Fault-free longest paths in star networks with conditional link faults

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Theoretical Computer Science 410 (2009) 766–775

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Theoretical Computer Science

journal homepage:www.elsevier.com/locate/tcs

Fault-free longest paths in star networks with conditional link faults

Ping-Ying Tsai

a

, Jung-Sheng Fu

b

, Gen-Huey Chen

c,∗

a_{Department of Computer Science and Information Engineering, Hwa Hsia Institute of Technology, Taipei, Taiwan, ROC} b_{Department of Electronics Engineering, National United University, Miaoli, Taiwan, ROC}

c_{Department of Computer Science and Information Engineering, National Taiwan University, Taipei, 10764, Taiwan, ROC}

a r t i c l e i n f o Article history:

Received 26 May 2007

Received in revised form 15 July 2008 Accepted 6 November 2008 Communicated by D. Peleg

Keywords:

Conditional fault model Embedding

Fault tolerance Hamiltonian laceability Random fault model Star network

a b s t r a c t

The star network, which belongs to the class of Cayley graphs, is one of the most versatile interconnection networks for parallel and distributed computing. In this paper, adopting the conditional fault model in which each node is assumed to be incident with two or more fault-free links, we show that an n-dimensional star network can tolerate up to 2n−7 link faults, and be strongly (fault-free) Hamiltonian laceable, where n≥4. In other words, we can embed a fault-free linear array of length n!−1 (n!−2) in an n-dimensional star network with up to 2n−7 link faults, if the two end nodes belong to different partite sets (the same partite set). The result is optimal with respect to the number of link faults tolerated. It is already known that under the random fault model, an n-dimensional star network can tolerate up to n−3 faulty links and be strongly Hamiltonian laceable, for n≥3.

1. Introduction

The star network [1], which belongs to the class of Cayley graphs [2], has been recognized as an attractive alternative to the hypercube. It possesses many favorable topological properties such as recursiveness, symmetry, maximal fault tolerance, sublogarithmic degree and diameter, and strong resilience (see [1]). They are all desirable when we are building an interconnection topology for a parallel and distributed system. Besides, the star network can embed rings [28], grids [18], trees [5], and hypercubes [27]. Efficient communication algorithms for shortest-path routing [28], multiple-path routing [9], broadcasting [26] and scattering [13] are also available.

A linear array, which is one of the most fundamental networks for parallel and distributed computation, is suitable for developing simple and efficient algorithms. Numerous algorithms that were designed on linear arrays for solving various algebraic problems and graph problems can be found in [21]. A linear array can be also used as a control/data flow structure for distributed computation in a network (refer to [3] for an example).

Since node faults and/or link faults may occur in networks, it is important to consider faulty networks. Previously, communication problems (e.g., routing [5,12], broadcasting [32], multicasting [25], and gossiping [10]), embedding problems [4,6,16,17,24,30], and fault diameters [8,19,29] were studied on various faulty networks. Among them, two fault models were adopted; one is the random fault model [5,8,10,12,16,17,24,25,32], and the other is the conditional fault model [4,6,19,29,30].

The random fault model assumed that the faults might occur anywhere without any restriction, whereas the conditional

fault model assumed that the fault distribution must be subject to some constraint, e.g., that two or more fault-free links

are incident to each node. As a consequence of the constraint, it is in general more difficult to solve problems under the conditional fault model than the random fault model.

∗_{Corresponding author. Fax: +886 2 23628167.}

E-mail address:ghchen@csie.ntu.edu.tw(G.-H. Chen).

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Fig. 1. S4and four embedded S3’s.

In this paper, under the conditional fault model and with the assumption of at least two fault-free links incident to each node, we show that an n-dimensional star network can tolerate up to 2n

−

7 link faults, while retaining strongly (fault-free) Hamiltonian laceability, where n

≥

4. The result is optimal with respect to the number of link faults tolerated. For the same problem, at most n

−

3 link faults can be tolerated if the random fault model is adopted [24]. With our results, all parallel algorithms developed on a linear array of length n

! −

1 or n

! −

2 can be executed as well on an n-dimensional star network with up to 2n

−

7 link faults.

Previous results under the conditional fault model are described as follows. With the same assumption as ours, an n-dimensional hypercube (n-cube for short) with 2n

−

5 link faults is strongly (fault-free) Hamiltonian laceable [30], and an m-ary n-cube with 4n

−

5 link faults has a fault-free Hamiltonian cycle [4]. On the other hand, with the assumption of each node having at least k fault-free neighbors, the minimum number of node faults whose removal may disconnect an n-cube is

(

n

−

k

)

2k_{, where 1}

_≤

_k

_{≤ b}

_n

_/

₂

_c

_[₂₀_{]. Such a minimum number was named the restricted-node-connectivity}

and denoted by Rk-node-connectivity [20]. The node-connectivity of an n-cube is known to be n. There is a lower bound of

md

₍₍

_n

₋

_d

₋

₁

₎₍

_m

₋

₁

₎₍

_s

₊

₁

_{) + (}

_m

₋

_s

₋

₁

₎₎

_{on the R}

k-node-connectivity of an m-ary n-cube [33], where d

= b

k

/(

m

−

1

)c

and s

=

k mod

(

m

−

1

)

.

When k

=

1, the R1-node-connectivity of an n-cube (an m-ary n-cube) is 2n

−

2 (4n

−

2 if m

≥

4, and 4n

−

3 if m

=

3)

[11] ([7]), and the R1-node-connectivities of cube-connected cycles, undirected binary de Bruijn networks and Kautz graphs

are all greater by one at most than their node-connectivities [23]. Besides, the maximal diameters of an n-cube with 2n

−

3 node faults and an n-dimensional star network with 2n

−

5 node faults are n

+

2 [19] and

b

3

(

n

−

1

)/

2

c +

2 [29], respectively. When they are fault-free, their diameters are n and

b

3

(

n

−

1

)/

2

c

, respectively.

In the next section, the structure of the star network is reviewed. Some necessary definitions, notations and previous results are also introduced. In Section3, some new properties of the star network are derived in order to prove the main result. The proof of the main result is shown in Section4. Finally, this paper concludes with some remarks in Section5.

2. Preliminaries

It is convenient to represent a network as a graph G, where each vertex (edge) of G uniquely represents a node (link) of the network. We use V

(

G

)

and E

(

G

)

to denote the vertex set and edge set of G, respectively. Throughout this paper, we use network and graph, node and vertex, link and edge, interchangeably. The following is a definition of star networks, in terms of graph theory.

Definition 1 ([1]). An n-dimensional star network, denoted by Sn, has the node set V

(

Sn) = {a1a2

· · ·

an

|

a1a2

· · ·

anis a

permutation of 1

,

2

, . . . ,

n

}

and the link set E

(

Sn) = {(a1a2

· · ·

an,aia2

· · ·

ai−1a1ai+1

· · ·

an)|a1a2

· · ·

an

∈

V

(

Sn)and 2

≤

i

≤

n

}

.

Snhas n

!

nodes, each of degree n

−

1. S1is a node, S2is a link, and S3is a cycle of length six. S4is shown inFig. 1. The

link (a1a2

· · ·

an, aia2

· · ·

ai−1a1ai+1

· · ·

an) is referred to as an i-dimensional link. We use e(i)

(v)

to denote the i-dimensional

link that is incident to node

v

, and let E(i)

₍

_S_{n) =}_{e(i)

_{(v)|v ∈}

_V

₍

_S_n)_{} be the set of all i-dimensional links in S}

n. Snis both node

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768 P.-Y. Tsai et al. / Theoretical Computer Science 410 (2009) 766–775

It can be observed fromFig. 1that S4contains four embedded S3’s, denoted by

h∗ ∗ ∗

1

i

3,

h∗ ∗ ∗

2

i

3,

h∗ ∗ ∗

3

i

3and

h∗ ∗ ∗

4

i

3,

respectively (

∗ ∗ ∗

1, for example, represents any permutation of 1, 2, 3, 4 ending with 1). In general, Sncontains n embedded

Sn−1’s.

For 1

≤

r

≤

n

−

1, an embedded Srin Snis denoted by

h

s1s2

· · ·

sn

i

r, where s1

= ∗

and there are exactly r

−

1 occurrences

of

∗

in s2s3

· · ·

sn. For example,

h∗

4

∗

2

i

2 denotes an embedded S2 in S4. When Sn is partitioned into

h∗

d−11

∗

n−d

i

n−1,

h∗

d−1₂

_∗

n−d

_i

n−1,

. . .

,

h∗

d−1n

∗

n−d

i

n−1, Sn is said to be partitioned along dimension d, where 1

<

d

≤

n and

∗

d−1 (

∗

n−d)

represents d

−

1 (n

−

d) consecutive

∗

’s. These n embedded Sn−1’s are connected by d-dimensional links. When d

=

n,

we useE

˜

p(n,q() Sn)to represent the set of those n-dimensional links in Snthat connect

h∗

n−1p

i

n−1and

h∗

n−1q

i

n−1, where p

6=

q.

Clearly, we have

| ˜

E(pn,q() Sn)| = (n

−

2

)!

.

A path from vertex u (

=

v

₀) to vertex

v

(

=

vk

) in a graph G, represented as

h

v

₀,

v

1,

. . .

,

vk

i

, is referred to as a u–v path. A

path (cycle) in G is a Hamiltonian path (Hamiltonian cycle) if it contains every vertex of G exactly once. G is bipartite if there is a partition of V

(

G

)

into V0

(

G

)

and V1

(

G

)

such that every

(

u

, v) ∈

E

(

G

)

has either u

∈

V0

(

G

)

and

v ∈

V1

(

G

)

or u

∈

V1

(

G

)

and

v ∈

V0

(

G

)

. The two subsets V0

(

G

)

and V1

(

G

)

are referred to as partite sets of G. Star networks are known to be bipartite [1].

A bipartite graph G with

|

V0

(

G

)| = |

V1

(

G

)|

is Hamiltonian laceable if it has a u–

v

Hamiltonian path for every u

∈

V0

(

G

)

and

every

v ∈

V1

(

G

)

[31], and strongly Hamiltonian laceable if it additionally has a longest u–

v

path of length

|

V0

(

G

)|+|

V1

(

G

)|−

2

for all u,

v ∈

V0

(

G

)

or u,

v ∈

V1

(

G

)

[15]. A Hamiltonian laceable graph G is hyper Hamiltonian laceable if for every

w ∈

V0

(

G

)(w ∈

V1

(

G

))

, G–

w

contains a u–

v

Hamiltonian path for all u,

v ∈

V1

(

G

)

(u,

v ∈

V0

(

G

)

) [22]. Clearly, a hyper

Hamiltonian laceable graph is strongly Hamiltonian laceable.

Given a vertex u in G, we define N

(

u

) = {v|(

u

, v) ∈

E

(

G

)}

to be the neighborhood of u, which is the set of vertices that are adjacent to u in G. The size of N

(

u

)

, i.e.,

|

N

(

u

)|

, is the degree of u. The minimum vertex degree of G is denoted by

δ(

G

) =

min

{|

N

(

u

)||

u

∈

V

(

G

)}

. Let V0_{be a vertex subset of G. We define N}

₍

_V0

_{) = Su}

∈V 0N

(

u

) −

V

0_{to be the neighborhood} of V0. Besides, we use G

[

V0

]

to denote the subgraph of G induced by V0. Throughout this paper, we use F (

⊆

E

(

Sn)) to denote the set of link faults in Sn.

Lemma 1 ([24]). Sn

−

F is strongly Hamiltonian laceable if

|

F

| ≤

n

−

3, and hyper Hamiltonian laceable if

|

F

| ≤

n

−

4, where

n

≥

4.

Lemma 2 ([14]). If

|

F

| ≤

2n

−

7 and

δ(

Sn

−

F

) ≥

2, then there exists 1

<

d

≤

n such that

|

E(d)

(

Sn) ∩F

| ≥

1 and

δ(h∗

d−1_q

_∗

n−d

_i

n−1

−

F

) ≥

2 for all 1

≤

q

≤

n, where n

≥

4. 3. Properties and main result

In this section, we first introduce some properties of Sn. Then we present our main result.

Lemma 3. Suppose u

=

u1u2

· · ·

un

∈

V

(h∗

n−1q

i

n−1

)

, where un

=

q and n

≥

3. For every r

∈ {

1

,

2

, . . . ,

n

} − {

u1

,

q

}

, there

exists

w ∈

V

(h∗

n−1_r

_i

n−1

)

and

v ∈

N

(

u

) ∩

V

(h∗

n−1q

i

n−1

)

such that

(w, v) =

e(n)

(v)

.

Proof. We assume u_c

=

r, where 1

<

c

<

n. Select

v =

u_cu₂

· · ·

u_c−1u1uc+1

· · ·

un(∈ N

(

u

) ∩

V

(h∗

n−1q

i

n−1

))

and

w =

unu2

· · ·

uc−1u1uc+1

· · ·

uc

(∈

V

(h∗

n−1r

i

n−1

))

. Clearly,

(w, v) =

e(n)

(v)

.

Lemma 4. Suppose that P is a path in

h∗

n−1_q

_i

n−1

−

F , where n

≥

4 and 1

≤

q

≤

n. If

|

F

| ≤

2n

−

7 and P is Hamiltonian or of

length

(

n

−

1

)! −

2, then P has a link

(

u

, v)

with e(n)

₍

_u

_),

_e(n)

_{(v) /∈}

_{F .}

Proof. Suppose conversely that no such (u,

v

) can be found in P. Then,

|

E(n)

₍

_S_{n) ∩}_F

_{| ≥}

₍₍

_n

₋

₁

_{)! −}

₂

_)/

_{2, which is greater}

than 2n

−

7 as n

≥

4, a contradiction.

In subsequent discussion, we let VA

=

S

r∈AV

(h∗

n

−1_r

_i

n−1

)

, where A

⊆ {

1

,

2

, . . . ,

n

}

. Lemma 5. Suppose that A

⊆ {

1

,

2

, . . . ,

n

}

and n

≥

5. For any s

∈

V

(h∗

n−1_p

_i

n−1

)

and t

∈

V

(h∗

n−1q

i

n−1

)

, there exists a longest

s–t path of length

|

A

| ×

(

n

−

1

)! −

1 or

|

A

| ×

(

n

−

1

)! −

2 in Sn

[

VA

] −

F , where p, q

∈

A and p

6=

q, provided the following two

conditions hold:

(1)

| ˜

E_i(_,n_j)

(

Sn) ∩F

|

< (

n

−

2

)!/

2 for all i

,

j

∈

A and i

6=

j;

(2)

h∗

n−1_r

_i

n−1

−

F is strongly Hamiltonian laceable for every r

∈

A. Proof. Since the distance between node ja₂

∗

n−4_a

n−1i and node jan−1

∗

n−4a2i is three, they belong to different partite

sets of

h∗

n−1i

i

_n−1. For each link (ja2

∗

n

−4_a

n−1i, ia2

∗

n−4an−1j) connecting

h∗

n−1i

i

n−1 with

h∗

n−1j

i

n−1, there exists another

link (ja_n−1

∗

n−4a2i, ian−1

∗

n−4a2j

)

connecting

h∗

n−1i

i

n−1with

h∗

n−1j

i

n−1. It is implied that there are an equal number (i.e.,

(

n

−

2

)!/

2) of nodes in V0

(h∗

n−1i

i

n−1

)

and V1

(h∗

n−1i

i

n−1

)

, respectively, that are connected to

h∗

n−1j

i

n−1.

Suppose A

= {

a1

,

a2

, . . . ,

a|A|

}

. A longest s–t path in Sn

[

VA

] −

F is shown inFig. 2, where a1

=

p and a|A|

=

q are assumed. Since

| ˜

Ea(n1),a2

(

Sn)| = (n

−

2

)!

, two links inE

˜

(n)

a1,a2

(

Sn) −F , one incident to V0

(h∗

n−1_a

1

i

n−1

)

and the other incident

to V1

(h∗

n−1a1

i

n−1

)

, can be found, as a consequence of (1). Hence, a link

(

x1

,

y2

) ∈ ˜

E( n)

a1,a2

(

Sn) −F can be determined such

that x1and s belong to different partite sets of

h∗

n−1a1

i

n−1and y2

∈

V

(h∗

n−1a2

i

n−1

)

. As a consequence of (2), there exists a

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Fig. 2. A longest s–t path in Sn[VA] −F .

Fig. 3. A longest s–t path in Snthat contains (u,v), where s, t∈V(h∗k_q_i k). Similarly, by the aid of (1) and (2), links

(

xk,yk+1

) ∈ ˜

E(

n)

ak,ak+1

(

Sn) −F and Hamiltonian yk–xkpaths in

h∗

n−1_a

k

i

n−1

−

F can

be obtained, where 2

≤

k

≤ |

A

| −

1. All these Hamiltonian paths together with a longest y|A|–t path in

h∗

n−1a|A|

i

n−1

−

F

constitute a longest s–t path in Sn

[

VA

] −

F . If the longest y|A|–t path has length

(

n

−

1

)! −

1, the length of the longest s–t path is computed as

(|

A

| −

1

) × ((

n

−

1

)! −

1

) + (|

A

| −

1

) + ((

n

−

1

)! −

1

) = |

A

| ×

(

n

−

1

)! −

1. Similarly, if the longest

y|A|–t path has length

(

n

−

1

)! −

2, the longest s–t path has length

|

A

| ×

(

n

−

1

)! −

2.

Lemma 6. Suppose that s, t are two distinct nodes of Snand

(

u

, v) ∈

E

(

Sn), where n

≥

4. If

{

s

,

t

} 6= {

u

, v}

, then there exists a

Hamiltonian s-t path or an s-t path of length

(

n

−

1

)! −

2 in Snthat contains

(

u

, v)

.

Proof. We prove this lemma by induction on n. This lemma holds for S4, which can be verified by exhaustive search (see

[34]). So, we assume that this lemma holds for Sk, and then consider Sk+1below, where k

≥

4.

Without loss of generality, suppose u,

v ∈

V

(h∗

k_q

_i

_k)_{for some 1}

_≤

_q

_≤

_k

₊

_{1. We also assume}

₍

_u

_{, v) ∈}

_E(l)

₍

_S

k+1

)

, where

2

≤

l

≤

k. With the following three cases, we show a longest s–t path of length

(

k

+

1

)! −

1 or

(

k

+

1

)! −

2 in Sk+1that

contains (u,

v

).

Case 1. s

∈

V

(h∗

k_q

_i

_k)_{and t}

_/∈

_V

_(h∗

k_q

_i

_k)_{or s}

_/∈

_V

_(h∗

k_q

_i

_k)_{and t}

_∈

_V

_(h∗

k_q

_i

_k)_{. We consider the situation of s}

_∈

_V

_(h∗

k_q

_i

_k)

and t

/∈

V

(h∗

k_q

_i

_k)_{. The discussion for the situation of s}

_/∈

_V

_(h∗

k_q

_i

_k)_{and t}

_∈

_V

_(h∗

k_q

_i

_k)_{is very similar. A desired s–t path can}

be obtained using the construction method ofFig. 2. We only need to change n

−

1 to k and

|

A

|

to k

+

1, and set a1

=

q.

Besides, the node x1is selected with

{

s

,

x1

} 6= {

u,

v}

. The induction hypothesis assures a Hamiltonian s–x1path in

h∗

ka1

i

k

that contains (u,

v

). The desired s–t path has length

(

k

+

1

) ×

k

! −

1

=

(

k

+

1

)! −

1 or

(

k

+

1

) ×

k

! −

2

=

(

k

+

1

)! −

2. Case 2. s, t

∈

V

(h∗

kq

i

k). A desired s–t path can be obtained as shown inFig. 3, where a1

=

q is assumed. The induction

hypothesis assures a longest s–t path in

h∗

k_a

1

i

kthat contains (u,

v

). A link (

w

,

w

0

) 6= (

u

, v)

can be selected from the path.

Let z

=

e(k+1)

_(w)

_{and z}0

₌

_e(k+1)

_(w

0

₎

_{. A Hamiltonian z–z}0_{path in S}

k+1

− h∗

ka1

i

kcan be obtained using the construction

method ofFig. 2(changing s to z, t to z0, n

−

1 to k,

|

A

|

to k, and

h∗

n−1ar

i

n−1to

h∗

kar+1

i

kfor all 1

≤

r

≤ |

A

|

)

. The Hamiltonian

z–z0_{path, combining with (}

_w

_{, z), (}

_w

0_{, z}0_{), and the s–}

_w

_and

_w

0_{–t paths in}

_h∗

k_a

1

i

k, forms a desired s–t path in Sk+1. The desired

s–t path has length

(

k

×

k

! −

1

) +

2

+

((

k

! −

1

) −

1

) = (

k

+

1

)! −

1 or

(

k

×

k

! −

1

) +

2

+

((

k

! −

2

) −

1

) = (

k

+

1

)! −

2. Case 3. s, t

/∈

V

(h∗

kq

i

k). Suppose s

∈

V

(h∗

kg

i

k)and t

∈

V

(h∗

kh

i

k), where g, h

∈ {

1

,

2

, . . . ,

k

+

1

} − {

q

}

. First we assume

g

6=

h. A desired s–t path can be obtained using the construction method ofFig. 2(changing n

−

1 to k and

|

A

|

to k

+

1, and letting a1

=

g, a2

=

q, and ak+1

=

h). The node x2is selected with

{

x2

,

y2

} 6= {

u

, v}

. The induction hypothesis assures a

Hamiltonian y2–x2path in

h∗

ka2

i

v

). The desired s–t path has length

(

k

+

1

) ×

k

! −

1

=

(

k

+

1

)! −

1 or

(

k

+

1

) ×

k

! −

2

=

(

k

+

1

)! −

2.

Next we assume g

=

h. A desired s–t path can be obtained by slightly modifying the construction method ofFig. 3. We only need to set a1

=

g

=

h and a2

=

q. The node x2is selected with

{

z

,

x2

} 6= {

u

, v}

. The induction hypothesis assures a

Hamiltonian z–x2path in

h∗

ka2

i

v

). A Hamiltonian y3–z0path in Sk+1

− h∗

ka1

i

k

− h∗

ka2

i

kcan be obtained,

similarly, using the construction method ofFig. 2. The desired s–t path, which consists of the s–

w

and

w

0_{–t paths in}

_h∗

k_a 1

i

k,

the Hamiltonian z–x2path, the Hamiltonian y3–z0path, and three links (

w

, z), (

w

0, z0), (x2, y3), has length

(

k

+

1

)! −

1 or

(

k

+

1

)! −

2.

Lemma 7. For any two distinct links (s, t), (u,

v

) in Sn, there exists a Hamiltonian cycle in Snthat contains both of them, where

n

≥

3.

Proof. Since S₃is a cycle of length 6, this lemma holds for S3. When n

≥

4, since s and t belong to different partite sets of

(5)

Fig. 4. A distribution of 2n−6 link faults over Sn.

The main result of this paper is presented in the following theorem whose proof is shown in the next section.

Theorem 1. With the assumption of two or more fault-free links incident to each node, an n-dimensional star network can tolerate

up to 2n

−

7 link faults, and be strongly (fault-free) Hamiltonian laceable, where n

≥

4.

Theorem 1is optimal with respect to the number of link faults tolerated.Fig. 4shows a distribution of 2n

−

6 link faults over Sn, where

h

s, u,

v

, t

i

is an s–t path and (s, u), (u,

v

) ((u,

v

), (

v

, t)) are the only two fault-free links incident to u (

v

). It is

easy to see that no fault-free Hamiltonian s–t path exists in the faulty Sn.

4. Proof of Theorem 1

With

|

F

| ≤

2n

−

7 and

δ(

Sn

−

F

) ≥

2, we show by induction that there exists a longest s–t path of length n

! −

1 or n

! −

2

between every two distinct nodes s, t of Sn

−

F . ByLemma 1(n

−

3

=

2n

−

7 as n

=

4), the theorem holds for S4. So, we

assume that this theorem holds for Sk, and then consider Sk+1in the rest of this section, where k

≥

4.

ByLemma 2, we can partition Sk+1along some dimension d such that

|

E(d)

(

Sk+1

)∩

F

| ≥

1 and

δ(h∗

d−1q

∗

k+1−d

i

k

−

F

) ≥

2

for all 1

≤

q

≤

k

+

1, where 1

<

d

≤

k

+

1. Without loss of generality, we assume d

=

k

+

1. Now that

|

E(k+1)

(

Sk+1

)∩

F

| ≥

1,

we have

|

E

(h∗

k_q

_i

_{k) ∩}_F

_{| ≤}

_2k

₋

_{6 for all 1}

_≤

_q

_≤

_k

₊

_{1. A desired s–t path in S}

k+1

−

F is constructed in Section4.1if

|

E

(h∗

k_q

_i

_{k) ∩} _F

_{| ≤}

_2k

₋

_{7 for all 1}

_≤

_q

_≤

_k

₊

_{1, and constructed in Section}_4.2_else.

4.1.

|

E

(h∗

k_q

_i

_{k) ∩}_F

_{| ≤}

_2k

₋

_{7 for all 1}

_≤

_q

_≤

_k

₊

₁

Suppose that

|

E

(h∗

k_q

_i

_{k) ∩} _F

_|

_≤

_2k

₋

_{7 for all 1}

_≤

_q

_≤

_k

₊

_{1. Since}

_|

_F

_|

_≤

_2k

₋

_{5, we have}

P

i,j∈{1,2,...,k+1} and i6=j

| ˜

E

(k+1)

i,j

(

Sk+1

) ∩

F

| ≤

2k

−

5, where k

≥

4. Notice that 2k

−

5

=

((

k

+

1

) −

2

)!/

2 when k

=

4,

and 2k

−

5

< ((

k

+

1

) −

2

)!/

2 when k

>

4. Two cases are discussed below.

Case 1. k

>

4. We have

| ˜

E_i(_,k_j+1)

(

Sk+1

)∩

F

|

< ((

k

+

1

)−

2

)!/

2 for all i, j

∈ {

1

,

2

, . . . ,

k

+

1

}

and i

6=

j. The induction hypothesis

assures that

h∗

k_q

_i

k

−

F is strongly Hamiltonian laceable for all 1

≤

q

≤

k

+

1. If s

∈

V

(h∗

kg

i

k)and t

∈

V

(h∗

kh

i

k)for some

g, h

∈ {

1

,

2

, . . . ,

k

+

1

}

and g

6=

h, then byLemma 5, Sk+1

−

F is strongly Hamiltonian laceable. If s, t

∈

V

(h∗

kg

i

k)for some

1

≤

g

≤

k

+

1, then a desired s–t path in Sk+1

−

F can be obtained by slightly modifying the construction method ofFig. 3.

We only need to set a1

=

g. The induction hypothesis assures a longest s–t path in

h∗

ka1

i

k

−

F . ByLemma 4, a link (

w

,

w

0)

with

(w,

z

) ∈ ˜

Ea(k1+,a12)

−

F and

(w

0

_,

z0

) ∈ ˜

Ea(k1+,a1k+1)

−

F can be selected from the path. ByLemma 5, a Hamiltonian z–z

0 path in

Sk+1

− h∗

ka1

i

k

−

F can be obtained. The resulting longest s–t path in Sk+1

−

F has length

(

k

+

1

)! −

1 or

(

k

+

1

)! −

2.

Case 2. k

=

4. If

| ˜

E_i(_,5_j)

(

S5

) ∩

F

|

< ((

4

+

1

) −

2

)!/

2

=

3 for all i, j

∈ {

1

,

2

, . . . ,

5

}

and i

6=

j, then the discussion is the same

as Case 1. So, we consider

| ˜

E_i0(5_,j0() S5

) ∩

F

| =

3 for i0, j0

∈ {

1

,

2

, . . . ,

5

}

and i0

6=

j0. Since

|

F

| ≤

3 as k

=

4, all link faults are

inE

˜

_i0(5_,j0() S5

)

. Assume that s

∈

V

(h∗

4g

i

4

)

and t

∈

V

(h∗

4h

i

4

)

, where g, h

∈ {

1

,

2

, . . . ,

5

}

. When g

6=

h, a desired s–t path in

S5

−

F can be obtained using the construction method ofFig. 2. We only need to change n

−

1 to 4 and

|

A

|

to 5, set a1

=

g

and a5

=

h, and set a2, a3

,

a4with

{

i0

,

j0

} 6= {

ar

,

ar+1

}

for all 1

≤

r

≤

4. The induction hypothesis assures a Hamiltonian

s–x1path in

h∗

4a1

i

4

−

F , a Hamiltonian yr–xr path in

h∗

4ar

i

4

−

F for all 2

≤

r

≤

4, and a longest y5–t path in

h∗

4a5

i

4

−

F .

The desired s–t path has length 5

! −

1 or 5

! −

2.

When g

=

h, a desired s–t path in S5

−

F can be obtained using the construction method ofFig. 3. We only need to change

k

+

1 to 5, set a1

=

g

=

h, and set a2, a3, a4, a5with

{

i0

,

j0

} 6= {

ar

,

a(rmod5)+1

}

for all 1

≤

r

≤

5. The induction hypothesis

assures a longest s–t path in

h∗

4_a

1

i

4

−

F . ByLemma 4, a link (

w

,

w

0) with

(w,

z

) ∈ ˜

Ea(51),a2

−

F and

(w

0

_,

_z0

_{) ∈ ˜}

_E(5)

a1,a5

−

F can be

selected from the path. The induction hypothesis assures a Hamiltonian yr–xrpath in

h∗

4ar

i

4

−

F for all 2

≤

r

≤

5, where

y2

=

z and x5

=

z0. The desired s–t path has length 5

! −

1 or 5

! −

2.

4.2.

|

E

(h∗

k

_{αik) ∩}

_F

_{| =}

_2k

₋

_{6 for some 1}

_≤

_{α ≤}

_k

₊

₁

Suppose that

|

E

(h∗

k

_{αik) ∩}

_F

_{| =}

_2k

₋

_{6, where 1}

_≤

_{α ≤}

_k

₊

_{1. Now that}

_|

_F

_{| ≤}

_2k

₋

_{5 and}

_|

_E(k+1)

₍

_S

k+1

) ∩

F

| ≥

1,

we have

|

E(k+1)

₍

_S

k+1

) ∩

F

| =

1. Besides, we have

|

E

(h∗

kq0

i

k) ∩F

| =

0 for all q0

∈ {

1

,

2

, . . . ,

k

+

1

} − {

α}

, and

| ˜

E_i(_,k_j+1)

(

Sk+1

) ∩

F

| ≤

1

(< ((

k

+

1

) −

2

)!/

2

)

for any i, j

∈ {

1

,

2

, . . . ,

k

+

1

}

and i

6=

j, where k

≥

4.

If s, t

∈

V

(h∗

k

_αik)

_{, then a link fault, say (h, h}0_{), is chosen from E}

_(h∗

k

_{αik) ∩}

_{F such that}

_{

_s

_,

_t

_{} 6= {}

_h

_,

_h0

_}

_{and e}(k+1)

₍

_h

₎

_,

(6)

Fig. 5. A longest s–t path in Sk+1−F when h, h0/∈N(V(h∗kak+1ik)).

of length k

! −

1 or k

! −

2 in

h∗

k

αik

−

(

F

− {

(

h

,

h0

)})

. A desired s–t path in Sk+1

−

F can be obtained using the construction

method ofFig. 3. We only need to set a1

=

α

, and

(w, w

0

) = (

h

,

h0

)

if P contains (h, h0). The desired s–t path has length

(

k

+

1

)! −

1 or

(

k

+

1

)! −

2.

If s

/∈

V

(h∗

k

αik)

or t

/∈

V

(h∗

k

αik)

, then a desired s–t path in Sk+1

−

F is constructed in Sections4.2.1and4.2.2, where

we use a1, a2, . . . , ak+1to denote the k

+

1 distinct integers from 1 to k

+

1 (i.e.,

{

a1

,

a2

, . . . ,

ak+1

} = {

1

,

2

, . . . ,

k

+

1

}

).

4.2.1. s

∈

V

(h∗

k

_αik)

_{and t}

_/∈

_V

_(h∗

k

_αik)

_{or s}

_/∈

_V

_(h∗

k

_αik)

_{and t}

_∈

_V

_(h∗

k

_αik)

Suppose that s

∈

V

(h∗

k

_αik)

_{and t}

_/∈

_V

_(h∗

k

_αik)

_{. The discussion for s}

_/∈

_V

_(h∗

k

_αik)

_{and t}

_∈

_V

_(h∗

k

_αik)

_{is similar. We}

assume t

∈

V

(h∗

k

βik)

, where

β 6= α

. A link fault (h, h0) is chosen from E

(h∗

k

αik) ∩

F such that s

/∈ {

h

,

h0

}

and e(k+1)

(

h

)

,

e(k+1)

₍

_h0

_{) /∈}

_{F . Set a}

1

=

α

and ak+1

=

β

.

We first consider the situation of h, h0

_/∈

_N

₍

_V

_(h∗

k_a

k+1

i

k)). Imagine that (h, h0) is fault-free. A vertex

v

1

∈

V

(h∗

ka1

i

k)is

determined such that

v

1and s belong to different partite sets of Sk+1, e(k+1)

(v

1

) /∈

F , and

v

1

/∈

N

(

V

(h∗

kak+1

i

k). The induction

hypothesis assures a longest s–

v

1path of length k

! −

1 in

h∗

ka1

i

k

−

(

F

− {

(

h

,

h0

)})

. If the longest s–

v

1path does not contain

(h, h0_{), then a desired s–t path of length}

₍

_k

₊

₁

_{)! −}

_{1 or}

₍

_k

₊

₁

_{)! −}

_{2 in S}

k+1

−

F can be obtained using the construction

method ofFig. 2. We only need to set x1

=

v

1, and change n

−

1 and

|

A

|

to k and k

+

1, respectively.

If the longest s–

v

1path contains (h, h0), then a desired s–t path in Sk+1

−

F can be obtained as shown inFig. 5. We set a2

and a3such that h

∈

N

(

V

(h∗

ka2

i

k))and h0

∈

N

(

V

(h∗

ka3

i

k)). There is an additional restriction to

v

1

:

v

1

/∈

N

(

V

(h∗

ka2

i

k))

and

v

1

/∈

N

(

V

(h∗

ka3

i

k)). Also we set a4such that

v

1

∈

N

(

V

(h∗

ka4

i

k)). Then, three vertices u2

∈

V

(h∗

ka2

i

k),

v

3

∈

V

(h∗

ka3

i

k),

and u4

∈

V

(h∗

ka4

i

k)are determined such that

(

u2

,

h

) =

e(k+1)

(

h

)

,

(v

3

,

h0

) =

e(k+1)

(

h0

)

, and

(

u4

, v

1

) =

e(k+1)

(v

1

)

.

ByLemma 5, there is a longest u4–t path of length

(

k

−

2

) ×

k

! −

1 or

(

k

−

2

) ×

k

! −

2 in Sk+1

[

VA

] −

F , where

A

= {

1

,

2

, . . . ,

k

+

1

} − {

a1

,

a2

,

a3

}

. Again, byLemma 5, there is a longest u2–

v

3path of length 2

×

k

! −

1 in Sk+1

[

VA

] −

F ,

where A

= {

a2

,

a3

}

. The desired s–t path has length

((

k

! −

1

) −

1

) + (

2

×

k

! −

1

) + ((

k

−

2

) ×

k

! −

1

) +

3

=

(

k

+

1

)! −

1

or

((

k

! −

1

) −

1

) + (

2

×

k

! −

1

) + ((

k

−

2

) ×

k

! −

2

) +

3

=

(

k

+

1

)! −

2.

Then we consider the situation of h

∈

N

(

V

(h∗

kak+1

i

k)), without loss of generality. Notice that at most one of h and h0

belongs to N

(

V

(h∗

k_a

k+1

i

k)). Imagine that (h, h0) is fault-free. A vertex u1

∈

V

(h∗

ka1

i

k)is determined such that u1and s

belong to different partite sets of Sk+1, e(k+1)

(

u1

) /∈

F , and u1

/∈

N

(

V

(h∗

kak+1

i

k). The induction hypothesis assures a longest

s–u1path of length k

! −

1 in

h∗

ka1

i

k

−

(

F

− {

(

h

,

h0

)})

. If the longest s–u1path does not contain (h, h0), a desired s–t path in

Sk+1

−

F can be obtained using the construction method ofFig. 2. We only need to set x1

=

u1, and change n

−

1 and

|

A

|

to

k and k

+

1, respectively. If the longest s–u1path contains (h, h0), two cases: h

∈

N

(

t

)

or h

/∈

N

(

t

)

, are discussed below.

Case 1. h

∈

N

(

t

)

−

F can be obtained as shown inFig. 6(a). We set a2with h0

∈

N

(

V

(h∗

ka2

i

k)). There

is an additional restriction to u1

:

u1

/∈

N

(

V

(h∗

ka2

i

k))and u1

/∈

N

(

V

(h∗

kak+1

i

k)). Also we set akwith u1

∈

N

(

V

(h∗

kak

i

k)).

Then, six vertices u2,

v

2

∈

V

(h∗

ka2

i

k), u3

∈

V

(h∗

ka3

i

k),

vk

∈

V

(h∗

kak

i

k), and uk+1,

vk

+1

∈

V

(h∗

kak+1

i

k)are determined

such that

(

u2

,

h0

) =

e(k+1)

(

h0

)

,

v

2and u2belong to different partite sets of Sk+1, e(k+1)

(v

2

) /∈

F ,

(

uk+1

, v

2

) =

e(k+1)

(v

2

)

,

vk

+1and uk+1belong to the same partite set of Sk+1, e(k+1)

(vk

+1

) /∈

F ,

(

u3

, vk

+1

) =

e(k+1)

(vk

+1

)

, and

(vk,

u1

) =

e(k+1)

(

u1

)

.

ByLemma 1, there are a longest u2–

v

2path of length k

! −

1 in

h∗

ka2

i

kand a longest uk+1–

vk

+1path of length k

! −

2 in

h∗

k_a

k+1

i

k

− {

t

}

. ByLemma 5, there is a longest u3–

vk

path of length

(

k

−

2

)×

k

! −

1 or

(

k

−

2

)×

k

! −

2 in Sk+1

[

VA

] −

F , where

A

= {

1

,

2

, . . . ,

k

+

1

}−{

a1

,

a2

,

ak+1

}

((

k

!−

1

)−

1

)+(

k

!−

1

)+(

k

!−

2

)+((

k

−

2

)×

k

!−

1

)+

4

=

(

k

+

1

)! −

1 or

((

k

! −

1

) −

1

) + (

k

! −

1

) + (

k

! −

2

) + ((

k

−

2

) ×

k

! −

2

) +

4

=

(

k

+

1

)! −

2.

Case 2. h

/∈

N

(

t

)

−

F is shown inFig. 6(b) when

w

0

6=

t and

(v

2

, w

0

) /∈

F , and shown inFig. 6(c) when

w

0

₌

t or

(v

2

, w

0

) ∈

F . We set a2such that h0

∈

N

(

V

(h∗

ka2

i

k)). There is an additional restriction to u1

:

u1

/∈

N

(

V

(h∗

ka2

i

k))

and u1

/∈

N

(

V

(h∗

kak+1

i

k)). Also we set akwith u1

∈

N

(

V

(h∗

kak

i

k)). Then, three vertices u2

∈

V

(h∗

ka2

i

k),

vk

∈

V

(h∗

kak

i

k),

and

w ∈

V

(h∗

k_a

k+1

i

k)are first determined such that

(

u2

,

h0

) =

e(k+1)

(

h0

)

,

(w,

h

) =

e(k+1)

(

h

)

, and

(vk,

u1

) =

e(k+1)

(

u1

)

. By Lemma 3, there exist

v

2

∈

V

(h∗

ka2

i

k)and

w

0

∈

N

(w) ∩

V

(h∗

kak+1

i

k)such that

(v

2

, w

0

) =

e(k+1)

(w

0

)

(seeFig. 6(b)).

If

w

0

₆₌

_{t and}

_(v

2

, w

0

) /∈

F , then, again byLemma 3, there exist u3

∈

V

(h∗

ka3

i

k)and

vk

+1

∈

N

(

t

)∩

V

(h∗

kak+1

i

k)such that

(

u3

, vk

+1

) =

e(k+1)

(vk

+1

)

. Besides,

(

u3

, vk

+1

) /∈

F can be satisfied, because

|

E(k+1)

(

Sk+1

) ∩

F

| =

1. ByLemma 7, there exists a

Hamiltonian cycle in

h∗

kak+1

i

kthat contains (t,

vk

+1) and (

w

,

w

0). ByLemma 1, there is a longest u2–

v

2path of length k

! −

1

in

h∗

k_a

2

i

k. ByLemma 5, there is a longest u3–

vk

path of length

(

k

−

2

) ×

k

! −

1 or

(

k

−

2

) ×

k

! −

2 in Sk+1

[

VA

] −

F , where

A

= {

1

,

2

, . . . ,

k

+

1

}−{

a1

,

a2

,

ak+1

}

((

k

!−

1

)−

1

)+(

k

!−

1

)+(

k

!−

2

)+((

k

−

2

)×

k

!−

1

)+

4

=

(

k

+

1

)! −

1 or

((

k

! −

1

) −

1

) + (

k

! −

1

) + (

k

! −

2

) + ((

k

−

2

) ×

k

! −

2

) +

4

=

(

k

+

1

)! −

2.

(7)

Fig. 6. A longest s–t path in Sk+1−F when h∈N(V(h∗kak+1ik)). (a) h∈N(t). (b) h/∈N(t)and (w06=t and(v2, w0) /∈F ). (c) h/∈N(t)and (w0=t or(v2, w0_{) ∈}

F ).

If

w

0

₌

_{t or}

_(v

2

, w

0

) ∈

F , then byLemma 3, there exist

v

3

∈

V

(h∗

ka3

i

k) and

w

00

∈

N

(w) ∩

V

(h∗

kak+1

i

k)such

that

(v

3

, w

00

) =

e(k+1)

(w

00

)

. Again byLemma 3, there exist u4

∈

V

(h∗

ka4

i

k) and

vk

+1

∈

N

(

t

) ∩

V

(h∗

kak+1

i

k)with

(

u4

, vk

+1

) =

e(k+1)

(vk

+1

)

. Besides, (

v

3,

w

00),

(

u4

, vk

+1

) /∈

F and

w

00

6=

t can be satisfied. ByLemma 7, there exists a

Hamiltonian cycle in

h∗

k_a

k+1

i

kthat contains (t,

vk

+1) and (

w

,

w

00). ByLemma 5, there is a longest u2–

v

3path of length

2

×

k

! −

1 in Sk+1

[

VA

] −

F , where A

= {

a2

,

a3

}

. If k

=

4,Lemma 1assures a longest u4–

v

4path of length 4

! −

1 or 4

! −

2 in

h∗

4_a

4

i

4. If k

>

4,Lemma 5assures a longest u4–

vk

path of length

(

k

−

3

) ×

k

! −

1 or

(

k

−

3

) ×

k

! −

2 in Sk+1

[

VA

] −

F , where

A

= {

1

,

2

, . . . ,

k

+

1

} − {

a1

,

a2

,

a3

,

ak+1

}

(

k

+

1

)! −

1 or

(

k

+

1

)! −

2.

4.2.2. s

,

t

/∈

V

(h∗

k

αik)

Suppose that s

∈

V

(h∗

k

_βik)

_{and t}

_∈

_V

_(h∗

k

_{γ ik)}

_{, where}

_{β, γ ∈ {}

₁

_,

₂

_{, . . . ,}

_k

₊

₁

_{} − {}

_α}

_{. We first consider the situation of}

β = γ

. A link fault (h, h0_{) is chosen from E}

_(h∗

k

_{αik) ∩}

_{F such that e}(k+1)

₍

_h

_),

_e(k+1)

₍

_h0

_{) /∈}

_{F . Set a}

1

=

β = γ

, and set a3

=

α

if

h, h0

_/∈

_N

₍

_V

_(h∗

k_a

1

i

k)), and a2

=

α

else. A desired s–t path in Sk+1

−

F can be constructed in a way similar toFig. 3. We only

explain below the construction for

α =

a3. The construction for

α =

a2is similar.

Refer toFig. 3again. For the purpose of our construction, we let A

= {

1

,

2

, . . . ,

k

+

1

}

,

(

x3

,

y3

) = (

h

,

h0

)(∈

E

(h∗

ka3

i

k)),

and select

(w, w

0

)(∈

E

(h∗

ka1

i

k))such that

{

w, w

0

} 6= {

s

,

t

}

,

w

,

w

0

/∈

N

(

V

(h∗

ka3

i

k)),

w

and y3belong to different partite

sets of Sk+1, and e(k+1)

(w)

, e(k+1)

(w

0

) /∈

F . Four vertices z, x2

∈

V

(h∗

ka2

i

k), y4

∈

V

(h∗

ka4

i

k), and z0

∈

V

(h∗

kak+1

i

k)are

determined such that

(

z

, w) =

e(k+1)

_(w)

_,

₍

_x

2

,

y3

) =

e(k+1)

(

y3

)

,

(

y4

,

x3

) =

e(k+1)

(

x3

)

, and

(

z0

, w

0

) =

e(k+1)

(w

0

)

.

There is a longest z–x2path of length k

! −

1 in

h∗

ka2

i

k. ByLemma 6, there exists a longest s–t path of length k

! −

1 or

k

! −

2 in

h∗

k_a

1

i

kthat contains (

w

,

w

0). Imagine that (x3, y3) is fault-free. The induction hypothesis assures a longest y3–x3

path of length k

! −

1 in

h∗

ka3

i

k

−

(

F

− {

(

x3

,

y3

)})

. ByLemma 5, there is a longest y4–z0path of length

(

k

−

2

) ×

k

! −

1 in

Sk+1

[

VA

] −

F , where A

= {

1

,

2

, . . . ,

k

+

1

} − {

a1

,

a2

,

a3

}

(

k

! −

1

) + ((

k

! −

1

) −

1

) + (

k

! −

1

) + ((

k

−

2

) ×

k

! −

1

) +

4

=

(

k

+

1

)! −

1 or

(

k

! −

1

) + ((

k

! −

1

) −

2

) + (

k

! −

1

) + ((

k

−

2

) ×

k

! −

1

) +

4

=

(

k

+

1

)! −

2. Then we consider the situation of

β 6= γ

. Set a1

=

β

and ak+1

=

γ

. A link fault (h, h0) is chosen from E

(h∗

k

αik) ∩

F such

that e(k+1)

₍

_h

_),

_e(k+1)

₍

_h0

_{) /∈}

_{F . Three cases are further discussed below.} Case 1. Neither of h and h0_{belongs to N}

₍

_V

_(h∗

k_a

1

i

k)) ∪N

(

V

(h∗

kak+1

i

k)). Set a3

=

α

−

F can be

constructed in a way similar toFig. 2. Refer toFig. 2again. For the purpose of our construction, we let A

= {

1

,

2

, . . . ,

k

+

1

}

and

(

x3

,

y3

) = (

h

,

h0

)(∈

E

(h∗

ka3

i

k)). We assume that s and y3belong to the same partite set of Sk+1. In case s and y3belong

to different partite sets of Sk+1, we have s and x3in the same partite set of Sk+1and exchange x3and y3inFig. 2.

Four vertices x1

∈

V

(h∗

ka1

i

k) − {s

}

,

y2

,

x2

∈

V

(h∗

ka2

i

k), and y4

∈

V

(h∗

ka4

i

(

x2

,

y3

) =

e(k+1)

₍

_y

3

)

, y2and x2belong to different partite sets of Sk+1

,

e(k+1)

(

y2

) /∈

F ,

(

x1

,

y2

) =

e(k+1)

(

y2

)

, and

(

y4

,

x3

) =

e(k+1)

(

x3

)

.

There are a longest s–x1path in

h∗

ka1

i

kand a longest y2–x2path in

h∗

ka2

i

k, each of length k

! −

1. Imagine that (x3, y3) is

(8)

Fig. 7. A longest s–t path in Sk+1−F whenβ 6= γ, h∈N(V(h∗ka1ik)), and h0 /∈N(V(h∗ka1ik)) ∪N(V(h∗kak+1ik)).

Fig. 8. A longest s–t path in Sk+1−F whenβ 6= γ, h∈N(V(h∗ka1ik)), and h0∈N(V(h∗kak+1ik)). (a) z6=t. (b) z=t.

a longest y4

−

t path of length (k

−

2

)×

k

! −

1 or

(

k

−

2

)×

k

! −

2 in Sk+1

[

VA

] −

F , where A

= {

1

,

2

, . . . ,

k

+

1

} − {

a1

,

a2

,

a3

}

. The

desired s–t path has length 3

×

(

k

!−

1

)+((

k

−

2

)×

k

!−

1

)+

3

=

(

k

+

1

)!−

1 or 3

×

(

k

!−

1

)+((

k

−

2

)×

k

!−

2

)+

3

=

(

k

+

1

)!−

2. Case 2. One of h and h0_{belongs to N}

₍

_V

_(h∗

k_a

1

i

k)) ∪ N

(

V

(h∗

kak+1

i

k)). Without loss of generality, we assume that h

∈

N

(

V

(h∗

ka1

i

k))and h0

/∈

N

(

V

(h∗

ka1

i

k))∪N

(

V

(h∗

kak+1

i

k)). A desired s–t path in Sk+1

−

F can be constructed as shown inFig. 7.

Set a2

=

α

and

(

u2

, v

2

) = (

h

,

h0

) ∈

E

(h∗

ka2

i

k). Six vertices

v

1

∈

V

(h∗

ka1

i

k),u1

∈

N

(v

1

) ∩

V

(h∗

ka1

i

k) − {s

}

,

w ∈

V

(h∗

ka1

i

k),

u3

∈

V

(h∗

ka3

i

k),

vk

∈

V

(h∗

kak

i

k), and uk+1

∈

V

(h∗

kak+1

i

k) − {t

}

are determined such that

(v

1

,

u2

) =

e(k+1)

(

u2

)

,

(

u3

, v

2

) =

e(k+1)

(v

2

)

, e(k+1)

(

u1

) /∈

F ,

(vk,

u1

) =

e(k+1)

(

u1

)

,

w

and s belong to different partite sets of Sk+1

,

e(k+1)

(w) /∈

F ,

and

(

uk+1

, w) =

e(k+1)

(w)

.

Imagine that (u2,

v

2) is fault-free. The induction hypothesis assures a longest u2–

v

2path of length k

! −

1 in

h∗

ka2

i

k

−

(

F

−

{

(

u2

, v

2

)})

. ByLemma 6, there exists a longest s–

w

path of length k

! −

1 in

h∗

ka1

i

kthat contains (u1,

v

1). ByLemma 1,

there is a longest uk+1–t path of length k

! −

1 or k

! −

2 in

h∗

kak+1

i

k. ByLemma 5, there is a longest u3–

vk

path of

length

(

k

−

2

) ×

k

! −

1 in Sk+1

[

VA

] −

F , where A

= {

1

,

2

, . . . ,

k

+

1

} − {

a1

,

a2

,

ak+1

}

(

k

!−

1

)+((

k

!−

1

)−

1

)+(

k

!−

1

)+((

k

−

2

)×

k

!−

1

)+

4

=

(

k

+

1

)!−

1 or

(

k

!−

1

)+((

k

!−

1

)−

1

)+(

k

!−

2

)+((

k

−

2

)×

k

!−

1

)+

4

=

(

k

+

1

)! −

2.

Case 3. Both of h0and h belong to N

(

V

(h∗

ka1

i

k)) ∪N

(

V

(h∗

kak+1

i

k)). If h

∈

N

(

V

(h∗

ka1

i

k))(∈ N

(

V

(h∗

kak+1

i

k))), then h0

∈

N

(

V

(h∗

kak+1

i

k))(∈N

(

V

(h∗

ka1

i

k))). Without loss of generality, we assume that h

∈

N

(

V

(h∗

ka1

i

k))and h0

∈

N

(

V

(h∗

kak+1

i

k)).

Set a2

=

α

. Two vertices

w ∈

V

(h∗

ka1

i

k) and z

∈

V

(h∗

kak+1

i

(w,

h

) =

e(k+1)

(

h

)

and

(

z

,

h0

_{) =}

_e(k+1)

₍

_h0

₎

_{. A desired s–t path in S}

k+1

−

F can be constructed as shown inFig. 8(a) if z

6=

t, and as shown in Fig. 8(b) if z

=

t.

If z

6=

t, then ten vertices u1

∈

N

(w) ∩

V

(h∗

ka1

i

k) − {s

}

,

v

1

∈

V

(h∗

ka1

i

k) − {s

}

, u2,

v

2

∈

V

(h∗

ka2

i

k), u3,

v

3

∈

V

(h∗

ka3

i

k),

u4

∈

V

(h∗

ka4

i

k),

vk

∈

V

(h∗

kak

i

k), uk+1

∈

V

(h∗

kak+1

i

k), and

vk

+1

∈

N

(

z

) ∩

V

(h∗

kak+1

i

k) − {t

}

are further determined

such that e(k+1)

(

u1

) /∈

F , (

v

3, u1

) =

e(k+1)

(

u1

)

, u3 and

v

3 belong to different partite sets of Sk+1, e(k+1)

(

u3

) /∈

F ,

(v

2

,

u3

) =

e(k+1)

(

u3

)

, u2and

v

2belong to different partite sets of Sk+1, e(k+1)

(

u2

) /∈

F ,

(vk,

u2

) =

e(k+1)

(

u2

)

,

v

1and t belong

to the same partite set of Sk+1, e(k+1)

(v

1

) /∈

F ,

(

uk+1

, v

1

) =

e(k+1)

(v

1

)

, e(k+1)

(vk

+1

) /∈

F , and

(

u4

, vk

+1

) =

e(k+1)

(vk

+1

)

.

There is a longest u3–

v

3path of length k

!−

1 in

h∗

ka3

i

k. Imagine that (h, h0) is fault-free. The induction hypothesis assures a

longest u2–

v

2path of length k

!−

1 in

h∗

ka2

i

k

−

(

F

−{

(

h

,

h0

)})

. ByLemma 6, there exist a longest uk+1–t path of length k

!−

1 in

h∗

kak+1

i

kthat contains (z,

vk

+1) and a longest s–

v

1path of length k

!−

1 or k

!−

2 in

h∗

ka1

i

kthat contains (u1,

w

). Since u4and

vk