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IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS,
VOL. 16, NO. 12,
DECEMBER 2005
Optimal Path Embedding in Crossed Cubes
Jianxi Fan, Xiaola Lin, and Xiaohua Jia, Senior Member, IEEE
Abstract—The crossed cube is an important variant of the hypercube. The n-dimensional crossed cube has only about half diameter,
wide diameter, and fault diameter of those of the n-dimensional hypercube. Embeddings of trees, cycles, shortest paths, and
Hamiltonian paths in crossed cubes have been studied in literature. Little work has been done on the embedding of paths except
shortest paths, and Hamiltonian paths in crossed cubes. In this paper, we study optimal embedding of paths of different lengths
n
between any two nodes in crossed cubes. We prove that paths of all lengths between dnþ1
2 e þ 1 and 2 1 can be embedded between
any two distinct nodes with a dilation of 1 in the n-dimensional crossed cube. The embedding of paths is optimal in the sense that the
dilation of the embedding is 1. We also prove that dnþ1
2 e þ 1 is the shortest possible length that can be embedded between arbitrary two
distinct nodes with dilation 1 in the n-dimensional crossed cube.
Index Terms—Crossed cube, graph embedding, optimal embedding, interconnection network, parallel computing system.
æ
1
I
INTRODUCTION
networks play an important role in
parallel computing systems. An interconnection network
can be represented by a graph G ¼ ðV ; EÞ, where V
represents the node set and E represents the edge set. In
this paper, we use graphs and interconnection networks
(networks for short) interchangeably.
Graph embedding is a technique in parallel computing
that maps a guest graph into a host graph (usually an
interconnection network). There are many applications of
graph embedding, such as architecture simulation, processor
allocation, VLSI chip design, etc. Architecture simulation is
the simulation of one architecture by another. This can be
modeled as a graph embedding, which embeds the guest
architecture (represented as a graph) into the host architecture, where the nodes of the graph represent the processors
and the edges of the graph represent the communication links
between the processors [22], [25], [33], [34]. In parallel
computing, a large process is often decomposed into a set of
small subprocesses that can execute in parallel with communications among these subprocesses. According to these
communication relations among these subprocesses, a graph
can be obtained, in which the nodes in the graph represent the
subprocesses and the edges of the graph represent the
communication links between these subprocesses [8]. Thus,
the problem of allocating these subprocesses into a parallel
computing systems can be again modeled as a graph
embedding problem. The problem of laying out circuits on
VLSI chips can also be reduced to graph embedding problems
[3], [30], [31].
Graph embedding can be formally defined as: Given two
graphs G1 ¼ ðV1 ; E1 Þ and G2 ¼ ðV2 ; E2 Þ, an embedding from
NTERCONNECTION
. J. Fan and X. Jia are with the Department of Computer Science, City
University of Hong Kong, 83 Tat Chee Avenue, Kowloon, Hong Kong.
E-mail: {Fanort, csjia}@cityu.edu.hk.
. X. Lin is with the College of Information Science and Technology, Sun Yatsen University, Guangzhou, China. E-mail: issxlin@zsu.edu.cn.
Manuscript received 26 Aug. 2004; revised 17 Dec. 2004; accepted 20 Mar.
2005; published online 20 Oct. 2005.
For information on obtaining reprints of this article, please send e-mail to:
tpds@computer.org, and reference IEEECS Log Number TPDS-0213-0804.
1045-9219/05/$20.00 ß 2005 IEEE
G1 to G2 is an injective mapping : V1 ! V2 . We call G1 the
guest graph and G2 the host graph. An important performance metric of embedding is dilation. The dilation of
embedding is defined as
dilðG1 ; G2 ; Þ ¼ maxfdistðG2 ; ðuÞ; ðvÞÞjðu; vÞ 2 E1 g;
where distðG2 ; ðuÞ; ðvÞÞ denotes the distance between the
two nodes ðuÞ and ðvÞ in G2 . The smaller the dilation of
an embedding is, the shorter the communication delay that
the graph G2 simulates the graph G1 .
has the
Embedding
from G1 to G2 is optimal if
smallest dilation among all the embeddings from G1 to G2 .
Clearly, dilðG1 ; G2 ; Þ 1. When dilðG1 ; G2 ; Þ ¼ 1,
is
surely optimal and G1 is a subgraph of G2 in this case.
Finding the optimal embedding of graphs is NP-hard.
Most of the works on graph embedding consider paths,
trees, meshes, and cycles as guest graphs because they are
the architectures widely used in parallel computing systems
[1], [2], [6], [16], [19], [20], [21], [22], [23], [25], [32], [33], [34].
In this paper, we will study the path embedding problem.
Some special paths take important roles in parallel computing. A shortest path (routing) between two nodes in an
interconnection network is an optimal communication path
in terms of delay. Edge-disjoint paths between two nodes
are fundamental of routing in high-speed networks [5], [26],
while node-disjoint paths are significant for fault-tolerant
routing [17], [18]. A longest path—Hamiltonian path can be
used in dual-path and multipath multicast routing algorithms to alleviate congestion or avoid deadlock incurred by
traditionally tree-based multicast algorithms in parallel
computing systems [4], [27], [28]. This paper addresses the
issue of path embedding in crossed cubes, which takes
paths as guest graphs and crossed cubes as host graphs.
The crossed cube is an important variant of the
hypercube [9]. It has drawn a great deal of attention [7],
[9], [10], [11], [12], [14], [15], [23], [24], [25], [33]. It has the
same node number and edge number as the hypercube with
the same dimension, but has only about half the diameter
[7], [9], wide diameter [7], and fault diameter [7] of those of
the hypercube. Embedding of shortest paths, Hamiltonian
paths, cycles, and trees as guest graphs into crossed cubes
Published by the IEEE Computer Society
FAN ET AL.: OPTIMAL PATH EMBEDDING IN CROSSED CUBES
were studied in literature [7], [9], [23], [25], [33]. In [7], [9],
Efe and Chang et al., respectively, provided an embedding
of shortest paths (i.e., shortest routing algorithms) between
any two nodes in the n-dimensional crossed cube. They also
showed that cycles of all lengths from 4 to 2n can be
embedded into the n-dimensional crossed cube. Huang et al.
gave fault-tolerant Hamiltonian path embedding in [23].
Yang et al. further provided fault-tolerant cycle embedding
in the n-dimensional crossed cube [33]. In the case of tree
embedding, Kulasinghe and Bettayeb gave a perfect result.
They proved that a complete binary tree of 2n 1 nodes can
be embedded into the n-dimensional crossed cube [25]. All
these embeddings in crossed cubes have dilations 1 and,
thus, they are all optimal.
There is little work reported on path embedding in
crossed cubes in literature. The most notable work on path
embedding in crossed cubes is the shortest path embedding
(shortest routing) [7], [9] and the Hamiltonian path
embedding [23]. In this paper, we will discuss the optimal
embedding of paths of different lengths between any two
nodes in crossed cubes. The original contributions of this
paper are as follows:
For any two distinct nodes x and y in the n-dimensional
crossed cube, and any integer l with dnþ1
2 eþ1l
2n 1, a path of length l can be embedded between x and y
with a dilation of 1 in the n-dimensional crossed cube. The
embedding is optimal in the sense that its dilation is 1.
2. There exist two distinct nodes x and y in the
n-dimensional crossed cube such that no path of length
dnþ1
2 e can be embedded between x and y with a dilation of
1 in the n-dimensional crossed cube. This conclusion
demonstrates that dnþ1
2 e þ 1 is the shortest possible length
that can be embedded between arbitrary two distinct nodes
with a dilation of 1 in the n-dimensional crossed cube.
In interconnection networks, the Hamilton-connectivity
(i.e., any two distinct nodes are connected by a Hamiltonian
path) is an important property. The results obtained in this
paper show a stronger connectivity for crossed cubes. We
prove that, in the n-dimensional crossed cube, any two
distinct nodes are connected not only by a Hamiltonian
path, but also by many other paths of consecutive lengths.
This property is not well-known in the existing interconnection networks.
The rest of this paper is organized as follows: Section 2
provides the preliminaries. Section 3 discusses embedding
n
paths of lengths from dnþ1
2 e þ 1 through 2 1 in the
n-dimensional crossed cube. Section 4 proves that the
existence of two nodes that no path of length dnþ1
2 e can be
embedded between these two nodes with a dilation of 1 in the
n-dimensional crossed cube. In Section 5, we give conclusions.
1.
2
PRELIMINARIES
Let V ðGÞ and EðGÞ denote the node set and the edge set of a
graph G, respectively. Given V 0 V ðGÞ, the subgraph
induced by V 0 in G is denoted by G½V 0 . A path P between
node u and node v in G is denoted by P : u ¼ uð0Þ ;
uð1Þ ; . . . ; uðkÞ ¼ v. Nodes u and v are the two end nodes of
path P . The length of path P is denoted by lenðP Þ. Path P can
also be denoted by: u ¼ uð0Þ ; uð1Þ ; . . . ; uði1Þ ; P1 ; uðjþ1Þ ; uðjþ2Þ ;
. . . ; uðkÞ ¼ v, where the path P1 is called a subpath of P between
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uðiÞ and uðjÞ , i.e., uðiÞ ; uðiþ1Þ ; . . . ; uðjÞ (i j). The path P1 ,
starting from uðiÞ and ending with uðjÞ , can be denoted by
pathðP ; uðiÞ ; uðjÞ Þ. For the path P , if u ¼ v (k 3), then P is a
cycle. If ðx; yÞ is an edge in a cycle C, we use C ðx; yÞ to
denote the path after deleting the edge ðu; vÞ in C.
Let distðG; x; yÞ denote the distance between two nodes x
and y of G. The diameter of G is defined as diamðGÞ ¼
maxfdistðG; x; yÞjx; y 2 V ðGÞ; x 6¼ yg.
If x is a node in the path P , then we denote it as x 2 P .
Otherwise, it is denoted as x 2
= P . Similarly, if ðx; yÞ is an
edge in the path P , then we denote it as ðx; yÞ 2 P .
Otherwise, it is denoted as ðx; yÞ 2
= P . IfTV 0 V ðGÞ and no
0
node in V is in P , then we write as V 0 P ¼ ;.
A binary string x of length n is denoted by xn1 xn2
. . . x1 x0 , where xn1 is the most significant bit and x0 is the
least significant bit. The ith bit xi of x can also be written as
bitðx; iÞ. The complement of xi is denoted by xi . Letting z be
a binary string of length k, the binary string of length ik
obtained by concatenating one by one i strings z is denoted
by zi .
The n-dimensional crossed cube (denoted by CQn ) is an
n-regular graph that contains 2n nodes and n2n1 edges.
Every node of CQn is identified by a unique binary string,
which is also called address, of length n. In this paper, we
would not distinguish between nodes and their binary
addresses. The n-dimensional crossed cube can be recursively defined as follows [9], [25].
Definition 1. Two binary strings, x ¼ x1 x0 and y ¼ y1 y0 , of
length two are said to be pair related (denoted by x y) if and
only if ðx; yÞ 2 fð00; 00Þ; ð10; 10Þ; ð01; 11Þ; ð11; 01Þg.
Definition 2. CQ1 is the complete graph on two nodes whose
addresses are 0 and 1. CQn consists of two subcubes CQ0n1
and CQ1n1 . The most significant bit of the addresses of the
nodes of CQ0n1 and CQ1n1 are 0 and 1, respectively. The
nodes u ¼ un1 un2 . . . u1 u0 and v ¼ vn1 vn2 . . . v1 v0 , where
un1 ¼ 0 and vn1 ¼ 1, are joined by an edge in CQn if and
only if
1.
2.
un2 ¼ vn2 if n is even, and
u2iþ1 u2i v2iþ1 v2i , for 0 i < bn1
2 c.
Given x; y 2 V ðCQn Þ, 0 i n 1 and n 1, if xj ¼ yj
for all j 2 fi þ 1; i þ 2; . . . ; n 1g, bitðx; iÞ ¼ bitðy; iÞ and
ðx; yÞ 2 EðCQn Þ, then we write as x i y. By Definition 2, if
ðx; yÞ 2 EðCQn Þ, then there exists an i such that 0 i n 1
and x i y, and for any j with 0 j n 1, there exists w 2
V ðCQn Þ such that x j w. If k 1, n 2, and y is a binary string
of length k, let V 0 ¼ fyxjx is a binary string of length n kg.
The subgraph induced by V 0 in CQn , i.e., CQn ½V 0 , is written as
CQynk . Obviously, by Definition 2, CQynk is isomorphic to
CQnk .
In Fig. 1, Fig. 1a, and Fig. 1b are two different drawings
of CQ3 ; Fig. 1c is CQ4 . We can easily find the symmetric
property of CQ3 .
In [9], Efe gave an Oðn2 Þ algorithm to find a shortest path
between any two distinct nodes in CQn . In [7], Chang et al.
improved this algorithm. They gave an OðnÞ algorithm,
which we call CSH algorithm, to get more shortest paths
between any two distinct nodes in CQn . CSH algorithm
introduces definitions of distance-preserving pair related and
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IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS,
VOL. 16, NO. 12,
DECEMBER 2005
Furthermore, Chang et al. proved an important result as
follows:
distðCQn ; u; vÞ ¼ ðu; vÞ:
This result will be used in the proofs of this paper.
For details on the CSH algorithm, see [7].
3
EMBEDDING PATHS OF LENGTHS FROM dnþ1
2 eþ1
n
TO 2 1
In this section, we will prove that paths of lengths from
n
dnþ1
2 e þ 1 through 2 1 can be embedded between any two
distinct nodes with a dilation of 1 in CQn for n 3. In other
words, there exists paths of all lengths from dnþ1
2 e þ 1 through
2n 1 between any two distinct nodes in CQn for n 3. The
following lemma shows the diameter of CQn .
Lemma 1 [7], [9]. If n 1, then diamðCQn Þ ¼ dnþ1
2 e.
Fig. 1. (a) and (b) Two different drawings of CQ3 . (c) CQ4 .
pair related distance. These two definitions are iterated as
follows [7]:
Let u and v be two distinct nodes in CQn . The ith double
bit of nodes u is defined as a 2-bit string u2iþ1 u2i for
0 i bn2 c 1, and as simply a single bit u2i for i ¼ bn2 c and
n odd. Bit l is called the most significant differing bit between
u and v if us ¼ vs for all s with l þ 1 s n 1 and ul 6¼ vl .
Let i ¼ b2l c be called the most significant differing double bit. A
function on u; v is defined as follows:
j ðu; vÞ ¼ 0 for all j i þ 1;
2 if u2i þ1 u2i ¼ v2i þ1 v2i ;
i ðu; vÞ ¼
1 otherwise:
Further, for j i 1, j ðu; vÞ can be defined using the
notion of distance-preserving pair related (abbreviated as d.p.
pair related) as follows:
In fact, arbitrarily selecting two distinct nodes in CQn , we
n
. Thus, in this problem, we have to
have jV ðCQ
2n n Þjn ¼ 2nþ1
consider 2 ð2 d 2 e 1Þ paths of all lengths from dnþ1
2 eþ
1 through 2n 1 between x and y in CQn . Even given a little
integer n ¼ 10, one has to consider more than 108 paths
between x and y. This is not realistic. In order to simplify the
proof, we adopt the induction on the dimension n of the
crossed cube to prove this result in Theorem 1. Before
beginning this proof, we need to give some preliminary
lemmas. The following lemma is on the existence of a special
cycle of length 4.
Lemma 2. If n 3 and x n1 y for x; y 2 V ðCQn Þ, letting x 1 u
and y 1 v, then ðu; vÞ 2 EðCQn Þ and, thus, C : x; u; v; y; x is a
cycle of length 4 that contains the edge ðx; yÞ in CQn .
Proof. Let x ¼ 0xn2 . . . x1 x0 , y ¼ 1yn2 . . . y1 y0 . Then, by the
definition of the crossed cube, we have ðx1 x0 ; y1 y0 Þ 2
fð00; 00Þ; ð10; 10Þ; ð01; 11Þ; ð11; 01Þg. In fact, we need only
verify the truth of the lemma for ðx1 x0 ; y1 y0 Þ 2 fð00;
00Þ; ð01; 11Þg. That is easy and, therefore, omitted.
u
t
Definition 3. u2jþ1 u2j and v2jþ1 v2j , for j i 1, are distancepreserving pair related if one of the following conditions holds:
Pbn1
2 c
1. ðu2jþ1 u2j ; v2jþ1 v2j Þ 2 fð01; 01Þ; ð11; 11Þg and k¼jþ1
k ðu; vÞ is even,
Pbn1
2 c
2. ðu2jþ1 u2j ; v2jþ1 v2j Þ 2 fð01; 11Þ; ð11; 01Þg and k¼jþ1
k ðu; vÞ is odd, and
3. ðu2jþ1 u2j ; v2jþ1 v2j Þ 2 fð00; 00Þ; ð10; 10Þg.
To find the embedding of paths of lengths from dnþ1
2 eþ1
to 2n 1, three special cases need to be dealt with
separately. The three special cases are when n is odd and:
1) distðCQn ; x; yÞ ¼ dnþ1
2 e and n 5; 2) distðCQn ; x; yÞ ¼
e
1
and
n
5;
3) distðCQn ; x; yÞ ¼ dnþ1
dnþ1
2
2 e 2 and
n 7. The results for the three cases are given, respectively,
in Lemmas 3, 4, and 7. They will be used in the proof of
Theorem 1.
We write u2jþ1 u2j d:p:
v2jþ1 v2j if u2jþ1 u2j and v2jþ1 v2j are
d.p. pair related, and u2jþ1 u2j d:p:
6 v2jþ1 v2j otherwise.
Then, j ðu; vÞ for j i 1 is recursively defined as
follows:
0 if u2jþ1 u2j d:p:
v2jþ1 v2j ;
j ðu; vÞ ¼
1 otherwise:
Lemma 3. If n 5 and n is odd, x ¼ 0xn2 . . . x1 x0 2 V ðCQ0n1 Þ
and y ¼ 1yn2 . . . y1 y0 2 V ðCQ1n1 Þ with xn2 xn3 d:p:
yn2
yn3 and xn2 xn3 d:p:
y
y
,
and
distðCQ
;
x;
yÞ
¼
dnþ1
n2
n3
n
2 e,
nþ1
then there exists a path of length d 2 e þ 1 between x and y in
CQn .
The pair related distance between u and v is defined,
denoted by ðu; vÞ, as
bn1
2 c
ðu; vÞ ¼
X
j¼0
j ðu; vÞ:
n1
n
Proof. Notice that dnþ1
2 e 1 ¼ b 2 c ¼ b 2 c. By the definition of
function i ðu; vÞ, we have i ðx; yÞ ¼ 1, i ¼ 0; 1; . . . ; dnþ1
2 e
1. For x1 x0 and y1 y0 , we need only deal with the cases that
x1 x0 ¼ y1 y0 , x1 x0 ¼ y1 y0 , x1 x0 ¼ y1 y0 , and x1 x0 ¼ y1 y0 .
Case 1. x1 x0 ¼ y1 y0 . We have the cases as below.
Case 1.1. x1 x0 2 f01; 11g. Then, y1 y0 2 f10; 00g. Let
x 0 u 1 v. Then, bitðv; 1Þbitðv; 0Þ ¼ y1 y0 2 f10; 00g. Therefore, 0 ðv; yÞ ¼ 0. Obviously, i ðv; yÞ ¼ i ðx; yÞ ¼ 1 for
FAN ET AL.: OPTIMAL PATH EMBEDDING IN CROSSED CUBES
nþ1
i 2 f1; 2; . . . ; dnþ1
2 e 1g and, thus, distðCQn ; v; yÞ ¼ d 2 e
1. By using the CSH algorithm, we can get a shortest
path P between v and y in CQn and distðCQn ; x; yÞ ¼
dist ðCQn ; u; yÞ 6¼ distðCQn ; z; yÞ for every node z in P .
T
So, fx; ug P ¼ ;. Then,
x; u; P
is a path of length dnþ1
2 e þ 1 between x and y in CQn .
Case 1.2. x1 x0 2 f10; 00g. Then, y1 y0 2 f01; 11g. Exchanging the position of x and y, this case is actually
reduced to Case 1.1.
Case 2. x1 x0 ¼ y1 y0 . We have the following cases.
Case 2.1. x1 x0 2 f00; 10g. Then, y1 y0 2 f10; 00g. Let x 0 u
n2 0
v w. Then, bitðw; 1Þbitðw; 0Þ ¼ y1 y0 2 f00; 10g and 0
ðw; yÞ ¼ dnþ1
ðw; yÞ ¼ 0. Clearly, i ðw; yÞ ¼ i ðx; yÞ ¼ 1
2 e2
nþ1
for i 2 f0; 1; . . . ; dnþ1
2 e 1g f0; d 2 e 2g. As a result,
distðCQn ; w; yÞ ¼ dnþ1
2 e 2. By using CSH algorithm, we
can get a shortest path P , whose length is dnþ1
2 e 2,
between w and y in CQn . We can easily verify that for any
z 2 P and z0 2 fx; u; vg, distðCQn ; z; yÞ 6¼ distðCQn ; z0 ; yÞ.
T
Therefore, fx; u; vg P ¼ ;. Then,
x; u; v; P
is a path of length dnþ1
2 e þ 1 between x and y in CQn .
Case 2.2. x1 x0 2 f01; 11g. Then, y1 y0 2 f11; 01g. Let
x n2 u 0 v and w 0 y. Then,
bitðv; 1Þbitðv; 0Þ ¼ bitðw; 1Þbitðw; 0Þ
¼ x1 x0 ¼ y1 y0 2 f10; 00g:
ðv; wÞ ¼ 0 and
It can be easily verify that 0 ðv; wÞ ¼ dnþ1
2 e2
nþ1
i ðv; wÞ ¼ i ðx; yÞ ¼ 1 f o r i 2 f0; 1; . . . ; d 2 e 1g f0;
dnþ1
2 e 2g. By using the CSH algorithm, we can construct
a shortest path P 0 , whose length is dnþ1
2 e 2, between v
and w in CQn . By the CSH algorithm, for any z 2 P 0 , we
have
bitðz; 1Þbitðz; 0Þ ¼ bitðu; 1Þbitðu; 0Þ ¼ bitðv; 1Þbitðv; 0Þ
¼ bitðw; 1Þbitðw; 0Þ ¼ x1 x0 ¼ y1 y0 2 f10; 00g:
T
Therefore, z 2
= fx; y; ug. That is, fx; y; ug P 0 ¼ ;. Then,
x; u; P 0 ; y
is a path of length dnþ1
2 e þ 1 between x and y in CQn .
Case 3. x1 x0 ¼ y1 y0 . If x1 x0 2 f00; 10g, then let
y 1 u n2 v 0 w. Otherwise, x1 x0 2 f01; 11g, let x 1 u0 n2
v0 0 w0 . Similar to Case 2.1, we can construct a path of
length dnþ1
2 e þ 1 between x and y in CQn .
Case 4. x1 x0 ¼ y1 y0 . Then, x1 x0 2 f01; 11g; ðx1 x0 ; y1 y0 Þ
Pdnþ1
2 e1
2 fð01; 01Þ; ð11; 11Þg, and k¼1
k ðx; yÞ is odd. Let x 0 u,
0
y v. Then, bitðu; 1Þbitðu; 0Þ ¼ bitðv; 1Þ bitðv; 0Þ 2 f00; 10g,
0 ðu; vÞ ¼ 0, and distðCQn ; u; vÞ ¼ dnþ1
2 e 1. By using the
CSH algorithm, we can get a shortest path P , whose length
is dnþ1
2 e 1, between u and v in CQn . By the CSH algorithm,
for any z 2 P , we have bitðz; 1Þbitðz; 0Þ ¼ bitðu; 1Þbitðu; 0Þ
2 f00; 10g and, thus, x1 x0 ¼ y1 y0 6¼ bitðz; 1Þbitðz; 0Þ. As a
T
sequence, fx; yg P ¼ ;. Then,
1193
x; P ; y
is a path of length dnþ1
u
2 e þ 1 between x and y in CQn . t
Lemma 4. If n 5 and n is odd, x ¼ 0xn2 . . . x1 x0 2 V ðCQ0n1 Þ
and y ¼ 1yn2 . . .y1 y0 2 V ðCQ1n1 Þ with xn2 xn3 d:p:
yn2 yn3
nþ1
and xn2 xn3 d:p:
yn2 yn3 , and distðCQn ; x; yÞ ¼ d 2 e 1,
nþ1
then there is a path of length d 2 e þ 1 between x and y in CQn .
Proof. We separately consider the two cases that 0 ðx; yÞ ¼ 1
and 0 ðx; yÞ ¼ 0.
Case 1. 0 ðx; yÞ ¼ 1. Since i ðx; yÞ ¼ 1 for i 2 fdnþ1
e1;
2P
bn1
nþ1
2 c
d 2 e 2g, dnþ1
e
¼
diamðCQ
Þ
>
distðCQ
;
x;
yÞ
¼
n
n
k¼0
2
k ðx; yÞ 3. Considering that n is odd, n 7. We consider
the cases as below.
Case 1.1. x1 x0 2 f00; 10g. Then, y1 y0 2 fx1 0; x1 1; x1 1g.
Case 1.1.1. y1 y0 ¼ x1 0. Let y 0 u 1 v 0 w. Then, 0 ðx; wÞ ¼
0 and i ðx; wÞ ¼ i ðx; yÞ for i 2 f1; 2; . . . ; dnþ1
2 e 1g. Thus,
e
2.
By
using
the
CSH
algorithm,
distðCQn ; x; wÞ ¼ dnþ1
2
we can get a shortest path P between x and w in CQn .
Similar T
to the proof in Lemma 3, we can deduce that
fy; u; vg P ¼ ;. Then,
P ; v; u; y
is a path of length dnþ1
2 e þ 1 between x and y in CQn .
Case 1.1.2. y1 y0 ¼ x1 1. Let y 1 u 0 v 1 w. Then, similar to
Case 1.1.1, we can construct a shortest path P 0 between x
and w in CQn and by using the path P 0 we can get a path
of length dnþ1
2 e þ 1 between x and y in CQn .
Case 1.1.3. y1 y0 ¼ x1 1. Let y 1 u n2 v 0 z 1 w. Then,
0 ðx; wÞ ¼ dnþ1
ðx; wÞ ¼ 0 and i ðx; wÞ ¼ i ðx; yÞ for i 2
2 e2
f0; 1; . . . ; dnþ1
e1gf0;
dnþ1
2
2 e2g. Thus, distðCQn ; x; wÞ
nþ1
¼ d 2 e 3. By using CSH algorithm, we can get a
shortest path P 00 between x and w in CQn and we can
T
deduce that fy; u; v; zg P 00 ¼ ;. Then,
P ; z; v; u; y
dnþ1
2 e
þ 1 between x and y in CQn .
is a path of length
Case 1.2. x1 x0 2 f01; 11g. Then, y1 y0 2 f00; 10; 01; 11g.
In fact, exchanging the position of x and y, the case for
y1 y0 2 f00; 10g can be reduced to Case 1.1. So, we only
consider the cases for y1 y0 2 f01; 11g. For y1 y0 2 f01; 11g,
let y 0 u 1 v 0 w. Similar to the proof in Lemma 3, we can
construct a path of length dnþ1
2 e þ 1 between x and y in CQn .
C as e 2 . 0 ðx; yÞ ¼ 0. T h e n, i ðx; yÞ ¼ 1 f o r i 2
f1; 2; . . . ; dnþ1
2 e 1g. We deal with the following cases.
Case 2.1. x1 x0 2 f00; 10g. Let y 1 u n2 v 1 w. Then,
0 ðx; wÞ ¼ dnþ1
ðx; wÞ ¼ 0 and i ðx; wÞ ¼ 1 for i 2 f0; 1;
2 e2
nþ1
. . . ; d 2 e 1g f0; dnþ1
2 e 2g. Thus, distðCQn ; x; wÞ ¼
dnþ1
e
2.
By
using
the
CSH algorithm, we can get a
2
shortest path P between x and w in CQn and we can
T
deduce that fy; u; vg P ¼ ;. Then,
P ; v; u; y
is a path of length dnþ1
2 e þ 1 between x and y in CQn .
Case 2.2. x1 x0 2 f01; 11g. Let x 1 u, y 1 v. Then,
0 ðu; vÞ ¼ 0 ðx; yÞ ¼ 0. Let P 0 : u ¼ z0 ; z1 ; . . . ; zk ¼ v be a
shortest path got by using CSH algorithm between u and v
in CQn . Supposing that zi1 ji zi for i 2 f1; 2; . . . ; kg, we will
1194
IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS,
VOL. 16, NO. 12,
DECEMBER 2005
Fig. 2. A path of length dnþ1
2 e þ 1 between x and y in CQ , where a straight line represents an edge and a curve line represents a path between two
nodes.
prove x 2
= P 0 . Since 0 ðu; vÞ ¼ 0, by the CSH algorithm, j16¼
1 and, thus, x 6¼ z1 . Further, for 2 i k, by the CSH
algorithm, distðCQn ; u; zi Þ ¼ distðCQn ; z0 ; zi Þ ¼ i 2 >
1 ¼ distðCQn ; z0 ; xÞ. Therefore, x 6¼ zi for i 2 f2; 3; . . . ; kg.
= fz0 ; z1 ;
And, it is clear that x 6¼ z0 ¼ u. To sum up, x 2
= P 0 . Similarly, we can prove y 2
= P 0.
. . . ; zk g. That is, x 2
Thus,
x; P 0 ; y
u
is a path of length dnþ1
2 e þ 1 between x and y in CQn . t
The proof of Lemma 7 need to use the following lemma:
Lemma 5. If n 3, x n1 y for x; y 2 V ðCQn Þ, letting
x 0 u n1 v 0 w 1 z and y 0 u0 n1 v0 0 w0 1 z0 , then z ¼ y, z0 ¼ x,
and C : x; u; v; w; y; x and C 0 : y; u0 ; v0 ; w0 ; x; y are the two
cycles of length 5 that contain the edge ðx; yÞ in CQn .
= fv; wg.
Furthermore, u 2
= fv0 ; w0 g and u0 2
Proof. Le t x ¼ 0xn2 . . . x1 x0 , y ¼ 1yn2 . . . y1 y0 . Then,
ðx1 x0 ; y1 y0 Þ 2 fðx1 0; x1 0Þ; ðx1 1; x1 1Þg. By Definition 2, we
can verify that z ¼ y, z0 ¼ x, and C : x; u; v; w; y; x and C 0 :
y; u0 ; v0 ; w0 ; x; y are the two cycles of length 5 that contain
the edge ðx; yÞ in CQn . Further, if ðx1 x0 ; y1 y0 Þ ¼ ðx1 0; x1 0Þ,
then bitðu; 1Þbitðu; 0Þ ¼ x1 1, bitðv0 ; 1Þbitðv0 ; 0Þ ¼ x1 1, and
bitðw0 ; 1Þbitðw0 ; 0Þ ¼ x1 0, while if ðx1 x0 ; y1 y0 Þ ¼ ðx1 1; x1 1Þ,
then bitðu; 1Þbitðu; 0Þ ¼ x1 0, bitðv0 ; 1Þbitðv0 ; 0Þ ¼ x1 0, and
bitðw0 ; 1Þbitðw0 ; 0Þ ¼ x1 1. Thus, u 2
= fv0 ; w0 g. Similarly, we
= fv; wg.
u
t
can also verify that u0 2
The following lemma will also be used in the proof of
Lemma 7 and Theorem 1.
Lemma 6 [15]. If n 2, for any ðx; yÞ 2 EðCQn Þ and any
integer l with 4 l 2n , there exists a cycle C of length l in
CQn such that ðx; yÞ is in C.
Lemma 7. If n 7 and n is odd, x ¼ 0xn2 . . . x1 x0 2
V ðCQ0n1 Þ and y ¼ 1yn2 . . .y1 y0 2 V ðCQ1n1 Þ with xn2 xn3
d:p:
d:p:
yn2 yn3 and xn2 xn3 yn2 yn3 , and distðCQn ; x; yÞ
nþ1
¼ d 2 e 2, then there is a path of length dnþ1
2 e þ 1 between x
and y in CQn .
Proof. Let x n1 z. Then, z 2 V ðCQ1n1 Þ, dnþ1e1 ðz; yÞ ¼ 0, and
2
1
i ðz; yÞ ¼ i ðx; yÞ, i ¼ 0; 1; . . . ; dnþ1
2 e 2. Hence, distðCQn1 ;
nþ1
z; yÞ ¼ d 2 e 3 1 and, thus, y 6¼ z. Without loss of
generality, we assume that x 2 V ðCQ000
n3 Þ (similarly prove
in other cases ). Then z 2 V ðCQ100
Þ
and
y 2 V ðCQ110
n3
n3 Þ. Let
0 n1 0 1 0
0 0 n1 0 0 0 1 0
z u
v w x and x u
v w z . By the conditions in
the lemma, we can easily verify that yn2 ¼ bitðu; n 2Þ
¼ bitðv0 ; n 2Þ ¼ bitðw0 ; n 2Þ. Therefore, y 2
= fu; v0 ; w0 g.
0
0
By Lemma 5, we have x ¼ x and z ¼ z and, thus, C :
z; u; v; w; x; z and C 0 : x; u0 ; v0 ; w0 ; z; x are two cycles of
length 5 in CQn such that u 2
= fv0 ; w0 g and u0 2
= fv; wg (see
1
Fig. 2a). Since distðCQn1 ; z; yÞ ¼ dnþ1
e
3
1,
we may let
2
P be a path of length dnþ1
e
3
between
z
and
y
in CQ1n1 .
2
Note that P is also a shortest path between z and y in
CQ1n1 . We deal with the following cases.
Case 1. u 2
= P . Then,
x; w; v; u; P
dnþ1
2 e
þ 1 between x and y in CQn (see
is a path of length
Fig. 2b).
Case 2. u 2 P . Since ðz; uÞ 2 EðCQ1n1 Þ and P is a
shortest path between z and y in CQ1n1 , P must be:
= P . Otherwise, P
z; u; . . . ; y. We can further claim that w0 2
must be z; u; . . . ; w0 ; . . . ; y. Thus, lenðpathðP ; z; w0 ÞÞ > 1.
On the other hand, since P is a shortest path between z
and y in CQ1n1 , the subpath between z and w0 in P must
be a shortest path between z and w0 in CQ1n1 .
FAN ET AL.: OPTIMAL PATH EMBEDDING IN CROSSED CUBES
Considering that ðz; w0 Þ 2 EðCQ1n1 Þ, the length of the
subpath between z and w0 in P should be 1. So, we obtain
a contradiction. Thus, we have the following cases:
= P . Then,
Case 2.1. v0 2
x; u0 ; v0 ; w0 ; P
is a path of length dnþ1
2 e þ 1 between x and y in CQn (see
Fig. 2c).
Case 2.2. v0 2 P . Then, P is z; u; . . . ; v0 ; . . . ; y. Since
= EðCQ1n1 Þ (Otherwise, z; v0 ; w0 ; z is a cycle of
ðz; v0 Þ 2
length 3 in CQ1n1 , but there is no cycle of length 3 in CQn
for all n 1, which can be proved similar to the proof of
Lemma 5 in [13]), z; w0 ; v0 is a shortest path between z and
v0 in CQ1n1 . Thus, distðCQ1n1 ; z; v0 Þ ¼ 2. Considering that
P is a shortest path between z and y in CQ1n1 , the
subpath between z and v0 in P must be a shortest path
between z and v0 in CQ1n1 . Therefore, lenðpathðP ; z; v0 ÞÞ
¼ 2. Since ðz; uÞ 2 EðCQ1n1 Þ, ðu; v0 Þ 2 V ðCQ1n1 Þ (See
Fig. 2d). Let P 0 ¼ pathðP ; v0 ; yÞ. Then, lenðP 0 Þ ¼ dnþ1
2 e
5. Thus,
x; w; v; u; z; w0 ; P 0
is a path of length dnþ1
u
2 e þ 1 between x and y in CQn . t
Further, Lemma 8 shows a property on distancepreserving pair related, which will be used in the proof
of Theorem 1.
Lemma 8. If n 3 and n is odd, x ¼ 0xn2 . . . x1 x0 2
V ðCQ0n1 Þ and y ¼ 1yn2 . . . y1 y0 2 V ðCQ1n1 Þ, u n1 x, and
v n1 y such that bitðv; n2Þ 6¼ bitðx; n 2Þ and bitðu; n 2Þ
d:p:
6¼ bitðy; n2Þ, then xn2 xn3 d:p:
yn2 yn3 and xn2 xn3 yn2 yn3 .
Proof. According to the definition of function , we can
easily claim that ðxn2 xn3 ; yn2 yn3 Þ 2 fð00; 10Þ; ð10; 00Þ;
ð01; 01Þ; ð11; 11Þg. We can easily verify xn2 xn3 d:p:
yn2 yn3 and xn2 xn3 d:p:
u
t
yn2 yn3 .
So far, we have introduced a series of lemmas. Now, let us
give a proof that there exists paths of all lengths from dnþ1
2 eþ1
through 2n 1 between any two distinct nodes in CQn for
n 3. Even with the the help of these lemmas, we point out,
we still have to deal with many cases in this proof.
Theorem 1. If n 3, for any x; y 2 V ðCQn Þ, x 6¼ y, and any
n
integer l, dnþ1
2 e þ 1 l 2 1, there exists a path of length
l between x and y in CQn .
Proof. we use induction on n.
We can easily verify the truth of the theorem for n ¼ 3.
Suppose that the theorem holds for 3 n 1, we
will prove that the theorem holds for n ¼ ( 4).
When n ¼ , for any x; y 2 V ðCQ Þ, x 6¼ y, and any
integer l with dþ1
2 e þ 1 l 2 1, without loss generality, we separately identify the two cases that both x
and y are in CQ01 , and x and y are, respectively, in
CQ01 and CQ11 to construct a path of length l in CQ .
First, we consider the case that both x and y are in
CQ01 . In this case, we further deal with the following
cases according to the value of l:
1195
1
1. By the induction
Case 1. dþ1
2 eþ1l2
hypothesis, there exist paths of lengths d2e þ 1, d2e þ 2,
. . . , 21 1 between x and y in CQ01 . Noticing that
d2e þ 1 dþ1
2 e þ 1, there exist paths of lengths l between
x and y in CQ01 and, thus, in CQ .
l1
Case 2. 21 l 2 1. Let l1 ¼ dl1
2 e, l2 ¼ b 2 c.
Then, d2e þ 1 22 1 l2 l1 21 1 and l1 þ l2 ¼
l 1. By the induction hypothesis, there exists a path P1
of length l1 between x and y in CQ01 . Select an edge
ðu; vÞ in P1 (see Fig. 3a). Let u0 1 u and v0 1 v. Then,
u0 ; v0 2 V ðCQ11 Þ and u0 6¼ v0 . By the induction hypothesis, there exists a path P2 of length l2 between u0 and v0
in CQ11 . Then,
pathðP1 ; x; uÞ; P2 ; pathðP1 ; v; yÞ
is a path of length l1 þ l2 þ 1 ¼ l between x and y in CQ .
Next, we consider the case that x in CQ01 and y in
CQ11 . In this case, we further deal with two cases
according to whether x and y are adjacent as follows:
Case 1. ðx; yÞ 2 EðCQ Þ. By Lemma 6, for any l0 with
4 l0 2 , there exists a cycle C of length l0 that contains
the edge ðx; yÞ in CQ . So, for any l00 with 3 l00 2 1,
there is a path C ðx; yÞ of length l00 between x and y in CQ .
Since 4, dþ1
2 e þ 1 4. Thus, there exists a path of
length l with dþ1
2 e þ 1 l 2 1 between x and y in CQ .
Case 2. ðx; yÞ 2
= EðCQ Þ. We have the following cases
according to the range of l.
1
. Let z 1 x. Then, z 2
Case 2.1. dþ1
2 eþ1l2
1
0
V ðCQ1 Þ and z 6¼ y. Let l ¼ l 1. We deal with the
following cases:
0
Case 2.1.1. is even. Then, d2e þ 1 ¼ dþ1
2 el 1
1. By the induction hypothesis, there exists a path
2
P of length l0 between z and y in CQ11 . Thus,
x; P
1
is a path of length l0 þ 1 ¼ l with dþ1
2 eþ1l2
between x and y in CQ .
1
;
Case 2.1.2. is odd. Then, 5. For dþ1
2 eþ2 l 2
þ1
0
1
d 2e þ 1 ¼ d 2 e þ 1 l 2
1. Similar to Case 2.1.1,
we can get a path of length l0 þ 1 ¼ l between x and y in
CQ . In order to show that there exists a path of length
1
y. Then, v 2
l ¼ dþ1
2 e þ 1 between x and y in CQ , let v
0
V ðCQ1 Þ and v 6¼ x. We consider the following cases:
Case 2.1.2.1. bitðv; 2Þ ¼ bitðx; 2Þ or bitðz; 2Þ
¼ bitðy; 2Þ. Then, there exists an i 2 f0; 1g such that
1i
fv; xg V ðCQ0i
2 Þ or fz; yg V ðCQ2 Þ. Without loss of
generality, we assume that there exists an i 2 f0; 1g such
that fv; xg V ðCQ0i
2 Þ. By the induction hypothesis, there
exists a path P of length d1
2 e þ 1 between x and v in
CQ0i
.
Thus,
2
P; y
þ1
is a path of length d1
2 e þ 2 ¼ d 2 e þ 1 between x and y
in CQ .
Case 2.1.2.2. bitðv; 2Þ 6¼ bitðx; 2Þ and bitðz; 2Þ
6¼ bitðy; 2Þ. By Lemma 8, we have x2 x3 d:p
y2 y3
and x2 x3 d:p:
y
y
.
Considering
that
diamðCQ
Þ ¼
2 3
e
for
1,
we
deal
with
the
following
cases:
dþ1
2
1196
IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS,
VOL. 16, NO. 12,
DECEMBER 2005
Fig. 3. A path of length l between x and y in CQ , where a straight line represents an edge and a curve line represents a path between two nodes.
þ1
Case 2.1.2.2.1. dþ1
2 e 2 distðCQ ; x; yÞ d 2 e. If
þ1
distðCQ ; x; yÞ ¼ d 2 e 2, we can deduce 6¼ 5. Supposing on the contrary that ¼ 5, distðCQ ; x; yÞ ¼ d5þ1
2 e
=
2 ¼ 1. Hence, ðx; yÞ 2 EðCQ Þ, contradictory to ðx; yÞ 2
EðCQ Þ. So, we have 7 if distðCQ ; x; yÞ ¼ dþ1
2 e 2.
By Lemma 7, there exists a path of length dþ1
2 eþ1
between x and y in CQ for distðCQ ; x; yÞ ¼ dþ1
2 e 2.
þ1
e;
d
e
1g,
by
Lemma
3 and
For ðCQ ; x; yÞ 2 fdþ1
2
2
e
þ
1
between
Lemma 4, there exists a path of length dþ1
2
x and y in CQ .
Case 2.1.2.2.2. 2 distðCQ ; x; yÞ dþ1
2 e 3. Notice
þ1
1
that b1
c
¼
d
e
1.
Since
z
x,
we
have
d þ1
ðz; yÞ
2
2
2 e1
¼ 0; dþ1e2 ðz; yÞ ¼ dþ1e2 ðx; yÞ ¼ 1. Therefore, i ðz; yÞ ¼
2
2
1
i ðx; yÞ for i ¼ 0; 1; . . . ; dþ1
2 e 3 and 1 distðCQ1 ; z; yÞ
þ1
d 2 e 4. Let P be a shortest path obtained using the
CSH algorithm between z and y in CQ11 . Then,
lenðP Þ ¼ distðCQ11 ; z; yÞ. By Lemma 2, there exists a
cycle x; u; w; z; x of length 4 that contains the edge ðx; zÞ
in CQ . The following proof will separately discuss
according to whether w 2 P .
1
If w 2
= P . Let m ¼ dþ1
2 e distðCQ1 ; z; yÞ. Then,
þ1
1
1. By Lemma 6, there exists
4md 2 e12
a cycle C of length m that contains the edge ðx; uÞ in
CQ01 (see Fig. 3b). Thus,
C ðx; uÞ; w; P
is a path of length ðm 1Þ þ 2 þ distðCQ11 ; z; yÞ ¼
dþ1
2 e þ 1 between x and y in CQ .
Otherwise, since P is a shortest path between z and y in
CQ11 , pathðP ; z; wÞ is a shortest path in CQ11 . And,
considering that ðz; wÞ 2 V ðCQ11 Þ, P must be z; w; . . . ; y.
Let P 0 ¼ pathðP ; w; yÞ. Then, lenðP 0 Þ ¼ distðCQ11 ; z; yÞ 1.
1
0
Let m0 ¼ dþ1
2 e distðCQ1 ; z; yÞ þ 2. Then, 6 m þ1
1
1. By Lemma 6, there exists a cycle C 0
d 2 e þ1 2
of length m0 that contains the edge ðx; uÞ in CQ01 (see
Fig. 3c). Thus,
C 0 ðx; uÞ; P 0
is a path of length ðm0 1Þ þ 1 þ ðdistðCQ11 ; z; yÞ 1Þ ¼
dþ1
2 e þ 1 between x and y in CQ .
Case 2.2. 21 þ 1 l 2 1. By Definition 2, we can
select u 2 V ðCQ01 Þ fxg, v 2 V ðCQ11 Þ fyg such that
l1
ðu; vÞ 2 EðCQ Þ. Let l1 ¼ dl1
2 e, l2 ¼ b 2 c. Then, d2e þ 1 2
1
l2 l1 2
1 and l1 þ l2 ¼ l 1. By the in2
duction hypothesis, there exist a path P1 of length l1
between x and u in CQ01 and a path P2 of length l2
between v and y in CQ11 (see Fig. 3d). Then,
P1 ; P2
is a path of length l1 þ l2 þ 1 ¼ l between x and y in CQ .t
u
The similar result was obtained independently by
Yang et al. in [35]. However, the result in [35] deduces the
existence of paths of lengths greater than or equal to
nþ1
dnþ1
2 e þ 2, but does not include d 2 e þ 1.
4
NONEXISTENCE OF EMBEDDING PATHS
LENGTHS dnþ1
2 e
OF
In Section 3, we have given the embedding of paths of
n
length from dnþ1
2 e þ 1 through 2 1 between any two
distinct nodes with a dilation of 1 in CQn . The following
theorem will prove that dnþ1
2 e þ 1 is the shortest possible
length that can be embedded between arbitrary two distinct
nodes with a dilation of 1 in CQn .
Theorem 2. If n 2, then there exists two nodes x, y 2 V ðCQn Þ
such that x 6¼ y and there is no path of length dnþ1
2 e between x
and y in CQn .
FAN ET AL.: OPTIMAL PATH EMBEDDING IN CROSSED CUBES
1197
Proof. The theorem obviously holds for n ¼ 2. Selecting
x ¼ 001, y ¼ 111 for n ¼ 3 and selecting x ¼ 0001, y ¼
1101 for n ¼ 4, we can easily verify the truth of the
theorem for n ¼ 3; 4 also. For n 5, we separately deal
with two cases as follows:
Case 1. dnþ1
. . . x1 x0 ¼
2 e is even. Select x ¼ xn1 xn2
Pbn1
n1
n
2 c
0 1, y ¼ yn1 yn2 . . . y1 y0 ¼ 1 . Then,
i¼1 i ðx; yÞ ¼
dnþ1
e1
is
odd.
By
the
definition
of
function
,
0 ðx; yÞ ¼ 0.
2
Suppose that there exists a path P of length dnþ1
2 e
between x and y in CQn . Let P be x ¼ xð0Þ , xð1Þ , . . . ,
nþ1
ðkÞ ðkÞ
ðkÞ ðkÞ
xðd 2 eÞ ¼ y, where xðkÞ ¼ xn1 xn2 . . . x1 x0 k ¼ 0; 1; for
ð0Þ j1 ð1Þ
. . . ; dnþ1
x . We can claim that j1 2
= f0; 1g.
2 e, and x
Supposing on the contrary that j1 2 f0; 1g, then we have
0 ðxð1Þ ; yÞ ¼ 1 and i ðxð1Þ ; yÞ ¼ i ðx; yÞ for i ¼ 1; 2; . . . ; bn1
2 c
Pbn1
Pbn1
ð1Þ
nþ1
2 c
2 c
and, thus,
ðx
;
yÞ
¼
ðx;
yÞ
¼
d
e
1.
i
i
i¼1
i¼1
2
So, the length of P is greater than
ð1Þ
distðCQn ; x ; yÞ ¼
n1
bX
2 c
i ðxð1Þ ; yÞ
i¼0
¼
n1
bX
2 c
i ðxð1Þ ; yÞ þ 0 ðxð1Þ ; yÞ ¼
i¼1
nþ1
;
2
contradictory to that the length of P is dnþ1
2 e.
Further, if n is odd, then j1 6¼ n 1. Otherwise, we have
ðxð1Þ ; yÞ ¼ 0, bn1
ðxð1Þ ; yÞ ¼ 2, and i ðxð1Þ ; yÞ ¼ 1 for
bn1
2 c
2 c1
n1
i ¼ 0; 1; . . . ; b 2 c 2. Similar to the above discussion, we
can deduce that the length of P is greater than dnþ1
2 e, a
contradiction.
Moreover, if j1 2 f2; 3; . . . ; 2bn1
we can claim
2 c 1g,
Pbn1
2 c
that j1 is not odd and if j1 is even, then
i ðxð0Þ ; yÞ is
j
i¼ 21 þ1
odd. Otherwise, we have i ðxð1Þ ; yÞ ¼ i ðx; yÞ for i ¼ 1; 2;
ð1Þ
. . . ; bn1
2 c and 0 ðx ; yÞ ¼ 1. Thus, we can get the same
contradiction as the above discussion.
In summary, j1 satisfies one of the two conditions as
follows:
Pbn1
2 c
1. j1 is even with 2 j1 2bn1
j
2 c 1 and
i¼ 21 þ1
ð0Þ
i ðx ; yÞ is odd;
2. j1 2 fn 1; n 2g and n is even.
By the definition of function , we have the following
conclusions:
If j1 satisfies the condition 1, then j1 ðxð1Þ ; yÞ ¼
ð1Þ 2 ð1Þ
0 ðxð1Þ ; yÞ ¼ 0; bn1
ðxð1Þ ; yÞ ¼ 2 with xn1 xn2 ¼ 00
2 c
ð1Þ
if n is even, bn1
ðxð1Þ ; yÞ ¼ 1 with xn1 ¼ 0 if n is
2 c
ð1Þ
ð1Þ
odd, and i ðxð1Þ ; yÞ ¼ 1 with x2iþ1 x2i ¼ 00 for
j1
i 2 f0; 1; 2; . . . ; bn1
2 c 1g f0; 2 g.
b. If j1 satisfies the condition 2, then 0 ðxð1Þ ; yÞ ¼ 0,
ð1Þ ð1Þ
bj1 cðxð1Þ ; yÞ ¼ 1 with xn1 xn2 2 f01;10g, and i ðxð1Þ ;
2
ð1Þ
ð1Þ
yÞ ¼ 1 with x2iþ1 x2i ¼ 00 for i 2 f0; 1; 2; . . . ; bn1
2 c
1g f0g.
Now, we separately deal with the following cases:
a.
n1
nþ1
Case 1.1. n is odd. Then, bn1
2 c ¼ 2 ¼ d 2 e 1. We
assume that k ðx
ðmÞ
; yÞ ¼ 0 for k ¼
0; j21 ; j22
; . . . ; j2m
with
Fig. 4. k0 ¼ maxfi < bn1
cji ðxðmÞ ; yÞ ¼ 1g and k00 ¼ maxfi < k0 ji ðxðmÞ ;
Pbn1c 2
yÞ ¼ 0g, where i¼k2 00 þ1 i ðxðmÞ ; yÞ is odd.
1mb
n1
c1
2
and for the even integers jl with 2 jl n 2 and
ðmÞ
1 l m, bn1
ðxðmÞ ; yÞ ¼ 1 with xn1 ¼ 0, and i0 ðxðmÞ ; yÞ
2 c
ðmÞ
ðmÞ
¼ 1 with x2i0 þ1 x2i0 ¼ 00 for i0 2 f0; 1; . . . ; bn1
2 c1g
f0; j21 ; j22 ; . . . ; j2m g. Let xðmÞ jmþ1 xðmþ1Þ , A ¼ f0; 1; j1 ; j1 þ
1; j2 ; j2 þ 1; . . . ; jm ; jm þ 1g and B ¼ fjjj is odd with 2 S
j n 2 and j1 ðxðmÞ ; yÞ ¼ 1g fjjj is even with 2 j 2
Pbn1c
n 2, j ðxðmÞ ; yÞ ¼ 1; and i¼2jþ1 i ðxðmÞ ; yÞ is eveng. We
2
S S2
will prove that jmþ1 2
= A B fn 1g as follows:
S S
Supposing on the contrary that jmþ1 2 A B fn 1g,
we have the following cases:
Case 1.1.1. jmþ1 2 A. By the definition of function , we
have bjmþ1 c ðxðmþ1Þ ; yÞ ¼ 1, i0 ðxðmþ1Þ ; yÞ ¼ 1 for i0 2 f0; 1;
2
j1 j2
jm
ðmþ1Þ
; yÞ ¼ 0 for k 2
. . . ; bn1
2 cg f0; 2 ; 2 ; . . . ; 2 g, and k ðx
ðmþ1Þ
; yÞ ¼
f0; j21 ; j22 ; . . . ; j2m g fbjmþ1
2 cg. Then, distðCQn ; x
Pbn1
c
ðmþ1Þ
nþ1
2
; yÞ ¼ d 2 e m. Thus, the length of P is
i¼0 i ðx
greater than or equal to lenðpathðP ; xð0Þ ; xðmþ1Þ ÞÞ þdist
nþ1
ðCQn ; xðmþ1Þ ; yÞ ¼ ðm þ 1Þ þ ðdnþ1
2 e mÞ ¼ d 2 e þ 1, con-
tradictory to that the length of P is dnþ1
2 e.
Case 1.1.2. jmþ1 2 B. By the definition of function , we
j1
have i0 ðxðmþ1Þ ; yÞ ¼ 1 for i0 2 f0; 1; . . . ; bn1
2 cg f0; 2 ,
j2
jm
j
0
ðmÞ
; yÞ ¼ 0, and 0 i < b mþ1
2 ; . . . ; 2 g. Let k ¼ maxfiji ðx
2 cg.
Pbn1
j1 j2
jm
0
ðmþ1Þ
2 c
Then, k 2 f0; 2 ; 2 ; . . . ; 2 g and
; yÞ ¼
i¼k0 þ1 i ðx
Pbn1
ðmÞ
ðmþ1Þ
ðmþ1Þ
2 c
0
ðx
;
yÞ.
Thus,
ðx
;yÞ
¼
1
and
ðx
; yÞ
k
k
i¼k0 þ1 i
j1 j2
jm
0
¼ 0 for k 2 f0; 2 ; 2 ; . . . ; 2 g fk g. Similar to Case 1.1.1, we
can conclude that the length of P is greater than or equal to
nþ1
dnþ1
2 e þ 1, contradictory to that the length of P is d 2 e.
Case 1.1.3. jmþ1 ¼ n 1. Let k0 ¼ maxfi < bn1
2 cj
ðmÞ
ðmÞ
i ðxðmÞ ; yÞ ¼ 1g (see Fig. 4). Then, x2k0 þ1 x2k0 ¼ 00 and
ðmÞ
ðmÞ
k ðxðmþ1Þ ; yÞ ¼ k ðxðmÞ ; yÞ ¼ 0 with x2kþ1 x2k ¼ 01 for k ¼
ðmþ1Þ
0
; yÞ ¼ 2.
k0 þ 1; k0 þ 2; . . . ; bn1
2 c 1 and, thus, k ðx
Further, let k00 ¼ maxfi < k0 ji ðxðmÞ ;yÞ ¼ 0g. Then,i0ðxðmþ1Þ ;
ðmÞ
ðmÞ
yÞ ¼ i0 ðxðmÞ ; yÞ ¼ 1 with x2i0 þ1 x2i0 ¼ 00 for i0 ¼ k00 þ 1;
Pbn1c
Pbn1c
k00 þ2; . . . ; k0 1 and, thus, i¼k2 00 þ1 i ðxðmþ1Þ ; yÞ ¼ i¼k2 00 þ1
Pbn1c
i ðxðmÞ ; yÞ. Obviously, if i¼k2 00 þ1 i ðxðmÞ ; yÞ is odd, then
ðmÞ
ðmÞ
ðmÞ
ðmÞ
x2k00 þ1 x2k00 ¼ 01; otherwise, x2k00 þ1 x2k00 ¼ 11. Without loss
Pbn1c
of generality, supposing that i¼k2 00 þ1 i ðxðmÞ ; yÞ is odd,
ðmÞ
ðmÞ
then x2k00 þ1 x2k00 ¼ 01. So, k00 ðxðmþ1Þ ; yÞ ¼ 1. Further, we
have i ðxðmþ1Þ ; yÞ ¼ i ðxðmÞ ; yÞ for i ¼ 0; 1; . . . ; k00 1.
Similar to the discussion in Case 1.1.1, the length of P is
1198
IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS,
greater than or equal to lenðP ; xð0Þ ; xðmþ1Þ Þ þ distðCQn ;
nþ1
xðmþ1Þ ; yÞ ¼ ðm þ 1Þ þ ðdnþ1
2 e mÞ¼ d 2 e þ 1, contradicnþ1
tory to that the length of P is d 2 e.
To sum up, jmþ1 is such an even integer that 2 jmþ1 Pbn1
ðmÞ
ðmÞ
2 c
i ðxðmÞ ; yÞ is odd with xjmþ1 þ1 xjmþ1 ¼ 00.
n 2 and
jmþ1
i¼
2
þ1
According to the definition of function , we have
ðmþ1Þ
n1
; yÞ
k ðxðmþ1Þ ; yÞ ¼ 0 for k 2 f0; j21 ; j22 ; . . . ; jmþ1
2 g; b 2 c ðx
ðmþ1Þ
¼ 1 with xn1
ðmþ1Þ ðmþ1Þ
with x2iþ1 x2i
¼ 0, and i ðxðmþ1Þ ; yÞ ¼ i ðxðmÞ ; yÞ ¼ 1
¼ 00 for i 2 f0; 1; . . . ; bn1
2 c 1g f0;
j1 j2
2 ; 2
; . . . ; jmþ1
2 g.
According to the above discussion, we have
n1
n1
n1
xðb 2 c1Þ ¼ 0ð01Þb 2 c and, thus, ðxðb 2 c1Þ ; yÞ 2 EðCQn Þ.
nþ1
ðdnþ1
2 e2Þ ; yÞ
Noticing that bn1
2 c ¼ d 2 e 1, we have x
nþ1
ðdnþ1
e2Þ
ðd
e1Þ
2 EðCQn Þ. Further, since ðx 2
;x 2
Þ 2 P and
nþ1
nþ1
nþ1
xðd 2 e1Þ ; yÞ 2 P , there exists a cycle y, xðd 2 e2Þ , xðd 2 e1Þ ,
y in CQn , whose length is 3, which contradicts to that there
is no cycle of length 3 in CQn (the method of proof is similar
to that of proof for Lemma 5 in [13]).
nþ1
Case 1.2. n is even. Then, bn1
2 c ¼ d 2 e 2. We
assume that one of the following two conditions holds:
1.
k ðxðmÞ ; yÞ ¼ 0 for k 2 f0; j21 ; j22 ; . . . ; j2m g with 1 ðmþ1Þ ðmþ1Þ
x2iþ1 x2i
jl n 3 and 1 l m and l 6¼ k , bn1
ðx
2 c
2.
ðmÞ
; yÞ ¼
ðmÞ ðmÞ
2 with xn1 xn2 ¼ 00, and i0 ðxðmÞ ; yÞ ¼ 1 with
ðmÞ
ðmÞ
x2i0 þ1 x2i0 ¼ 00 for i0 2 f0; 1; . . . , bn1
2 c 1g f0;
j1 j2
jm
2 , 2 , . . . , 2 g.
j
j
k ðxðmÞ ; yÞ ¼ 0 for k 2 f0; j21 ; j22 ; . . . ; jk20 1 ; k20 þ1 ; k20 þ2 ;
ðx
ðmÞ
and i0 ðxðmÞ ; yÞ ¼ 1 with x2i0 þ1 x2i0 ¼ 00 for
n1
c 1g
i0 2 f0; 1; . . . ; b
2
j1 j2
jk0 1 jk0 þ1 jk0 þ2
jm
0; ; ; . . . ;
;
;
;...;
;
2 2
2
2
2
2
where 1 k0 m.
Let xðmÞ jmþ1 xðmþ1Þ . We deal with the following
cases.
Case 1.2.1. Case 1 holds. Let A0 ¼ f0; 1; j1 ; j1 þ 1; j2 ;
0
j2 þ 1; . . . ; jm ; jm þ 1g and B ¼ fjjj is odd with 2 j S
n 3 and j1 ðxðmÞ ; yÞ ¼ 1g fjjj is even with 2 j n 2
Pbn1c
3; j ðxðmÞ ; yÞ ¼ 1; and i¼2jþ1 i ðxðmÞ ; yÞis eveng. Then, si2
milar to the discussion in Case 1.1.1 and Case 1.1.2, we
S
= A0 B0 . So, jmþ1 satisfies one of the following
have jmþ1 2
two cases.
Case 1.2.1.1. jmþ1 is such an even integer that 2 Pbn1
ðmÞ
2 c
i ðxðmÞ ; yÞ is odd with xjm þ1
jmþ1 n 3 and
jmþ1
i¼
2
þ1
xjm ¼ 00. Then, we have k ðxðmþ1Þ ; yÞ ¼ 0 for k 2 f0;
j1 j2
2 ; 2
j2
2
; . . . ; jmþ1
2 g.
2
ðmþ1Þ ðmþ1Þ
; yÞ ¼ 1 with xn1 xn2 2 f10; 01g, k ðxðmþ1Þ ; yÞ ¼
ðmþ1Þ ðmþ1Þ
with x2iþ1 x2i
1g f0; j21 ,
j2
2
ðmÞ
ðmÞ
¼ x2iþ1 x2i ¼ 00 for i 2 f0; 1; . . . ; bn1
2 c
, . . . ; j2m g.
Case 1.2.2. Case 2 holds. Let A0 ¼ f0; 1; j1 ; j1 þ 1; . . . ;
jk0 1 ; jk0 1 þ1; jk0þ1 ; jk0 þ1 þ 1; jk0 þ2 ; jk0 þ2 þ 1; . . . ; jm ; jm þ 1g
and B0 ¼ fjjj is odd with 2 j n 3 and j1 ðxðmÞ ; yÞ
2
S
¼ 1g fjjj is even with 2 j n 3, j ðxðmÞ ; yÞ ¼ 1; and
2
Pbn1
2 c
ðxðmÞ ; yÞ is even g. Then, similar to the discussion
i¼2j þ1 i
S
= A0 B0 .
in Case 1.1.1 and Case 1.1.2, we have jmþ1 2
= fn 1; n 2g. SupposFurther, we can show that jmþ1 2
ing on the contrary that jmþ1 2 fn 1; n 2g, we have
ðmþ1Þ ðmþ1Þ
xn1 xn2
ðmþ1Þ ðmþ1Þ
2 f00; 11g. If xn1 xn2
¼ 11, similar to the
discuss in Case 1.1.3, we can obtain a contradiction. If
ðmþ1Þ ðmþ1Þ
xn1 xn2
¼ 00, we have bn1
ðxðmþ1Þ ; yÞ ¼ 2 and
2 c
i ðxðmþ1Þ ; yÞ¼ i ðxðmÞ ; yÞ for i ¼ 0; 1 . . . ; bn1
2 c 1. Hence,
bn1
2 c
distðCQn ; x
ðmþ1Þ
; yÞ ¼
X
bn1
2 c
i ðx
ðmþ1Þ
; yÞ ¼
i¼0
¼d
X
i ðxðmÞ ; yÞ þ 1
i¼0
nþ1
e m:
2
As a result, the length of P is larger than or equal to
lenðP ; xð0Þ ; xðmþ1Þ Þ þ distðCQn ; xðmþ1Þ ; yÞ
nþ1
nþ1
e mÞ ¼ d
e þ 1;
¼ ðm þ 1Þ þ ðd
2
2
2
ðmÞ
,
ðmÞ
with jk0 2 fn1; n2g and xn1 xn2 2 f10; 01g,
2
j1
2
0 for k 2 f0; j21 ; j22 ; . . . ; j2m g, and i ðxðmþ1Þ ; yÞ ¼ i ðxðmÞ ; yÞ
integers jk with 2 jk n3, bjk0 c ðxðmÞ ; yÞ ¼ 1
ðmÞ
ðmÞ
Case 1.2.1.2. jmþ1 2 fn1; n 2g. Then, we have bjmþ1 c
ðmþ1Þ
. . . ; j2m g with 1 m bn1
2 c 1 and the even
ðmÞ
ðmÞ
DECEMBER 2005
¼ x2iþ1 x2i ¼ 00
for i 2 f0; 1; . . . , bn1
2 cg f0;
m bn1
2 c 1 and the even integers jl with 2 0
VOL. 16, NO. 12,
ðmþ1Þ
; . . . ; jmþ1
; yÞ¼ i ðxðmÞ ; yÞ with
2 g and i ðx
contradictory to that the length of P is dnþ1
2 e. Then, jmþ1
must be such an even integer that 2 jmþ1 n 3 and
Pbn1
ðmÞ
ðmÞ
2 c
i ðxðmÞ ; yÞ is odd with xjm þ1 xjm ¼ 00. Further,
jmþ1
i¼
2
þ1
we have k ðxðmþ1Þ ; yÞ ¼ 0 for k 2 f0; j21 ; j22 ; jk20 1 ;
jk0 þ1
2
;
jk0 þ2
2
;
ðmþ1Þ
j
; yÞ
. . . ; jmþ1
2 g; b k0 c ðx
¼ 1 with jk0 2 fn 1; n 2g and
ðmþ1Þ ðmþ1Þ
xn1 xn2
2 f10; 01g , and i ðxðmþ1Þ ; yÞ ¼
2
¼
ðmÞ ðmÞ
xn1 xn2
ðmþ1Þ ðmþ1Þ
i ðxðmÞ ; yÞ ¼ 1 with x2iþ1 x2i
ðmÞ
ðmÞ
¼ x2iþ1 x2i ¼ 00 for i 2
j
j
jmþ1
j1 j2 jk0 1
k0 þ1
k0 þ2
f0; 1; . . . ; bn1
2 c 1g f0; 2 ; 2 ; 2 , 2 ; 2 ; . . . ; 2 g.
According to the above discussion, we have
n1
n1
n1
ðbn1
x 2 cÞ 2 f01ð01Þb 2 cþ1 ; 10ð01Þb 2 cþ1 g and, thus, ðxðb 2 cÞ ;
nþ1
yÞ 2 EðCQn Þ. Noticing that bn1
2 c ¼ d 2 e2, we have
nþ1
nþ1
nþ1
ðxðd 2 e2Þ ; yÞ 2 EðCQn Þ. Further, since ðxðd 2 e2Þ ; xðd 2 e1Þ Þ
nþ1
nþ1
2 P and ðxðd 2 e1Þ ; yÞ 2 P , we get a cycle y, xðd 2 e2Þ ,
nþ1
xðd 2 e1Þ , y in CQn , whose is length 3, a contradiction.
Case 2. dnþ1
2 e is odd. Select x ¼ xn1 xn2 . . . x1 x0 ¼
0n1 1, y ¼ yn1 yn2 . . . y1 y0 ¼ 1n2 01. Similar to Case 1, we
can prove that there is no a path of length dnþ1
2 e between x
u
and y in CQn when n 5. Thus, the proof is omitted. t
FAN ET AL.: OPTIMAL PATH EMBEDDING IN CROSSED CUBES
This theorem shows that dnþ1
2 e þ 1 is the shortest possible
length that can be embedded between arbitrary two distinct
nodes with a dilation of 1 in the n-dimensional crossed cube,
which implies that this range of path lengths is optimal.
5
CONCLUSIONS
In this paper, we have studied embedding of paths of
different lengths between any two nodes in crossed cubes.
We have proven that paths of all lengths between dnþ1
2 eþ1
and 2n 1 can be embedded between any two distinct
nodes with a dilation of 1 in the n-dimensional crossed
cubes. We have also proven that dnþ1
2 e þ 1 is the shortest
possible length that can be embedded between arbitrary
two distinct nodes with a dilation of 1 in the n-dimensional
crossed cube. The embedding of paths is optimal in the
sense that its dilation is 1.
ACKNOWLEDGMENTS
The authors would like to thank the anonymous referees for
their valuable comments. Their suggestions are very helpful
to revise the paper. This research is partially supported by
Hong Kong grants CERG under No. 9040816 and No. 9040909.
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manuscript.
Jianxi Fan received the BS and MS degrees in
computer science from Shandong Normal University and Shandong University, China, in 1988
and 1991, respectively. He became a lecturer,
an associate professor, and a professor in 1994,
1997, and 2002, respectively, in the College of
Information Engineering at Qingdao University,
China. He visited the Department of Computer
Science, City University of Hong Kong, Hong
Kong, from April 2003 to October 2003. He is
currently a PhD candidate in the Department of Computer Science, City
University of Hong Kong, Hong Kong. His research interests include
interconnection networks, design and analysis of algorithms, graph
theory, etc.
1200
IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS,
Xiaola Lin received the BS and MS degrees in
computer science from Peking University, Beijing, China, and the PhD degree in computer
science from Michigan State University, East
Lansing, Michigan. He is currently a professor of
computer science at Sun Yat-sen University,
Guangzhou, China. His research interests include parallel and distributed computing and
computer networks.
VOL. 16, NO. 12,
DECEMBER 2005
Xiaohua Jia received the BS, BSc (1984), and
MEng (1987) degrees from the University of
Science and Technology of China, and received
the DSc (1991) degree in information science
from the University of Tokyo, Japan. He is
currently a professor in the Department of
Computer Science at the City University of Hong
Kong, adjunct with the School of Computing,
Wuhan University, China. His research interests
include distributed systems, computer networks,
WDM optical networks, Internet technologies, and mobile computing. He
is a senior member of the IEEE.
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