Fault-tolerant cycle-embedding of crossed cubes
✩Ming-Chien Yang
a, Tseng-Kuei Li
b,∗, Jimmy J.M. Tan
a, Lih-Hsing Hsu
aaDepartment of Computer and Information Science, National Chiao Tung University, Hsinchu, Taiwan 30050, R.O.C. bDepartment of Computer Science and Information Engineering, Ching Yun University, JungLi, Taiwan, 320, R.O.C.
Received 22 March 2003; received in revised form 22 July 2003 Communicated by M. Yamashita
Abstract
The crossed cube CQn introduced by Efe has many properties similar to those of the popular hypercube. However, the
diameter of CQn is about one half of that of the hypercube. Failures of links and nodes in an interconnection network are
inevitable. Hence, in this paper, we consider the hybrid fault-tolerant capability of the crossed cube. Letting feand fvbe the
numbers of faulty edges and vertices in CQn, we show that a cycle of length l, for any 4 l |V (CQn)|− fv, can be embedded
into a wounded crossed cube as long as the total number of faults (fv+ fe) is no more than n− 2, and we say that CQnis
(n− 2)-fault-tolerant pancyclic. This result is optimal in the sense that if there are n − 1 faults, there is no guarantee of having
a cycle of a certain length in it. 2003 Published by Elsevier B.V.
Keywords: Cycle embedding; Crossed cube; Pancyclic; Hamiltonian; Fault tolerance
1. Introduction
Network topology is essential for parallel and distributed computation, and many topologies have been proposed, for example, hypercubes, butterfly graphs and star graphs. The hypercube is one of the most popular networks since it has a simple structure and is easy to implement. However, there are still some different points of view to construct new topologies; for example, a new topology having smaller diameter.
✩
This work was supported in part by the National Science Council of the Republic of China under Contract NSC 91-2218-E-231-002.
* Corresponding author.
E-mail address: [email protected] (T.-K. Li).
To lower the diameter, we may change some links of the hypercube. Some variations of the hypercube have been studied in the literature. In [2], Efe first studied the crossed cube CQn, which has a structure similar to that of the hypercube, including recursive structure, the same number of vertices, and the same number of edges. However, the diameter of CQn is only about one half of that of the hypercube, and the diameter is an important factor for parallel computing speed. Other studies have been done to explore more properties of the crossed cube CQn, such as edge
congestion of CQn, as studied in [1]. Furthermore, embedding of binary trees, hamiltonian paths, and hamiltonian cycles into CQnwere discussed in [5,6].
The graph embedding problem asks if a guest graph is a subgraph of a host graph, and an important benefit of graph embedding is that we can apply
0020-0190/$ – see front matter 2003 Published by Elsevier B.V. doi:10.1016/j.ipl.2003.08.007
existing algorithms for guest graphs to host graphs. This problem has attracted a burst of studies over the years.
The pancycle problem asks if a cycle of length l is a subgraph of a given graph with a given positive integer l. Hwang [4] and Fan [3] et al. studied this problem on butterfly graphs and Möbius cubes, respectively. But they did not consider the possibilities of failures of nodes and/or links.
Failures are inevitable when a network is put in use. Therefore, the fault-tolerant capacity of an interconnection network is a crucial issue in parallel computing. Furthermore, both nodes and links may simultaneously be faulty in a network. Hence we study the hybrid fault tolerance of CQnin this paper. Letting fvand febe the numbers of faulty vertices and edges
in CQn, respectively, we show that a cycle of length l, for any 4 l |V (CQn)| − fv, is a subgraph of a
wounded crossed cube with (fv+ fe) (n − 2). That
is, CQnis (n− 2)-fault-tolerant pancyclic. In addition,
this result is optimal, and the reason is explained as follows: The n-dimensional crossed cube is n-regular. As a result, if there are (n− 1) faulty edges incident to a single node, a hamiltonian cycle cannot be embedded into a wounded CQn.
The rest of this paper is organized as follows: Section 2 includes the definition of the crossed cubes and some basic notation and terminologies. Then, the proof of the pancyclicity of CQnis given in Section 3. For the case n= 4, the proof is a little tedious, and we leave some parts of it to Appendix A.
2. Definitions and notation
Given a simple graph G, we use V (G) and E(G) to denote the vertex and edge sets of G, respectively. In order to define the crossed cube CQn, as proposed by Efe [2], the pair related set R is introduced. Let R = {(00, 00), (10, 10), (11, 01), (01, 11)}. Two binary strings a1a2 and b1b2 of length 2 are pair
related, denoted by a1a2∼ b1b2, if (a1a2, b1b2)∈ R.
The following is the recursive definition of the n-dimensional crossed cube CQn. CQn has 2n vertices, each labeled by a binary string of length n. CQ1 is a complete graph with two vertices labeled 0 and 1, respectively. For n 2, CQn is obtained by taking
two copies of CQn−1, denoted by CQ0n−1 and CQ1n−1, respectively, and adding 2n−1edges as follows:
Let
VCQ0n−1= {0xn−2. . . x1x0: xi= 0 or 1}
and
VCQ1n−1= {1yn−2. . . y1y0: yi= 0 or 1}.
A vertex 0xn−2. . . x1x0 ∈ V (CQ0n−1) and a vertex
1yn−2. . . y1y0∈ V (CQ1n−1) are adjacent if
(1) xn−2= yn−2 if n is even, and
(2) x2i+1x2i ∼ y2i+1y2ifor 0 i < (n − 1)/2 .
We take CQ3 and CQ4 as examples and display them in Fig. 1 (a) and (b), respectively. In Fig. 1(c), we use a different way to draw CQ3in order to see its vertex-symmetry.
We now introduce some basic terminologies and notation needed for later discussion. A path is a se-quence of vertices with any two consecutive vertices being adjacent in G. We use u1, u2, . . . , ul to denote
a path that begins with u1and ends with ul. In
addi-tion, u1, u2, . . . , ul is a cycle if u1= ul. A
hamil-tonian path is defined as a path which contains all the vertices of G exactly once. A graph G is hamiltonian connected if, for any two vertices of G, there exists a hamiltonian path between them. We say that a graph G is pancyclic if G contains a cycle of length l as a sub-graph, for every 4 l |V (G)|. A cycle is a hamil-tonian cycle if it traverses all the vertices of G exactly once. A graph G is hamiltonian if G contains a hamil-tonian cycle.
To consider a wounded graph, we give the fol-lowing terminologies and notation. Given a graph G, let Fv⊆ V (G) and Fe ⊆ E(G); and F = Fv∪ Fe.
Let G be the graph obtained from G by deleting all the edges in Fe. We use G− F to denote the
subgraph of G induced by V (G)− Fv. We call a
graph G k-fault-tolerant hamiltonian connected (ab-breviated as k-hamiltonian connected) if G− F is hamiltonian connected for any F with|F | k. We call a graph G k-fault-tolerant hamiltonian (abbreviated as k-hamiltonian) if G− F is hamiltonian for any F with
|F | k. A graph G is called k-fault-tolerant pancyclic (abbreviated as k-pancyclic) if G− F is pancyclic for any F with|F | k.
Fig. 1. (a) CQ3, (b) CQ4, and (c) CQ3drawn in a different way.
3. Main result
We use CQijn−2 to denote an (n− 2)-dimensional crossed cube which is a subgraph of CQn induced by the vertices labeled ij xn−3. . . x0. We say that an edge
is a critical edge of CQnif it is an edge in CQin−1with
one endpoint in CQi0n−2 and the other in CQi1n−2 for i∈ {0, 1}.
Lemma 1. Let (u1, u2) be a critical edge of CQn
which is in CQ0n−1, and v1, v2be the neighbors of u1
and u2in CQ1n−1, respectively, for n 4. Then (v1, v2)
is also a critical edge of CQnin CQ1n−1.
Proof. We discuss two cases: (1) n is even, and (2) n
is odd.
Case 1. n is even. Without loss of generality, we assume that u1= 00xn−3xn−4. . . x1x0 and u2=
01yn−3yn−4. . . y1y0, where x2i+1x2i ∼ y2i+1y2i for
0 i (n − 3)/2 . Then v1= 10yn−3yn−4. . . y1y0,
and v2= 11xn−3xn−4. . . x1x0. By definition, v1 and
v2 are adjacent, and (v1, v2) is a critical edge in
CQ1n−1.
Case 2. n is odd. Without loss of generality, we assume that u1= 00xn−3xn−4xn−5. . . x1x0. Suppose
that xn−3= 0. Then u1= 000xn−4xn−5. . . x1x0, u2=
010yn−4yn−5. . . y1y0, where x2i+1x2i ∼ y2i+1y2i for
0 i (n − 4)/2 , v1 = 100yn−4yn−5. . . y1y0,
and v2= 110xn−4xn−5. . . x1x0. Thus, v1 and v2 are
adjacent, and (v1, v2) is a critical edge in CQ1n−1. It
can be checked that the statement is also true for the case xn−3= 1. ✷
It is observed that vertices u1, u2, v1, v2 in the
above lemma form a 4-cycle. We call this cycle a crossed 4-cycle in CQn. It is clear that, for each vertex 00xn−3· · · x0, there is exactly one crossed
4-cycle corresponding to the vertex. Thus, there are 2n−2 disjoint crossed 4-cycles in CQn. We note that a crossed 4-cycle contains two critical edges.
Huang et al. [5] showed the validity of the follow-ing theorem. Based on this theorem, we show the pan-cyclicity of the crossed cube by induction.
Theorem 1 [5]. The crossed cube CQn is (n− 2)-hamiltonian and (n− 3)-hamiltonian connected for n 3.
The base case is n= 3, and the proof is given in the following.
Theorem 2. CQ3is 1-pancyclic.
Proof. Note that CQ3can be redrawn as Fig. 1(c), and it is vertex-transitive. We consider two cases (1) one faulty vertex, and (2) one faulty edge as follows:
Case 1. One faulty vertex. Without loss of gener-ality, we assume that vertex x = 000 is faulty. We list cycles of lengths from 4 to 7 as follows: 001, 111, 101, 011, 001 , 001, 111, 110, 010, 011, 001 , 001, 111, 110, 100, 101, 011, 001 , and 001, 111, 101, 100, 110, 010, 011, 001 .
Case 2. One faulty edge. Without loss of generality, we assume that the faulty edge e is incident to 000. By case 1, there are cycles of lengths from 4 to 7 in the faulty CQ3. For a cycle of length 8, suppose that e=
Fig. 2. Cases of Theorem 3.
100, 000 is the desired one. Suppose that e = (000, 001). Then 000, 010, 110, 111, 001, 011, 101, 100, 000 is a cycle of length 8. If e = (000, 100), the case is symmetric to the case e= (000, 001). ✷
Let F be a set of faults in CQn. We say that a vertex
u in one subcube of CQn is a safe crossing-point in
CQn− F if u still connects to the neighbor in the other
subcube in CQn− F , i.e., the corresponding neighbor
uin the other subcube of u is fault-free, and the edge (u, u) is also fault-free. The main result is as follows.
Theorem 3. The crossed cube CQn is (n −
2)-pancyclic for n 3.
Proof. We prove this by induction on n. It follows
from Theorem 2 that CQ3 is 1-pancyclic. Now we proceed to the induction step. Suppose that CQn−1 is (n− 3)-pancyclic for some n 4. We will show that CQnis (n−2)-pancyclic. Let F ⊆ V (CQn)∪E(CQn) be the set of faults. We divide F into five disjoint parts: Fv0= F ∩ VCQ0n−1, Fe0= F ∩ ECQ0n−1, Fv1= F ∩ VCQ1n−1, Fe1= F ∩ ECQ1n−1, Fec= F ∩(u, v)| (u, v) is an edge
between CQ0n−1and CQ1n−1. Let f = |F |, fv0= |Fv0|, fe0= |Fe0|, fv1= |Fv1|, fe1= |F1
e|, and fec= |Fec|. For convenience of discussion,
we define the following subsets of F : Fv = F ∩
V (CQn), Fe= F ∩E(CQn), F0= Fv0∪Fe0, and F1=
Fv1∪ Fe1. And let fv= |Fv|, fe= |Fe|, f0= |F0|, and
f1= |F1|. Note that f0+ f1= f − fec.
Case 1. There is a subcube containing all the (n−2) faults. Without loss of generality, we assume that f0=
n− 2. Thus, there is no fault outside CQ0n−1, i.e., f1= fec= 0. We discuss the existence of cycles of lengths from 4 to 2n− fv according to the following
cases.
Case 1.1. Cycles of lengths from 4 to 2n−1. Since CQn−1is (n− 3)-pancyclic, CQ1n−1contains cycles of lengths from 4 to 2n−1 for n 4. Clearly, CQn− F
also contains cycles of these lengths.
Case 1.2. A cycle of length 2n−1+1. (See Fig. 2(a).) We want to construct a cycle containing 2n−1 − 1 vertices in CQ1n−1 and two vertices in CQ0n−1. To avoid faults in CQ0n−1, we introduce a term called the shadows of the faults. Let u1, u2, v2, v1, u1 be
a crossed 4-cycle with u1, u2in CQ0n−1 and v1, v2 in
CQ1n−1, respectively. If there is a fault on this cycle but the fault is not in CQ1n−1, we call edge (v1, v2)
a shadow fault of F on CQ1n−1. (Similarly, we may define a shadow fault on CQ0n−1.) Let Fs= {e | edge e is a shadow fault of F on CQ1n−1}. Since all crossed 4-cycles are vertex disjoint, |Fs| n − 2. If |Fs| = n− 2, we arbitrarily pick an edge e1in Fs , and let
F= Fs − e1, or else F= Fs. Then |F| n − 3
and CQ1n−1− Fis still pancyclic. So there is a cycle C of length 2n−1− 1 in CQ1
n−1− F. Clearly, there
are two critical edges on C. Let (a, b) = e1 be a
critical edge on C, so (a, b) /∈ Fs. Let a, b be the neighbors of a and b in CQ0n−1, respectively. Then a, a, b, b, a is a fault-free crossed 4-cycle. Suppose that C= a, Q, b, a . Then a, a, Q, b, b, a forms a cycle of length 2n−1+ 1 in CQn− F .
Case 1.3. Cycles of lengths from 2n−1+ 2 to 2n− fv. (See Fig. 2(b).) By Theorem 1, CQ0n−1 is (n−
a hamiltonian path, say P = u1, u2, . . . , u2n−1−f0 v , where fv0= fv. Let 2 l 2n−1− fv. We construct
a cycle of length 2n−1+ l as follows: Suppose that the neighbors of u1 and ul in CQ1n−1 are v1 and
vl, respectively. Since CQn−1 is (n− 4)-hamiltonian
connected and n 4, there is a hamiltonian path Q in CQ1n−1between v1and vlcontaining 2n−1 vertices.
So u1, . . . , ul, vl, Q, v1, u1 forms a cycle of length
2n−1+ l.
Case 2. Both f0 and f1 are at most n− 3. By induction hypothesis, CQ0n−1− F0and CQ1n−1− F1 are still pancyclic. We discuss the existence of cycles of all lengths from 4 to 2n− fvin the following cases.
Case 2.1. Cycles of lengths from 4 to 2n−1− fv1. By induction hypothesis, CQ1n−1 is (n− 3)-pancyclic. Thus, we have cycles of lengths from 4 to 2n−1− fv1 in CQ1n−1− F1.
Case 2.2. A cycle of length 2n−1− fv1+ 1. (See Fig. 2(c).) We construct the cycle using a similar way used in Case 1.2. Let Fs = {e | edge e is a shadow fault of F on CQ1n−1}. Then |Fs∪ F1| n − 2. If |Fs ∪ F1| = n − 2, we arbitrarily choose an edge
e1 in Fs, and let F= Fs ∪ F1− e1, or else F=
Fs ∪ F1. Then|F| n − 3 and CQ1n−1− F is still pancyclic. Since F∩ V (CQ1n−1)= Fv1, there is a cy-cle C of length 2n−1− fv1− 1 in CQ1n−1− F. Since 2n−1− fv1− 1 > 2n−2 for n 4, C contains two crit-ical edges. Let (a, b)= e1 be a critical edge on C,
so (a, b) /∈ Fs. Let a, b be the neighbors of a and b in CQ0n−1, respectively. Then a, a, b, b, a is a fault-free crossed 4-cycle. Suppose that C= a, Q, b, a . Then a, a, Q, b, b, a forms a cycle of length 2n−1− fv1+ 1 in CQn− F .
Case 2.3. Cycles of lengths from 2n−1− fv1+ 2 to 2n− fv. (See Fig. 2(d).) Without loss of generality,
we assume that n− 3 f0 f1. If f1= n − 3, then f0= n − 3, and 2n − 6 n − 2 = f , which implies n 4. Thus, we need to discuss the case f1= n − 3 just for n = 4. We leave this particular case to Appendix A, and assume that f1 n − 4 in the following discussion.
By Theorem 1, CQ1n−1 − F1 is still hamiltonian connected, i.e., there is a path of length 2n−1− fv1− 1 between any two vertices in CQ1n−1− F1. As a result, if we can find a path of length l in CQ0n−1− F0with the two endpoints being safe crossing-points, then we find a cycle of length l+ 2 + (2n−1− fv1− 1).
Since we want to construct cycles of lengths from 2n−1 − fv1+ 2 to 2n− fv, 1 l 2n−1 − (fv−
fv1)− 1 = 2n−1 − fv0− 1. Now we construct a path of length l in CQ0n−1 for each l, 1 l 2n−1 − fv0− 1. By Theorem 1, CQ0n−1is (n− 3)-hamiltonian. Thus we have a hamiltonian cycle C = u0, u1, . . . ,
u2n−1−f0
v−1, u0 of length 2
n−1− f0
v in CQ0n−1− F 0.
We claim that there exist two safe crossing-points ui and uj on C such that (j − i)(mod 2n−1−f0
v)= l. Suppose on the contrary that there do not exist such ui
and uj. Then there are at least(2n−1− fv0)/2 faults
outside CQ0n−1. However, (2n−1− fv0)/2 + fv0 2n−2 > n− 2 for n 2. We obtain a contradiction. Thus, there exist such two vertices uiand uj. And then
we find a path of length l on C. Hence, the theorem follows. ✷
Appendix A
In the following, we construct cycles of lengths from 2n−1− fv1+ 2 to 2n− fvfor the case f0= f1=
n− 3. Since f0+ f1= 2n − 6 n − 2, n 4. Thus,
we need only to discuss the case f0= f1= 1 for n= 4 here. We shall use some symmetric properties of CQ3to reduce the cases.
For convenience of discussion (see Fig. 3(a)), we call (000, 010), (001, 011), (111, 101), (110, 100) as inner edges of CQ3 and (000, 001), (001, 111),
(111, 110), (110, 010), (010, 011), (011, 101), (101, 100), (100, 000) as outer edges of CQ3, respectively.
Let x be a vertex of CQ3. An inner edge e is said to be an N -edge of x if x connects to one of the endpoints of e. Hence CQ3has two N -edges of x. An inner edge e is said to be an H -edge of x if x is not incident to e, and e is not an N -edge of x. Therefore, CQ3has one H -edge of x.
To explore the pancyclicity of CQ4−F , we need an
observation, and it is stated in the following lemma.
Lemma 2. Let x be a faulty vertex in CQ3. Then the two N -edges of x, e1and e2, are on cycles of lengths
from 4 to 7 in CQ3− x. And the H -edge of x is on
cycles of lengths 4, 5, and 7 in CQ3− x.
Proof. Without loss of generality, we may assume that
x = 000. Then the two N-edges of x are (001, 011) and (110, 100), and the H -edge of x is (111, 101). We
Fig. 3. (a) N -edge and H -edge of x and (b) a cycle of length 12 with one faulty vertex 0000 and one faulty edge (1100, 1110).
list all the cycles as follows: 001, 111, 101, 011, 001 , 111, 110, 100, 101, 111 , 001, 111, 110, 010, 011, 001 , 111, 110, 010, 011, 101, 111 , 110, 010, 011, 101, 100, 110 , 001, 111, 110, 100, 101, 011, 001 , and 001, 111, 101, 100, 110, 010, 011, 001 . ✷
We continue to discuss cycles in CQ4− F , and consider two situations: (1) cycles of lengths from 8 to 14 and (2) cycles of lengths 15 and 16.
Case 1. Cycles of lengths from 8 to 14. Suppose that there is a faulty vertex x in CQ03 or a faulty edge e1
which is incident to x. Let (u1, u2) and (u3, u4) be the
two N -edges of x. By Lemma 2, (u1, u2) and (u3, u4)
are on cycles of lengths from 4 to 7 in CQ03− F0. Let v1, v2, v3, and v4 be the neighbors of u1, u2, u3,
and u4in CQ13, respectively. It is not difficult to check
that both (v1, v2) and (v3, v4) are inner edges of CQ13.
(See Fig. 3(b).) And (v1, v2) can not reach (v3, v4)
via exactly one edge of CQ13. Suppose that the other fault is a faulty vertex y in CQ13, or a faulty edge e2
which is incident to y. Then (v1, v2) or (v3, v4), say
(v1, v2) is an N -edge or H -edge of y in CQ13− F1.
By Lemma 2, (v1, v2) is on cycles of lengths 4, 5,
and 7 in CQ13− F1. We use Ci to denote a cycle of
length i. Let Ci and Cj be cycles containing (u1, u2)
and (v1, v2), respectively, 4 i 7, j = 4, 5, or 7.
Then we can construct a cycle Cl from Ci and Cj by
adding (u1, v1) and (u2, v2), and deleting (u1, u2) and
(v1, v2) for 8 l 14. For example, Fig. 3(b) shows a
cycle of length 12 in CQ4with one faulty vertex 0000
and one faulty edge (1100, 1110).
Case 2. Cycles of lengths from 15 to 16. If fv= 2,
we need only to find cycles of lengths from 4 to 14 which we did in the previous cases. If fv= 1 and
fe = 1, we have to find a cycle of length 15. By
Theorem 1, CQ4is 2-hamiltonian, so there is a cycle of length 15 in CQ4− F . If fe= 2, since CQ4 is
2-hamiltonian, there is also a cycle of length 16. Suppose that F = {(x1, y1), (x2, y2)}. Let F= {(x1, y1), x2}.
Then there is a cycle of length 15 in CQ4− F. This
cycle is also fault-free in CQ4− F .
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