Fault-tolerant hamiltonian connectedness of cycle composition networks

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Fault-tolerant hamiltonian connectedness of cycle

composition networks

Tz-Liang Kueng

a

, Cheng-Kuan Lin

a

, Tyne Liang

a,*,1

,

Jimmy J.M. Tan

a

, Lih-Hsing Hsu

b,2

a_{Department of Computer Science, National Chiao Tung University, Hsinchu 30050, Taiwan, ROC} b_{Department of Computer Science and Information Engineering, Providence University, Taichung 43301, Taiwan, ROC}

Abstract

It is important for a network to tolerate as many faults as possible. With the graph representation of an interconnection network, a k-regular hamiltonian and hamiltonian connected network is super fault-tolerant hamiltonian if it remains hamiltonian after removing up to k 2 vertices and/or edges and remains hamiltonian connected after removing up to k 3 vertices and/or edges. Super fault-tolerant hamiltonian networks have an optimal ﬂavor with regard to the fault-tol-erant hamiltonicity and fault-tolfault-tol-erant hamiltonian connectivity. For this reason, a cycle composition framework was pro-posed to construct a (k + 2)-regular super fault-tolerant hamiltonian network based on a collection of n k-regular super fault-tolerant hamiltonian networks containing the same number of vertices for n P 3 and k P 5. This paper is aimed to emphasize that the cycle composition framework can be still applied even when k = 4.

Keywords: Hamiltonian; Hamiltonian connected; Fault tolerance; Super fault-tolerant hamiltonian

1. Introduction

The architecture of an interconnection network is usually represented by a graph whose vertices and edges represent processors and communication links, respectively. Thus, we use the terms graph and network inter-changeably. Throughout this paper, we concentrate on loopless undirected graphs. For the graph deﬁnitions

and notations we follow the ones given by Bondy and Murty[1]. A graph G consists of a nonempty set V(G)

and a subset E(G) of {(u, v)j(u, v) is an unordered pair of V(G)}. The set V(G) is called the vertex set of G and

E(G) is called the edge set. Two vertices u and v of G are adjacent if (u, v)2 E(G). A graph H is a subgraph of G

if V(H) V(G) and E(H) E(G). Let S be a nonempty subset of V(G). The subgraph induced by S is the

subgraph of G with its vertex set S and with its edge set which consists of those edges joining any two vertices

* _{Corresponding author.}

E-mail address:[email protected](T. Liang).

1

The author was supported in part by the National Science Council of the Republic of China, under Contract NSC 95-2221-E-009-194.

2

This work was supported in part by the National Science Council of the Republic of China, under Contract NSC 95-2221-E-233-002. Available online at www.sciencedirect.com

Applied Mathematics and Computation 196 (2008) 245–256

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in S. We use G S to denote the subgraph of G induced by V(G) S. Analogously, the subgraph generated

by a nonempty subset F E(G) is the subgraph of G with its edge set F and its vertex set consisting of those

vertices of G incident with at least one edge of F. We use G F to denote the subgraph of G with vertex set

V(G) and edge set E(G) F. The degree of a vertex u in G, denoted by degG(u), is the number of edges incident

to u. A graph G is k-regular if all its vertices have the same degree k. A matching of size k in a graph G is a set of k edges with no shared endpoints. The vertices belonging to the edges of a matching are saturated by the matching; the others are unsaturated. A perfect matching is a matching that saturates every vertex of G.

A path P of length k from a vertex x to a vertex y in a graph G is a sequence of distinct vertices

hv0; v1; v2; . . . ; vki such that x = v0, y = vk, and ðvi1; viÞ 2 EðGÞ for every 1 6 i 6 k. More precisely, we write

P¼ hv0; e1; v1; e2; v2; . . . vk1; ek; vki, in which ei¼ ðvi1; viÞ 2 EðGÞ for every i. For convenience, we write P

ashv0; . . . ; vi; Q; vj; . . . ; vki where Q ¼ hvi; . . . ; vji. Note that we allow Q to be a path of length zero. Moreover,

we use P1to denote the pathhvk; vk1; . . . ; v1; v0i. To emphasize the beginning and ending vertices of P, we

also write P as P[x, y]. A path of a graph G is a hamiltonian path if it spans G. A cycle is a path with at least three vertices such that the ﬁrst vertex is the same as the last one. A cycle of G is a hamiltonian cycle if it tra-verses all vertices of G. A graph G is hamiltonian if it has a hamiltonian cycle, and G is hamiltonian connected if there exists a hamiltonian path joining any two vertices of G.

A suitable network is generally designed to satisfy some speciﬁed requirements. For example, the hamilto-nian property is one of the major concerns for designing the network topology and fault tolerance is desirable in massive parallel systems. So these two properties can be concerned in the network topology as follows. A graph G is called l-fault-tolerant hamiltonian (resp. l-fault-tolerant hamiltonian connected) if it remains ham-iltonian (resp. hamham-iltonian connected) after removing at most l vertices and/or edges. The fault-tolerant

hamiltonicity of G, HfðGÞ, is deﬁned to be the maximum integer l such that G F remains hamiltonian

for every F V(G) [ E(G) with jFj 6 l if G is hamiltonian, and undeﬁned otherwise. Obviously, HfðGÞ 6

dðGÞ 2, where d(G) = min{degG(v)jv 2 V(G)}. A regular graph G is optimal fault-tolerant hamiltonian if

HfðGÞ ¼ dðGÞ 2. The fault-tolerant hamiltonian connectivity of G, HjfðGÞ, is deﬁned to be the maximum

integer l such that G F remains hamiltonian connected for every F V(G) [ E(G) with jFj 6 l if G is

ham-iltonian connected, and undeﬁned otherwise. Obviously, HjfðGÞ 6 dðGÞ 3. A regular graph G is optimal

fault-tolerant hamiltonian connected if HjfðGÞ ¼ dðGÞ 3. A regular graph is super fault-tolerant hamiltonian

if HfðGÞ ¼ dðGÞ 2 and HjfðGÞ ¼ dðGÞ 3. For instance, twisted-cubes, crossed-cubes, mo´bius cubes and

recursive circulant graphs are all super fault-tolerant hamiltonian[2,4–6,8].

A network will have higher fault tolerance if it is super fault-tolerant hamiltonian. For this reason, Chen

et al.[3]proposed a systematic framework to recursively construct super fault-tolerant hamiltonian graphs as

follows. Let G0, G1, . . ., Gn 1be n k-regular super fault-tolerant hamiltonian graphs with the same number of

vertices. The cycle composition network H ¼ GðG0; G1; . . . ; Gn1; M0;1; M1;2; . . . ; Mn2;n1; Mn1;0Þ is deﬁned to

G0 G1 G2 Gn-1 a perfect matching a perfect matching a perfect matching

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be the graph with the vertex set VðH Þ ¼Sn1_i¼0VðGiÞ and the edge set EðH Þ ¼

Sn1

i¼0ðEðGiÞ [ Mi;iþ1Þ where Mi;jis

an arbitrary perfect matching between the vertices of Giand those of Gj. SeeFig. 1. Then Chen et al.[3]proved

that GðG0; G1; . . . ; Gn1; M0;1; M1;2; . . . ; Mn2;n1; Mn1;0Þ, abbreviated as Gh0;1;...;n1;0i, is super fault-tolerant

hamiltonian for n P 3 and k P 5.

Theorem 1 [3]. Assume n P 3 and k P 5. Let G0; G1; . . . ; Gn1 be n k-regular super fault-tolerant hamiltonian

graphs with the same number of vertices. For any 0 6 i 6 n 1, let Mi;iþ1 be a perfect matching between the

vertices of Giand those of Gi +1. Then Gh0;1;...;n1;0i is (k + 2)-regular super fault-tolerant hamiltonian.

For instance, the recursive circulant graph, which was proposed by Park and Chwa[7], is essentially

con-structed as a special case in this way, and it is shown to be super fault-tolerant hamiltonian under a certain

condition [8]. Similarly, k-ary n-cubes are also recursively constructed using the same framwork [9]. In this

paper, we shall extendTheorem 1by showing that Gh 0, 1, . . ., n1, 0 iis still super fault-tolerant hamiltonian even

when k = 4. Such an extension is signiﬁcant because only the remaining case of k = 3 needs to be concerned carefully or to be checked by computer while the topological properties of cycle composition networks are investigated.

2. Fault-tolerant hamiltonicity

For the ease of exposition, the notations we used in this paper are described as follows. We denote the

graph GðGi; Giþ1; . . . ; Gj; Mi;iþ1; Miþ1;iþ2; . . . ; Mj1;jÞ by Ghi;iþ1;...;ji. Let u be a vertex of Giwith some i. We use

(u)to denote the vertex of Gi1such that ((u), u)2 Mi1, i, and use (u)+to denote the vertex of Gi + 1such

that ðu; ðuÞþÞ 2 Mi;iþ1. Hence, we have u¼ ððuÞÞþ¼ ððuÞþÞ. Moreover, all additions and subtractions are

considered modulo n. In order to prove the main results, we need the following lemmas.

Lemma 1. Assume n P 1. Let G0; G1; . . . ; Gn1 be n 4-regular super fault-tolerant hamiltonian graphs with the

same number of vertices. For any 0 6 i 6 n 2, let Mi, i+1be a perfect matching between the vertices of Giand

those of Gi + 1. Moreover, let Fi V(Gi)[ E(Gi) withjFij 6 1 for every 0 6 i 6 n 1 and let Xi;iþ1 Mi;iþ1with

jXi;iþ1j 6 1 such that jFij þ jFiþ1j þ jXi;iþ1j 6 2 is satisfied for all 0 6 i 6 n 2. Let u and v be two vertices of

G0 F0. Then there is a hamiltonian path of Gh0;...;n1i ððSi¼0n1FiÞ [ ðSn2_i¼0Xi;iþ1ÞÞ joining u to v.

Proof. For convenience, let F ¼ ðSn1_i¼0FiÞ [ ð

Sn2

i¼0Xi;iþ1Þ. We prove this lemma by induction on n. Obviously,

the statement is trivial when n = 1. For any n P 2, we suppose that the statement holds for n 1. Depending

on jV(G0)j, two cases are distinguished.

Case 1: Suppose thatjV(G0)j = 5. Thus, G0is isomorphic to the complete graph K5. First assumejF0j = 0.

Since jF0j þ jF1j þ jX0;1j 6 2, we can choose two vertices x, y of G0 such that j{x, y} \ {u, v}j 6 1 and

jF \ fðxÞþ;ðyÞþ;ðx; ðxÞþÞ; ðy; ðyÞþÞgj ¼ 0. Accordingly, we can construct a hamiltonian path P ¼ hu; P1;

x; y; P2; vi of G0, in which P1or P2may be a path of length zero. On the other hand, assume thatjF0j = 1. Since

G0is 4-regular super fault-tolerant hamiltonian, there is a hamiltonian path P of G0 F0joining u to v. Since

jF0j þ jF1j þ jX0;1j 6 2 and jF0j = 1, there exists an edge (x, y) on P such that jF \ fðxÞþ;ðyÞþ;ðx; ðxÞþÞ;

ðy; ðyÞþÞgj ¼ 0. Accordingly, we write P ¼ hu; P1; x; y; P2; vi, in which P1or P2may be a path of length zero. By

induction hypothesis, there is a hamiltonian path T of Gh1;...;n1i ððSn1i¼1FiÞ [ ðSn2_i¼1Xi;iþ1ÞÞ joining (x)+to

(y)+. Thenhu; P1; x;ðxÞþ; T ;ðyÞþ; y; P2; vi is a hamiltonian path of Gh0;1;...;n1i F joining u to v. SeeFig. 2for

illustration. x y u v G<1,...,n-1> T G0 P1 P2 + (x) + (y) G1 Gn-1

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Case 2: Suppose that jV(G0)j P 6. Since G0 is super fault-tolerant hamiltonian, there is a hamiltonian

path P of G0 F0 joining u to v. Since jF0j þ jF1j þ jX0;1j 6 2, there exists an edge (x, y) on P such

thatjF \ fðxÞþ;ðyÞþ;ðx; ðxÞþÞ; ðy; ðyÞþÞgj ¼ 0. Accordingly, we write P ¼ hu; P1; x; y; P2; vi, in which P1or P2

may be a path of length zero. By induction hypothesis, there is a hamiltonian path T of G_h1;...;n1i ððSn1i¼1FiÞ [ ðSn2i¼1Xi;iþ1ÞÞ joining (x)+to (y)+. Thenhu; P1; x;ðxÞþ; T ;ðyÞþ; y; P2; vi is a hamiltonian

path of Gh0;1;...;n1i F joining u to v. h

same number of vertices. For any 0 6 i 6 n 2, let Mi;iþ1 be a perfect matching between the vertices of Giand

those of Gi + 1. Moreover, let Fi V(Gi)[ E(Gi) withjFij 6 1 for every 0 6 i 6 n 1 and let Xi;iþ1 Mi;iþ1 with

jXi;iþ1j 6 1 for every 0 6 i 6 n 2 such that jFij þ jFiþ1j þ jXi;iþ1j 6 2 is satisfied for all 0 6 i 6 n 2. Let u

be a vertex of G0 F0 and v be a vertex of Gt Ft with t P 0. Then there is a hamiltonian path of

Gh0;...;n1i ððSn1i¼0FiÞ [ ðSn2i¼0Xi;iþ1ÞÞ joining u to v.

Proof. For convenience, let F ¼ ðSn1_i¼0FiÞ [ ð

Sn2

i¼0Xi;iþ1Þ. When t = 0, the statement follows from Lemma 1.

Hence, we suppose t > 0 in the following. Since Gt is 4-regular, we have jV(Gt)j P 5. Moreover, since

jFt1j + jFtj + jXt1, tj 6 2, we can choose a vertex w of Gt (Ft[ {v}) such that jF \ fw; ðwÞ;

ðw; ðwÞÞgj ¼ 0 and (w)5u.

Let y0= u and xt1= (w). Since every Gi, 0 6 i 6 t 1, is 4-regular and jFij þ jFiþ1j þ jXi;iþ1j 6 2, we

sequentially choose a vertex xi of Gi Fi and denote (xi)+ by yi + 1 such that xi5yi and

jF \ fxi; yiþ1;ðxi; yiþ1Þgj ¼ 0 from i = 0 to i = t 3 while t P 3. Next, we choose a vertex xt2 of

Gt2 (Ft2[ {yt2}) and denote (xt2)+by yt1such thatjF \ {xt2, yt1, (xt2, yt1)}j = 0 and yt15xt1

while t P 2. Since every Gi, 0 6 i 6 t 1, is super fault-tolerant hamiltonian, there is a hamiltonian path Piof

Gi Fijoining yito xi. ByLemma 1, there is a hamiltonian path T of Ght;...;n1i ððSn1i¼tFiÞ [ ðSn2i¼t Xi;iþ1ÞÞ

joining w to v. Thenhu ¼ y₀; P0; x0;ðx0Þþ¼ y1; . . . ; xt2;ðxt2Þþ¼ yt1; Pt1; xt1¼ ðwÞ; w; T ; vi is a

hamilto-nian path of Gh0;...;n1i F joining u to v. SeeFig. 3for illustration. h

UsingLemma 2, we prove the following result.

Theorem 2. Assume n P 3. Let G0; G1; . . . ; Gn1be n 4-regular super fault-tolerant hamiltonian graphs with the

same number of vertices. For any 0 6 i 6 n 1, let Mi;iþ1 be a perfect matching between the vertices of Giand

those of Gi+1. Then Gh0;1;...;n1;0i is optimal fault-tolerant hamiltonian.

Proof. Obviously, Gh0;1;...;n1;0iis 6-regular. Thus, we are going to show that it is 4-fault-tolerant hamiltonian.

Let F be a faulty set of Gh0;1;...;n1;0iwithjFj 6 4. For convenience, let Fi= F\ (V(Gi)[ E(Gi)) for 0 6 i 6 n 1.

Without loss of generality, we assume thatjF0j P jFij for all 1 6 i 6 n 1. Depending on jF0j, ﬁve cases are

distinguished.

Case 1: Suppose thatjF0j = 4. Let F0¼ ff1; f2; f3; f4g. Since G0is 2-fault-tolerant hamiltonian, there is a

hamiltonian cycle C in G0 ff3; f4g.

Subcase 1.1: Suppose that both f1 and f2 are on C but they are not adjacent. Thus, we can write

C¼ hx1; f1; y1; H1; x2; f2; y2; H2; x1i, in which H1or H2may be a path of length zero. ByLemma 2, there is a

hamiltonian path S1½ðx1Þ;ðy1Þ

in Gn 1and there is a hamiltonian path S2½ðx2Þþ;ðy2Þ

þ in Gh1;...;n2i. Then u G0 G1 Gt-1 x0 P0 P1 x1 + (x0) + (x1) Pt-1 v w Gn-1 Gt T xt-2(xt-2) + xt-1 = yt-1 = y1 = y0 _{= y}₂ -(w) G<t,...,n-1>

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hx1;ðx1Þ; S1;ðy1Þ

; y₁; H1; x2,ðx2Þþ; S2,ðy2Þ þ

; y₂; H2; x1i is a hamiltonian cycle of Gh0;1;...;n1;0i F . SeeFig. 4a

for illustration.

Subcase 1.2: Suppose that both f1 and f2 are on C and they are adjacent. Thus, we write C¼

hx; R; y; f1; f2; xi. By Lemma 2, there is a hamiltonian path H of Gh1;...;n1i joining (y)+ to (x)+. Then

hx; R; y; ðyÞþ; H ;ðxÞþ; xi is a hamiltonian cycle of Gh0;1;...;n1;0i F . SeeFig. 4b for illustration.

Subcase 1.3: Suppose that either f1or f2is on C. Without loss of generality, we assume that f1is on C. Thus,

we write C ashx; R; y; f1; xi. Then a hamiltonian cycle of Gh0;1;...;n1;0i F can be formed in the same way as

that used in Subcase 1.2.

Subcase 1.4: Suppose that neither f1 nor f2 is on C. Thus, we write C as hx; R; y; xi with any edge

(x, y)2 E(C). Then a hamiltonian cycle of Gh0;1;...;n1;0i F can be formed in the same way as that used in

Subcase 1.2.

Case 2: Suppose that jF0j = 3. Let F0¼ ff1; f2; f3g. Since G0 is 2-fault-tolerant hamiltonian, there is a

hamiltonian cycle C in G0 ff2; f3g. Thus, we have either f162 V(C) [ E(C) or f12 V(C) [ E(C). Accordingly,

we write C ¼ hx; R; y; xi by picking any edge (x, y) on C if f162 V(C) [ E(C); we write C = hx, R, y, f1,xi if f1is

on C. Let F0_{= F}_F

0. Since jFj 6 4 and jF0j = 3, jF0j 6 1. Moreover, one can see either

jfðxÞþ;ðyÞþ;ðx; ðxÞþÞ; ðy; ðyÞþÞg \ F j ¼ 0 or jfðxÞ;ðyÞ;ðx; ðxÞÞ; ðy; ðyÞÞg \ F j ¼ 0. With symmetry, we

assume that jfðxÞþ;ðyÞþ;ðx; ðxÞþÞ; ðy; ðyÞþÞg \ F j ¼ 0. By Lemma 2, there is a hamiltonian path H of

Gh1;...;n1i F0 joining (y)+to (x)+. Thenhx; R; y; ðyÞþ; H ;ðxÞþ; xi is a hamiltonian cycle of Gh0;1;...;n1;0i F .

Case 3: Suppose thatjF0j = 2 and jFij = 2 with any 1 6 i 6 n 1. Since both G0and Giare 2-fault-tolerant

hamiltonian, there is a hamiltonian cycle C in G0 F0and there is a hamiltonian cycle T in Gi Fi. Since

every Gj, 0 6 j 6 n 1, is 4-regular, jV(Gj)j P 5.

Subcase 3.1: Suppose that i2 {1,n 1}. With symmetry, we assume that i = 1. Apparently, there is a vertex

u in G0 F0 such that (u)+ is in G1 F1. Without loss of generality, we write C =hu, R1, x, ui and

T =h(u)+

, y, R2, (u)+i so that (y)+ is diﬀerent from (x). By Lemma 2, there is a hamiltonian path H of

Gh2;...;n1i F joining (x) to (y)+. Then hu, R1, x, (x), H, (y)+, y, R2, (u)+, ui is a hamiltonian cycle of

Gh0;1;...;n1;0i F . See Fig. 5a.

Subcase 3.2: Suppose that i62 {1, n 1}. Obviously, there is a vertex u in G0 F0and a vertex v in Gi Fi

such that (u)+5(v). Without loss of generality, we write C =hu, x, R1, ui and T = hv, R2, y, vi so that (y)+is

diﬀerent from (x). ByLemma 2, there is a hamiltonian path P1of Gh1;...;i1i joining (u)+to (v). Similarly,

there is a hamiltonian path P2 of Ghiþ1;...;n1i joining (y)+ to (x). Then hu, (u)+, P1, (v), v, R2, y, (y)

+

,

P2, (x), x, R1, ui is a hamiltonian cycle of Gh0;1;...;n1;0i F . See Fig. 5b for illustration.

Case 4: Suppose thatjF0j = 2 and jFij 6 1 for every 1 6 i 6 n 1. Since G0is 2-fault-tolerant hamiltonian,

there is a hamiltonian cycle C in G0 F0. Since G0is 4-regular, we havejV(G0 F0)j P 3. For convenience, let

m =jV(G0 F0)j. Accordingly, we write C ¼ hu0; u1; u2; . . . ; um1; u0i. Without loss of generality, we assume

thatjF \ {(u0)+, (u1), (u0, (u0)+), (u1, (u1))}j = 0. Let F0= F F0. ByLemma 2, there is a hamiltonian path T

of Gh1;...;n1i F0 joining (u0)+ to (u1). Then hu0;ðu0Þþ; T ;ðu1Þ; u1; . . . ; um1; u0i is a hamiltonian cycle of G_{h0;1;...;n1;0i} F . See Fig. 6a for illustration.

Case 5: Suppose thatjF0j 6 1. That is, jFij 6 1 for all 0 6 i 6 n 1. For convenience, let Xi, i+1= F\ Mi, i+1

for 0 6 i 6 n 1. Suppose that there exists an integer t of f0; 1; . . . ; n 1g such that jFtj + jFt+1j +

jXt, t+1j P 3. Without loss of generality, t can be assumed to be n 1. Otherwise, t is ﬁxed to be n 1.

Accordingly, we havejFij + jFi+1j + jXi, i+1j 6 2 for 0 6 i 6 n 2. Since jFn1j + jF0j + jXn1, 0j 6 4, we can

choose a vertex x of Gn1 Fn1such thatjF \ {(x)+, (x, (x)+)}j = 0. Let F0= F Xn1, 0. ByLemma 2, there

x y R H -(x1) -(y1) G0 G1 Gn- 1 G0 G1 H2 H1 S1 S2 x1 y1 x2 y2 + (x2) + (y2) G<1,... ,n-1> Gn- 1 + (x) + (y) Gn-2 G<1,... ,n-2>

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is a hamiltonian path T of Gh0;1;...;n1i F0 joining x to (x)+. Then hx, T, (x)+, xi is a hamiltonian cycle of Gh0;1;...;n1;0i F . SeeFig. 6b for illustration. h

3. Fault-tolerant hamiltonian connectedness

In this section, we are going to show that the cycle composition network is optimal fault-tolerant hamilto-nian connected. This result is divided into three propositions.

Proposition 1. Assume n P 1. Let G0; G1; . . . ; Gn1 be n 4-regular super fault-tolerant hamiltonian graphs with

the same number of vertices. For any 0 6 i 6 n 1, let Mi, i+1be a perfect matching between the vertices of Giand

those of Gi+1. Let F be a subset of V(G0)[ E(G0) withjFj = 3. Then Gh0;1;...;n1;0i F is hamiltonian connected.

Proof. Let F = {f1, f2, f3}. Since G0 is 2-fault-tolerant hamiltonian, there is a hamiltonian cycle C in

G0 {f2, f3}. Since G0is 4-regular, jV(C)j P 3. Let u and v be two vertices of Gh0;1;...;n1;0i F . Then we have

to construct a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. The following cases are distinguished.

Case 1: Suppose that u and v are in G0 F. Since G0is 1-fault-tolerant hamiltonian connected, there is a

hamiltonian path H of G0 {f3} joining u to v. Suppose that f1and f2are exclusive from H. Thus, we write

H =hu, P1, x, y, P2, vi with any edge (x, y) 2 E(H). Suppose that either f1or f2is exclusive from H. Without loss

of generality, we assume that f2is exclusive from H. Thus, we may write H =hu, P1, x, f1, y, P2, vi. Suppose that

both f1and f2are on H and they are adjacent. Thus, we may write H =hu, P1, x, f1, f2, y, P2, vi. ByLemma 2,

there is a hamiltonian path T of Gh1;...;n1i joining (x)+ to (y)+. Then hu, P1, x, (x)

+

, T, (y)+, y, P2, vi is a

hamiltonian path of Gh0;1;...;n1;0i F joining u to v. SeeFig. 7a for illustration.

Suppose that both f1 and f2 are on H but they are not adjacent. Thus, we may write H =hu, A1,

x1, f1, y1, A2, x2, f2, y2, A3, vi. UsingLemma 2, we can ﬁnd a hamiltonian path D1of Gh1;...;n2ijoining (x1)+to

(y1)+. Similarly, there is a hamiltonian path D2 of Gn1 joining (x2) to (y2). Hence, hu, A1, x1,

xu

a

b

y x u y v G0 G1 Gn-1 G2 + (y) -(x) + (u) R1 R2 G<2,...,n-1> G<i+1,...,n-1> G0 Gn-1 Gi Gi+1 P1 P2 (u)+ -(x) _(y)+ -(v) G1 G<1,...,i-1> Gi-1 R1 _R₂ H

Fig. 5. Illustration for case 3 ofTheorem 2.

G0 G1 Gn-1 + (u0) T (u1) -u1 u2 um -1 u0 G<1,...,n-1> T + (x) Gn-1 G1 G0 x

a

b

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(x1)+, D1, (y1)+, y1, A2, x2, (x2), D2, (y2), y2, A3, vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. See

Fig. 7b for illustration.

Case 2: Suppose that u and v are in Gifor some 1 6 i 6 n 1. With symmetry, we assume that i 5 n 1.

Suppose that f1is on the hamiltonian cycle C of G0 {f2, f3}. SincejV(C)j P 3, we write C = hx, P, y, f1, xi.

Otherwise, we write C =hx, P, y, xi with any edge (x, y) 2 E(C).

Subcase 2.1: Suppose that (x)+5u and (x)+5v. Thus, either (y)5(u)+or (y)5(v)+. Without loss of

generality, we assume that (y)5(v)+. ByLemma 2, there is a hamiltonian path T1of Gh1;...;ii fvg joining u

to (x)+. Similarly, there is a hamiltonian path T2 of Ghiþ1;...;n1i joining (y) to (v)+. Then

hu, T1, (x)+, x, P, y, (y), T2, (v)+, vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. See Fig. 8a for

illustration.

Subcase 2.2: Suppose that (x)+= u or (x)+= v. Without loss of generality, we assume that (x)+= u. By

Lemma 2, there is a hamiltonian path T of Gh1;...;n1i fug joining (y)to v. Thenhu = (x)+, x, P, y, (y), T, vi is

a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. SeeFig. 8b for illustration.

Case 3: Suppose that u is in G0 F and v is in Giwith any i > 0. Since i 5 1 or i 5 n 1, we may assume

that i 5 1. Since jV(C)j P 3, we write C = hu, T, z, ui with z 5 u. Moreover, T can be written as

hu, P1, x, f1, y, P2, zi if f1is on T, or T can be written ashu, P1, x, y, P2, zi otherwise.

Subcase 3.1: Suppose that (z)5v. Since G1 is 1-fault-tolerant hamiltonian connected, there is a

hamiltonian path H of G1joining (x)+to (y)+. ByLemma 2, there is a hamiltonian path R of Gh2;...;n1ijoining

(z)to v. Thenhu, P1, x, (x)

+

, H, (y)+, y, P2, z, (z), R, vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v.

See Fig. 9a.

Subcase 3.2: Suppose that (z)= v. ByLemma 2, there is a hamiltonian path H of Gh1;...;n1i fvg joining

(x)+to (y)+. Thenh u, P1, x, (x)+, H, (y)+, y, P2, z, (z)= vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to

v. SeeFig. 9b for illustration.

Case 4: Suppose that u is in Giand v is in Gjfor any 1 6 i < j 6 n 1. Suppose that f1is on C. Then we write

C =hx, P, y, f1, xi. Otherwise, we write C = h x, P, y, xi with any (x, y) 2 E(C). Since (x)+5u or (y)+5u, we

may assume that (x)+5u.

T F

-a

b

x y u v + (x) + (y) P1 P2 G0 G1 G<1,...,n-1> G<1,...,n-2> Gn- 1 Gn-1 G0-F G1 D1 D2 Gn-2 A1 A2 A3 u v + (x1) + (y1) -(x2) -(y2) x1 y1 y2 x2

Fig. 7. Illustration for case 1 ofProposition 1.

P yx

a

b

v u P y x T G1 G0 G_i Gn-1 T₁ T2 + (x) + (v) G<i+1,...,n-1> G<1,...,n-1> G1 G0 Gn-1 -(y) (y)- u (x)= + Gi+1 G<1,...,i> v

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Subcase 4.1: Suppose that (y)5v. ByLemma 2, there is a hamiltonian path T1of Gh1;...;iijoining u to (x)+.

Similarly, there is a hamiltonian path T2of Ghiþ1;...;n1ijoining (y)to v. Thenhu, T1, (x)+, x, P, y, (y), T2, vi is a

Subcase 4.2: Suppose that (y)= v. ByLemma 2, there is a hamiltonian path H of Gh1;...;n1i fvg joining u

to (x)+. Thenhu, H, (x)+, x, P, y, (y)= vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. SeeFig. 10(b)

for illustration. h

the same number of vertices. For any 0 6 i 6 n 1, let Mi,i+1be a perfect matching between the vertices of Gi

and those of Gi+1. Let F be a faulty set of Gh0;1;...;n1;0i such that jFj = 3 and jF \ (V(G0)[ E(G0))j = 2. Then

Gh0;1;...;n1;0i F is hamiltonian connected.

Proof. For convenience, let Fi= F\ (V(Gi)[ E(Gi)) and Xi,i+1= F\ Mi,i+1 for every 0 6 i 6 n 1.

More-over, let F0_{= F}_F

0. Obviously, we havejF0j = 2, jF0j = 1, and jFij 6 1 for all 1 6 i 6 n 1. Since G0is

4-reg-ular, jV(G0)j P 5 and jV(G0 F0)j P 3. Moreover, since G0 is 2-fault-tolerant hamiltonian, there is a

hamiltonian cycle C in G0 F0. Let u and v be any two vertices of Gh0;1;...;n1;0i F . Then we have to construct

a hamiltonian path of Gh0;1;...;n1;0i F joining u to v.

Case 1: Suppose that u and v are in G0 F0. SincejV(G0 F0)j P 3, we may write C = hu, P, y, ui in which

y 5 u. Moreover, we may write P =hu, H1, x, v, H2, yi. Note that the length of H1becomes zero if u = x. Since

jF0_{j = 1, we have jX}

0,1j + jF1j = 0 or jXn1,0j + jFn1j = 0. With symmetry, we assume that jX0,1j + jF1j = 0.

By Lemma 2, there is a hamiltonian path T of Gh1;...;n1i F0 joining (x)+ to (y)+. Then

hu; H1; x;ðxÞþ; T ;ðyÞþ; y; H12 ; vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. See Fig. 11 for

illustration.

Case 2: Suppose that u and v are in either G1 F1or Gn1 Fn1. With symmetry, we assume that u and v

are in G1 F1. y x H u z y u z v Gn-1 G2 G<2,...,n-1> Gi (z)- _x _(x)+ + (y) P2 P1 G1 G0 R H P1 + (x) + (y) -v (z)= Gn-1 G1 G0 G<1,...,n-1> P2

P yx v u P H u + (x) -(y) T1 T2 G1 G0 Gi Gj Gi+1 Gn-1 G<i+1,...,n-1> G<1,...,i> y x + (x) Gj Gn-1 G1 G0 v=(y) -G<1,...,n-1>

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Subcase 2.1: Suppose thatjX0,1j + jF1j = 1. Since jV(G0 F0)j P 3, we choose a vertex x of the hamiltonian

cycle C such thatjF0_{\ {(x)}+

, (x, (x)+)}j = 0. Hence, C can be written as C = hy, x, z, P, yi. Since (x)+₅

u or

(x)+5v, we assume that (x)+5v. Since G1is 1-fault-tolerant hamiltonian connected, there is a hamiltonian

path Q[u, v] of G1 F1. Since (x)

+

5v, we write Q =hu, T1, (x)

+

, w, T2, vi. Note that T1or T2may be a path of

length zero. Moreover, we select a vertex from {y, z}, say y, such that (y)5(w)+. ByLemma 2, there is a

hamiltonian path H of Gh2;...;n1i joining (y) to (w)+. Then hu, T1, (x)+, x, z, P, y, (y), H, (w)+, w, T2, vi is a

Subcase 2.2: Suppose that jX0,1j + jF1j = 0. Thus, we can choose a vertex x of C such that jF0\

{(x)+, (x, (x)+)}j = 0 and (x)+62 {u, v}. Hence, the hamiltonian cycle C of G0 F0 can be written as

C =hy, x, z, P, yi.

Subcase 2.2.1: Suppose thatj{(y)+, (z)+}\ {u, v}j P 1. Without loss of generality, we assume that (z)+= u.

By Lemma 2, there is a hamiltonian path T of Gh1;...;n1i ðF0[ fugÞ joining (x)+ to v. Then hu =

(z)+, z, P, y, x, (x)+, T, vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. SeeFig. 12b for illustration.

Subcase 2.2.2: Suppose that j{(y)+

, (z)+}\ {u, v}j = 0. Since jF0_{\ {(y)}_{, (y, (y)}_{)}j = 0 or jF}0_{\ {(z)}_,

(z, (z))}j = 0, we assume that jF0_{\ {(y)}

, (y, (y))}j = 0. Since G1is 1-fault-tolerant hamiltonian connected,

there is a hamiltonian path Q of G1 {((x)+, ((y)))}. Since (x)+62 {u, v}, Q can be represented by

h u, T1, w1, (x)+, w2, T2, vi. Note that T1 or T2 may be a path of length zero. Accordingly, we have that

jF0_{\ {(w}

1)+, (w1, (w1)+)}j = 0 or jF0\ {(w2)+, (w2, (w2)+)}j = 0. Without loss of generality, we assume that

jF0_{\ {(w}

2) +

, (w2, (w2) +

)}j = 0. ByLemma 2, there is a hamiltonian path H of Gh2;...;n1i F0 joining (y)to

(w2) + . Thenh u, T1, w1, (x) + , x, z, P, y, (y), H, (w2) +

, w2, T2, vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u

to v. SeeFig. 12c.

Case 3: Suppose that u and v are in Gi Fiwith any 1 < i < n 1. Without loss of generality, we assume

that Pi1_j¼1jFjj þPi1j¼0jXj;jþ1j ¼ 0. Since jV(G0 F0)j P 3, we ﬁrst choose a vertex x of C such that

jF0_{\ {(x)}_{, (x, (x)}_{)}j = 0. Thus, we can write C = hz, x, y, P, zi. Next, we choose a vertex t of G}

i (Fi[ {u})

such thatjF0_{\ {(t)}+

,(t, (t)+)}j = 0 and (t)+5(x). Since Giis 1-fault-tolerant hamiltonian connected, there is

a hamiltonian path H in Gi Fijoining u to t. Then H can be represented byhu, R1, w, v, R2, ti, in which R1or

R2may be a path of length zero. Since (y)+5(w)or (z)+5(w), we assume that (y)+5(w). ByLemma 2,

there is a hamiltonian path T1of Gh0;...;i1i F0joining (w)to (y)+. Similarly, there is a hamiltonian path T2

of Ghiþ1;...;n1i F0joining (x)to (t)+. As a result,hu; R1; w;ðwÞ; T1;ðyÞþ; y; P ; z; x;ðxÞ; T2;ðtÞþ; t; R12 ; vi is a

u v y x v z u v w P H

a

b

C

+ (x) + u (z)= G0 G1 Gn-1 G<1,...,n-1> P T G0 G1 G2 Gn-1 G<2,...,n-1> y x z + (x) + (w) T1 T2 -(y) H G2 Gn-1 G<2,...,n-1> -(y) y x z P T1 T2 + (w2) w1 w2 + (x) G0 G1

x v u y T H1 H2 (x) (y) Gn-1 G1 G0 G< 1,...,n-1>

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Case 4: Suppose that u is in G0 F0and v is in Gi Fiwith any i > 0. SincejV(G0 F0)j P 3, we can write

C =hx, u, y, P, xi. Since jF0_{j = 1, we have jX}

0,1j + jF1j = 0 or jXn1,0j + jFn1j = 0. Without loss of generality,

we assume jX0,1j + jF1j = 0. Hence, we have (x)+5v or (y)+5v. Without loss of generality, we assume

(x)+5v. By Lemma 2, there is a hamiltonian path H of Gh1;...;n1i F0 joining (x)+ to v. Then

hu, y, P, x, (x)+, H, vi is a hamiltonian path of Gh0;1;...;n1;0i F joining u to v. SeeFig. 13b for illustration.

Case 5: Suppose that u is in Gi Fiand v is in Gj Fj for any 1 6 i < j 6 n 1. Since jF0j = 1, we have

jX0,1j + jF1j = 0 or jXn1,0j + jFn1j = 0. Without loss of generality, we assume jXn1,0j + jFn1j = 0. Since

jV(G0 F0)j P 3, we can choose a vertex x of C such that (x)+5u andjF0\ {(x)+, (x, (x)+)}j = 0. Moreover,

at least one neighbor of x on C, namely y, satisﬁes (y)5v. Accordingly, we can write C =h x, P, y, xi. By

Lemma 2, there is a hamiltonian path T1of Gh1;...;ii F0joining u to (x)+. Similarly, there is a hamiltonian path

T2 of Ghiþ1;...;n1i F0 joining (y) to v. Then hu, T1, (x)+, x, P, y, (y), T2, vi is a hamiltonian path of

Gh0;1;...;n1;0i F joining u to v. SeeFig. 13c for illustration. h

same number of vertices. For any 0 6 i 6 n 2, let Mi, i+1 be a perfect matching between the vertices of Gi

and those of Gi+1. Moreover, let Fi V(Gi)[ E(Gi) withjFij 6 1 for every 0 6 i 6 n 1 and let Xi,i+1 Mi,i+1

withjXi,i+1j 6 1 for every 0 6 i 6 n 2 such that jFij + jFi+1j + jFi+2j + jXi,i+1j + jXi+1,i+2j 6 2 is satisfied for all

0 6 i 6 n 3. Let u and v be two vertices of Gt Ft with 0 < t < n 1. Then there is a hamiltonian path of

Gh0;...;n1i ððSn1i¼0FiÞ [ ð

Sn2

i¼0Xi;iþ1ÞÞ joining u to v.

Proof. For convenience, let F ¼ ðSn1_i¼0FiÞ [ ðSn2_i¼0Xi;iþ1Þ. Since Gt is 4-regular super fault-tolerant

hamilto-nian, there is a hamiltonian path P of Gt Ftjoining u to v. Depending onjFtj, we distinguish the following

two cases.

Case 1: Suppose that jFtj = 1. Thus, one can see that jV(Gt Ft)j P 4. Let w1= u. Thus, we write P as

hu = w1, w2, w3, w4, R, vi. Since jFtj = 1, one can see that jFt1j + jFt+1j + jXt1,tj + jXt,t+1j 6 1. Hence, we

select a vertex wifrom {w2, w3} such thatjF \ {(wi), (wi)+, (wi, (wi)), (wi, (wi)+)}j = 0. Accordingly, one can

see that either jF \ {(wi1)+, (wi+1),(wi1, (wi1)+), (wi+1, (wi+1))}j = 0 or jF \ {(wi1), (wi+1)+, (wi1,

(wi1)), (wi+1, (wi+1)+)}j = 0. Without loss of generality, we assume jF \ {(wi1)+, (wi+1), (wi1, (wi1)+),

(wi+1, (wi+1))}j = 0. Hence, we can further write P as h u = w1, P1, wi1, wi, wi+1, P2, vi. ByLemma 2, there is a

hamiltonian path T of Gh0;...;t1i ððSt1i¼0FiÞ [ ðS_i¼0t2Xi;iþ1ÞÞ joining (wi) to (wi+1). Similarly, there is a

hamiltonian path Q of Ghtþ1;...;n1i ððSn1i¼tþ1FiÞ [ ðSn2_i¼tþ1Xi;iþ1ÞÞ joining (wi1)+ to (wi)+. Then hu = w1,

P1,wi1,(wi1)+, Q, (wi) +

, wi, (wi), T, (wi+1), wi+1, P2, vi is a hamiltonian path of Gh0;...;n1i F joining u to v.

P xy t w z v u x u y v P H G1 G0 Gi Gi+1 Gn-1 + (y) R1 R2 + (t) Gi-1 T1 T2 G<i+1,...,n-1> -(x) -(w) G<1,...,i-1> G1 G0 + (x) Gi Gn-1 G<1,...,n-1> P yx v u G1 G0 + (x) Gn-1 -(y) Gj Gi T1 T2 Gi+1 G<1,...,i> G<i+1,...,n-1>

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Case 2: Suppose thatjFtj = 0. First, assume that jV(Gt)j P 6. Hence, we can select two adjacent edges (x, y),

(y, z)2 E(P) such that jF \ {(x)+, (y)+, (y), (z), (x, (x)+), (y, (y)+), (y, (y)), (z, (z))}j = 0 or jF \ {(x),

(x, (x)), (y), (y, (y)), (y, (y)+), (z, (z)+), (y)+, (z)+}j = 0. Without loss of generality, we assume that jF \ {(x)+,

(y)+, (y), (z), (x, (x)+), (y, (y)+), (y, (y)), (z, (z))}j = 0. Accordingly, P can be written as hu, P1, x, y, z, P2, vi,

in which P1 or P2 may be a path of length zero. By Lemma 2, there is a hamiltonian path T of

G_h0;...;t1i ððSt1i¼0FiÞ [ ðSt2i¼0Xi;iþ1ÞÞ joining (y) to (z). Similarly, there is a hamiltonian path Q of G_{htþ1;...;n1i} ððSn1i¼tþ1FiÞ [ ðSn2i¼tþ1Xi;iþ1ÞÞ joining (x)+ to (y)+. Then hu, P1, x, (x)+, Q, (y)+, y, (y), T, (z),

z, P2, vi is a hamiltonian path of Gh0;...;n1i F joining u to v.

Next, assume thatjV(Gt)j = 5. Thus, Gtis isomorphic to the complete graph K5. Let V(Gt) = {u = w1, w2,

w3, w4, w5= v}. First of all, we choose a vertex from {w2,w3,w4}, say w2, such that jF \ {(w2),

(w2)+, (w2, (w2)), (w2, (w2)+)}j = 0. Secondly, we choose two vertices x, y from {w3, w4, w5} such that

jF \ {(x)+

, (x, (x)+), (y), (y, (y))}j = 0. Accordingly, a hamiltonian path of Gtcan be written ash u = w1, P1,

x, w2, y, P2, w5= vi. Then a hamiltonian path of Gh0;...;n1i F joining u to v can be formed in a way similar to

that mentioned above. h

same number of vertices. For any 0 6 i 6 n 2, let Mi,i+1be a perfect matching between the vertices of Giand

those of Gi+1. Moreover, let Fi V(Gi)[ E(Gi) withjFij 6 1 for every 0 6 i 6 n 1 and let Xi,i+1 Mi,i+1with

jXi,i+1j 6 1 for every 0 6 i 6 n 2 such that jFij + jFi+1j + jFi+2j + jXi,i+1j + jXi+1,i+2j 6 2 is satisfied for all

0 6 i 6 n 3. Let u be a vertex of Gs Fs and v be a vertex of Gt Ftwith 0 6 s 6 t 6 n 1. Then there is

a hamiltonian path of Gh0;...;n1i ððSn1i¼0FiÞ [ ðSn2_i¼0Xi;iþ1ÞÞ joining u to v.

Proof. For convenience, let F ¼ ðSn1_i¼0FiÞ [ ðSn2_i¼0Xi;iþ1Þ. When s = 0, the statement follows fromLemma 2.

When 0 < s = t < n 1, the statement follows from Lemma 3. So, we consider the case when 0 < s < t in

the following. Since Gs is 4-regular, we have jV(Gs)j P 5. Moreover, since jFsj + jFs+1j + jXs,s+1j 6 2, we

can choose a vertex x of Gs (Fs[ {u}) such that jF \ {x, (x)+, (x, (x)+)}j = 0 and (x)+5v. By Lemma 2,

there is a hamiltonian path P of Gh0;...;si ððSsi¼0FiÞ [ ðSs1i¼0Xi;iþ1ÞÞ joining u to x. Similarly, there is a

hamil-tonian path T of Ghsþ1;...;n1i ððSn1i¼sþ1FiÞ [ ðSn2i¼sþ1Xi;iþ1ÞÞ joining (x)+to v. Thenhu, P, x, (x)+, T, vi is a

ham-iltonian path of Gh0;...;n1i F joining u to v. h

the same number of vertices. For any 0 6 i 6 n 1, let Mi,i+1 be a perfect matching between the vertices of Gi

and those of Gi+1. Let F be a faulty set of Gh0;1;...;n1;0i such that jFj = 3 and jF \ (V(Gi)[ E(Gi))j 6 1 for

0 6 i 6 n 1. Then Gh0;1;...;n1;0i F is hamiltonian connected.

Proof. Let u be a vertex of Ga Faand let v be a vertex of Gb Fbfor any 0 6 a 6 b 6 n 1. For

conve-nience, let Fi= F\ (V(Gi)[ E(Gi)) and Xi,i+1= F\ Mi,i+1 for every 0 6 i 6 n 1. Obviously, we have

jFij 6 1. Moreover, let t be the integer such that jXt,t+1j = max{jXi,i+1j j0 6 i 6 n 1}. Depending on jXt,t+1j,

two cases are distinguished.

Case 1: Suppose thatjXt,t+1j P 1. Without loss of generality, t can be assumed to be n 1. Accordingly, we

have jXi,i+1j 6 1 for every 0 6 i 6 n 2. Let F0= F Xn1,0. Hence, we have jF0j 6 2 and jFij + jFi+1j +

jFi+2j + jXi,i+1j + jXi+1,i+2j 6 2 for all 0 6 i 6 n 3. ByLemma 4, Gh0;1;...;n1i F0is hamiltonian connected.

Case 2: Suppose thatjXt,t+1j = 0. Then we set t to be a 1. Obviously, we have jFij + jFi+1j + jXi,i+1j 6 2

for all 0 6 i 6 n 2. ByLemma 2, Gha;aþ1;...;n1;0;...;a1i F is hamiltonian connected.

Finally, Gh0;1;...;n1;0i F is concluded to be hamiltonian connected. h

Theorem 3. Assume n P 3. Let G0; G1; . . . ; Gn1 be n 4-regular super fault-tolerant hamiltonian graphs with the

same number of vertices. For any 0 6 i 6 n 1, let Mi,i+1be a perfect matching between the vertices of Giand

those of Gi+1. Then Gh0;1;...;n1;0i is optimal fault-tolerant hamiltonian connected.

Proof. Obviously, Gh0;1;...;n1;0i is 6-regular. Thus, we are going to show that Gh0;1;...;n1;0i is 3-fault-tolerant

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F\ (V(Gi)[ E(Gi)) for 0 6 i 6 n 1. Without loss of generality, we assume that jF0j P jFij for all 1 6

i 6 n 1. Depending on jF0j, three cases are distinguished. The ﬁrst case that jF0j = 3 is proved byProposition

1. The second case whenjF0j = 2 is proved byProposition 2. Finally, the case forjF0j 6 1 follows from

Prop-osition 3. h

According toTheorem 1–3, we have the following corollary.

Corollary 1. Assume n P 3 and k P 4. Let G₀; G₁; . . . ; Gn1 be n k-regular super fault-tolerant hamiltonian

graphs with the same number of vertices. For any 0 6 i 6 n 1, let Mi,i+1 be a perfect matching between the

vertices of Giand those of Gi+1. Then Gh0;1;...;n1;0i is (k + 2)-regular super fault-tolerant hamiltonian.

4. Conclusion

In this paper, we improve the result of Chen et al.[3]by showing that on the basis of n 4-regular super

fault-tolerant hamiltonian networks G0; . . . ; Gn1, n P 3, the cycle composition network Gh0;1;...;n1;0iis super

fault-tolerant hamiltonian. However, we conjecture that this result may not be true based on n cubic networks. Therefore, such an extension is signiﬁcant because only the remaining case for 3-regular graphs needs to be checked with brute force while the topological properties of the cycle composition network is investigated. References

[1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North Holland, New York, 1980.

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[3] Y.-C. Chen, L.-H. Hsu, J.J.M. Tan, A recursively construction scheme for super fault-tolerant hamiltonian graphs, Applied Mathematics and Computation 177 (2006) 465–481.

[4] W.-T. Huang, Y.-C. Chuang, J.J.M. Tan, L.-H. Hsu, Fault-free hamiltonian cycle in faulty mo´bius cubes, Journal of Computing and Systems 4 (2000) 106–114.

[5] W.-T. Huang, Y.-C. Chuang, L.-H. Hsu, J.J.M. Tan, On the fault-tolerant hamiltonicity of crossed cubes, IEICE Transaction on Fundamentals E85-A (6) (2002) 1359–1371.

[6] W.-T. Huang, J.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.

[7] J.-H. Park, K.-Y. Chwa, Recursive circulant: a new topology for multicomputer networks, in: Proceedings of the International Symposium on Parallel Architectures, Algorithms and Networks (I-SPAN’94), IEEE Computer Society Press, New York , 1994, pp. 73–80.

[8] C.-H. Tsai, J.J.M. Tan, Y.-C. Chuang, L.-H. Hsu, Hamiltonian properties of faulty recursive circulant graphs, Journal of Interconnection Networks 3 (2002) 273–289.

[9] M.-C. Yang, J.J.M. Tan, L.-H. Hsu, Hamiltonian circuit and linear array embedding in faulty k-ary n-cubes, Journal of Parallel and Distributed Computing 4 (2007) 362–368.