Fault-tolerant hamiltonian laceability of hypercubes

Fault-tolerant hamiltonian laceability of hypercubes

Chang-Hsiung Tsai, Jimmy J.M. Tan, Tyne Liang, Lih-Hsing Hsu

Department of Computer and Information Science, National Chiao Tung University, Hsinchu, 30050 Taiwan, ROC

Received 16 January 2001; received in revised form 27 December 2001 Communicated by F.Y.L. Chin

Abstract

It is known that every hypercube Qnis a bipartite graph. Assume that n
2 and F is a subset of edges with |F |  n − 2. We prove that there exists a hamiltonian path in Qn− F between any two vertices of different partite sets. Moreover, there exists a path of length 2n− 2 between any two vertices of the same partite set. Assume that n
3 and F is a subset of edges with |F |  n − 3. We prove that there exists a hamiltonian path in Qn− {v} − F between any two vertices in the partite set without v. Furthermore, all bounds are tight.2002 Elsevier Science B.V. All rights reserved.

Keywords: Hamiltonian laceable; Hypercube; Fault tolerance

1. Introduction

In this paper, a network is represented as an undi-rected graph. For the graph definition and notation we follow [1]. G= (V, E) is a graph if V is a fi-nite set and E is a subset of{(a, b) | (a, b) is an or-dered pair of V}. We say that V is the vertex set and

E is the edge set. Two vertices a and b are

adja-cent if (a, b)∈ E. A path is a sequence of adjacent vertices, written asv0, v1, v2, . . . , vm , in which all the vertices v0, v1, . . . , vmare distinct except possibly v0= vm. We also write the path v0, P , vm , where P = v1, . . . , vm−1 . A path is a hamiltonian path if its vertices are distinct and they span on V . A cycle is a path with at least three vertices such that the first

✩

This work was supported in part by the National Science Council of the Republic of China under Contract NSC 89-2115-M-009-020.

* _{Corresponding author.}

E-mail address: [email protected] (L.-H. Hsu).

vertex is the same as the last one. A cycle is a hamil-tonian cycle if it traverses every vertex of G exactly once. A graph is hamiltonian if it has a hamiltonian cycle. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two vertices of

G. A graph G= (V0∪ V1, E) is bipartite if V (G) is

the union of two disjoint sets V0and V1such that each edge consists of one vertex from each set.

As the hamiltonicity of a graph G is concerned, it is an important issue to investigate if G is hamiltonian or hamiltonian connected. However, any hamiltonian bipartite graph G = (V0 ∪ V1, E) satisfies |V0| =

|V1|. Since the colors of the bipartite path alternates, all hamiltonian bipartite graphs are not hamiltonian-connected. Simmons [8] introduces the concept of hamiltonian laceability for those hamiltonian bipartite graphs. A hamiltonian bipartite graph G = (V0 ∪

V1, E) is hamiltonian laceable if there is a hamiltonian path between any two vertices x and y with x∈ V0and

y∈ V1. Hsieh et al. [3] further extend this concept into 0020-0190/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved.

strongly hamiltonian laceable. A hamiltonian laceable graph G= (V0 ∪ V1, E) is strongly if there is a

simple path of length|V0∪ V1| − 2 between any two vertices of the same partite set. Lewinter et al. [7] also introduce the concept of hyper-hamiltonian laceable. A hamiltonian laceable graph G= (V0∪ V1, E) is

hyper-hamiltonian laceable if for any vertex v∈ Vi, i= 0, 1, there is a hamiltonian path of G − v between

any two vertices of V1−i.

The edge fault-tolerant hamiltonicity proposed by Hsieh, Chen, and Ho [4], measures the performance of the hamiltonian property in the faulty networks. A hamiltonian graph G is k edge fault-tolerant hamil-tonian if G− F remains hamiltonian for every F ⊂

E(G) with |F |  k. The edge fault-tolerant

hamil-tonicity,He(G), is defined to be the maximum integer k such that G is k edge fault-tolerant hamiltonian if G

is hamiltonian, and undefined if otherwise. It is easy to see thatHe(G) δ(G) − 2 for any hamiltonian graph G where

δ(G)= min
deg(v)| v ∈ V (G).

We can further study other fault-hamiltonicity. A ham-iltonian laceable graph G is k edge fault-tolerant hamiltonian laceable if G− F remains hamiltonian laceable for every F⊂ E(G) with |F |  k. The edge fault-tolerant hamiltonian laceability,HL_e(G), is

de-fined to be the maximum integer k such that G is

k edge fault-tolerant hamiltonian laceable, and

unde-fined if otherwise. A strongly hamiltonian laceable graph G is k edge fault-tolerant strongly hamiltonian laceable if G− F remains strongly hamiltonian lace-able for every F⊂ E(G) with |F |  k. The edge fault-tolerant strongly hamiltonian laceability,H_eSL(G), is

defined to be the maximum integer k such that G is k edge fault-tolerant strongly hamiltonian laceable, and undefined if otherwise. A hyper-hamiltonian laceable graph G is k edge fault-tolerant hyper-hamiltonian laceable if G− F remains hyper-hamiltonian laceable for every F ⊂ E(G) with |F |  k. The edge fault-tolerant hyper-hamiltonian laceability,H_eh(G), is

de-fined to be the maximum integer k such that G is

k edge fault-tolerant hyper-hamiltonian laceable, and

undefined if otherwise.

Network topology is usually represented by a graph where nodes represent processors and edges represent links between processors. The binary hypercube, Qn, is one of the most popular topologies [6]. Let u=

un−1un−2. . . u1u0and v= vn−1vn−2. . . v1v0 be two

n-bit binary strings. The Hamming distance h(u, v)

between two vertices u and v is the number of different bits in the corresponding strings of both vertices. The

n-dimensional hypercube consists of all n-bit binary

strings as its vertices and two vertices u and v are adjacent if and only if h(u, v)= 1. Latifi et al. [5] proved thatHe(Qn)= n − 2 if n
2. Harary et al. [2] proved that Qnis strongly hamiltonian laceable if and only if n
2. Lewinter et al. [7] proved that Qn is hyper-hamiltonian laceable if and only if n
3. In this paper, we prove that

HL

e(Qn)= HSLe (Qn)= n − 2 if n
2 and

Hh

e(Qn)= n − 3 if n
3.

Using our approach, we can easily prove thatHe(Qn)

= n − 2 if n
2.

2. Fault-tolerant hamiltonian laceability of hypercubes

Let Qn be the n-dimensional hypercube. In this section, we will prove that HSL_e (Qn)
n − 2 and Hh

e(Qn)
n − 3. It is clear that HSLe (G) n − 2 and Hh

e(G) n − 3, for any n-regular bipartite graph G. So H_eSL(Qn)= n − 2, Hhe(Qn)= n − 3, and these results are optimal. Qncan be divided into two copies of Qn−1, denoted by Q0_n−1 and Q1_n−1. Let Ec be the set of crossing edges, i.e.,

Ec=

(u, u)| (u, u)∈ E(Qn), u∈ V (Q0_n−1) and u∈ V (Q1_n−1).

Let F be the set of faulty edges of Qn, F0= F ∩

E(Q0_n−1), F1= F ∩E(Q1_n−1), and Fc= F ∩Ec. Also let f0= |F0|, f1= |F1|, and fc= |Fc|. Since Qn is edge symmetric, it suffices to consider only the case that fc
1. That means, Qn can be split into Q0_n−1 and Q1_n−1 using any dimension d , where 1 d  n. Therefore, given a faulty edge set F , Qncan be split into Q0_n−1 and Q1_n−1such that Fcis not an empty set.

Lemma 1. Q3 is 1 edge fault-tolerant hamiltonian laceable,HL_e(Q3)= 1.

Proof. Let e be a faulty edge in Q3. Q3 can be split into Q0₂and Q1₂such that e is neither in E(Q0₂) nor in E(Q1

2). Suppose that x and y are with different colors. Case 1: x, y∈ V (Q0₂) or x, y∈ V (Q1₂). Without

loss of generality, we may assume that x, y∈ V (Q0₂). Q2 is a cycle of length four, so x and y are adja-cent. Let P = x, x0, x1, y be a path in Q02. Since |F | = 1, there exists an edge, denoted by (u, v), in this path such that (u, u) and (v, v) are fault-free

and u, v ∈ Q1₂. Obviously, u and v are adjacent. Let R= u, w, z, v be a path in Q1₂. Therefore,

E(P )∪ E(R) − {(u, v)} forms a hamiltonian path in Q3joining x and y.

Case 2: x∈ V (Q0₂) and y∈ V (Q1₂), or x∈ V (Q1₂)

and y∈ V (Q0₂). Without loss of generality, we may

assume that x∈ V (Q0₂) and y ∈ V (Q1₂). Since there

are two vertices in Q0₂ adjacent to x, we may choose a fault-free edge (u, v) such that u∈ V (Q0₂), u is

adjacent to x and v∈ V (Q1₂). Obviously, v and y

are adjacent. Let x, x0, x1, u be a path in Q0₂ and y, y0, y1, v be a path in Q1₂, respectively. Combining these two paths, we have a hamiltonian path in Q3 joining x and y. ✷

Lemma 2. Q3is 1 edge fault-tolerant strongly hamil-tonian laceable, i.e.,HSL_e (Q3)= 1.

Proof. Let e be a faulty edge in Q3. Q3 can be split into Q0₂and Q1₂such that e is neither in E(Q0₂) nor in E(Q1₂). Suppose that x and y are with the same color.

In order to prove this lemma, we will construct a fault-free path of length 6 joining x and y.

Case 1: x, y∈ V (Q0₂) or x, y∈ V (Q1₂). Without

loss of generality, we may assume that x, y∈ V (Q0₂).

Letx, u, y be a path in Q0₂such that (u, u) is

fault-free and u∈ Q1₂. There is a vertex v in {x, y} such that (v, v) is fault free and v ∈ Q1₂. Without loss of generality, we assume that v= x. Obviously, u and v are adjacent. Letu, w, z, v be a path in Q1₂. Therefore,x, v, z, w, u, u, y forms a path of length

6 in Q3joining x and y.

Case 2: x∈ V (Q0₂) and y∈ V (Q1₂), or x∈ V (Q1₂)

and y∈ V (Q0₂). Without loss of generality, we may

assume that x∈ V (Q0₂) and y ∈ V (Q1₂). Let u be

another vertex in V (Q0₂) having the same color as x.

Also let v∈ V (Q1₂), v and y have the same color. So

these four vertices x, y, u, and v are all with the same color. And at least one of u and v is not an endpoint of the faulty edge e. We may assume that (u, u) is

a fault-free edge such that u∈ V (Q1₂). Obviously, u

and y are adjacent. Let y, y0, v, u be a path in Q1₂

andx, x0, u be a path in Q0₂, respectively. Therefore, x, x0, u, u, v, y0, y forms a path of length 6 joining

x and y. ✷

Lemma 3. The hypercube Qn, n
2, is (n − 2) edge fault-tolerant hamiltonian laceable, i.e., H_eL(Qn)= n− 2.

Proof. We prove this lemma by induction on n. First,

we observe that the lemma holds for n = 2. By Lemma 1, the lemma holds if n= 3. For n
4, we assume that the lemma is true for every integer k < n. Let x and y be two vertices with different colors in

Qn, i.e., x and y are in different partite sets. In order to prove that HL_e(Qn)
n − 2, we must construct a fault-free hamiltonian path joining x and y for any given faulty edge set F with|F | = n−2. Since fc
1, f0 n − 3 and f1 n − 3.

Case 1: x, y∈ V (Q0_n−1) or x, y∈ V (Q1_n−1). (See

Fig. 1(a).) Without loss of generality, we may as-sume that x and y are in Q0_n−1. By induction hypoth-esis, HL_e(Q0_n−1)= n − 3, there exists a hamiltonian

path x, P0, y with 2n−1− 1 edges. We claim that

there exists an edge (u, v)∈ E(x, P0, y ) such that

both crossing edges (u, u) and (v, v) are fault-free.

Since |E(x, P0, y )| = 2n−1− 1, we have 2n−1− 1

choices. If none of the edges of x, P0, y meets the

requirements of such an edge (u, v), then there are at least (2n−1− 1)/2 faults in Fc. (Because a sin-gle fault in Fc eliminates 2 edges ofx, P0, y .) And

(2n−1_{−1)/2 > n−2 for n
4, this contradicts with} the fact that |F |  n − 2. Therefore, we can always find such an edge (u, v). Obviously, uand vare with different colors. SinceHL_e(Q1_n−1)= n − 3, there

ex-ists a fault-free hamiltonian pathu, P1, v in Q1n−1. Therefore, Ex, P0, y ∪ Eu, P1, v  ∪
(u, u), (v, v)−
(u, v)

forms a hamiltonian path in Qn− F joining x and y. Case 2: x ∈ V (Q0_n−1) and y∈ V (Q1_n−1), or x ∈ V (Q1_n−1) and y∈ V (Q0_n−1). (See Fig. 1(b).) Without

Fig. 1. (a) Case 1: x, y∈ V (Q0_n−1). (b) Case 2: x∈ V (Q0_n−1) and y∈ V (Q1_n−1). loss of generality, we assume that x∈ V (Q0_n−1) and

y∈ V (Q1_n−1). We claim that we can choose a vertex u in Q0_n−1 such that u and x are with different colors, and the crossing edge (u, u) is fault-free,

where u∈ V (Q1_n−1). Since |V (Q0_n−1)| = 2n−1, we have 2n−2 choices. (Because there are 2n−2 vertices in Q0_n−1 which have different colors from x.) If none of the vertices in Q0_n−1 meets the requirements of such vertex u, then there are at least 2n−2 faults in

Fc. This contradicts with the fact that |F |  n − 2 for n
2. Therefore, we can always find such a vertex u. Then, u and y are with different colors. SinceHL_e(Q0_n−1)= HL_e(Q1_n−1)= n − 3, there exists

a hamiltonian pathx, P0, u and u, P1, y in Q0_n−1 and in Q1_n−1, respectively. Therefore,

Ex, P0, u ∪ Eu, P1, y 

∪
(u, u)

forms a hamiltonian path in Qn− F joining x and y. This completes the induction. ✷

Given any faulty edge set F with|F | = n − 2, we can chose an edge (u, v) in Qn − F . By the proof above, there exists a hamiltonian path u, P, v in

Qn− F joining u and v. So it is easy to see that Eu, P, v ∪
(u, v)

forms a hamiltonian cycle in Qn− F . Hence He(Qn)

= n − 2 if n
2.

Theorem 1. The hypercube Qn, n
2, is (n − 2) edge fault-tolerant strongly hamiltonian laceable, i.e.,

HSL

e (Qn)= n − 2.

Proof. By Lemma 3, we haveHL_e(Qn)= n − 2. So all we have to show is the following. Let x and y be

two vertices with the same color in Qn. We must find a fault-free pathx, P, y of length 2n− 2 for any given faulty edge set F with|F | = n − 2. This theorem can be proved using by the same way of Lemma 3 and hence the detail proof is omitted. ✷

Theorem 2. The hypercube Qn, n
3, is (n − 3) edge fault-tolerant hyper-hamiltonian laceable, i.e.,

Hh

e(Qn)= n − 3.

Proof. The proof is again by using induction on n.

Lewinter et al. [7] proved that Qn, n
3, is hyper-hamiltonian laceable. So the lemma holds for the case n= 3. For n
4, we assume the theorem is true for every integer k < n. By induction hypothe-sis, Hh_e(Qn−1)= n − 4. Now, we consider Qn. By Lemma 3, we haveHl_e(Qn)= n − 2. Obviously, Qn is hamiltonian laceable after removing n− 3 edges. In order to prove thatH_eh(Qn)
n−3, it suffices to show the following. After deleting a given vertex w from

Qn, let x and y be any two vertices in the larger par-tite set of Qn. We must construct a hamiltonian path of Qn− F − w joining x and y for any given faulty edge set F with|F | = n − 3. Without loss of general-ity, we may assume that w∈ V (Q0_n−1). Since fc
1, f0 n − 4 and f1 n − 4.

Case 1: x, y∈ V (Q0_n−1). (See Fig. 2(a).) By

in-duction hypothesis, Hh_e(Q0_n−1)= n − 4, there exists

a fault-free hamiltonian pathx, P0, y with 2n−1− 2

edges in Q0_n−1 − w. We now show that there ex-ists an edge (u, v) ∈ E(x, P0, y ) such that both

crossing edges (u, u) and (v, v) are fault-free. Since

|E(x, P0, y )| = 2n−1− 2, we have 2n−1− 2 choices.

If none of the edges of x, P0, y meets the

Fig. 2. (a) Case 1: x, y∈ V (Q0_n−1). (b) Case 2: x, y∈ V (Q1_n−1). (c) Case 3: x∈ V (Q0_n−1) and y∈ V (Q1_n−1). 2n−2 − 1 faults in Fc. (Because a single fault in Fc

eliminates 2 edges of x, P0, y .) And 2n−2 _{− 1 >} n− 3 for n
4, this contradicts with the fact that

|F |  n − 3. Therefore, we can always find such an edge (u, v). Obviously, uand vare with different col-ors. SinceHL_e(Q1_n−1)= n − 3, there exists a fault-free

hamiltonian pathu, P1, v in Q1_n−1. Therefore,

Ex, P0, y ∪ Eu, P1, v 

∪

(u, u), (v, v)−
(u, v)

forms a hamiltonian path in Qn− F − w joining x and y.

Case 2: x, y∈ V (Q1_n−1). (See Fig. 2(b).) First, we

will choose a vertex u in Q1_n−1 such that u and x are with different colors, and the crossing edge (u, u) is

fault-free, where u∈ V (Q0_n−1). Since|V (Q1_n−1)| =

2n−1, we have 2n−2choices. (Because there are 2n−2 vertices in Q1_n−1which have different colors from x.) If none of the vertices in Q1_n−1meets the requirements of such vertex u, then there are at least 2n−2 faults in Fc. This contradicts with the fact that|F |  n − 3 for n
2. Therefore, we can always find such a vertex u. SinceHh_e(Q1_n−1)= n−4, there exists a

fault-free hamiltonian path x, P, y in Q1_n−1 − u joining

x and y. We claim that there exists a vertex t such

that the edge (u, t) is fault-free in Q1_n−1, the edge

(t, v)∈ E(x, P, y ), and the crossing edge (v, v) is

fault-free in Qn where v∈ Q0_n−1. Since the number of neighboring vertices of u in Q1_n−1is n− 1, we have

n− 1 choices. If none of the vertices in Q1_n−1 meets the requirements of such a vertex t , then there are at least n− 1 faults in Fc, making it contradictory to the fact that|F |  n − 3. Therefore, we can always find such a vertex t .

We then divide the pathx, P, y into two sections x, P0, v and t, P1, y , or x, P0, t and v, P1, y .

Without loss of generality, we assume the casex, P0, t

andv, P1, y . Thus, we have two sections as x, P0, u

andv, P1, y . Let (u, u) and (v, v) be two crossing

edges incident to vertices u and v, respectively. Then,

u and v are with the same color in Q0_n−1. Since u and x are with different colors, u and v are in the larger partite set of Q0_n−1− w. By induction hypothe-sis,HL_e(Q0_n−1)= n−4, there exists a fault-free

hamil-tonian pathu, R, v in Q0_n−1. Therefore,

Ex, P0, t ∪ Eu, R, v ∪

Ev, P1, y ∪
(u, u), (v, v), (t, u)

forms a hamiltonian path in Qn− F − w joining x and y.

Case 3: x ∈ V (Q0_n−1) and y∈ V (Q1_n−1), or x ∈ V (Q1_n−1) and y∈ V (Q0_n−1). (See Fig. 2(c).) Without

loss of generality, we assume that x∈ V (Q0_n−1) and y∈ V (Q1_n−1). First, we will choose a vertex u, u= x,

in Q0_n−1 such that u and x are with the same color, and the crossing edge (u, u) is fault-free, where u∈ V (Q1_n−1). Since|V (Q0_n−1)| = 2n−1, we have 2n−2−1 choices. (Because there are 2n−2−1 fault-free vertices in Q0_n−1 which have the same color as x.) If none of the vertices in Q0_n−1 meets the requirements of such vertex u, then there are at least 2n−2− 1 faults in

Fc. This contradicts with the fact that|F |  n − 3 for n
3. Therefore, we can always find such a vertex u.

Then, uand y are with different colors. By induction hypothesis, H_eh(Q0_n−1)= n − 4, there exists a

fault-free hamiltonian path x, P0, u in Q0_n−1 joining x and u. Also, sinceHL_e(Q1_n−1)= n − 3, there exists a

fault-free hamiltonian pathu, P1, y in Q1n−1joining uand y. Therefore,

Ex, P0, u ∪ Eu, P1, y 

∪
(u, u)

forms a hamiltonian path in Qn− F − w joining x and y. The proof is complete. ✷

References

[1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North-Holland, New York, 1980.

[2] F. Harary, M. Lewinter, Hypercubes and other recursively defined hamilton laceable graphs, Congr. Numer. 60 (1987) 81– 84.

[3] S.Y. Hsieh, G.H. Chen, C.W. Ho, Hamiltonian-laceability of star graphs, Networks 36 (2000) 225–232.

[4] S.Y. Hsieh, G.H. Chen, C.W. Ho, Fault-free hamiltonian cycles in faulty arrangement graphs, IEEE Trans. Parallel Distributed Systems 10 (32) (1999) 223–237.

[5] S. Latifi, S. Zheng, N. Bagherzadeh, Optimal ring embedding in hypercubes with faulty links, in: Fault-Tolerant Computing Symp., 1992, pp. 178–184.

[6] F.T. Leighton, Introduction to Parallel Algorithms and Architec-ture: Arrays, Trees, Hypercubes, Morgan Kaufmann, San Ma-teo, CA, 1992.

[7] M. Lewinter, W. Widulski, Hyper-Hamilton laceable and caterpillar-spannable product graphs, Comput. Math. Appl. 34 (1997) 99–104.

[8] G. Simmons, Almost all n-dimensional rectangular lattices are Hamilton laceable, Congr. Numer. 21 (1978) 103–108.