Fault-free Hamiltonian cycles in crossed cubes with conditional link faults

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Fault-free Hamiltonian cycles in crossed cubes

with conditional link faults

Hao-Shun Hung

a

, Jung-Sheng Fu

b

, Gen-Huey Chen

a,*

a

Department of Computer Science and Information Engineering, National Taiwan University, Taipei 10764, Taiwan b

Department of Electronics Engineering, National United University, Miaoli, Taiwan Received 16 August 2006; received in revised form 23 April 2007; accepted 19 May 2007

Abstract

The crossed cube, which is a variation of the hypercube, possesses some properties superior to the hypercube. In this paper, assuming that each node is incident with at least two fault-free links, we show that an n-dimensional crossed cube contains a fault-free Hamiltonian cycle, even if there are up to 2n 5 link faults. The result is optimal with respect to the number of link faults tolerated. We also verify that the assumption is practically meaningful by evaluating its occurrence probability, which is very close to 1.

Keywords: Conditional link fault; Crossed cube; Fault-tolerant embedding; Forbidden faulty set model; Hamiltonian cycle; Hypercube; Interconnection network

1. Introduction

The hypercube is a popular interconnection network (network for short) with many attractive properties such as regularity, symmetry, small diameter, strong connectivity, recursive construction, partition ability, and relatively low link complexity[25]. On the other hand, the crossed cube[6,7], which can result by changing some connections of the hypercube, is superior to the hypercube in diameter and mean distance. The diameter of an n-dimensional crossed cube isd(n + 1)/2e, which is about one half of the diameter of an n-dimensional hypercube. The mean distance of the crossed cube is smaller than the hypercube.

The crossed cube has been studied extensively in the literature [3,4,8,11–13,17,19,20,29]. It was demon-strated in[19] that a (2n 1)-node complete binary tree can be embedded into a 2n

-node crossed cube with dilation one. The dilation will go up to two if the same tree is embedded into a 2n-node hypercube (see

[28]). Path embedding in the crossed cube can be found in [11,12]. The connectivity of an n-dimensional crossed cube is n (see[20]). It was shown in[3]that both the n-wide diameter and the (n 1)-fault diameter of an n-dimensional crossed cube weredn/2e + 2. In[29], the crossed cube was shown to be pancyclic. It was

* _{Corresponding author. Tel.: +886 2 23625336x427; fax: +886 2 23628167.} E-mail address:[email protected](G.-H. Chen).

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further shown in[17,29]that an n-dimensional crossed cube is (n 2)-Hamiltonian, (n 3)-Hamiltonian-con-nected, and (n 2)-fault-tolerant pancyclic. Recently, fault diagnosis of the crossed cube was explored in

[4,13].

The random fault model, which was considered in [15,17,29,30], assumed that faults might happen any-where in a network without any restriction. On the other hand, Harary introduced in[16]the concept of con-ditional faults. Let P represent a property of a graph G and S be a vertex subset of G. The P-connectivity of G was deﬁned to be the minimum |S| so that G S is disconnected and every component of G S satisﬁes the property P. Considering P the property that each node is incident with at least one fault-free node, P-connec-tivities were computed for hypercubes[9], k-ary n-cubes[5], cubes-connected cycles[23], undirected de Bruijn networks[23], and Kautz networks[23]. In addition, under the same assumption (i.e., the property P), the diameters of hypercubes and star graphs were computed in[21,24], respectively.

In the past 20 years, few Hamiltonian properties were derived on networks with conditional faults, although P-connectivities were obtained for some networks. It appears that Hamiltonicity problems on net-works become more diﬃcult, when conditional faults are assumed. Previous results were obtained only on hypercubes and k-ary n-cubes. Under the assumption that each node is incident with two or more fault-free links, it was shown in [1,2] that a k-ary n-dimensional hypercube (an n-dimensional hypercube) contains a fault-free Hamiltonian cycle, even if there are up to 4n 5 (2n 5) link faults. Besides, it was shown in

[26]that an n-dimensional hypercube with 2n 5 link faults is strongly (fault-free) Hamiltonian laceable. In this paper, with the same assumption as[1,2], we show that an n-dimensional crossed cube contains a fault-free Hamiltonian cycle, even if there are up to 2n 5 link faults. The result is optimal with respect to the number of link faults tolerated. The rest of this paper is organized as follows. In Section2, the structure of the crossed cube is reviewed. Some necessary deﬁnitions and notations are also introduced. In Section3, some favorable properties of the crossed cube are derived. These properties will help the derivation of the main result. In Section4, the main result and its optimality are shown. In Section5, the probability that the assump-tion holds when there are 2n 5 link faults is analyzed. In Section6, this paper concludes with some remarks.

2. Preliminaries

A network is conveniently represented with an undirected graph, where the vertices (edges) of the graph denote the nodes (links) of the network. Throughout this paper, vertex and node, edge and link, and graph and network are used interchangeably. Moreover, we use CQnto denote an n-dimension crossed cube. CQn

can be constructed recursively. Initially, CQ1contains a link whose two end nodes are labeled with 0 and

1, respectively, and CQ2 is isomorphic to a two-dimensional hypercube. For n P 3, CQn can be obtained

by joining two CQn1’s, denoted by CQ0n1 and CQ 1

n1, with 2 n1

links, as described below.

Each node label of CQ0_n1 CQ1_n1 is preceded with a bit 0 (1). There is a link joining a node u = 0un2un3. . . u0 in CQ0_n1 with another node v = 1vn2vn3. . . v0 in CQ1_n1 if and only if

(u2i+1u2i, v2i+1v2i)2 {(0 0, 0 0), (1 0, 1 0), (0 1, 1 1), (1 1, 0 1)} for all 0 6 i 6 b(n 1)/2c 1 and un2= vn2 if n

is even. In subsequent discussion (u, v) is referred to as a crossing link of CQnand u(v) is referred to as the

crossing node of v(u) with respect to CQn. For convenience, we use cn(v) to denote the crossing node of v with

respect to CQn. So, we have cn(v) = u and cn(u) = v.Fig. 1depicts CQ3and CQ4, where (0 0 0, 1 0 0), (0 1 0, 1 1 0),

(0 0 1, 1 1 1) and (0 1 1, 1 0 1) are four crossing links of CQ3.

Formally, CQn can be deﬁned as follows, where we use u2i+1u2i v2i+1v2i to denote (u2i+1u2i, v2i+1v2i)2

{(0 0, 0 0), (1 0, 1 0), (0 1, 1 1), (1 1, 0 1)}.

Deﬁnition 1 [6]. The node set of CQn is {vn1vn2. . . v0| vi2 {0, 1} for all 0 6 i 6 n 1}. Two nodes

u = un1un2. . . u0and v = vn1vn2. . . v0of CQnare adjacent if and only if there exists 0 6 d 6 n 1 so that

the following four conditions are satisﬁed:

(1) un1un2. . . ud+1= vn1vn2. . . vd+1, if d < n 1;

(2) ud¼ vd (vd is the complement of vd);

(3) ud1= vd1, if d is odd;

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Notice that (u, v) above is a crossing link of CQd+1, which joins CQ0dwith CQ1d. Such a link is called a d-link. Every node in CQnis incident with n links, which are 0-link, 1-link, 2-link, . . ., (n 1)-link. A path (cycle) in a

graph G is called a Hamiltonian path (cycle) if it contains every vertex of G exactly once. The following two lemmas were shown in[17].

Lemma 1 [17]. A CQnwith at most n 2 link faults contains a fault-free Hamiltonian cycle, where n P 3.

Lemma 2 [17]. Suppose that there are felink faults and fvnode faults in a CQn. If fe+ fv6n 3, then for every

two fault-free nodes u, v of the CQn, there is a fault-free path of length 2n fv 1 joining u and v in the CQn,

where n P 3.

In fact, the fault-free path of length 2n fv 1 inLemma 2is a Hamiltonian path in the graph that results

by removing the fvfaulty nodes from the CQn.

3. Properties

In this section, some properties of the crossed cube are derived. They will be used when we construct a fault-free Hamiltonian cycle in a faulty crossed cube, which was described in the next section.

Lemma 3 [20]. Suppose that (u, v) is a d-link in CQn, where d is odd or d = n 2. Then, (cn(u), cn(v)) is also a

d-link.

For a graph G, we use V(G) (E(G)) to denote the vertex set (edge set) of G. Also we use Px,yto denote a path

between two nodes x, y of the crossed cube.

Lemma 4. Suppose that u, v, x and y are four distinct nodes in CQn, where n P 4. There exist Pu,vand Px,yso

that we have V(Pu,v)\ V(Px,y) = B and V(Pu,v)[ V(Px,y) = V(CQn).

Proof. We prove this lemma by induction on n. For n = 4, the lemma can be easily veriﬁed by exhaustive search (by the aid of a computer program)[33]. We assume that the lemma holds for n = k P 4. The situation of n = k + 1 is discussed with the following four cases.

Case 1: u is in V CQ _k0and x is in V CQ 1_k. If v is in V CQ 0_kand y is in V CQ 1_k, then byLemma 2, there are a Hamiltonian path between u and v in CQ0_k and a Hamiltonian path between x and y in CQ1_k, which are a desired Pu,vand a desired Px,y, respectively.

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If v is in V CQ1 k and y is in V CQ0 k

, then there is a Hamiltonian path between u and y in CQ0

k similarly. A

link (s, t) in the path can be found so that we have ck+1(s), ck+1(t)62 {x,v}. Refer to Fig. 2a, where

s0_{= c}

k+1(s) and t0= ck+1(t). By the induction hypothesis, there are Ps0_,v and P_x;t0 so that we have VðPs0_;vÞ \ V ðP_x;t0Þ ¼ £ and V ðP_s0_;vÞ [ V ðP_x;t0Þ ¼ V CQ 1_k. Hence a desired P_u,vand a desired P_x,ycan be con-structed as the two bold paths inFig. 2a.

If v and y are in V CQ1 k

, then we select a node s in CQ0

k so that we have s62 {u, v, y} and ck+1(s) 5 x. A

desired Pu,v and a desired Px,y can be constructed as shown in Fig. 2b, where we have

s0_{¼ c}

kþ1ðsÞ; V ðPu;vÞ \ V ðPy;sÞ ¼ £; V ðPu;vÞ [ V ðPy; sÞ ¼ V CQ0k

, and VðPx;s0Þ ¼ V CQ1

k

(i.e., Px;s0 is a Hamiltonian path in CQ1_k). The discussion for the situation that v and y are in V CQ 1_kis similar. Case 2: u is in V CQ 1_kand x is in V CQ 1_k. The construction of a desired Pu,vand a desired Px,yis similar to

Case 1.

Case 3: u and x are in V CQ 0_k. If v is in V CQ 0_kand y is in V CQ _k1or v is in V CQ 1_kand y is in V CQ 0_k, then a desired Pu,vand a desired Px,ycan be obtained by the construction method ofFig. 2b.

If v and y are in V CQ 0_k, then we arbitrarily select a link (s, t) in Pu,v. A desired Pu,vand a desired Px,ycan

be constructed as shown in Fig. 3a, where we have s0_{¼ c}

kþ1ðsÞ; t0¼ ckþ1ðtÞ; V ðPu;vÞ \ V ðPx;yÞ ¼ £; VðPu;vÞ [ V ðPx;yÞ ¼ V CQ0k , and VðPs0_;t0Þ ¼ V CQ1 k . If v and y are in V CQ1 k

, then we arbitrarily select nodes s,t from V CQ0 k

fx; ug so that we have ck+1(s),

ck+1(t)62 {v, y}. A desired Pu,v and a desired Px,y can be constructed as shown in Fig. 3b, where we

have s0_{¼ c}

kþ1ðsÞ; t0¼ ckþ1ðtÞ; V ðPu;sÞ \ V ðPx;tÞ ¼ £; V ðPu;sÞ [ V ðPx;tÞ ¼ V CQ0k

; VðPy;t0Þ \ V ðP_v;s0Þ ¼ £, andðPy;t0Þ [ V ðP_v;s0Þ ¼ V CQ 1_k.

Case 4: u and x are in V CQ1 k

. The construction of a desired Pu,vand a desired Px,yis similar to Case 3. h

Lemma 5. Suppose that (u, v) is an arbitrary fault-free (n 1)-link in a CQnwith at most n 2 link faults, where

n P 4. There exists a fault-free Hamiltonian cycle in the CQnthat contains (u, v).

Proof. Let fc be the number of faulty crossing links of the CQn and f0(f1) be the number of link faults in

CQ0_n1

CQ1_n1

. We have fc+ f0+ f16n 2. Without loss of generality, we assume f0P f1. We also

assume that u is in V CQ 0_n1and v is in V CQ 1_n1. For n = 4, the lemma can be easily veriﬁed by exhaustive search. For n P 5, three cases are discussed below:

Case 1: f0< n 3. There is a fault-free (n 1)-link (x, x0) 5 (u, v), where x is in V CQ0_n1

and x0_{= c} n(x).

ByLemma 2, there are a fault-free Hamiltonian path between u and x in and a fault-free Hamiltonian path

Fig. 2. Construction of Pu,vand Px,ywhen u is in V CQ0k and x is in V CQ1 k . (a) v is in V CQ0 k and y is in V CQ1 k . (b) v and y are in V CQ0k .

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between u and x0_{in CQ}0

n1. A desired fault-free Hamiltonian cycle can be constructed as the bold cycle in

Fig. 4a.

Case 2: f0= n 3. ByLemma 1, there is a fault-free Hamiltonian cycle in CQ0_n1. Since f0+ f1+ fc6n 2,

we have fc61. There is a fault-free (n 1)-link (x, x0) 5 (u, v), where x is neighboring to u in the

Hamil-tonian cycle and x0_{= c}

n(x). ByLemma 2, there is a fault-free Hamiltonian path between x0and v in CQ1_n1.

A desired fault-free Hamiltonian cycle can be obtained by the construction method ofFig. 4a.

Case 3: f0= n 2. We have fc= f1= 0. Suppose that (x, y) is a faulty link in CQ0_n1. There are n 3 link

faults in CQ0_n1, exclusive of (x, y). ByLemma 1(imagining that (x, y) is fault-free), there exists a Hamilto-nian cycle in that can avoid the n 3 link faults. If (x, y) is not contained in the Hamiltonian cycle, then we select a fault-free (n 1)-link (t, t0_{) 5 (u, v), where t is neighboring to u in the Hamiltonian cycle and}

t0_{= c}

n(t). A desired fault-free Hamiltonian cycle can be obtained by the construction method ofFig. 4a.

Otherwise ((x, y) is contained in the Hamiltonian cycle), if x = u, then a desired fault-free Hamiltonian cycle can be obtained by the construction method ofFig. 4a (replacing (x, x0_{) with (y, y}0_{), where y}0_{= c}

n(y)). If

x 5 u, then a desired fault-free Hamiltonian cycle can be constructed as shown inFig. 4b, where s62 {x, y}

Fig. 3. Construction of Pu,vand Px,ywhen u and x are in V CQ0k

. (a) v and y are in V CQ0 k . (b) v and y are in V CQ1 k .

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is neighboring to u in the Hamiltonian cycle of CQ0

n1 and (x, x0), (y, y0), (s, s0) are fault-free (n

1)-links. h

Lemma 6. Suppose that (u, v) is an arbitrary fault-free d-link in a CQnwith at most n 2 link faults, where

1 6 d 6 n 2 is odd and n P 4. There exists a fault-free Hamiltonian cycle in the CQnthat contains (u, v).

Proof. We prove this lemma by induction on n. For n = 4, the lemma can be easily veriﬁed by exhaustive search. We assume that the lemma holds for n = k P 4. Then we consider the situation of n = k + 1. Let fc

be the number of faulty crossing links of the CQk+1and f0(f1) be the number of link faults in CQ0_k CQ1_k

. We have f0+ f1+ fc6k 1. Without loss of generality, we assume f0P f1. Two cases are discussed below.

Case 1: f06k 2. If (u, v) is in E CQ0k

, then by the induction hypothesis, there exists a fault-free Hamil-tonian cycle in CQ0_k that contains (u, v). Moreover, there exists a link (s, t) 5 (u, v) in the Hamiltonian cycle so that (s, s0_{) and (t, t}0_{) are fault-free, where s}0_{= c}

k+1(s) and t0= ck+1(t). Since there are 2k(>2(k 1) + 1)

links contained in the Hamiltonian cycle, such a link (s, t) exists. On the other hand, we have f16b(k 1)/

2c < k 3, as a consequence of f0P f1and f0+ f1+ fc6k 1. Then, byLemma 2, there is a fault-free

Hamiltonian path between s0 _{and t}0_{in CQ}1

k. A desired fault-free Hamiltonian cycle can be constructed as

shown inFig. 5a. The discussion for the situation that (u, v) is in E CQ 1_kis similar.

Case 2: f0= k 1. We have fc= f1= 0. Suppose that (x, y) is a faulty link in CQ0k such that we have

{x0_{, y}0_{} 5 {u, v}, where x}0_{= c}

k+1(x) and y0= ck+1(y). There are k 2 link faults in CQ0k, exclusive of

(x, y). We ﬁrst assume (u, v) is in E CQ 0_k. By the induction hypothesis (imagining that (x, y) is fault-free) there exists a Hamiltonian cycle, denoted by C, in CQ0

k that contains (u, v), but does not contain the

k 2 link faults. A desired fault-free Hamiltonian cycle can be obtained by the construction method of

Fig. 5a (replacing (s, t) with (x, y) if (x, y) is contained in C, and replacing (s, t) with another link in C if (x, y) is not contained in C).

Then we assume that (u, v) is in E CQ1 k

. By Lemma 1(imagining that (x, y) is fault-free), there exists a Hamiltonian cycle, denoted by C0_{, in CQ}0

k that can avoid the k 2 link faults. If (x, y) is not contained

in C0_{, then we select a link (s, t) from C}0_{so that we have {s}0_{, t}0_}_{\ {u, v} = B, where s}0_c

k+1(s) and t0ck+1(t).

By Lemma 4, there are two paths Pu,s0 and P_v,t0 that satisfy VðP_u;s0Þ \ V ðP_v;t0Þ ¼ £ and V ðP_u;s0Þ[ VðPv;t0Þ ¼ V CQ1

k

. Besides, they are fault-free because f1= 0. A desired fault-free Hamiltonian cycle can

be constructed as shown inFig. 5b.

If (x, y) is contained in C0 _{and {x}0_{, y}0_}_{\ {u,v} = B, then a desired fault-free Hamiltonian cycle can be}

obtained by the construction method of Fig. 5b (replacing (s, t) with (x, y)). If (x, y) is contained in C0

and {x0_{, y}0_}_{\ {u, v} = B, then a desired fault-free Hamiltonian cycle can be obtained as shown in}

Fig. 5c. InFig. 5c, x0_{= u is assumed, and the fault-free path between v and y}0 _{that contains all nodes in}

V CQ 1_k fug can be assured byLemma 2(imagining that u is faulty). h

In the next section, we show the main result of this paper, which is stated below.

Fig. 5. Construction of a fault-free Hamiltonian cycle in CQk+1that contains (u, v). (a) f06k 2. (b) (x, y) is not contained in C0. (c) (x, y) is contained in C0_{and {x}0_{, y}0_}_{\ {u, v} 5 B.}

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Theorem 1. With the assumption that each node is incident with at least two fault-free links, there is a fault-free Hamilton cycle in a CQn with up to 2n 5 link faults, where n P 3. Moreover, the result is optimal with the

respect to the number of link faults tolerated. 4. Proof ofTheorem 1

For n = 3, the correctness of the theorem can be assured byLemma 1. So we assume n P 4. We prove the theorem by induction. When n = 4, the theorem can be easily veriﬁed by exhaustive search. We assume that the theorem holds for n = k P 4. In the rest of this section, the situation of n = k + 1 is considered.

Let fc be the number of faulty crossing links of the CQk+1 and f0(f1) be the number of link faults in

CQ0_k CQ1_k. We have f0+ f1+ fc62k 3. Without loss of generality, we assume f0P f1. Three cases are

dis-cussed below:

Case 1 : f062k 5. There is at most one node in CQ0k that is incident with only one fault-free link in CQ 0 k

(the other fault-free link is a crossing link of the CQk+1), for otherwise there are at least 2k 3 link faults in

CQ0_k, which is a contradiction.

We ﬁrst assume that every node in CQ0

k is incident with at least two fault-free links in CQ 0

k. By the induction

hypothesis, there exists a fault-free Hamiltonian cycle in CQ0

k. A link (s, t) can be found in the Hamiltonian

cycle so that (s, s0_{) and (t, t}0_{) are fault-free, where s}0_{= c}

k+1(s) and t0= ck+1(t). In addition, (s, t) should be

selected to be a (k 1)-link if f1= k 2 (it surely exists because fc61 as f1= k 2). If f16k 3, then by

Lemma 2there is a fault-free Hamiltonian path between s0_{and t}0_{in CQ}1

k. If f1= k 2, then byLemma 3

(s0_,t0_{) is a (k}_{1)-link, and by}_{Lemma 5}_{there is a fault-free Hamiltonian cycle in CQ}1

kthat contains (s0, t0). A

desired fault-free Hamiltonian cycle can be constructed as shown inFig. 6a.

Then we assume that a node p in CQ0_k is incident with only one fault-free link in CQ0_k. Hence, we have f0P k 1 (thus f1+ fc6k 2) and (p, p0) is fault-free, where p0= ck+1(p). A faulty link (p, q) with

(q, q0_{) fault-free can be found in CQ}0

k, where q0= ck+1(q). In addition, (p, q) should be selected to be a d-link

for some odd d if f1= k 2 (thus fc= 0). By the induction hypothesis (imagining that (p, q) is fault-free

because p is incident with only one fault-free link in CQ0_k), there is a Hamiltonian cycle that can avoid all faulty links but (p, q) in CQ0_k. Notice that (p, q) should be included in the Hamiltonian cycle.

If f16k 3, then by Lemma 2 there exists a fault-free Hamiltonian path between p0 and q0 in CQ1_k. If

f1= k 2, then byLemma 3(p0, q0) is a d-link, and byLemma 6there exists a fault-free Hamiltonian cycle

in CQ1

k that contains (p0, q0). A desired fault-free Hamiltonian cycle can be obtained by the construction

method ofFig. 6a (replacing (s, t) with (p, q)).

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Case 2 : f0= 2k 4. We have f1+ fc61. Similarly, there is at most one node in CQ0_k that is incident with

only one fault-free link in CQ0

k. If every node in CQ 0

k is incident with at least two fault-free links in CQ 0 k, then

we arbitrarily select a faulty link (x, y) in CQ0

k with (x, x0) and (y, y0) fault-free, where x0= ck+1(x) and

y0_{= c}

k+1(y). By the induction hypothesis (imagining that (x, y) is fault-free), there exists a Hamiltonian

cycle, denoted by C, in CQ0

k that can avoid all faulty links but (x, y) in CQ 0

k. A desired fault-free

Hamilto-nian cycle can be obtained by the construction method ofFig. 6a (replacing (s, t) with (x, y) if (x, y) is con-tained in C, and replacing (s, t) with another link in C if (x, y) is not concon-tained in C). On the other hand, if there is a node p in CQ0_k that is incident with only one fault-free link in CQ0_k, then we arbitrarily select a faulty link (p, q) in CQ0_k with (q, q0_{) fault-free, where q}0_{= c}

k+1(q). By the induction hypothesis (imaging that

(p, q) is fault-free), there is a Hamiltonian cycle that can avoid all faulty links but (p, q) in CQ0_k. A desired fault-free Hamiltonian cycle can be obtained by the construction method ofFig. 6a (replacing (s, t) with (p, q)).

Case 3 : f0= 2k 3. We have f1= fc= 0. There are at most two nodes in CQ0k that are incident with only

one fault-free link in CQ0_k. We ﬁrst assume that every node in CQ0_k is incident with at least two fault-free links in CQ0

k. There are two faulty links (x, y) and (g, h) in CQ 0

k so that we have {x, y}\ {g, h} = B. By

the induction hypothesis (imagining that (x, y) and (g, h) are fault-free), there exists a Hamiltonian cycle, denoted by C0_{, in CQ}0

k that can avoid all faulty links but (x, y) and (g, h).

If both (x, y) and (g, h) are contained in C0_{, then a desired fault-free Hamiltonian cycle can be constructed as}

shown in Fig. 6b, where we have x0_{= c}

k+1(x), y0= ck+1(y), g0= ck+1(g), h0= ck+1(h), and the two paths

Py0_;g0and P_x0_;h0 with VðP_y0_;g0Þ \ V ðP_x0_;h0Þ ¼ £ and V ðP_y0_;g0Þ [ V ðP_x0_;h0Þ ¼ V CQ0

k

can be assured byLemma 4. Otherwise, if (x, y) or (g, h) are not contained in C0_{, then a desired fault-free Hamiltonian cycle can be}

obtained by the construction method ofFig. 6a (replacing (s, t) with (x, y) (or (g, h)) if (x, y) (or (g, h)) is contained in C0_{, and replacing (s, t) with any link in C}0 _{if neither (x, y) nor (g, h) is contained in C}0_).

Then we assume that there is exactly one node p in CQ0_k that is incident with only one fault-free link in CQ0_k. Suppose that (p, q) is an arbitrary faulty link in CQ0_k. There are at least (2k 3) (k 3) (k 1) = 1 faulty link in CQ0_k whose end nodes are neither p nor q. Let (x, y) be such a faulty link in CQ0_k, i.e., {p, q}\ {x, y} = B. By the induction hypothesis (imagining that (p, q) and (x, y) are fault-free), there exists a Hamiltonian cycle in CQ0_k that can avoid all faulty links but (p, q) and (x, y). Notice that (p, q) should be included in the Hamiltonian cycle.

If (x, y) is contained in the Hamiltonian cycle, then a desired fault-free Hamiltonian cycle can be obtained by the construction method ofFig. 6b (replacing (g, h) with (p, q)). Otherwise, a desired fault-free Hamil-tonian cycle can be obtained by the construction method ofFig. 6a (replacing (s, t) with (p, q)).

Next, we assume that there are two nodes p and w in CQ0

kthat each are incident with only one fault-free link

in CQ0

k. Notice that (p, w) is a faulty link, for otherwise there are more than 2k 3 faulty links in CQ 0 k,

which is a contradiction. Suppose that (p, q) and (w, z) are two faulty links with {p, q}\ {w, z} = B. By the induction hypothesis (imagining that (p, q) and (w, z) are fault-free), there exists a Hamiltonian cycle in CQ0_k that can avoid all faulty links but (p, q) and (w, z). Notice that (p, q) and (w, z) should be included

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in the Hamiltonian cycle. A desired fault-free Hamiltonian cycle can be obtained by the construction method ofFig. 6b (replacing (x, y) and (g, h) with (p, q) and (w, z), respectively).

Finally, we show a distribution of 2n 4 link faults over a CQnsuch that no fault-free Hamiltonian cycle

can be found in the faulty CQn. Let us consider two nodes u = 0n(n consecutive 0’s) and v = 0n212of the

CQn. Refer toFig. 7where u and v have their 0-links and 1-links fault-free and the other n 2 links faulty.

It is easy to see that no fault-free Hamiltonian cycle exists in the faulty CQn.

5. The probability

Recall that we have assumed inTheorem 1 that each node in CQnis incident with at least two fault-free

links, while there are up to 2n 5 link faults in the CQn, where n P 3. In this section, we further show that

the assumption is practically meaningful by analyzing its probability when the CQncontains 2n 5 link faults.

By pnwe denote the probability that each node in such a faulty CQnis incident with at least two fault-free

links. Since there are a total of n· 2n1 links contained in CQn, there are

n 2n1

2n 5

ways to distribute the 2n 5 link faults. All these fault distributions are assumed having equal probability of occurrence.

Clearly, p3= 1 and p4is computed as 1

16 4 3 32 3 ¼ 0:987097, where 16 4₃

is the number of fault

distributions having some node incident with three link faults. When n P 5, there are 2n n 2n1_n n 5 or 2n n n 1 n 2n1 n n 4

fault distributions having some node incident with n(or n 1) link faults. It is not diﬃcult to check that n 2

n1_n n 4 > n 2 n1_n n 5 for n P 5. Thus, we have p_n¼ 1 2n n 2 n1_n n 5 ! þ 2n n n 1 n 2 n1_n n 4 ! n 2n1 2n 5 ! Table 1 Values of formula(1) n Values of(1) 5 >0.999400 6 >0.999995 7 >(1–8· 109 8 >(1–4)· 1012 9 >(1–4)· 1016 10 >(1–9)· 1021 11 >(1–6)· 1026 12 >(1–9)· 1032 13 >(1–4)· 1038 14 >(1–3)· 1045 15 >(1–7)· 1053 16 >(1–5)· 1061 17 >(1–6)· 1070 18 >(1–3)· 1079 19 >(1–2)· 1089 20 >(1–5)· 10100

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>1 2n ðn þ 1Þ n 2 n1_n n 4 n 2n1 2n 5 ¼ 1 2n ðn þ 1Þ n 2 n1 ð2n 5Þ Q2n5 i¼n3i Qn1 i¼0 n 2n1 i : ð1Þ

Table 1shows the values of formula(1)for n = 5, 6, . . ., 20. When n increases, the values of(1)go up rapidly and toward 1. Consequently, the probability that each node of CQnwith at most 2n 5 link faults is incident

with at least two fault-free links is very close to 1, even if n is small. 6. Discussion and conclusion

In this paper, with the assumption of at least two fault-free links incident to each node, we have shown that there exists a fault-free Hamiltonian cycle in an n-dimensional crossed cube (CQn) with up to 2n 5 link

faults. We also veriﬁed that the assumption is practically meaningful by evaluating its occurrence probability, which is very close to 1, even if n is small. Besides, the result is optimal with respect to the number of link faults tolerated. A recursive algorithm for constructing a fault-free Hamiltonian cycle can be easily obtained from the proof ofTheorem 1.

With the same assumption, a previous work by Chan and Lee[2]constructed a fault-free Hamiltonian cycle in an n-dimensional hypercube with 2n 5 link faults. The hypercube is highly symmetric, and it can be parti-tioned at any dimension into two smaller hypercubes. With this favorable property, fault-free Hamiltonian cycles were successfully obtained for some intractable distributions of link faults. On the other hand, the crossed cube is not symmetric and it can be partitioned into two smaller crossed cubes at only two dimensions. It was the major diﬃculty encountered when we tried to extend the result of[2]to the crossed cube. In order to achieve the extension, some new properties, as described in Section3, on the crossed cube were therefore derived.

It was shown in[22]that each member in the class of hypercube-like networks[27]contains a Hamiltonian cycle. Both the hypercube and crossed cube belong to the class. With the same assumption, the hypercube can tolerate up to 2n 5 link faults, while retaining a fault-free Hamiltonian cycle. We are wondering how many link faults the other hypercube-like networks can tolerate, while retaining a fault-free Hamiltonian cycle.

Moreover, it was shown in[29]that a CQnwith at most n 3 (link or node) faults contains fault-free cycles

of every possible length, except three. In[18], the authors showed, with the same assumption, that a CQnwith

at most 2n 5 link faults contains fault-free cycles of lengths ranging from 4 to 2n

. We are interested in explor-ing other topological properties, such as connectivity, diameter and Hamiltonian-connectedness, of the crossed cube under the assumption of conditional faults.

In[14], the h-extraconnectivity (h-edge-extraconnectivity) problem was deﬁned for a graph G, which, assum-ing that S was a set of faulty vertices (edges) in G, was required to compute the minimum |S| so that G S was disconnected and every component of G S had at least h + 1 nodes, where h P 0 was an integer. When h = 0, the h-extraconnectivity (h-edge-extraconnectivity) of G is equal to the vertex (edge) connectivity of G. When h = 1, the h-extraconnectivity (h-edge-extraconnectivity) of G is equal to the conditional vertex (edge) connectivity[9,10]of G. Recently, the two problems were further solved in[31,32] for h = 2. In[31], the 2-edge-extraconnectivities of hypercubes, twisted cubes, crossed cubes and Mo¨bius cubes were computed. In[32], the 2-extraconnectivity and 2-edge-extraconnectivity of folded hypercubes were computed. We are also interested in studying the two problems on hypercube-like networks.

Acknowledgements

The authors would like to express their gratitude to the anonymous reviewers for their valuable comments and suggestions which improve the paper a lot. The authors are also grateful to the National Science Council of the Republic of China, Taiwan for ﬁnancially supporting this research under Contract No. NSC 94-2213-E-239-014.

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