j6 1, there exists an index i such that ðui ,uhi Þ and ðvi ,vhi Þ are F-fault free. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path Q1 in AQ1n1 F ðAQ1n1 Þ joining uhi and vhi . Thus, hui ,uhi ,Q1 ,vhi ,vi ,P i ,ui i forms a hamiltonian cycle in AQn F. See Fig. 4(c) for an illustration. Case 4: j F ðAQ0n1 j¼ 2n 3. Thus, j F F ðAQ0n1 Þ j¼ 0. Let f1 and f2 be any two elements in F ðAQ0n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle C ¼ hu,f10 ,v,P 1 ,w,f20 ,x,P 2 ,ui in AQ0n1 ðF ðAQ0n1 Þ ff1 ,f2 gÞ where f10 ¼ f1 if f1 is on C or f10 is an arbitrary edge of C otherwise and f20 ¼ f2 if f2 is on C or f20 is an arbitrary edge of C otherwise. Without loss of generality, we assume that l(P2) 6 l(P1). Subcase 4.1: l(P2) P 1. Obviously, uh, vh, wh and xh are distinct. Since AQn1 has property 2H, there exist two disjoint spanning paths huh, P3, vhi and hwh, P4, xhi in AQ1n1 . Thus, hu, uh, P3, vh, v, P1, w, wh, P4, xh, x, P2, ui is a hamiltonian cycle of AQn F. See Fig. 4(d) for an illustration. Subcase 4.2: l(P2) = 0. Then u = x. Moreover w, u, and v are distinct. Obviously, one of the following cases holds: (1) {wh, wc, uh, uc h c , v , v } are all distinct, (2) (wh = vc and wc = vh), (3) h (w = uc and wc = uh), and (4) (vh = uc and vc = uh). Suppose that (1) {wh, wc, uh, uc, vh, vc} are all distinct or (2) (wh = vc and wc = vh). Since AQn 1 has property 2H, there exist two disjoint spanning paths hwh, P5, uhi and huc, P6, vhi in AQ1n1 . Then hu, uc, P6, vh, v, P1, w, wh, P5, uh, ui forms a hamiltonian cycle of AQn F. See Fig. 4(e) for an illustration. Suppose that (3) (wh = uc and wc = uh) or (4) (vh = uc and vc = uh). Without loss of generality, we can assume that wh = uc and wc = uh. Since AQn1 is (2n 6)-fault
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hamiltonian connected, there is a hamiltonian path huh, P7, vhi in AQ1n1 fwh g. Hence, hu, uh, P7, vh, v, P1, w, wh, ui forms a hamiltonian cycle of AQn F. See Fig. 4(f) for an illustration. This completes the proof of the lemma. h Proof of Lemma 9. Let F be any subset of V(AQn) [ E(AQn) with jFj < 2n 4. Without loss of generality, we assume that j F ðAQ0n1 Þ jPj F ðAQ1n1 Þ j. We want to construct a hamiltonian path joining u and v in AQn F for any two different vertices u and v 2 V(AQn) F. Case 1: jF(AQ0n1)j 6 2n 6. Since AQn1 is (2n 6)-fault hamiltonian connected, both AQ0n1 F ðAQ0n1 Þ and AQ1n1 F ðAQ1n1 Þ are hamiltonian connected. Subcase 1.1: u,v 2 V ðAQ0n1 Þ or u,v 2 V ðAQ1n1 Þ. Without loss of generality, we assume that u, v 2 V(AQ0n1). Since AQ0n1 F ðAQ0n1 Þ is hamiltonian connected, there exists a hamiltonian path P of AQ0n1 F ðAQ0n1 Þ joining u and v. We claim that there exists an edge (w, x) on P such that both (w, wh) and (x, xh) are F-fault free. Suppose that there is no such edge. Then at least [l(P)/2] edges in Ehn [ Ecn are F-fault. Since lðP Þ P 2n1 j F ðAQ0n1 Þ j , j F j¼j F ðAQ0n1 Þ j þ ð2n1 j F ðAQ0n1 Þ jÞ=2 P 2n2 þ j F ðAQ0n1 Þ j =2 > 2n 4. We have a contradiction. Hence, such edge exists as claimed. We can write P as hu, P1, w, x, P2, vi. Since AQ1n1 F ðAQ1n1 Þ is also hamiltonian connected, there is a hamiltonian path hwh, P3, xhi in AQ1n1 F ðAQ1n1 Þ. Thus, hu, P1, w, wh, P3, xh, x, P2, vi is a hamiltonian path of AQn F joining u and v. See Fig. 5(a) for an illustration. Subcase 1.2: u 2 AQ0n1 and v 2 AQ0n1 . Since jFj 6 2n 4 < 2n1, there exists an F-fault free edge (w, wh) such that w 2 V ðAQ0n1 Þ, w 5 u, and wh 5 v. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path hu, P1, wi in AQ0n1 ðAQ0n1 Þ and there is a hamiltonian path hwh, P2, vi in AQ1n1 F ðAQ1n1 Þ. Then hu, P1, w, wh, P2, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(b) for an illustration. Case 2: j F ðAQ0n1 Þ j¼ 2n 5. Hence, j F F ðAQ0n1 Þ j6 1. Subcase 2.1: u,v 2 V ðAQ0n1 Þ. Suppose that there exists a hamiltonian path P of AQ0n1 F ðAQ0n1 Þ joining u and v. Since ðlðpÞ j F ðAQ0n1 Þ jÞ=2 P 2n 4,, there exists an edge (w, x) in P such that (w, wh) and (x, xh) are F-fault free. We can write P as hu, P1, w, x, P2, vi. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P3 of AQ1n1 F ðAQ1n1 Þ joining wh and xh. Thus, hu, P1, w, wh, P3, xh,
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Fig. 5. Illustrations for Lemma 9.
x, P2, vi forms a hamiltonian path of AQn F joining u and v. Suppose that there is no hamiltonian path of AQ0n1 F ðAQ0n1 Þ joining u and v. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path Qi of AQ0n1 ðF ðAQ0n1 Þ ffi gÞ joining u and v for every fi 2 F ðAQ0n1 Þ. We can write Qi as hu,P 1i ,wi ,fi ,xj ,P 2i ,vi. Since j F ðAQ0n1 Þ j¼ 2n 5 P 5 and j F F ðAQ0n1 Þ j6 1, there exists an index i such that ðwi ,whi Þ and ðxi ,xhi Þ are F-fault free. Since AQ1n1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path R1 of AQ1n1 F ðAQ1n1 Þ joining whi and xhi . Then hu,P 1i ,wi ,whi ,R1 ,xhi , xi ,P 2i ,vi forms a hamiltonian path of AQnF joining u and v. See Fig. 5(c) for an illustration.
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Subcase 2.2: u 2 V ðAQ0n1 Þ and v 2 V ðAQ1n1 Þ.Let A = {xj(x, v) 2 E(AQn) and x 2 V ðAQ1n1 Þg. Obviously, jAj = 2n 3.Let B = {yhjy 2 A} [ {ycjy 2 A}. By Lemma 3, jBj = 4n 8.Since jFj 6 2n 4, there exists a y 2 A such that either (y, yh) or (y, yc), say (y, y*), is F-fault free and w* 5 u. Suppose that there exists a hamiltonian path P of AQ0n1 F ðAQ0n1 Þ joining u and y*. Since ðlðpÞ j F ðAQ0n1 Þ jÞ=2 P 2n 4, there exists an edge (w, x) in P such that (w, wh) and (x, xh) are F-fault free and wh, xh and v are distinct. We can write P as hu, P1, w, x, P2, y*i. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P3 of AQ1n1 F ðAQ1n1 Þ fv,yg joining wh and xh. Thus, hu, P1, w, wh, P3, xh, x, P2, y*, y, vi forms a hamiltonian path of AQn F joining u and v. Suppose that there is no hamiltonian path of AQ0n1 F ðAQ0n1 Þ joining u and y*. Since AQn1 is (2n 6)fault hamiltonian connected, there exists a hamiltonian path Qi of AQ0n1 ðF ðAQ0n1 Þ ffi gÞ joining u and y* for every fi 2 F ðAQ0n1 Þ. We can write Qi as hu,P 1i ,ri ,fi ,si ,P 2i ,y i. Since j F ðAQ0n1 Þ j¼ 2n 5 P 5 andj F F ðAQ0n1 Þ j6 1, there exists an index i such that ðri ,rhi Þ and ðsi ,shi Þ are F-fault free. Since AQ1n1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path R1 of AQ1n1 ðF ðAQ1n1 Þ fy,vgÞ joining rhi and shi . Then hu,P 1i ,ri ,rhi ,R1 ,shi ,si , P 2i ,y ,y,vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(d) for an illustration. Subcase 2.3: u,v 2 V ðAQ1n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there is a hamiltonian cycle C of AQ0n1 F . Let A = {1x1 . . . xnj there exists some i and j in (2, . . . , n) such that xi ¼ ui , xj ¼ uj , and xk ¼ uk if k 5 {i, j} and let B = {1x2 . . . xnjthere exists some k 2 {2, . . . , n} such that xk = xk and xi ¼ xi if i 5 k}. Obviously, A [ B is the subset of x 2 V ðAQ1n1 Þ j d AQ1 ðx,uÞ ¼ 2g. n1 n n1 n1 Moreover, j A [ B j¼ þ ¼ . Further2 1 2 more, every vertex in A [ B has two common neighbors with n 1 u in AQn1 . Since ð2n 4Þ P 4, there are at least four 2 vertices in A [ B, say {xij1 6 i 6 4}, such that xhi 2 C. Let wi be a common neighbor of u and xi. Moreover, let yhi and zhi be the two neighbors of xhi in C. Obviously, fyhi j 1 6 i 6 4g [ fzhi j 1 6 i 6 4g P 6. There exists some i such that one of yhi and zhi , say yhi , satisfying yi 62 {u,v, wi} and yi is F-fault free. We can write C as hxhi ,zhi ,P 1 ,yhi ,xhi g.Since AQn1 is (2n 6)-fault hamiltonian, there exists a hamiltonian path P2
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of AQ1n1 F ðAQ1n1 Þ fu,wi ,xi g joining bi and v. Then hu,wi ,xi , xhi ,zhi ,P 1 ,yhi ,yi ,P 2 ,vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(e) for an illustration. Case 3: j F ðAQ0n1 Þ j¼ 2n 4. Hence, j F F ðAQ0n1 Þ j¼ 0. Subcase 3.1: u,v 2 V ðAQ0n1 Þ. Let f1 and f2 be any two elements in F ðAQ0n1 Þ. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P joining u and v in AQ0n1 ðF ðAQ0n1 Þ ff1 ,f2 gÞ which may or may not include f1 or f2 on it. Removing f1 and f2, path P is divided into three, two, or one pieces depending on the cases in which f1 and f2 are on P or not. For the last two cases, we may arbitrarily delete one or two more edges from P to make it into three subpaths. Therefore, without loss of generality, we may write these three subpaths as hu,P 1 ,w,f10 ,x,P 2 ,y,f20 ,z,P 3 ,vi. Suppose that l(P2) P 1. Hence, wh, xh, yh and zh are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hwh, P4, xhi and hyh, P5, zhi in AQ1n1 . Then hu, P1, w, wh, P4, xh, x, P2, y,yh, P5, zh, z, P3, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(f) for an illustration. Suppose that l(P2) = 0. Hence, x = y. Suppose that fðx,wÞ,ðx,zÞg \ C n ¼ ;. Then h x , xc, wh and zh are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hwh, P6, xhi and hxc, P7, zhi in AQ1n1 . Then hu, P1, w, wh, P6, xh, x, P2, y, yh, P7, zh, z, P3, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(g) for an illustration. Suppose that fðx,wÞ,ðx,zÞg \ C n 6¼ ;. Without loss of generality, we assume that (x, w) is in C n . Then xh = wc and xc = wh. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path P8 of AQ1n1 fxc g joining xh and zh. Then hu, P1, w, xc, x, xh, P8, zh, z, P3, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(h) for an illustration. Subcase 3.2: u 2 V ðAQ0n1 Þ and v 2 V ðAQ1n1 Þ. Let f1 be any element in F ðAQ0n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle C ¼ hu,P 1 ,w,f10 ,x,P 2 ,ui in AQ0n1 ðF ðAQ0n1 Þ f1 Þ where f10 ¼ f1 if f1 is on cycle C or f10 is any edge of C otherwise. Without loss of generality, we may assume that l(P1) 6 l(P2). Then we rewrite C as hu,P 1 ,w,f10 , x,P 3 ,y,ui. Suppose that l(P1) = 0. Hence, u = w. Thus, we can rewrite C as hu,f10 ,x,P 03 ,ui. Then there exists a vertex x 0 2 {xh, xc} such that x 0 5 v. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P4 of AQ1n1 joining x 0 and v. Then hu,P 03 ,x,x0 ,vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(l) for an illustration. Suppose that l(P1) P 1. Then we have the following three cases. (1) wh, xh, yh, and v are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hwh, P4, xhi and hyh, P5, vi in AQ1n1 . Then hu, P1, w, wh, P4, xh, x, P3, y, yh, P5, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(j) for an illustration.
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(2) yh = v. Since AQn1 is (2n6)-fault hamiltonian connected, there is a hamiltonian path hwh, P6, xhi of AQ1n1 fvg. Thus, hu, P1 w, wh, P6, xh,x, P2, y, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(k) for an illustration. (3) xh = v or wh = v. Without loss of generality, we assume that xh = v. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path hwh, P7, yhi of AQ1n1 fvg. Thus, hu, P1, w, wh, P7, yh, y, P2, x, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(1) for an illustration. Subcase 3.3: u,v 2 V ðAQ1n1 Þ. Suppose that F ðAQ0n1 Þ C n . By definition, the hypercube Qn1 is a spanning subgraph of AQ0n1 F ðAQ0n1 Þ. Thus, there is a hamiltonian cycle C in AQ0n1 F ðAQ0n1 Þ [9]. We claim that there is an edge (w, x) on C such that wh, xh, u, and v are distinct. Since l(C) = 2n1 and (2n1/ 2) 3 = 2n2 3 P 5, there are at least five edges we can choose. Hence, such edge exists. Since AQ1n1 has property 2H, there are two disjoint spanning paths hu, P2, whi and hxh, P3, vi of AQ1n1 . Then hu, P2, wh, w, P1, x, xh, P3, vi is a hamiltonian path of AQn F joining u and v. See Fig. 5(m) for an illustration. Suppose that F ðAQ0n1 Þ 6 C n . Let f1 be any element in F ðAQ0n1 Þ C n . Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle C ¼ hw,P 1 ,x, f10 ,wiin AQ0n1 ðF ðAQ0n1 Þ f1 Þ where f10 ¼ f1 if f1 is incident with C or f10 is any edge of C otherwise. Hence, wh, wc, xh, and xc are distinct. We have the following two cases. (1) wh, xh, u, and v are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hu, P4, whi and hxh, P5, vi of AQ1n1 . Then hu, P4, wh, w, P1, x, xh, P5, vi hamiltonian path of AQn F joining u and v. See Fig. 5(n) for an illustration. (2) wh, xh, u, and v are not distinct. Without loss of generality, we assume that wh = u. Let x 0 be a element in {xh, xc} such that x 0 5 v. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path P6 of AQ1n1 fug joining x 0 and v. Then hu, w, P1, x, x 0 , P6, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(o) for an illustration. Hence, this lemma is proved.
h
References [1] S. Abraham, K. Padmanabhan, The twisted cube topology for multiprocessors: A study in network asymmetry, Journal of Parallel and Distributed Computing 13 (1991) 104–110. [2] L.N. Bhuyan, D.P. Agrawal, Generalized hypercube and hyperbus structures for a computer network, IEEE Transaction on Computers C-33 (1984) 323–333. [3] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North Holland, New York, 1980.
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[4] C.P. Chang, J.N. Wang, L.H. Hsu, Topologial properties of twisted cubes, Information Sciences 113 (1999) 147–167. [5] F.B. Chedid, R.B. Chedid, A new variation on hypercubes with smaller diameter, Information Process Letters 46 (1993) 275–280. [6] S.A. Choudum, V. Sunitha, Augmented cubes, Networks 40 (2) (2002) 71–84. [7] T.Y. Feng, A survey of interconnection networks, IEEE Computer 14 (1981) 12–27. [8] W.T. Huang, J.M. Tan, C.N. Hung, L.H. Hsu, Fault-tolerant hamiltonicity of twisted cubes, Journal of Parallel and Distributed Computing 62 (2002) 591–604. [9] Y. Saad, M.H. Schultz, Topological properties of hypercubes, IEEE Transaction on Computers 34 (1988) 867–872. [10] D. Wang, Embedding hamiltonian cycles into folded hypercubes with faulty links, Journal of Parallel and Distributed Computing 61 (2001) 545–564.
Fault hamiltonicity of augmented cubes Hong-Chun Hsu a
c
a,*
, Liang-Chih Chiang b, Jimmy J.M. Tan b, Lih-Hsing Hsu c
Department of Computer Science and Information Management, Providence University, Shalu, Taichung, Taiwan 43301, ROC b Department of Computer and Information Science, National Chiao Tung University, Hsinchu, Taiwan 300, ROC Department of Computer and Information Engineering, Ta Hwa Institute of Technology, Hsinchu, Taiwan 300, ROC Received 11 December 2003; revised 25 September 2004; accepted 9 October 2004 Available online 23 December 2004
Abstract In this paper, we consider the fault hamiltonicity and the fault hamiltonian connectivity of the augmented cubes AQn. Assume that F V(AQn) [ E(AQn) and n P 4. We prove that AQn F is hamiltonian if jFj 6 2n 3 and that AQn F is hamiltonian connected if jFj 6 2n 4. Moreover, these bounds are tight. Ó 2004 Elsevier B.V. All rights reserved. Keywords: Fault-tolerant; Hamiltonian; Hamiltonian connected; Augmented cubes
1. Introduction Network topology is a crucial factor for the interconnection networks since it determines the performance of the networks. Many interconnection network topologies have been proposed in the literature for the purpose of connecting
*
Corresponding author. E-mail address: [email protected] (H.-C. Hsu).
0167-8191/$ - see front matter Ó 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.parco.2004.10.002
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thousands of processing elements. Network topology is usually represented by a graph where vertex represent processors and edges represent links between processors. For the graph definition and notation, we follow [3]. G = (V, E) is a graph if V is a finite set and E is a subset of {(u, v)j(u, v) is an unordered pair of V}. We say that V is the vertex set and E is the edge set. For any vertex x of V, degG(x) denotes its degree in G. We use d(G) to denote min{degG(x)jx 2 V(G)}. Two vertices u and v are adjacent if (u, v) 2 E. A path, denoted by hv0, v1, v2, . . . , vki, is a sequence of distinct vertices where vi and vi+1 are adjacent for all i, 0 6 i 6 k 1. The length of a path P, l(P), is the number of edges in P. We also write the path hv0, v1, v2, . . . , vki as hv0, P1, vi, vi+1, . . . , vj, P2, vt, . . . , vki, where P1 is the subpath hv0, v1, . . . , vii and P2 is the subpath hvj, vj+1, . . . , vti. It is possible to write a path as hv0, v1, P, v1, v2, . . . , vki if l(P) is 0. Sometimes, a path can also be represented by hv0, v1, . . . , vi, e, vi+1, . . . , vni to emphasize that e is the edge (vi, vi+1). We use d(u, v) to denote the distance between u and v, i.e., the length of a shortest path joining u and v. A path is a hamiltonian path if its vertices are distinct and span V. A cycle is a path with at least three vertices such that the first vertex is the same as the last vertex. There are a lot of interconnection network topologies proposed in literature [7]. Among these topologies, the n-dimensional hypercube, denoted by Qn, is a popular one. Several variations of the hypercubes have been investigated to improve the efficiency of the hypercubes [1,2,4,5,10]. The augmented cube AQn, recently proposed by Choudum and Sunitha [6], is one of such variations. For any positive integer n, AQn is a vertex transitive, (2n 1)-regular, and (2n 1)-connected graph with 2n vertices. In this paper, we consider two important parameters of the augmented cubes, fault hamiltonicity and fault hamiltonian connectivity. These two parameters for interconnection networks are proposed by Huang et. al. [8]. A hamiltonian cycle of G is a cycle that traverses every vertex of G exactly once. A graph is hamiltonian if it has a hamiltonian cycle. A hamiltonian graph G is k-fault hamiltonian if G F remains hamiltonian for every F V(G) [ E(G) with jFj 6 k. The fault hamiltonicity, Hf(G), is defined to be the maximum integer k such that G is k-fault hamiltonian if G is hamiltonian and is undefined otherwise. Clearly, Hf (G) 6 d(G) 2 if Hf (G) is defined. In this paper, we prove that Hf (AQn) = 2n 3 if n 62 {1, 3}, Hf(AQ3) = 2, and Hf(AQ1) is undefined. Since d(AQn) = 2n 1, the fault hamiltonicity of the augmented cube AQn is exactly 2n 3 for n P 4. Therefore, we obtain the optimal result. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two vertices of G. All hamiltonian connected graphs except the complete graphs K1 and K2 are hamiltonian. A graph G is k-fault hamiltonian connected if G F remains hamiltonian connected for every F V(G) [ E(G) with jFj 6 k. The fault hamiltonian connectivity, H kf ðGÞ, is defined to be the maximum integer k such that G is k-fault hamiltonian connected if G is hamiltonian connected and is undefined otherwise. Clearly, H kf ðGÞ 6 dðGÞ 3 if H kf ðGÞ is defined and jV(G)j P 4. In this paper, we prove that H kf f ðAQn Þ ¼ 2n 4 if n 62 1,3, H kf ðAQ3 Þ ¼ 1, and H kf ðAQ1 Þ ¼ 0. Again, this result concerning the fault hamiltonian connectivity of the augmented cube AQn is optimal because d(AQn) = 2n 1.
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In the following section, we give the definition of the augmented cubes and discuss some properties of AQn. Then we present the fault hamiltonian properties of AQ3 and AQ4 in Section 3. In the final section, we discuss Hf (AQn) and H kf ðAQn Þ. 2. Augmented cubes Let n P 1 be an integer. The graph of the n-dimensional augmented cube [6], denoted by AQn has 2n vertices, each labeled by an n-bit binary string V(AQn) = {u1u2 . . . unjui 2 {0, 1}}. AQ1 is the graph K2 with vertex set {0, 1}. For n P 2, AQn can be recursively constructed by two copies of AQn1, denoted by AQ0n1 and AQ1n1 and by adding 2n edges between AQ0n1 and AQ1n1 as follows: Let V ðAQ0n1 Þ ¼ fð0u2 u3 . . . un Þ j ui ¼ 0 or 1 for 2 6 i 6 n} and V ðAQ1n1 Þ ¼ fð1v2 v3 . . . vn Þ j vi ¼ 0 or 1 for 2 6 i 6 n}. A vertex u = (0u2u3 . . . un) of AQ0nl is joined to a vertex v = (1v2v3 . . . vn) of AQ1n1 if and only if either (i) ui = vi for 2 6 i 6 n; in this case, (u, v) is called a hypercube edge and we set v = uh, or (ii) ui ¼ vi for 2 6 i 6 n; in this case, (u, v) is called a hypercube edge and we set v = uc . The augmented cubes AQ1, AQ2, and AQ3 are illustrated in Fig. 1. It is proved in [6] that AQn is a vertex transitive, (2n 1)-regular, and (2n 1)-connected graph with 2n vertices for any positive integer n. Let Ehn ¼ fðu,uh Þ j u 2 V ðAQ0n1 Þg and Ecn ¼ fðu,uc Þ j u 2 V ðAQ0n1 Þg. Obviously, Ehn and Ecn are two perfect matchings between the vertices of AQ0n1 and AQ1n1 . Hence, j Ehn j¼j Ecn j¼ 2n1 . Let C n ¼ fðu,vÞ j u ¼ 0u1 . . . un and v ¼ 0 u1 . . . un g. In other words, C n is the set of all complement 0 edges in AQn1 . The following lemmas are derived directly from the definition. Lemma 1. Let x and y be any two vertices in AQ0n1 with n P 4. Suppose that ðx,yÞ 62 C n . Then xh, xc, yh and yc are all distinct. Lemma 2. Assume that ðu,vÞ 2 AQ0n1 . Then ðuh ,vh Þ 2 AQ1n1 and ðuc ,vc Þ 2 AQ1n1 .
Fig. 1. The argumented cubes: (a) AQ1; (b) AQ2 and (c) AQ3.
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Lemma 3. Assume there is a vertex u in A0n1 . Let A ¼ fv j ðu,vÞ 2 EðAQ0n1 Þg and B = {vh, vcjv 2 A}. Then jAj = 2n 3 and jBj = 4n 8. To discuss the fault hamiltonicity and the fault hamiltonian connectivity of AQn, we need to introduce the following two terms for graphs. A graph G has property 2H if it satisfies the following conditions: Let {w, x} and {y, z} be two pairs of four distinct vertices of G. There exist two disjoint paths P1 and P2 of G such that (1) P1 joins w to x, (2) P2 joins y to z, and (3) P1 [ P2 span G. A hamiltonian connected graph G = (V, E) is said to be 1-vertex hamiltonian connected if G {u} is still hamiltonian connected after removing any vertex u 2 V. Thus, any k-fault hamiltonian connected graph is 1-vertex hamiltonian connected for k P 1. Furthermore, any 1-vertex hamiltonian connected graph is hamiltonian connected. Let F V(AQn) [ E(AQn) be any faulty set of AQn. An edge (u, v) is called F -fault free if (u, v) 62 F, u 62 F, and v 62 F; otherwise it is called F-fault. Let H = (V 0 , E 0 ) be a subgraph of AQn. We use F(H) to denote the set (V 0 [ E 0 ) \ F. We need the following lemmas. The proof of these lemmas are put in Appendix. Lemma 4. Hf(AQ3) = 2. Lemma 5. H kf ðAQ3 Þ ¼ 1. Lemma 6. Hf(AQ4) = 5. and H kf ðAQ4 Þ ¼ 4. Moreover, AQ4 has property 2H. Lemma 7. Let n P 4. Suppose that AQn1 is 1-vertex hamiltonian connected and has property 2H. Then AQn also has property 2H. Lemma 8. Let n P 5. Suppose that AQn1 is (2n 5)-fault hamiltonian, (2n 6)fault hamiltonian connected, and has property 2H. Then AQn is (2n 3)-fault hamiltonian. Lemma 9. Let n P 5. Suppose that AQn1 is (2n 5)-fault hamiltonian, (2n 6)fault hamiltonian connected, and has property 2H. Then AQn is (2n 4)-fault hamiltonian connected.
3. Fault hamiltonian and fault hamiltonian connectivity of the augmented cubes Theorem 10. Let n be a positive integers with n P 4. Then AQn is (2n 3)-fault hamiltonian, (2n 4)-fault hamiltonian connected, and has property 2H. Proof. We prove this theorem by induction on n. The induction base is n = 4. By Lemmas 7 and 6, the theorem holds for n = 4. With Lemmas 7–9, we prove the induction step. h
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With Lemmas 4 and 5, we have the following corollary. Corollary 11. Let n be a positive integer. (1) Hf(AQn) = 2n 3 and H kf ðAQn Þ ¼ 2n 4 if n 62 {1, 3}; (2) Hf(AQ1) is undefined and H kf ðAQ1 Þ ¼ 0; and (3) Hf(AQ3) = 2 and H kf ðAQ3 Þ ¼ 1.
Appendix A Proof of Lemma 4. For the proof, we have the following observation. Let C1 and C2 be any two cycles of K4. Then E(C1) \ E(C2) 5 ;. Now, we prove that Hf (AQ3) < 3. Let F = {001, 010, 100}. Then AQ3 F is the graph in Fig. 2(a). Obviously, AQ3 F is non-hamiltonian. Hence, Hf(AQ3) < 3. Then we prove that Hf (AQ3) P 2. Let F V(AQ3) [ E(AQ3) with jFj 6 2. Without loss of generality, we assume that j F ðAQ02 Þ jPj F ðAQ12 Þ j. Case 1: j F ðAQ02 Þ j¼ 2. Since AQ2 is a complete graph K4, there is a hamiltonian path P0 of AQ02 F joining some u and v. Obviously, there is a hamiltonian path P1 of AQ12 joining uh and vh. Hence, hu, uh, P1 vh, v, P0, ui forms a hamiltonian cycle of AQ3 F. Case 2: j F ðAQ02 Þ j¼ 1. Suppose that j F ðAQ12 Þ j¼ 1. Hence, there is a hamiltonian cycle C0 in AQ02 F ðAQ02 Þ there is a hamiltonian cycle C1 in AQ12 F ðAQ12 Þ. By Lemma 3, there exists an edge (u, v) in C0 such that both (u, uh) and (v, vh) are F-fault free and the edge (uh, vh) is in C1. Then we may write C0 as hu, P1, v, ui and write C1 as huh, P2, vh, uhi. Then hu, uh, P2, vh, v, P1, ui forms a hamiltonian cycle of AQ3 F. Suppose that j F ðAQ12 Þ j¼ 0. Hence, AQ12 is hamiltonian connected. Let C0 be the hamiltonian cycle of AQ02 F ðAQ02 Þ. Then for any edge (u, v) in C0, there exists a hamiltonian path P2 of AQ12 joining uh and vh. Then hu, uh, P2, vh, v, P1, ui forms a hamiltonian cycle of AQ3 F.
Fig. 2. The graphs of AQ3 F.
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Case 3: j F ðAQ02 Þ j¼ 0. Hence, both AQ02 and AQ12 are hamiltonian connected. Then we choose any two distinct F-fault free edges in Eh [ Ec say (u, v) and (w, x), which u,w 2 AQ02 and v,x 2 AQ12 . Let P1 be a hamiltonian path of AQ02 joining u and w, and P2 be a hamiltonian path of AQ12 joining v and x. Thus, hu, v, P2, x, w, P1, ui forms a hamiltonian cycle of AQ3 F. h Proof of Lemma 5. Now, we prove that H kf ðAQ3 Þ < 2. Let F = {001, 111}. Then AQ3 F is the graph in Fig. 2(b). Obviously, there is no hamiltonian path of AQ3 F joining 010 to 100. Hence, H kf ðAQ3 Þ < 2. Then we prove that H kf ðAQ3 Þ ¼ 1. Let F V(AQ3) [ E(AQ3) with jFj 6 1. By the symmetric property of AQ3, we may assume that F is the following three cases: (1)F = {000}, (2) F = {(000, 100)}, and (3) F = {(000, 111)}. The corresponding hamiltonian paths are listed below. Hence, AQ3 F is hamiltonian connected. h
F = {000}
F = {(000, 001)}
F = {(000, 111)}
h001, 010, 110, 100, 101, 111, 011i h001, 011, 111, 101, 100, 110, 010i h001, 010, 011, 111, 101, 110, 100i h001, 011, 010, 110, 100, 111, 101i h001, 010, 011, 111, 101, 100, 110i h001, 011, 010, 110, 100, 101, 111i h010, 001, 101, 100, 110, 111, 011i h010, 001, 011, 111, 101, 110, 100i h010, 011, 001, 110, 100, 111, 101i h010, 001, 011, 111, 101, 100, 110i h010, 011, 001, 110, 100, 101, 111i h011, 010, 001, 110, 101, 111, 100i h011, 010, 001, 110, 100, 111, 101i h011, 010, 001, 101, 111, 100, 110i h011, 010, 001, 110, 100, 101, 111i h100, 111, 110, 010, 011, 001, 101i h100, 111, 101, 001, 011, 010, 110i h100, 110, 101, 001, 010, 011, 111i h101, 100, 111, 011, 001, 010, 110i h101, 100, 110, 010, 001, 011, 111i h110, 100, 101, 001, 010, 011, 111i
h000, 010, 011, 100, 101, 111, 110, 001i h000, 011, 001, 101, 100, 111, 110, 010i h000, 010, 001, 101, 100, 111, 110, 011i h000, 010, 011, 001, 101, 111, 110, 100i h000, 010, 011, 001, 110, 111, 100, 101i h000, 010, 011, 001, 101, 100, 111, 110i h000, 010, 011, 001, 101, 100, 110, 111i h001, 010, 000, 100, 110, 101, 111, 011i h001, 011, 000, 111, 101, 100, 110, 010i h001, 010, 011, 000, 111, 101, 110, 100i h001, 011, 000, 010, 110, 100, 111, 101i h001, 010, 000, 011, 111, 101, 100, 110i h001, 011, 000, 010, 110, 100, 101, 111i h010, 001, 101, 000, 100, 110, 111, 011i h010, 001, 011, 000, 111, 101, 110, 100i h010, 000, 011, 001, 110, 100, 111, 101i h010, 001, 011, 000, 111, 101, 100, 110i h010, 000, 011, 001, 110, 100, 101, 111i h011, 000, 010, 001, 110, 101, 111, 100i h011, 000, 010, 001, 110, 100, 111, 101i h011, 000, 010, 001, 101, 111, 100, 110i h011, 000, 010, 001, 110, 100, 101, 111i h100, 111, 110, 010, 000, 011, 001, 101i h100, 111, 101, 001, 011, 000, 010, 110i h100, 110, 101, 001, 010, 000, 011, 111i h101, 100, 111, 000, 011, 001, 010, 110i h110, 100, 101, 001, 010, 000, 011, 111i
h000, 010, 011, 100, 101, 111, 110, 001i h000, 011, 001, 101, 100, 111, 110, 010i h000, 010, 001, 101, 100, 111, 110, 011i h000, 010, 011, 001, 101, 111, 110, 100i h000, 010, 011, 001, 110, 111, 100, 101i h000, 010, 011, 001, 101, 100, 111, 110i h000, 010, 011, 001, 101, 100, 110, 111i h001, 010, 000, 100, 110, 101, 111, 011i h001, 011, 000, 100, 101, 111, 110, 010i h001, 010, 000, 011, 111, 101, 110, 100i h001, 011, 000, 010, 110, 100, 111, 101i h001, 010, 000, 011, 111, 101, 100, 110i h001, 011, 000, 010, 110, 100, 101, 111i h010, 001, 101, 000, 100, 110, 111, 011i h010, 001, 000, 011, 111, 101, 110, 100i h010, 000, 011, 001, 110, 100, 111, 101i h010, 001, 000, 011, 111, 101, 100, 110i h010, 000, 011, 001, 110, 100, 101, 111i h011, 000, 010, 001, 110, 101, 111, 100i h011, 000, 010, 001, 110, 100, 111, 101i h011, 000, 010, 001, 101, 111, 100, 110i h011, 000, 010, 001, 110, 100, 101, 111i h100, 111, 110, 010, 000, 011, 001, 101i h100, 111, 101, 001, 011, 000, 010, 110i h100, 110, 101, 001, 010, 000, 011, 111i h101, 111, 100, 000, 011, 001, 010, 110i h110, 100, 101, 001, 010, 000, 011, 111i
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Proof of Lemma 6. To prove this lemma is very tedious. With long and detail discussion, we have completed theoretical proof for this lemma. Nevertheless, we do not present it in this paper for reducing complexity. However, we can also verify this lemma using computer. Our computer program was coded by depth first search method (a kind of brute force method). h Proof of Lemma 7. Let {w, x} and {y, z} be two pairs of four distinct vertices of AQn. We are going to construct two disjoint spanning paths, R1 and R1, of AQn such that R1 joins w and x and R2 joins y and z. We consider the following four cases. Case 1: {w, x, y, z} are all in AQ0n1 . By the assumption of this lemma, AQn1 has property 2H. There exist two disjoint paths hw, P1, xi and hy, P2, zi spanning AQ0n1 . Without loss of generality, we assume that l(P1) 6 l(P2). Suppose that n = 3. Then AQ02 is isomorphic to the complete graph K4. Obviously, these two paths of AQ02 are hw, xi and hy, zi. Since AQ12 is K4, there is a hamiltonian path hwh, P3, xhi of AQ12 . Thus, hw, wh, P3, xh, xi and hy, zi form the desired disjoint spanning paths of AQ3. Suppose that n P 4. Then l(P2) P 4. Hence, P2 can be written as hy, P4, u, v, zi. By Lemma 2, ðuh ,vh Þ 2 EðAQ1n1 Þ. By the assumption of this lemma, AQ1n1 is 1-vertex hamiltonian connected. Hence, AQ1n1 is hamiltonian connected. Thus, there exists a hamiltonian path huh, P3, vhiin AQ1n1 . Then R1 = hw, P1, xi and R2 = hy, P4, u, uh, P3, vh,v, zi form the desired disjoint spanning paths of AQn. See Fig. 3(a) for an illustration. Case 2: {w, x, y} are in AQ0n1 and z is in AQ1n1 . Let y* be a vertex in {yh, yc} such that y* 5 z. By the assumption of this lemma, there exists a hamiltonian path hy*, P2, zi in AQ1n1 and there exists a hamiltonian path hw*, P1, xi in AQ0n1 fyg. Then R1 = hw, P1, xi and R2 = hy, y*, P2, zi form the desired disjoint spanning paths of AQn. See Fig. 3(b) for an illustration.
Fig. 3. Illustrations for Lemma 7.
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Case 3: {w, x} are in AQ0n1 and {y, z} are in AQ1n1 . By the assumption of this lemma, there is a hamiltonian path R1 = hw, P1, xiin AQ0n1 and there is a hamiltonian R2 = hy, P2, zi in AQ1n1 . Obviously, R1 and R2 form the desired disjoint spanning paths of AQn. See Fig. 3(c) for an illustration. Case 4: {w, y} are in AQ0n1 and {x, z} are in AQ1n1 . Subcase 4.1: {(w, x),(y, z)} \ E(AQn) 5 ;. Without loss of generality, we assume that (w, x) 2 E(AQn). Since n P 3, there exists a vertex u in AQ0n1 such that u 5 y and uh 5 z. By the assumption of this lemma, there exists a hamiltonian path hy, P1, ui of AQ0n1 fwg there exists a hamiltonian path huh, P2, zi of AQ1n1 fxg. Obviously, R1 = hw, xi and R2 = hy, P1, u,uh, P2, zi form the desired disjoint spanning paths of AQn. See Fig. 3(d) for an illustration. Subcase 4.2: {(w, x),(y, z)} \ E(AQn) = ;. Then one of wh and wc is not z, say w*, and one of zh and zc is not w, say z*. By the assumption of this lemma, there exists a hamiltonian path hw*, P1, xi in AQ1n1 fzg and there exists a hamiltonian path hy, P2, z*i in AQ0n1 fwg. Thus, R1 = hw, w*, P1, xi and R2 = hy, P2, z*, zi form the desired disjoint spanning paths of AQn. See Fig. 3(e) for an illustration. Hence, the lemma is proved. h Proof of Lemma 8. Let F be any subset of V(AQn) [ E(AQn) with jFj 6 2n 3. Without loss of generality, we may assume that j F ðAQ0n1 Þ jPj F ðAQ1n1 Þ j. We are going to construct a hamiltonian cycle of AQn F depending on the following cases. Case 1: j F ðAQ0n1 Þ j6 2n 6. In this case, AQ0n1 F ðAQ0n1 Þ and AQ1n1 F ðAQ1n1 Þ are hamiltonian connected. Since j Ehu [ Ecn j j F jP 2n ð2n 3Þ P 25, there are two distinct F-fault free edges (u, v) and (x, y) such that u and x are in AQ0n1 F ðAQ0n1 Þ, and v and y are in AQ1n1 F ðAQ1n1 Þ. By the assumption of this lemma, there exist hamiltonian paths hu, P1, xi in AQ0n1 F ðAQ0n1 Þ and hv, P2, yi in AQ1n1 F ðAQ1n1 Þ. Then hu, v, P2, y, x, P1, ui forms a hamiltonian cycle of AQn F. See Fig. 4(a) for an illustration. Case 2: j F ðAQ0n1 Þ j¼ 2n 5. Hence, j F F ðAQ0n1 Þ j6 2. Since AQn1 is (2n5)fault hamiltonian, there is a hamiltonian cycle C in AQ0n1 F ðAQ0n1 Þ. Since n1 lðCÞ=2 P 2 2nþ5 > 5, there exists an edge (u, v) in C such that (u, uh) and 2 h (v, v ) are F-fault free. We can write C as hu, P1, v, ui. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path huh, P2, vhi of AQ1n1 F ðAQ1n1 Þ. Then hu, uh, P2, vh, v, P1, ui is a hamiltonian cycle of AQn F. See Fig. 4(b) for an illustration. Case 3: j F ðAQ0n1 Þ j¼ 2n 4. Suppose that there exists a hamiltonian cycle C of AQ0n1 F ðAQ0n1 Þ. As in Case 2, there exists an edge (u, v) in C such that (u, uh) and (v, vh) are F-fault free. We can write C as hu, P1, v, ui. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path P2 of
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Fig. 4. Illustrations for Lemma 8.
AQ1n1 F ðAQ1n1 Þ. joining uh and vh. Thus, hu, uh, P2, vh, v, P1, ui forms a hamiltonian cycle of AQn F. Suppose that there is no hamiltonian cycle of AQ0n1 F ðAQ0n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle Ci of AQ0n1 ðF ðAQ0n1 Þ ffi gÞ for every fi 2 F ðAQ0n1 Þ, We can write Ci as hui, Pi, vi, fi, uii. Since j F ðAQ0n1 Þ jP 6 and j F F ðAQ0n1 Þ j6 1, there exists an index i such that ðui ,uhi Þ and ðvi ,vhi Þ are F-fault free. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path Q1 in AQ1n1 F ðAQ1n1 Þ joining uhi and vhi . Thus, hui ,uhi ,Q1 ,vhi ,vi ,P i ,ui i forms a hamiltonian cycle in AQn F. See Fig. 4(c) for an illustration. Case 4: j F ðAQ0n1 j¼ 2n 3. Thus, j F F ðAQ0n1 Þ j¼ 0. Let f1 and f2 be any two elements in F ðAQ0n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle C ¼ hu,f10 ,v,P 1 ,w,f20 ,x,P 2 ,ui in AQ0n1 ðF ðAQ0n1 Þ ff1 ,f2 gÞ where f10 ¼ f1 if f1 is on C or f10 is an arbitrary edge of C otherwise and f20 ¼ f2 if f2 is on C or f20 is an arbitrary edge of C otherwise. Without loss of generality, we assume that l(P2) 6 l(P1). Subcase 4.1: l(P2) P 1. Obviously, uh, vh, wh and xh are distinct. Since AQn1 has property 2H, there exist two disjoint spanning paths huh, P3, vhi and hwh, P4, xhi in AQ1n1 . Thus, hu, uh, P3, vh, v, P1, w, wh, P4, xh, x, P2, ui is a hamiltonian cycle of AQn F. See Fig. 4(d) for an illustration. Subcase 4.2: l(P2) = 0. Then u = x. Moreover w, u, and v are distinct. Obviously, one of the following cases holds: (1) {wh, wc, uh, uc h c , v , v } are all distinct, (2) (wh = vc and wc = vh), (3) h (w = uc and wc = uh), and (4) (vh = uc and vc = uh). Suppose that (1) {wh, wc, uh, uc, vh, vc} are all distinct or (2) (wh = vc and wc = vh). Since AQn 1 has property 2H, there exist two disjoint spanning paths hwh, P5, uhi and huc, P6, vhi in AQ1n1 . Then hu, uc, P6, vh, v, P1, w, wh, P5, uh, ui forms a hamiltonian cycle of AQn F. See Fig. 4(e) for an illustration. Suppose that (3) (wh = uc and wc = uh) or (4) (vh = uc and vc = uh). Without loss of generality, we can assume that wh = uc and wc = uh. Since AQn1 is (2n 6)-fault
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hamiltonian connected, there is a hamiltonian path huh, P7, vhi in AQ1n1 fwh g. Hence, hu, uh, P7, vh, v, P1, w, wh, ui forms a hamiltonian cycle of AQn F. See Fig. 4(f) for an illustration. This completes the proof of the lemma. h Proof of Lemma 9. Let F be any subset of V(AQn) [ E(AQn) with jFj < 2n 4. Without loss of generality, we assume that j F ðAQ0n1 Þ jPj F ðAQ1n1 Þ j. We want to construct a hamiltonian path joining u and v in AQn F for any two different vertices u and v 2 V(AQn) F. Case 1: jF(AQ0n1)j 6 2n 6. Since AQn1 is (2n 6)-fault hamiltonian connected, both AQ0n1 F ðAQ0n1 Þ and AQ1n1 F ðAQ1n1 Þ are hamiltonian connected. Subcase 1.1: u,v 2 V ðAQ0n1 Þ or u,v 2 V ðAQ1n1 Þ. Without loss of generality, we assume that u, v 2 V(AQ0n1). Since AQ0n1 F ðAQ0n1 Þ is hamiltonian connected, there exists a hamiltonian path P of AQ0n1 F ðAQ0n1 Þ joining u and v. We claim that there exists an edge (w, x) on P such that both (w, wh) and (x, xh) are F-fault free. Suppose that there is no such edge. Then at least [l(P)/2] edges in Ehn [ Ecn are F-fault. Since lðP Þ P 2n1 j F ðAQ0n1 Þ j , j F j¼j F ðAQ0n1 Þ j þ ð2n1 j F ðAQ0n1 Þ jÞ=2 P 2n2 þ j F ðAQ0n1 Þ j =2 > 2n 4. We have a contradiction. Hence, such edge exists as claimed. We can write P as hu, P1, w, x, P2, vi. Since AQ1n1 F ðAQ1n1 Þ is also hamiltonian connected, there is a hamiltonian path hwh, P3, xhi in AQ1n1 F ðAQ1n1 Þ. Thus, hu, P1, w, wh, P3, xh, x, P2, vi is a hamiltonian path of AQn F joining u and v. See Fig. 5(a) for an illustration. Subcase 1.2: u 2 AQ0n1 and v 2 AQ0n1 . Since jFj 6 2n 4 < 2n1, there exists an F-fault free edge (w, wh) such that w 2 V ðAQ0n1 Þ, w 5 u, and wh 5 v. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path hu, P1, wi in AQ0n1 ðAQ0n1 Þ and there is a hamiltonian path hwh, P2, vi in AQ1n1 F ðAQ1n1 Þ. Then hu, P1, w, wh, P2, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(b) for an illustration. Case 2: j F ðAQ0n1 Þ j¼ 2n 5. Hence, j F F ðAQ0n1 Þ j6 1. Subcase 2.1: u,v 2 V ðAQ0n1 Þ. Suppose that there exists a hamiltonian path P of AQ0n1 F ðAQ0n1 Þ joining u and v. Since ðlðpÞ j F ðAQ0n1 Þ jÞ=2 P 2n 4,, there exists an edge (w, x) in P such that (w, wh) and (x, xh) are F-fault free. We can write P as hu, P1, w, x, P2, vi. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P3 of AQ1n1 F ðAQ1n1 Þ joining wh and xh. Thus, hu, P1, w, wh, P3, xh,
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Fig. 5. Illustrations for Lemma 9.
x, P2, vi forms a hamiltonian path of AQn F joining u and v. Suppose that there is no hamiltonian path of AQ0n1 F ðAQ0n1 Þ joining u and v. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path Qi of AQ0n1 ðF ðAQ0n1 Þ ffi gÞ joining u and v for every fi 2 F ðAQ0n1 Þ. We can write Qi as hu,P 1i ,wi ,fi ,xj ,P 2i ,vi. Since j F ðAQ0n1 Þ j¼ 2n 5 P 5 and j F F ðAQ0n1 Þ j6 1, there exists an index i such that ðwi ,whi Þ and ðxi ,xhi Þ are F-fault free. Since AQ1n1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path R1 of AQ1n1 F ðAQ1n1 Þ joining whi and xhi . Then hu,P 1i ,wi ,whi ,R1 ,xhi , xi ,P 2i ,vi forms a hamiltonian path of AQnF joining u and v. See Fig. 5(c) for an illustration.
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Subcase 2.2: u 2 V ðAQ0n1 Þ and v 2 V ðAQ1n1 Þ.Let A = {xj(x, v) 2 E(AQn) and x 2 V ðAQ1n1 Þg. Obviously, jAj = 2n 3.Let B = {yhjy 2 A} [ {ycjy 2 A}. By Lemma 3, jBj = 4n 8.Since jFj 6 2n 4, there exists a y 2 A such that either (y, yh) or (y, yc), say (y, y*), is F-fault free and w* 5 u. Suppose that there exists a hamiltonian path P of AQ0n1 F ðAQ0n1 Þ joining u and y*. Since ðlðpÞ j F ðAQ0n1 Þ jÞ=2 P 2n 4, there exists an edge (w, x) in P such that (w, wh) and (x, xh) are F-fault free and wh, xh and v are distinct. We can write P as hu, P1, w, x, P2, y*i. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P3 of AQ1n1 F ðAQ1n1 Þ fv,yg joining wh and xh. Thus, hu, P1, w, wh, P3, xh, x, P2, y*, y, vi forms a hamiltonian path of AQn F joining u and v. Suppose that there is no hamiltonian path of AQ0n1 F ðAQ0n1 Þ joining u and y*. Since AQn1 is (2n 6)fault hamiltonian connected, there exists a hamiltonian path Qi of AQ0n1 ðF ðAQ0n1 Þ ffi gÞ joining u and y* for every fi 2 F ðAQ0n1 Þ. We can write Qi as hu,P 1i ,ri ,fi ,si ,P 2i ,y i. Since j F ðAQ0n1 Þ j¼ 2n 5 P 5 andj F F ðAQ0n1 Þ j6 1, there exists an index i such that ðri ,rhi Þ and ðsi ,shi Þ are F-fault free. Since AQ1n1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path R1 of AQ1n1 ðF ðAQ1n1 Þ fy,vgÞ joining rhi and shi . Then hu,P 1i ,ri ,rhi ,R1 ,shi ,si , P 2i ,y ,y,vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(d) for an illustration. Subcase 2.3: u,v 2 V ðAQ1n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there is a hamiltonian cycle C of AQ0n1 F . Let A = {1x1 . . . xnj there exists some i and j in (2, . . . , n) such that xi ¼ ui , xj ¼ uj , and xk ¼ uk if k 5 {i, j} and let B = {1x2 . . . xnjthere exists some k 2 {2, . . . , n} such that xk = xk and xi ¼ xi if i 5 k}. Obviously, A [ B is the subset of x 2 V ðAQ1n1 Þ j d AQ1 ðx,uÞ ¼ 2g. n1 n n1 n1 Moreover, j A [ B j¼ þ ¼ . Further2 1 2 more, every vertex in A [ B has two common neighbors with n 1 u in AQn1 . Since ð2n 4Þ P 4, there are at least four 2 vertices in A [ B, say {xij1 6 i 6 4}, such that xhi 2 C. Let wi be a common neighbor of u and xi. Moreover, let yhi and zhi be the two neighbors of xhi in C. Obviously, fyhi j 1 6 i 6 4g [ fzhi j 1 6 i 6 4g P 6. There exists some i such that one of yhi and zhi , say yhi , satisfying yi 62 {u,v, wi} and yi is F-fault free. We can write C as hxhi ,zhi ,P 1 ,yhi ,xhi g.Since AQn1 is (2n 6)-fault hamiltonian, there exists a hamiltonian path P2
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of AQ1n1 F ðAQ1n1 Þ fu,wi ,xi g joining bi and v. Then hu,wi ,xi , xhi ,zhi ,P 1 ,yhi ,yi ,P 2 ,vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(e) for an illustration. Case 3: j F ðAQ0n1 Þ j¼ 2n 4. Hence, j F F ðAQ0n1 Þ j¼ 0. Subcase 3.1: u,v 2 V ðAQ0n1 Þ. Let f1 and f2 be any two elements in F ðAQ0n1 Þ. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P joining u and v in AQ0n1 ðF ðAQ0n1 Þ ff1 ,f2 gÞ which may or may not include f1 or f2 on it. Removing f1 and f2, path P is divided into three, two, or one pieces depending on the cases in which f1 and f2 are on P or not. For the last two cases, we may arbitrarily delete one or two more edges from P to make it into three subpaths. Therefore, without loss of generality, we may write these three subpaths as hu,P 1 ,w,f10 ,x,P 2 ,y,f20 ,z,P 3 ,vi. Suppose that l(P2) P 1. Hence, wh, xh, yh and zh are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hwh, P4, xhi and hyh, P5, zhi in AQ1n1 . Then hu, P1, w, wh, P4, xh, x, P2, y,yh, P5, zh, z, P3, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(f) for an illustration. Suppose that l(P2) = 0. Hence, x = y. Suppose that fðx,wÞ,ðx,zÞg \ C n ¼ ;. Then h x , xc, wh and zh are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hwh, P6, xhi and hxc, P7, zhi in AQ1n1 . Then hu, P1, w, wh, P6, xh, x, P2, y, yh, P7, zh, z, P3, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(g) for an illustration. Suppose that fðx,wÞ,ðx,zÞg \ C n 6¼ ;. Without loss of generality, we assume that (x, w) is in C n . Then xh = wc and xc = wh. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path P8 of AQ1n1 fxc g joining xh and zh. Then hu, P1, w, xc, x, xh, P8, zh, z, P3, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(h) for an illustration. Subcase 3.2: u 2 V ðAQ0n1 Þ and v 2 V ðAQ1n1 Þ. Let f1 be any element in F ðAQ0n1 Þ. Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle C ¼ hu,P 1 ,w,f10 ,x,P 2 ,ui in AQ0n1 ðF ðAQ0n1 Þ f1 Þ where f10 ¼ f1 if f1 is on cycle C or f10 is any edge of C otherwise. Without loss of generality, we may assume that l(P1) 6 l(P2). Then we rewrite C as hu,P 1 ,w,f10 , x,P 3 ,y,ui. Suppose that l(P1) = 0. Hence, u = w. Thus, we can rewrite C as hu,f10 ,x,P 03 ,ui. Then there exists a vertex x 0 2 {xh, xc} such that x 0 5 v. Since AQn1 is (2n 6)-fault hamiltonian connected, there exists a hamiltonian path P4 of AQ1n1 joining x 0 and v. Then hu,P 03 ,x,x0 ,vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(l) for an illustration. Suppose that l(P1) P 1. Then we have the following three cases. (1) wh, xh, yh, and v are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hwh, P4, xhi and hyh, P5, vi in AQ1n1 . Then hu, P1, w, wh, P4, xh, x, P3, y, yh, P5, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(j) for an illustration.
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(2) yh = v. Since AQn1 is (2n6)-fault hamiltonian connected, there is a hamiltonian path hwh, P6, xhi of AQ1n1 fvg. Thus, hu, P1 w, wh, P6, xh,x, P2, y, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(k) for an illustration. (3) xh = v or wh = v. Without loss of generality, we assume that xh = v. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path hwh, P7, yhi of AQ1n1 fvg. Thus, hu, P1, w, wh, P7, yh, y, P2, x, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(1) for an illustration. Subcase 3.3: u,v 2 V ðAQ1n1 Þ. Suppose that F ðAQ0n1 Þ C n . By definition, the hypercube Qn1 is a spanning subgraph of AQ0n1 F ðAQ0n1 Þ. Thus, there is a hamiltonian cycle C in AQ0n1 F ðAQ0n1 Þ [9]. We claim that there is an edge (w, x) on C such that wh, xh, u, and v are distinct. Since l(C) = 2n1 and (2n1/ 2) 3 = 2n2 3 P 5, there are at least five edges we can choose. Hence, such edge exists. Since AQ1n1 has property 2H, there are two disjoint spanning paths hu, P2, whi and hxh, P3, vi of AQ1n1 . Then hu, P2, wh, w, P1, x, xh, P3, vi is a hamiltonian path of AQn F joining u and v. See Fig. 5(m) for an illustration. Suppose that F ðAQ0n1 Þ 6 C n . Let f1 be any element in F ðAQ0n1 Þ C n . Since AQn1 is (2n 5)-fault hamiltonian, there exists a hamiltonian cycle C ¼ hw,P 1 ,x, f10 ,wiin AQ0n1 ðF ðAQ0n1 Þ f1 Þ where f10 ¼ f1 if f1 is incident with C or f10 is any edge of C otherwise. Hence, wh, wc, xh, and xc are distinct. We have the following two cases. (1) wh, xh, u, and v are distinct. Since AQ1n1 has property 2H, there are two disjoint spanning paths hu, P4, whi and hxh, P5, vi of AQ1n1 . Then hu, P4, wh, w, P1, x, xh, P5, vi hamiltonian path of AQn F joining u and v. See Fig. 5(n) for an illustration. (2) wh, xh, u, and v are not distinct. Without loss of generality, we assume that wh = u. Let x 0 be a element in {xh, xc} such that x 0 5 v. Since AQn1 is (2n 6)-fault hamiltonian connected, there is a hamiltonian path P6 of AQ1n1 fug joining x 0 and v. Then hu, w, P1, x, x 0 , P6, vi forms a hamiltonian path of AQn F joining u and v. See Fig. 5(o) for an illustration. Hence, this lemma is proved.
h
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