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Topological Properties on the Wide and Fault

Diameters of Exchanged Hypercubes

Tsung-Han Tsai, Y-Chuang Chen, and Jimmy J.M. Tan

Abstract—The n-dimensional hypercube is one of the most popular topological structure for interconnection networks in parallel computing and communication systems. The exchanged hypercube EHðs; tÞ, a variant of the hypercube, retains several valuable and desirable properties of the hypercube such as a small diameter, bipancyclicity, and super connectivity. In this paper, we construct s þ 1 (or t þ 1) internally vertex-disjoint paths between any two vertices for parallel routes in the exchanged hypercube EHðs; tÞ for 3  s  t. We also show that both theðs þ 1Þ-wide diameter and s-fault diameter of the exchanged hypercube EHðs; tÞ are s þ t þ 3 for 3  s  t. Index Terms—Hypercube, exchanged hypercube, interconnection network, internally vertex-disjoint paths, wide diameter, fault diameter

Ç

1

I

NTRODUCTION

A

multiprocessor/multicomputer interconnection network is usually modeled as a graph, in which vertices corre-spond to processors/computers, and edges correcorre-spond to connections/communication links. Throughout this paper, the terms networks and graphs are interchangeable. A graph G is a two-tuple ðV; EÞ, where V is a nonempty vertex set, and E is a subset of fðu; vÞ j ðu; vÞ is an unordered pair of V g. V ðGÞ and EðGÞ denote the vertex set and the edge set of G, respectively. Two vertices, u and v, of a graph G are adjacent if ðu; vÞ 2 EðGÞ. The neighborhoods of a vertex v in graph G, denoted by NGðvÞ, is fx j ðv; xÞ 2 EðGÞg. A path P of length k

from vertex u to vertex v in a graph G is a sequence of dis-tinct vertices written as x0! x1! x2!    ! xk where

x0¼ u, xk¼ v, and ðxi; xiþ1Þ 2 EðGÞ for every 0  i  k  1

if k  1. The path P can be written as u ! P ! v to empha-size its first and last vertices. For convenience, P can also be written as x0!    ! xi! Q ! xj!    ! xk, where Q ¼

xi!    ! xj. Given a path P from u to v, all vertices in P

except u and v are called internal vertices of P . Two paths are called internally vertex-disjoint (abbreviated as internally dis-joint) if they share no internal vertex. A cycle is a path with at least three vertices such that the last vertex is adjacent to the first one. For clarity, a cycle of length k is represented by x1! x2!    ! xk! x1. The distance between two vertices

u and v in graph G, denoted by dGðu; vÞ, is the length of the

shortest path between u and v.

To design an interconnection network with desired topologies is an important issue [5]. The hypercube is one of the most popular interconnection network structures in par-allel computing and communication systems [7], [11], [19],

[23]. This is partly because of many attractive properties of the hypercube such as regularity, recursive structure, vertex and edge symmetry, and maximum connectivity, as well as the effective routing and broadcasting. An n-dimensional hypercube, denoted by Qn, is a graph with 2n vertices and n  2n1

edges. Each vertex is labeled by an n-bit binary string u ¼ un1un2   u0. Two vertices are adjacent if and only if

their strings differ exactly in one bit position. Let u ¼ un1un2   u0 and v ¼ vn1vn2   v0 be two n-bit

binary strings. The Hamming distance between two vertices u and v, denoted by Hðu; vÞ, is the number of different bits in the corresponding strings of both vertices. Thus, Hðu; vÞ ¼ dQnðu; vÞ. Note that Qnhas diameter n [23].

As a variant of the n-dimensional hypercube, the exchanged hypercube EHðs; tÞ, which was proposed by Loh et al. [13], is defined by removing some edges from the hypercube. To make EHðs; tÞ useful in reliable and critical applications, studies have been conducted, which have pro-duced some significant results. EHðs; tÞ retains several desirable properties of the hypercube such as a small diameter [13], bipancyclicity [16], and super connectivity [17] and this makes it even better than a hypercube. This is evi-dent in the fact that even though the number of edges of an exchanged hypercube is nearly half of that of a hypercube, their diameters are similar. Thus, exchanged hypercubes have lower link costs than hypercubes. To transfer informa-tion safely and quickly between any two vertices in exchanged hypercubes, we need to find as many as possible internally disjoint paths between the two vertices. This idea was proposed in Menger’s theorem [18], which states that there are k internally disjoint paths between any two verti-ces in an interconnection network if k is less than or equal to the connectivity of this network. Moreover, this intercon-nection network has many benefits such as parallel routing and fault tolerance. In recent years, many literature referen-ces discuss the topic of internally disjoint paths in some spe-cific networks, such as hypercubes [20], crossed cubes [8], ðn; kÞ-star graphs [12], folded hypercubes [15], hypercube-like graphs [19], hierarchical hypercubes [21], and aug-mented k-ary n-cubes [22]. Next, we discuss the fault and wide diameters of exchanged hypercubes. The fault diameter,

 T.-H. Tsai and J.J.M. Tan are with the Department of Computer Science, National Chiao Tung University, Hsinchu 300, Taiwan.

E-mail: {tsaich, jmtan}@cs.nctu.edu.tw.

 Y.-C. Chen is with the Department of Information Management, Minghsin University of Science and Technology, Xinfeng, Hsinchu 304, Taiwan. E-mail: cardy@must.edu.tw.

Manuscript received 5 Dec. 2013; accepted 2 Feb. 2014. Date of publication 23 Feb. 2014; date of current version 14 Nov. 2014.

Recommended for acceptance by J. Zhang.

For information on obtaining reprints of this article, please send e-mail to: reprints@ieee.org, and reference the Digital Object Identifier below.

Digital Object Identifier no. 10.1109/TPDS.2014.2307853

1045-9219ß 2014 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.

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which was first proposed in [9], is used to estimate the effects of faults on the diameter, while the wide diameter is used to measure the diameter of the connections with pre-scribed bandwidths, and it is a combination of both the diameter and connectivity. The fault and wide diameters have been discussed in [2], [3], [4], [6], [12], [15], [21], [22]. In this study, we construct s þ 1 (or t þ 1) internally disjoint paths between any two vertices for parallel routes in the exchanged hypercube EHðs; tÞ for 3  s  t. We also prove that both the ðs þ 1Þ-wide diameter and s-fault diameter are s þ t þ 3 for 3  s  t.

The rest of this paper is organized as follows. In the next section, we provide the definition for exchanged hyper-cubes and describe their properties. In Section 3, the main results are presented; the internally disjoint paths between any two vertices in EHðs; tÞ for 3  s  t are discussed and it is demonstrated that both ðs þ 1Þ-wide diameter and s-fault diameter are s þ t þ 3 for 3  s  t. In Section 4, con-cluding remarks are presented.

2

P

RELIMINARIES

The exchanged hypercube is defined as an undirected graph EHðs; tÞ ¼ GðV; EÞ, where s  1 and t  1. The definition of exchanged hypercubes is given as follows.

Definition 1. The vertex set V of exchanged hypercube EHðs; tÞ ðs  1; t  1Þ is the set

futþs   utþ1ut   u1u0j ui2 f0; 1g for 0  i  s þ tg:

Let u ¼ utþsutþs1   u0and v ¼ vtþsvtþs1   v0be two

ver-tices in EHðs; tÞ. There is an edge ðu; vÞ in EHðs; tÞ if and only if ðu; vÞ is in one of the following sets:

E1¼fðu; vÞ j u06¼ v0; ui¼ vifor 1  i  s þ tg;

E2¼fðu; vÞ j u0¼ v0¼ 0; Hðu; vÞ ¼ 1 with ui6¼ vi

for some t þ 1  i  s þ tg; and

E3¼fðu; vÞ j u0¼ v0¼ 1; Hðu; vÞ ¼ 1 with ui6¼ vi

for some 1  i  tg;

where Hðu; vÞ denotes the Hamming distance between two ver-tices u and v.

According to the definition of EHðs; tÞ, the number of vertices is 2sþtþ1 and the number of edges is

ðs þ t þ 2Þ2sþt1. For a vertex x with x

0¼ 0, the vertex

degree is s þ 1, whereas the vertex degree with x0¼ 1 is

t þ 1. EHðs; tÞ is a subgraph of the ðs þ t þ 1Þ-dimensional hypercube Qsþtþ1, and thus it is also a bipartite graph. Fig. 1

illustrates the exchanged hypercubes EHð1; 1Þ, EHð1; 2Þ, EHð2; 1Þ and EHð2; 2Þ. Dashed links correspond to the edge set E1, solid links correspond to the edge set E2, and bold

links correspond to the edge set E3.

Loh et al. [13] stated the following properties. Property 1. The diameter of EHðs; tÞ is s þ t þ 2. Property 2. EHðs; tÞ is isomorphic to EHðt; sÞ.

Property 3. EHðs; tÞ can be decomposed into two copies of EHðs  1; tÞ or EHðs; t  1Þ.

Property 4. The subgraphs induced by the vertices of the form     zfflffl}|fflffl{

s

utut1   u10 and utþsutþs1   utþ1    

zfflffl}|fflffl{t

1 in EHðs; tÞ are isomorphic to Qs and Qt, respectively, where

 2 f0; 1g.

The subgraphs induced by the vertex sets V ðQsÞ and

V ðQtÞ are denoted by S and T , respectively. Then, S ffi Qs

and T ffi Qt. Therefore, by Property 4, there are 2tand 2s

dis-tinct induced subgraphs Qsand Qt, respectively. Denote by

Qi

s (respectively, Q j

t) for 0  i  2t 1 (respectively,

0  j  2s 1) where i (respectively, j) with radix 10 is the

value of utut1   u1 (respectively, utþsutþs1   utþ1). Let

hsðu; vÞ (respectively, htðu; vÞ) denote the number of

differ-ent bits between u and v in dimensions t þ 1 to s þ t (respec-tively, 1 to t). When the context is clear, hsðu; vÞ and htðu; vÞ

are simply written as hs and hs, respectively. Moreover,

since EHðs; tÞ is isomorphic to EHðt; sÞ by Property 2, we may, without loss of generality, assume that s  t in this paper.

A vertex set S V ðGÞ is a separating set or a vertex cut if G  S is disconnected. The connectivity of G, written as kðGÞ, is the minimum size of a vertex cut. Let dðGÞ be the minimum degree of G, then it is clear that kðGÞ  dðGÞ. A graph G is k-connected if the connectivity kðGÞ is at least k. Moreover, a graph G has connectivity k if G is k-connected

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but not ðk þ 1Þ-connected. This follows from Menger’s theo-rem [18], which states that the connectivity of a graph is at least k if and only if there exist k internally disjoint paths between any two vertices.

Let a and b be two positive integers such that a  kðGÞ and b  kðGÞ  1. Given any two vertices u and v of G, let Daðu; vÞ denote the set of all a internally disjoint paths

between u and v. Each element of Daðu; vÞ consists of a

internally disjoint paths. jDaðu; vÞj denotes the number of

elements in Daðu; vÞ. Let liðu; vÞ denote the longest length

among the a paths of the i-th element of Daðu; vÞ. Thus,

lDaðu; vÞ and d

f

bðu; vÞ are defined as follows:

lDaðu; vÞ ¼ min

1ijDaðu;vÞj

liðu; vÞ;

dfbðu; vÞ ¼ max

F V;jF j¼bfdGFðu; vÞju; v =2 F g;

where G  F denotes the subgraph of G induced by V  F . In other words, dfbðu; vÞ denotes the longest distance

between u and v when any b faulty vertices occur.

Definition 2. [1] The a-wide diameter of G, denoted by DaðGÞ,

is defined as

DaðGÞ ¼ max

u;v2VflDaðu; vÞg:

In particular, DkðGÞðGÞ is the wide diameter of G and D1ðGÞ is

simply the diameter DðGÞ of G.

Definition 3. [1] The b-fault diameter of G, denoted by Df bðGÞ, is defined as DfbðGÞ ¼ maxu;v2V  dfbðu; vÞ 

In particular, DfkðGÞ1ðGÞ is the fault diameter of G.

Obviously, DðGÞ  DfkðGÞ1ðGÞ  DkðGÞðGÞ. For the

hyper-cubes Qn, Latifi [10] proved that DnðQnÞ ¼ Dfn1ðQnÞ ¼ n þ 1

for n  2. For the crossed cubes CQn, Chang et al. [1] proved

that DnðCQnÞ ¼ Dfn1ðCQnÞ ¼ dn2e þ 2 for n  2. In this

paper, we also discuss and prove the wide and the fault diameters of exchanged hypercubes, and proved that Dsþ1ðEHðs; tÞÞ ¼ DfsðEHðs; tÞÞ ¼ s þ t þ 3 for 3  s  t. The

following three theorems are needed in the proofs of our results.

Theorem 1. [14] The connectivity of the exchanged hypercubes EHðs; tÞ is s þ 1 for 1  s  t.

From Menger’s theorem, there exist s þ 1 internal disjoint paths between any two vertices in the exchanged hypercube EHðs; tÞ.

Theorem 2. [20] Let u, v be any two vertices of the n-dimensional hypercube Qnand assume that Hðu; vÞ ¼ d. Then there are n

internally disjoint paths between u and v such that d of them are of length d, and the remaining n  d paths are of length d þ 2.

Theorem 3. [16] In the exchanged hypercube EHðs; tÞ for 1  s  t, the vertices in the set Vc¼ futþs   u0j u0¼

c; ui 2 f0; 1g for 1  i  s þ tg are vertex-transitive, where

c 2 f0; 1g.

For convenience, consecutive i 0’s and 1’s are denoted by 0iand 1i, respectively. That is, 0i¼ 00    0zfflfflffl}|fflfflffl{

i

and 1i¼ 11    1zfflfflffl}|fflfflffl{ i

.

3

W

IDE AND

F

AULT

D

IAMETERS OF

E

XCHANGED

H

YPERCUBES

In this section, our goal is to prove that Dsþ1ðEHðs; tÞÞ ¼

Df

sðEHðs; tÞÞ ¼ s þ t þ 3 for 3  s  t.

Lemma 1. Df

sðEHðs; tÞÞ  s þ t þ 3 for 1  s  t.

Proof. Let u; u0; and v be three vertices of EHðs; tÞ. We

con-sider that u ¼ 0s0t0, u0¼ 0s0t1 and v ¼ 1s1t1. See Fig. 2

for illustration. Suppose that F is a faulty vertex set such that F ¼ NEHðs;tÞðuÞ  u0. The shortest path between u

and v, denoted by P , in EHðs; tÞ  F must pass through u0. Thus, P can be written as u ! u0! R ! v where R is

the shortest path from u0to v in EHðs; tÞ  F . The subpath R can be written as follows:

u0! H ! 0s1t1 ! 0s1t0 ! L ! 1s1t0 ! v:

Note that the length of subpath u0! H ! 0s1t1 is t and

all vertices of H are in Q0

t; moreover, the length of

sub-path 0s1t0 ! L ! 1s1t0 is s and all vertices of L are in

Q2t1

t . Thus, the length of the subpath u0! R ! v is

s þ t þ 2, and it follows that dEHðs;tÞFðu; vÞ ¼ 1 þ ðsþ

t þ 2Þ ¼ s þ t þ 3. Therefore, DfsðEHðs; tÞÞ  s þ t þ 3 for

1  s  t. tu Next, to show that Dsþ1ðEHðs; tÞÞ  s þ t þ 3 for

3  s  t, internally disjoint paths between any two verti-ces u and v of EHðs; tÞ are constructed in Lemmas 2-11. Table 1 illustrates the conditions of vertices u and v in Lemmas 2-11. For convenience, some symbols are used in the following proofs. Let u ! P ! v be a path from u to v in EHðs; tÞ. The predecessor vertex of v in P is denoted by preðP; u; iÞ if their i-th bits are different. Sim-ilarly, the successor vertex of u in P is denoted by sucðP; u; jÞ if their j-th bits are different. We use lðP Þ to denote the length of P .

Lemma 2. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u0¼ v0¼ 0, and htðu; vÞ ¼ 0, then there exist s þ 1

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internally disjoint paths Pi for 1  i  s þ 1 between u and v

such that hsof them are of length hs, s  hspaths are of length

hsþ 2, and one path is of length hsþ 6.

Proof. By Theorem 3, we may assume without loss of gen-erality that u ¼ 0s0t0 and v ¼ 0shs1hs0t0 are in Q0

s. See

Fig. 3 for illustration. By Theorem 2, in Q0

s, there exist s

internally disjoint paths between u and v such that hs

of them are of length hsand the remaining s  hspaths

are of length hsþ 2. Let u ! Pi! v for 1  i  s be

those s internally disjoint paths. The following sets of s þ 1 internally disjoint paths between u and v in EHðs; tÞ can be set:

We construct the paths Pi for 1  i  s from u to v as

follows:

u ! Pi! v:

Note that all the paths u ! Pi! v are in Q0s. Thus,

lðPiÞ ¼ hs for 1  i  hs and lðPiÞ ¼ hsþ 2 for hsþ 1 

i  s.

The path Psþ1 can be constructed from u to v as

fol-lows:

u ! 0s0t1 ! 0s0t111

! 0s0t110 ! L ! 0shs1hs0t110

! 0shs1hs0t111 ! 0shs1hs0t1 ! v:

Note that edge 0s0t1 ! 0s0t111 is in Q0

t, subpath

0s0t110 ! L ! 0shs1hs0t110 is in Q1

s and edge

0shs1hs0t111 ! 0shs1hs0t1 is in Q2hs1

t . It can be seen

that Psþ1 is also internally disjoint to Pi for 1  i  s.

Moreover, we have lðLÞ ¼ hs. Therefore, lðPsþ1Þ ¼ hsþ 6.

This completes the proof. tu Lemma 3. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u0¼ v0¼ 0 and htðu; vÞ 6¼ 0, then there exist s þ 1 internally

disjoint paths Pifor 1  i  s þ 1 between u and v such that

hsþ 1 of them are of length hsþ htþ 2 and s  hspaths are

of length hsþ htþ 4.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t0 and v ¼ 0shs1hs0tht1ht0 are in Q0

s

and Q2ht1

s , respectively. Depending on hs, two cases are

distinguished.

Case 1: hs¼ 0. Then, u ¼ 0s0t0 and v ¼ 0s0tht1ht0. See

Fig. 4 for illustration. Let ui ¼ u 0si10i10t0 ¼

0si10i10t0 and vi¼ v 0si10i10t0 ¼ 0si10i10tht1ht0

for 1  i  s where is the exclusive-or operation. To construct a path from ui to vi, we need the following

intermediate vertices: xi¼ ui 0s0t1 and yi¼ vi 0s0t1.

Accordingly, xi¼ 0si10i10t1 and yi¼ 0si10i10tht

1ht1. Now we construct path Pifor 1  i  s from u to v

as follows:

u ! ui! xi! R

i! yi! vi! v:

TABLE 1

The Conditions of Lemmas 2-11 in EHðs; tÞ

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Note that edge u ! uiis in Q0 s, subpath xi! Ri! yiis in Q2i1 t and edge vi! v is in Q2 ht1 s while ui! xi and

yi! vi are two edges in E

1. This can be confirmed that

those Pifor 1  i  s are internally disjoint.

It remains to construct the ðs þ 1Þ-th internally dis-joint path from u to v. Path Psþ1 can be constructed as

follows:

u ! u0! R ! v0! v:

Note that u0¼ 0s0t1 and v0¼ 0s0tht1ht1. We can find

that subpath u0! R ! v0 is in Q0

t while u ! u0 and

v0! v are two edges in E 1.

By inspection, the vertices in Q0

t are not in Pi for

1  i  s. Thus, path Psþ1is also internally disjoint to Pi

for 1  i  s. Since both xi and yi are in Q2i1

t ,

lðRiÞ ¼ Hðxi; yiÞ ¼ ht. Moreover, both u0 and v0 are in

Q0

t, and lðRÞ ¼ ht. Therefore, lðPiÞ ¼ htþ 4 for 1  i  s

and lðPsþ1Þ ¼ htþ 2.

Case 2: 1  hs s. Then, u ¼ 0s0t0 and v ¼

0shs1hs0tht1ht0. See Fig. 5 for illustration. Let

w ¼ 0shs1hs0t0 and z ¼ 0s0tht1ht0 be in Q0

s and Q2

ht1

s ,

respectively. By Theorem 2, in Q0

s, there exist s internally

disjoint paths between u and w such that hsof them are

of length hsand the remaining s  hspaths are of length

hsþ 2. Let u ! Hi! w for 1  i  s be those internally

disjoint paths. Similarly, in Q2ht1

s , there exist s internally

disjoint paths between z and v such that hsof them are of

length hs, and the remaining s  hs paths are of length

hsþ 2. We also denote z ! Li! v for 1  i  s are those

internally disjoint paths.

Now the path P1can be constructed as follows:

u ! H1! w ! w0! R0! v0! v:

Note that w0¼ w 0s0t1 ¼ 0shs1hs0t1 and v0¼ v

0s0t1 ¼ 0shs1hs0tht1ht1 where is the exclusive-or

operation. We can find that subpath u ! H1! w is in Q0s

and subpath w0! R0! v0is in Q2hs1

t while w ! w0and

v0! v are two edges in E

1. Since both w0and v0 are in

Q2ths1, lðR0Þ ¼ Hðw0; v0Þ ¼ ht. In addition, we have

lðH1Þ ¼ hs. Hence lðP1Þ ¼ hsþ htþ 2.

Based on Hi and Li for 2  i  s, we construct s  1

internally disjoint paths from u to v as follows. Let wi¼

preðHi; w; t þ iÞ and vi¼ preðLi; v; t þ iÞ where their

ðt þ iÞ-th bits are different. This is, wi¼ 0shs1hsi01i10t0

and vi¼ 0shs1hsi01i10tht1ht0 when 2  i  h

s, and

wi¼ 0si10ihs11hs0t0 and vi¼ 0si10ihs11hs0tht1ht0

when hsþ 1  i  s. Assume that H0iis the subpath of Hi

without containing w. Clearly, all u ! Hi0! wi for

2  i  s are internally disjoint. To construct a path from wi to vi, we need the following intermediate

verti-ces: xi¼ wi 0s0t1 and yi¼ vi 0s0t1 for 2  i  s

where is the exclusive-or operation. Accordingly, xi¼ 0shs1hsi01i10t1 and yi¼ 0shs1hsi01i1 0tht1ht1

when 2  i  hs, and xi¼ 0si10ihs11hs0t1 and yi¼

0si10ihs11hs0tht1ht1 when hsþ 1  i  s. We

con-struct a path from wito v as follows:

wi ! xi! R

i! yi! vi:

Combining the subpaths above, we can obtain paths Pi

for 2  i  s from u to v as follows:

u ! H0i! wi! xi! Ri! yi! vi ! v:

Note that, for 2  i  hs (respectively, hsþ 1  i  s),

subpath u ! Hi0! wi is in Q0s, subpath xi ! Ri! yi is in Q2hs2i11 t (respectively, Q2 hsþ2i11 t ) and edge vi ! v is in Q2ht1

s while wi! xi and yi! vi are two edges in

E1. Moreover, the vertices in Q2

hs1

t are not in Pi for

2  i  s. It is easy to verify that all those Pifor 2  i  s

are internally disjoint.

Since both xi and yi are in Q2hs2i11

t (or Q2

hsþ2i11

t ),

lðRiÞ ¼ Hðxi; yiÞ ¼ ht. Note that subpath u ! Hi0! wi is

of length hs 1 for 2  i  hs and hsþ 1 for hsþ

1  i  s. As a result, lðPiÞ ¼ hsþ htþ 2 for 2  i  hs

and lðPiÞ ¼ hsþ htþ 4 for hsþ 1  i  s.

Next, the path Psþ1can be constructed as follows:

u ! u0! R ! z0! z ! L 1! v:

Note that u0¼ 0s0t1 and z0¼ 0s0tht1ht1. We can find

that subpath u0! R ! z0 is in Q0

t and subpath

z ! L1! v is in Q2

ht1

s while u ! u0and z0! z are two

edges in E1. By inspection, the vertices in Q0tare not in Pi

for 1  i  s. Thus, all paths Pifor 1  i  s þ 1 are

inter-nally disjoint. Since both u0 and z0 are in Q0 t,

lðRÞ ¼ Hðw0; v0Þ ¼ h

t. In addition, both z and v are in

Q2ht1

s , and lðL1Þ ¼ hs. Therefore, lðPsþ1Þ ¼ hsþ htþ 2.

This completes the proof. tu Lemma 4. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u0¼ v0¼ 1, and hsðu; vÞ ¼ 0, then there exist t þ 1 internally

disjoint paths Pi for 1  i  t þ 1 between u and v such that

htof them are of length ht, t  htpaths are of length htþ 2,

and one path is of length htþ 6.

Proof. By Theorem 3, we may assume without loss of generality that u ¼ 0s0t1 and v ¼ 0s0tht1ht1 are in Q0

t.

By Theorem 2, in Q0

t, there exist t internally disjoint

paths between u and v such that ht of them are of

length ht, and the remaining t  ht paths are of length

htþ 2. Let u ! Hi! v for 1  i  t be those t

inter-nally disjoint paths. The following sets of t þ 1

(6)

internally disjoint paths between u and v in EHðs; tÞ can be set:

We construct the paths Pi for 1  i  t from u to v as

follows:

u ! Pi! v:

Note that all the paths u ! Pi! v are in Q0t. Thus,

lðPiÞ ¼ ht for 1  i  ht and lðPiÞ ¼ htþ 2 for

htþ 1  i  t.

The path Ptþ1 can be constructed from u to v as

fol-lows:

u ! 0s0t0 ! 0s110t0

! 0s110t1 ! L ! 0s110tht1ht1

! 0s110tht1ht0 ! 0s0tht1ht0 ! v:

Note that edge 0s0t0 ! 0s110t0 is in Q0

s, subpath 0s1

10t1 ! L ! 0s110tht1ht1 is in Q1

t and edge 0s1

10tht1ht0 ! 0s0tht1ht0 is in Q2ht1

s . It is easy to verify

that Ptþ1is also internally disjoint to Pifor 1  i  t.

Fur-thermore, we have lðLÞ ¼ ht. Therefore, lðPtþ1Þ ¼ htþ 6.

This completes the proof. tu Lemma 5. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u0¼ v0¼ 1, and hsðu; vÞ 6¼ 0, then there exist t þ 1 internally

disjoint paths Pifor 1  i  t þ 1 between u and v such that

htþ 1 of them are of length hsþ htþ 2 and t  htpaths are

of length hsþ htþ 4.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t1 and v ¼ 0shs1hs0tht1ht1 are in Q0

t

and Q2hs1

t , respectively. Depending on ht, two cases are

distinguished.

Case 1: ht¼ 0. Then, u ¼ 0s0t1 and v ¼ 0shs1hs0t1. Let

ui¼ u 0s0ti10i10 ¼ 0s0ti10i11 and vi¼ v 0s0ti

10i10 ¼ 0shs1hs0ti10i11 for 1  i  t where is

the exclusive-or operation. To construct a path from ui

to vi, we need the following intermediate vertices: xi¼

ui 0s0t1 ¼ 0s0ti10i10 and yi ¼ vi 0s0t1 ¼ 0shs

1hs0ti10i10. Now we construct path P

i for 1  i  t

from u to v as follows: u ! ui ! xi! R

i! yi! vi! v:

Note that edge u ! uiis in Q0

t, subpath xi! Ri! yiis in

Q2i1

s and edge vi! v is in Q2

hs1

t while ui! xi and

yi! viare two edges in E 1.

It remains to construct the ðt þ 1Þ-th internally disjoint path from u to v. Path Ptþ1can be constructed as follows:

u ! u0! R ! v0! v:

Note that u0¼ 0s0t0 and v0¼ 0shs1hs0t0. We can find

that subpath u0! R ! v0 is in Q0

s while u ! u0 and

v0! v are two edges in E

1. This can be confirmed that

those Pi for 1  i  t þ 1 are internally disjoint.

More-over, we have lðRiÞ ¼ lðRÞ ¼ hs. Therefore, lðPiÞ ¼ hsþ 4

for 1  i  t and lðPtþ1Þ ¼ hsþ 2.

Case 2: 1  ht t. Then, u ¼ 0s0t1 and v ¼ 0shs

1hs0tht1ht1. Let w ¼ 0s0tht1ht1 and z ¼ 0shs1hs0t1 be in

Q0

t and Q2

hs1

t , respectively. By Theorem 2, in Q0t, there

exist t internally disjoint paths between u and w such that htof them are of length ht, and the remaining t  ht

paths are of length htþ 2. Let u ! Hi! w for 1  i  t

be those internally disjoint paths. Similarly, there exist t internally disjoint paths between z and v such that htof

them are of length ht, and the remaining t  htpaths are

of length htþ 2. We also denote z ! Li ! v for 1  i  t

are those internally disjoint paths.

Now the path P1can be constructed as follows:

u ! H1! w ! w0! R0! v0! v:

Note that w0¼ 0s0tht1ht0 and v0¼ 0shs1hs0tht1ht0. We

can find that subpath u ! H1! w is in Q0t and subpath

w0! R0! v0 is in Q2ht1

s while w ! w0 and v0! v are

two edges in E1. Additionally, we have lðR0Þ ¼ hs and

lðH1Þ ¼ ht. Hence lðP1Þ ¼ hsþ htþ 2.

Based on Hi and Li for 2  i  t, we construct t  1

internally disjoint paths from u to v as follows. Let wi¼ preðH

i; w; iÞ and vi ¼ preðLi; v; iÞ where their i-th

bits are different. This is, wi¼ 0s0tht1hti01i11 and

vi¼ 0shs1hs0tht1hti01i11 when 2  i  h

t, and wi¼

0s0tht1hti01i11 and vi¼ 0shs1hs0ti10iht11ht1 when

htþ 1  i  t. Assume that Hi0is the subpath of Hi

with-out containing w. Clearly, all u ! Hi0! wi for 2  i  t

are internally disjoint. To construct a path from wi to

vi, we need the following intermediate vertices: xi¼ 0s0tht1hti01i10 and yi¼ 0shs1hs0tht1hti01i10

when 2  i  hs, and xi¼ 0s0ti10iht11ht0 and

yi¼ 0shs1hs0ti10iht11ht0 when hsþ 1  i  t. We

con-struct paths Pifor 2  i  t from u to v as follows:

u ! H0i! wi! xi! Ri! yi! vi ! v:

Note that, for 2  i  ht (respectively, hsþ 1  i  t),

subpath u ! H0 i! wi is in Q0t, subpath xi ! Ri! yi is in Q2ht2i11 s (respectively, Q2 htþ2i11 s ) and edge vi ! v is in Q2hs1

t while wi! xi and yi! vi are two edges in

E1. Additionally, we have lðHi0Þ ¼ ht 1 for 2  i  ht

and lðHi0Þ ¼ htþ 1 for htþ 1  i  t, and lðRiÞ ¼ hs for

2  i  t, As a result, lðPiÞ ¼ hsþ htþ 2 for 2  i  ht

and lðPiÞ ¼ hsþ htþ 4 for htþ 1  i  t.

Next, the path Ptþ1can be constructed as follows:

u ! u0! R ! z0! z ! L 1! v:

Note that u0¼ 0s0t0 and z0¼ 0shs1hs0t0. We can find

that subpath u0! R ! z0 is in Q0

s and subpath

z ! L1! v is in Q2

hs1

t while u ! u0and z0! z are two

edges in E1. Moreover, we have lðRÞ ¼ hsand lðL1Þ ¼ ht.

Therefore, lðPtþ1Þ ¼ hsþ htþ 2.

It is easy to verify that all those Pifor 1  i  t þ 1 are

internally disjoint. This completes the proof. tu tu Lemma 6. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u06¼ v0 and hsðu; vÞ ¼ htðu; vÞ ¼ 0, then there exist s þ 1

internally disjoint paths Pifor 1  i  s þ 1 between u and v

such that s of them are of length 7 and one path is of length 1. Proof. By Theorem 3, we may assume without loss of

gener-ality that u ¼ 0s0t0 and v ¼ 0s0t1 are in Q0

s and Q0t,

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vi¼ v 0s0ti10i10 ¼ 0s0ti10i11 for 1  i  t where

is exclusive-or operation. To construct a path from ui

to vi, we need the following intermediate vertices:

xi ¼ ui 0s0t1 ¼ 0si10i10t1, yi ¼ xi 0s0ti10i10 ¼

0si10i10ti10i11, zi¼ wi 0si 10i10t0 ¼ 0si10i10ti

10i10 and wi¼ vi 0s0t1 ¼ 0s 0ti10i10. Now we

con-struct path Pifor 1  i  t from u to v as follows:

u ! ui ! xi! yi! zi! wi! vi! v:

Note that, for 1  i  s, edge u ! ui is in Q0 s, edge xi! yi is in Q2i1 t , edge zi! wi is in Q2 i1 s and edge vi! v is in Q0

t while ui! xi, yi! zi and wi! vi are

three edges in E1. Thus, lðPiÞ ¼ 7 for 1  i  s. This can

be confirmed that all those Pi for 1  i  s are internally

disjoint.

Finally, the path Psþ1 is u ! v, and lðPsþ1Þ ¼ 1. This

completes the proof. tu Lemma 7. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u06¼ v0, hsðu; vÞ 6¼ 0 and htðu; vÞ ¼ 0, then there exist s þ 1

internally disjoint paths Pi for 1  i  s þ 1 between u and v

such that hs of them are of length hsþ 5, s  hs paths are of

length hsþ 7, and one path is of length hsþ 1.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t0 and v ¼ 0shs1hs0t1 are in Q0

s and

Q2hs1

t , respectively. Let w ¼ 0shs1hs0t0 be in Q0s. By

The-orem 2, in Q0

s, there exist s internally disjoint paths

between u and w such that hs of them are of length hs,

and the remaining s  hs paths are of length hsþ 2. Let

u ! Hi! w for 1  i  s be those internally disjoint

paths.

The path P1can be constructed as follows:

u ! H1! w ! v:

Note that subpath u ! H1! w is in Q0swhile w ! v is an

edge in E1. Thus, lðP1Þ ¼ hsþ 1.

Based on Hi for 2  i  s, we construct s  1

inter-nally disjoint paths from u to v as follows. Let wi¼ preðH

i; w; iÞ where their ðt þ iÞ-th bits are different.

This is, wi¼ 0shs1hsi01i10t0 when 2  i  h s and

wi¼ 0si10ihs11hs0t0 when h

sþ 1  i  s. Assume that

H0

iis the subpath of Hiwithout containing w. Clearly, all

u ! Hi0! wi for 2 ! i ! s are internally disjoint. Let

vi¼ v 0shs1hs0t0 ¼ 0shs1hs0ti10i11 where is the

exclusion-or operation. To construct a path from wito vi,

we need the following intermediate vertices: ai¼ 0shs

1hsi01i10t1, bi¼ 0shs1hsi01i10ti10i11 and ci¼ 0shs

1hsi01i10ti10i10 when 2  i  h

s, and ai ¼ 0si

10ihs11hs0t1, bi ¼ 0si10ihs11hs0ti10i11 and ci¼

0si10ihs11hs0ti10i10 when h

sþ 1  i  s. di¼ 0shs

1hs0ti10i11 when 2  i  s. We construct paths P i for

2  i  s from u to v as follows:

u ! Hi0! wi! ai! bi! ci! di! vi! v:

Note that, for 2  i  hs (respectively, hsþ 1  i  s),

subpath u ! H0 i! wi is in Q0s, edge ai! bi is in Q2hs2i11 t (respectively, Q2 hsþ2i11 t ), edge ci! diis in

Q2si1 and edge vi! v is in Q2

hs1

s while wi! ai, bi ! ci

and di! vi are three edges in E

1. Moreover, we have

lðH0

iÞ ¼ hs 1 for 2  i  hs and lðHi0Þ ¼ hsþ 1 for

hsþ 1  i  s. As a result, lðPiÞ ¼ hsþ 5 for 2  i  hs

and lðPiÞ ¼ hsþ 7 for hsþ 1  i  s.

The path Psþ1 can be constructed from u to v as

fol-lows:

u ! u0! x ! y ! L ! z ! v1! v:

Note that u0¼ 0s0t1, x ¼ 0s0t111, y ¼ 0s0t110 and

z ¼ 0shs1hs0t110. We can find that edge u0! x is in Q0 t,

subpath y ! L ! z is in Q1

s and edge v1! v is in Q2

hs1

s

while u ! u0, x ! y and z ! v1 are three edges in E 1.

Furthermore, we have lðLÞ ¼ hs. Therefore, lðPsþ1Þ ¼

hsþ 5.

It is easy to verify that all those Pifor 1  i  s þ 1 are

internally disjoint. This completes the proof. tu Lemma 8. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u06¼ v0, hsðu; vÞ ¼ 0 and htðu; vÞ 6¼ 0, then there exist s þ 1

internally disjoint paths Pifor 1  i  s þ 1 between u and v

such that the following two cases are distinguished.

1) If htðu; vÞ  s, then s of them are of length htþ 5 and

one path is of length htþ 1.

2) If htðu; vÞ  s  1, then ht of them are of length

htþ 5, s  htpaths are of length htþ 7, and one path

is of length htþ 1.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t0 and v ¼ 0s0tht1ht1 are in Q0

s and Q0t,

respectively. Let ui¼ u 0si10i10t0 ¼ 0si10i10t0 for

1  i  s where is the exclusive-or operation. Now the path P1can be constructed as follows:

u ! u1! x ! L ! y ! z ! v0! v:

Note that x ¼ 0s110t1, y ¼ 0s110tht1ht1, z ¼ 0s1

10tht1ht0 and v0¼ 0s0tht1ht0. We can find that edge

u ! u1is in Q0

s, subpath x ! L ! y is in Q1t, edge z ! v0

is in Q2ht1 while u1! x, y ! z and v0! v are three

edges in E1. Moreover, we have lðLÞ ¼ ht. Hence

lðP1Þ ¼ htþ 5.

Let w ¼ 0s0t1 be in Q0

t. By Theorem 2, in Q0t, there exist

t internally disjoint paths between w and v such that htof

them are of length ht, and the remaining t  htpaths are

of length htþ 2. Let w ! Hi! v for 1  i  t be those

internally disjoint paths. Based on Hi for 2  i  s, we

construct s  1 internally disjoint paths from u to v as fol-lows. Let wi¼ sucðH

i; w; iÞ where their i-th bits are

dif-ferent. This is, wi¼ 0s0ti10i11 when 2  i  s. Assume

that Hi0 is the subpath of Hi without containing w.

Clearly, all wi! H0

i ! v for 2  i  s are internally

dis-joint. To construct a path from ui to wi, we need the

fol-lowing intermediate vertices: ai¼ 0si10i10t1, bi¼ 0si

10i10ti10i11, ci¼ 0si10i10ti10i10 and di¼ 0s0ti

10i10 when 2  i  s. We construct paths P i for

2  i  s from u to v as follows:

u ! ui! ai! bi! ci! di! wi! H0 i! v:

Note that edge u ! ui is in Q0

s, edge ai! bi is in Q2

i1

t ,

edge ci! diis in Q2i1

(8)

while ui! ai, bi! ci and di ! wiare three edges in E 1.

In addition, if htðu; vÞ  s, then lðHi0Þ ¼ ht 1 for

2  i  s; otherwise, lðHi0Þ ¼ ht 1 for 2  i  ht and

lðHi0Þ ¼ htþ 1 for htþ 1  i  s. Thus, if htðu; vÞ  s, we

have lðPiÞ ¼ htþ 5 for 2  i  s; otherwise, lðPiÞ ¼ htþ 5

for 2  i  htand lðPiÞ ¼ htþ 7 for htþ 1  i  s.

The path Psþ1can be constructed as follows:

u ! w ! H1! v:

Note that subpath w ! H1! v is in Q0twhile u ! w is an

edge in E1. Moreover, we have lðH1Þ ¼ ht. Therefore,

lðP1Þ ¼ htþ 1.

It is easy to verify that all those Pifor 1  i  s þ 1 are

internally disjoint. This completes the proof. tu Lemma 9. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u06¼ v0, hsðu; vÞ ¼ s and htðu; vÞ 6¼ 0, then there exist s þ 1

internally disjoint paths Pi for 1  i  s þ 1 between u and v

such that the following two cases are distinguished.

1) If htðu; vÞ  s, then s of them are of length s þ htþ 3

and one path is of length s þ htþ 1.

2) If htðu; vÞ  s  1, then htþ 2 of them are of length

s þ htþ 3 and s  ht 1 paths are of length

s þ htþ 5.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t0 and v ¼ 1s0tht1ht1 are in Q0

s and

Q2s1

t , respectively. Let w ¼ 1s0t0 and x ¼ 1s0t0 be in Q0s

and Q2s1

t , respectively. By Theorem 2, in Q0s, there exist

s internally disjoint paths between u and w such that s of them are of length s. In addition, there exist t internally disjoint paths between x and v such that htof them are of

length ht, and the remaining t  ht paths are of length

htþ 2. Let u ! Hi! w for 1  i  s and x ! Li! v for

1  i  t be those internally disjoint paths. Based on Hi

for 1  i  s  1, we construct s  1 internally disjoint paths from u to v as follows. Let wi¼ preðH

i; w; iÞ where

their ðt þ iÞ-th bits are different. This is, wi¼ 1si01i10t0

when 1  i  s  1. Assume that Hi0is the subpath of Hi

without containing w. Clearly, all u ! H0

i! wi for

1  i  s  1 are internally disjoint. Similarly, based on Lifor 1  i  s  1, we construct s  1 internally disjoint

paths from u to v as follows. Let xi¼ sucðL

i; x; iÞ where

their i-th bits are different. This is, xi¼ 1s0ti10i11

when 1  i  s  1. Assume that L0

i is the subpath of Li

without containing x. Clearly, all wi ! L0

i! v for

1  i  s  1 are internally disjoint. To construct a path from wito xi, we need the following intermediate

verti-ces: ai¼ 1si01i10t1, bi¼ 1si01i10ti10i11, ci ¼ 1si

01i10ti10i10 and di ¼ 1s0ti10i10 when 1  i  s  1.

We construct paths Pi for 1  i  s  1 from u to v as

follows:

u ! Hi0! wi! ai! bi! ci! di! xi! L0i! v:

Note that subpath u ! H0

i! wi is in Q0s, edge ai ! bi is in Q2s2i11 t , edge ci! di is in Q2 i1 s , and subpath xi! L0 i! v is in Q2 s1 t while wi ! ai, bi! ci and

di ! xi are three edges in E

1. Furthermore, we have

lðHi0Þ ¼ s  1 for 1  i  s  1. In addition, if htðu; vÞ  s,

then lðL0iÞ ¼ ht 1 for 1  i  s  1; otherwise,

lðL0iÞ ¼ ht 1 for 1  i  ht and lðL0iÞ ¼ htþ 1 for

htþ 1  i  s  1. Thus, if htðu; vÞ  s, we have lðPiÞ ¼

s þ htþ 3 for 1  i  s  1; otherwise, lðPiÞ ¼ s þ htþ 3

for 1  i  ht and lðPiÞ ¼ s þ htþ 5 for htþ 1 

i  s  1.

The path Pscan be constructed as follows:

u ! Hs! w ! x ! Ls! v:

Note that subpath u ! Hs! w is in Q0s, subpath

x ! Ls! v is in Q2

s1

t while w ! x is an edge in E1.

Moreover, if htðu; vÞ  s, then lðLsÞ ¼ ht 1; otherwise,

lðLsÞ ¼ htþ 1. Hence, if htðu; vÞ  s, we have

lðPsÞ ¼ s þ htþ 1; otherwise, lðPsÞ ¼ s þ htþ 3.

It remains to construct the ðs þ 1Þ-th internally disjoint path from u to v. Path Psþ1can be constructed as follows:

u ! u0! R ! y ! z ! K ! v0! v:

Note that u0¼ 0s0t1, y ¼ 0s0tht1ht1, z ¼ 0s0tht1ht0 and

v0¼ 1s0tht1ht0. We can find that subpath u0! R ! y is

in Q0

t and z ! K ! v0 is in Q2

ht1

s while u ! u0, y ! z

and v0! v are three edges in E

1. Furthermore, we have

lðRÞ ¼ htand lðKÞ ¼ s. Therefore, lðPsþ1Þ ¼ s þ ht¼ 3.

It is easy to verify that all those Pifor 1  i  s þ 1 are

internally disjoint. This completes the proof. tu Lemma 10. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u06¼ v0, 1  hsðu; vÞ  s  1 and s  htðu; vÞ  t, then there

exist s þ 1 internally disjoint paths Pi for 1  i  s þ 1

between u and v such that hsþ 2 of them are of length

hsþ htþ 3 and s  hs 1 paths are of length hsþ htþ 5.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t0 and v ¼ 0shs1hs0tht1ht1 are in Q0

s

and Q2hs1

t , respectively. Let w ¼ 0shs1hs0t0 and

x ¼ 0shs1hs0t1 be in Q0

s and Q2

hs1

t , respectively. By

The-orem 2, in Q0

s, there exist s internally disjoint paths

between u and w such that hs of them are of length hs,

and the remaining s  hs paths are of length hsþ 2. In

addition, there exist t internally disjoint paths between x and v such that ht of them are of length ht, and the

remaining t  ht paths are of length htþ 2. Let

u ! Hi ! w for 1  i  s and x ! Li! v for 1  i  t

be those internally disjoint paths. Based on Hi for

1  i  s  1, we construct s  1 internally disjoint paths from u to w as follows. Let wi¼ preðH

i; w; iÞ where their

ðt þ iÞ-th bits are different. This is, wi¼ 0shs1hsi01i10t0

when 1  i  hs and wi¼ 0si10ihs11hs0t0 when

hsþ 1  i  s  1. Assume that H0i is the subpath of Hi

without containing w. Clearly, all u ! H0

i! wi for

1  i  s  1 are internally disjoint. Similarly, based on Li for 1  i  s  1, we construct s  1 internally disjoint

paths from x to v as follows. Let xi¼ sucðL

i; x; iÞ where

their i-th bits are different. This is, xi¼ 0shs1hs0ti10i11

when 1  i  s  1. Assume that L0

i is the subpath of Li

without containing x. Clearly, all xi! L0

i ! v for

1  i  s  1 are internally disjoint. To construct a path from wi to xi, we need the following intermediate

(9)

ci¼ 0shs1hsi01i10ti10i10 when 1  i  h s, and

ai¼ 0si10ihs11hs0t1, bi¼ 0si10ihs11hs0ti10i11, ci¼

0si10ihs11hs0ti10i10 when h

sþ 1  i  s  1. di¼

0shs1hs0ti10i10 when 1  i  s  1. We construct paths

Pifor 1  i  s  1 from u to v as follows:

u ! Hi0! wi! ai! bi! ci! di! xi! L0i! v:

Note that, for 1  i  hs(respectively, hsþ 1  i  s  1),

subpath u ! Hi0! wi is in Q0s, edge ai! bi is in Q2ths2i11 (respectively, Q2 hsþ2i11 t ), edge ci! di is in Q2i1 s , and subpath xi ! L0i! v is in Q2 hs1 t while

wi! ai, bi! ciand di! xiare three edges in E

1.

More-over, we have lðH0

iÞ ¼ hs 1 for 1  i  hs and lðHi0Þ ¼

hsþ 1 for hsþ 1  i  s  1, and lðL0iÞ ¼ hs 1 for 1 

i  s  1. Thus, we have lðPiÞ ¼ hsþ htþ 3 for 1  i  hs

and lðPiÞ ¼ hsþ htþ 5 for hsþ 1  i  s  1.

The path Pscan be constructed as follows:

u ! Hs! w ! x ! Ls! v:

Note that subpath u ! Hs! w is in Q0s and subpath

x ! Ls! v is in Q2

hs1

t while w ! x is an edge in E1.

Moreover, we have lðHsÞ ¼ hsþ 2 and lðLsÞ ¼ ht. Hence

lðPsÞ ¼ hsþ htþ 3.

It remains to construct the ðs þ 1Þ-th internally dis-joint path from u to v. Path Psþ1 can be constructed as

follows:

u ! u0! R ! y ! z ! K ! v0! v:

Note that u0¼ 0s0t1, y ¼ 0s0tht1ht1, z ¼ 0s0tht1ht0 and

v0¼ 0shs1hs0tht1ht0. We can find that subpath u0!

R ! y is in Q0

t and z ! K ! v0is in Q2

ht1

s while u ! u0,

y ! z and v0! v are three edges in E

1. Furthermore, we

have lðRÞ ¼ ht and lðKÞ ¼ hs. Therefore, lðPsþ1Þ ¼

hsþ ht¼ 3.

It is easy to verify that all those Pifor 1  i  s þ 1 are

internally disjoint. This completes the proof. tu Lemma 11. Let u and v be two vertices of EHðs; tÞ for 3  s  t with u ¼ utþsutþs1   u0 and v ¼ vtþsvtþs1   v0. If

u06¼ v0, 1  hsðu; vÞ  s  1 and 1  htðu; vÞ  s  1, then

there exist s þ 1 internally disjoint paths Pi for 1  i  s þ 1

between u and v such that the following three cases are distinguished.

1) If hsþ ht t þ 1, then s þ t  hs ht 1 of them

are of length hsþ htþ 5 and hsþ ht t þ 2 paths are

of length hsþ htþ 3.

2) If s  hsþ ht t, then s  1 of them are of length

hsþ htþ 5 and two paths are of length hsþ htþ 3.

3) If hsþ ht s  1, then hsþ htþ 1 of them are of

length hsþ htþ 5, s  hs ht 1 of them are of

length hsþ htþ 7, and one path is of length

hsþ htþ 3.

Proof. By Theorem 3, we may assume without loss of gener-ality that u ¼ 0s0t0 and v ¼ 0shs1hs0tht1ht1 are in Q0

s

and Q2ths1, respectively. Let w ¼ 0shs1hs0t0 and

x ¼ 0shs1hs0t1 be in Q0 s and Q2

hs1

t , respectively. By

The-orem 2, in Q0

s, there exist s internally disjoint paths

between u and w such that hs of them are of length hs,

and the remaining s  hs paths are of length hsþ 2. In

addition, there exist t internally disjoint paths between x and v such that ht of them are of length ht, and the

remaining t  ht paths are of length htþ 2. Let

u ! Hi ! w for 1  i  s and x ! Li! v for 1  i  t

be those internally disjoint paths. Based on Hi for

1  i  s  1, we construct s internally disjoint paths from u to w as follows. Let wi = preðH

i; w; iÞ where their

ðt þ iÞ-th bits are different. This is, wi¼ 0shs1hsi01i10t0

when 1  i  hs and wi ¼ 0si10ihs11hs0t0 when

hsþ 1  i  s. Assume that Hi0is the subpath of Hi

with-out containing w. Clearly, all u ! Hi0! wi for 1  i  s

are internally disjoint. Similarly, based on Lk for

1  k  t, we construct t internally disjoint paths from x to v as follows. Let xk¼ sucðL

k; x; kÞ where their k-th bits

are different. This is, xk¼ 0shs1hs0tk10k11 when

1  k  t. Assume that L0k is the subpath of Lk without

containing x. Clearly, all xk! L0

k! v for 1  k  t are

internally disjoint. To construct a path from wi to xi, we

need the intermediate vertices: ai¼ 0shs1hsi01i10t1,

bi¼ 0shs1hsi01i10tk10k11 and ci¼ 0shs1hsi01i1

0tk10k10 when 1  i  h

s, and ai¼ 0si10ihs11hs0t1,

bi¼ 0si10ihs11hs0tk10k11 and ci ¼ 0si10ihs11hs

0tk10k10 when h

sþ 1  i  s  1. dk¼ 0shs1hs0tk

10k10 when 1  i  s  1. Now we construct paths Pi

for 1  i  s  1 from u to v as follows: u ! Hi0! wi! ai! bi! ci! dk! xk! L

0 k! v:

Note that, for 1  i  hs(respectively, hsþ 1  i  s  1),

subpath u ! H0 i! wi is in Q0s, edge ai! bi is in Q2hs2i11 t (respectively, Q2 hsþ2i11 t ), edge ci! dk is in Q2k1 s , and subpath xi! L0k! v is in Q2 hs1 t while

wi! ai, bi! ciand dk! xkare three edges in E

1.

More-over, we have lðHi0Þ ¼ hs 1 for 1  i  hs and lðHi0Þ ¼

hsþ 1 for hsþ 1  i  s, and lðL0iÞ ¼ ht 1 for 1  i  ht

and lðL0

iÞ ¼ htþ 1 for htþ 1  i  t.

Then the path Pscan be constructed as follows:

u ! Hs! w ! x ! Lk! v:

Note that subpath u ! Hs! w is in Q0s, subpath

x ! Lk! v is in Q2ths1 while w ! x is an edge in E1.

Moreover, we have lðHsÞ ¼ hsþ 2 and lðLkÞ is equal to ht

or htþ 2 depending on k.

Next, we calculate the length of path Pifor 1  i  s.

Case 1: hsþ ht t þ 1.

For 1  i  t  ht, let k ¼ htþ i and then lðPiÞ ¼

hsþ htþ 5. For t  htþ 1  i  hs, let k ¼ htþ i  t and

then lðPiÞ ¼ hsþ htþ 3. For hsþ 1  i  s  1, let k ¼

htþ i  t and then lðPiÞ ¼ hsþ htþ 5. For i ¼ s, let

k ¼ htþ s  t and then lðPsÞ ¼ hsþ htþ 3.

Case 2: s  hsþ ht t.

For 1  i  hs, let k ¼ htþ i and then lðPiÞ ¼

hsþ htþ 5. For hsþ 1  i  s  1, let k ¼ i  hsand then

lðPiÞ ¼ hsþ htþ 5. For i ¼ s, let k ¼ s  hs and then

lðPsÞ ¼ hsþ htþ 3.

Case 3: hsþ ht s  1:

For 1  i  hs, let k ¼ htþ i and then lðPiÞ ¼ hsþ

htþ 5. For hsþ 1  i  hsþ ht, let k ¼ i  hs and then

(10)

and then lðPiÞ ¼ hsþ htþ 7. For i ¼ s, let k ¼ s and then

lðPsÞ ¼ hsþ htþ 5.

Finally, the path Psþ1can be constructed as follows:

u ! u0! R ! y ! z ! K ! v0! v:

Note that u0¼ 0s0t1, y ¼ 0s0tht1ht1, z ¼ 0s0tht1ht0

and v0¼ 0shs1hs0tht1ht0. We can find that subpath

u0! R ! y is in Q0

t and z ! K ! v0 is in Q2

ht1

s while

u ! u0, y ! z and v0! v are three edges in E

1.

Further-more, we have lðRÞ ¼ ht and lðKÞ ¼ hs. Therefore,

lðPsþ1Þ ¼ hsþ ht¼ 3.

It is easy to verify that all those Pifor 1  i  s þ 1 are

internally disjoint. This completes the proof. tu By Lemmas 2-11 above, s þ 1 or t þ 1 internally disjoint paths between any two vertices of the exchanged hyper-cubes EHðs; tÞ can be constructed, and it can be verified that the length of each the internally disjoint paths is at most s þ t þ 3. Take Lemma 2 for instance, one of the s þ 1 inter-nally disjoint paths is of length hsþ 6. Then, hsþ 6 

s þ t þ 3 since 3  s  t. Additionally, take Lemma 3 for instance, s  hs of the s þ 1 internally disjoint paths is of

length hsþ htþ 4. Suppose that hs s  1, then hsþ htþ

4  ðs  1Þ þ t þ 4 ¼ s þ t þ 3. Suppose that hs¼ s, then no

path is of length hsþ htþ 4 since s  hs¼ 0. Therefore, the

following corollary can be obtained.

Corollary 1. Dsþ1ðEHðs; tÞÞ  s þ t þ 3 for 3  s  t.

The wide diameter and fault diameter of the exchanged hypercubes EHðs; tÞ for 3  s  t are stated in Theorem 4.

Theorem 4. Dsþ1ðEHðs; tÞÞ ¼ DfsðEHðs; tÞÞ ¼ s þ t þ 3 for

3  s  t. Proof. Clearly, Df

sðEHðs; tÞÞ  Dsþ1ðEHðs; tÞÞ. Additionally,

by Lemma 1 and Corollary 1, we have that s þ t þ 3  DfsðEHðs; tÞÞ  Dsþ1ðEHðs; tÞÞ  s þ t þ 3 for

3  s  t. Therefore, Dsþ1ðEHðs; tÞÞ ¼ DfsðEHðs; tÞÞ ¼ s þ

t þ 3 for 3  s  t, and this completes the proof. tu For the cases of s ¼ 1; 2 on the wide diameter and fault diameter of the exchanged hypercubes EHðs; tÞ, the statement of Theorem 4 is not true. The following is a counterexample. Consider that u ¼ 0s0t1, u0¼ 0s0t0, and

v ¼ 0s1t1. Assume that F is a faulty vertex set such that

F ¼ NEHðs;tÞðuÞ  u0. The shortest path P between u and v

in EHðs; tÞ  F can be written as the following: P : u ¼ 0s0t1 ! u0! 0s110t0 ! 0s110t1 ! L

! 0s111t1 ! 0s111t0 ! 0s1t0

! v ¼ 0s1t1:

Note that the path L is of length t. It follows that dEHðs;tÞFðu; vÞ ¼ t þ 6. Therefore, we have DfsðEHðs; tÞÞ 

t þ 6 > s þ t þ 3 for s ¼ 1; 2.

4

C

ONCLUDING

R

EMARKS

The topology of a network is an important consideration in the design of interconnection networks since it affects many key properties such as efficiency and fault tolerance. The exchanged hypercube EHðs; tÞ, which is beneficial in parallel computing and communication systems, consti-tutes nearly half the number of edges in comparison with the hypercube Qsþtþ1 and yet retains the advantages of

many topologies; furthermore, it provides good applica-tion to support. In this paper, we focus on constructing s þ 1 internally disjoint paths between any two vertices u and v in the exchanged hypercube EHðs; tÞ. However, if u0¼ v0¼ 1, t þ 1 ( s þ 1) internally disjoint paths

between u and v in EHðs; tÞ can be constructed. We also discuss the wide and fault diameters of the exchanged hypercube EHðs; tÞ. We proved that Dsþ1ðEHðs; tÞÞ ¼

Df

sðEHðs; tÞÞ ¼ s þ t þ 3 for 3  s  t. These properties

demonstrate that interconnection networks modeled by the exchanged hypercube EHðs; tÞ are extremely robust. They have high fault tolerance and reliability on a topolog-ical structure for interconnection networks. Finally, Table 2 illustrates the comparison of some properties on the n-dimension hypercube Qn, crossed hypercube CQn and

exchanged hypercube EHðs; tÞ.

A

CKNOWLEDGMENTS

The authors would like to express their gratitude to the anonymous referees and the editors for their review. Their valuable comments and suggestions help them to improve the present manuscript significantly. This work was sup-ported in part by the National Science Council of the Repub-lic of China under Contract NSC 102-2221-E-159-007, NSC 102-2221-E-009-060, and in part by the Aiming for the Top University and Elite Research Center Development Plan.

R

EFERENCES

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Boca Raton, FL, USA: CRC, 2008.

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[10] S. Latifi, “Combinatorial Analysis of Fault-Diameter of the n-Cube,” IEEE Trans. Comput., vol. 42, no. 1, pp. 27-33, Jan. 1993. [11] F.T. Leighton, Introduction to Parallel Algorithms and Architectures:

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Tsung-Han Tsai received the BS degree in mathematics from the National Changhua Uni-versity of Education, Taiwan, in 2004, and the MS degree in applied mathematics from National Chiao Tung University (NCTU), Taiwan, in 2006. He is currently working toward the PhD degree in computer science at NCTU. His research inter-ests include interconnection networks, algo-rithms, fault-tolerant computing, and graph theory.

Y-Chuang Chen received the BS, MS, and PhD degrees in computer science from National Chiao Tung University, Taiwan, in 1998, 2000, and 2003, respectively. He has been on the faculty of the Department of Information Management, Minghsin University of Science and Technology, since 2002. His research interests include inter-connection networks, graph theory, reliability analysis, and algorithms.

Jimmy J.M. Tan received the BS and MS degrees in mathematics from National Taiwan University in 1970 and 1973, respectively, and the PhD degree from Carleton University, Ottawa, in 1981. He has been on the faculty of the Department of Computer Science, National Chiao Tung University, since 1983. His research interests include design and analysis of algo-rithms, combinatorial optimization, and intercon-nection networks.

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數據

Fig. 1. Four examples of the exchanged hypercubes EHð1; 1Þ, EHð1; 2Þ, EHð2; 1Þ and EHð2; 2Þ.
Fig. 2. An illustration for Lemma 1.
Fig. 3 for illustration. By Theorem 2, in Q 0
Fig. 5. An illustration for the Case 2 of Lemma 3.

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