Ring embedding in faulty pancake graphs

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Ring embedding in faulty pancake graphs

Chun-Nan Hung

a,∗

_{, Hong-Chun Hsu}

b

_{, Kao-Yung Liang}

a

_{, Lih-Hsing Hsu}

b a_{Department of Computer Science and Information Engineering, Da-Yet University, Changhua 51505, Taiwan, R.O.C.} b_{Department of Computer and Information Science, National Chiao Tung University, Hsinchu 300, Taiwan, R.O.C.}

Received 17 July 2002; received in revised form 29 November 2002 Communicated by M. Yamashita

Abstract

In this paper, we consider the fault hamiltonicity and the fault hamiltonian connectivity of the pancake graph Pn. Assume that F⊆ V (Pn)∪ E(Pn). For n 4, we prove that Pn− F is hamiltonian if |F | (n − 3) and Pn− F is hamiltonian connected if|F | (n − 4). Moreover, all the bounds are optimal.

Keywords: Fault tolerance; Hamiltonian; Hamiltonian connected; Pancake graphs

1. Introduction

Network topology is a crucial factor for the in-terconnection networks since it determines the per-formance of the networks. Many interconnection net-work topologies have been proposed in the litera-ture for the purpose of connecting hundreds or thou-sands of processing elements. Network topology is al-ways represented by a graph where nodes represent processors and edges represent links between proces-sors.

For the graph definition and notation we follow [2]. G= (V, E) is a graph if V is a finite set and E is a subset of {(u, v) | (u, v) is an unordered pair of V }. We say that V is the vertex set and E is the edge set. For any vertex x of V , deg_G(x) denotes its

* _{Corresponding author. Address: 112 Shan-Jiau Rd, Da-Tsuen,} Changhua 51505, Taiwan, R.O.C.

E-mail address: [email protected] (C.-N. Hung).

degree in G. We use δ(G) to denote min{degG(x)| x ∈ V (G)}. Two vertices u and v are adjacent if (u, v)∈ E. A path is represented by v0, v1, v2, . . . ,

vk. The length of a path Q is the number of edges in Q. We also write the path v0, v1, v2, . . . , vk as v0, Q1, vi, vi+1, . . . , vj, Q2, vt, . . . , vk, where Q1is

the pathv0, v1, . . . , vi and Q2is the pathvj, vj+1,

. . . , vt. Hence, it is possible to write a path as v0, v1, Q, v1, v2, . . . , vk if the length of Q is 0.

Sometimes, a path is also represented byv0, v1, . . . ,

vi, e, vi+1, . . . , vn to emphasize that e is the edge

(vi, vi+1). We use d(u, v) to denote the distance

be-tween u and v, i.e., the length of the shortest path join-ing u and v. A path is a hamiltonian path if its vertices span V . A cycle is a path with at least three vertices such that the first vertex is the same as the last vertex. Throughout this paper, we assume that n is a posi-tive integer. We usen to denote the set {1, 2, . . ., n}. The n-dimensional pancake graph, denoted by Pn, is

a graph with the vertex set

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Fig. 1. The pancake graphs P2, P3, and P4.

V (Pn)=

u1u2. . . un| ui∈ n and ui = uj

for i = j.

The adjacency is defined as follows: u1u2. . . ui. . . un

is adjacency to v1v2. . . vi. . . vn through an edge of

dimension i with 2 i n if vj= ui_−j+1if 1 j i and vj= uj if i < j n. The pancake graphs P2, P3,

and P4are illustrated in Fig. 1. By definition, Pnis an

(n− 1)-regular graph with n! vertices. Moreover, it is vertex transitive.

The pancake graphs are an important family of in-terconnection networks proposed by Akers and Krish-nameurthy [1]. Some interesting properties of pancake graphs are studied [3,10,4–7]. In particular, Gates and Papadimitriou [6] studied the diameter of the pancake graphs. Until now, we do not know the exact value of the diameter of the pancake graphs [7]. Kanevsky and Feng [10] proved that all cycles of length l where 6 l n! − 2 and l = n! can be embedded in the pan-cake graph Pnwith n 4.

In this paper, we consider two important properties of the pancake graphs, fault hamiltoniancity and fault hamiltonian connectivity. These two parameters for interconnection networks are proposed by Huang et al. [8,9].

A hamiltonian cycle of G is a cycle that traverses every vertex of G exactly once. A graph is hamiltonian if it has a hamiltonian cycle. A hamiltonian graph G is k-fault hamiltonian if G− F remains hamiltonian for every F⊂ V (G) ∪ E(G) with |F | k. The fault hamiltonicity,Hf(G), is defined to be the maximum

integer k such that G is k-fault hamiltonian if G

is hamiltonian and undefined if otherwise. Clearly, Hf(G) δ(G) − 2 if Hf(G) is defined. In this paper,

we prove thatHf(Pn)= n−3 if n 4. Since δ(Pn)=

n− 1, the fault hamiltonicity of the pancake graph Pn is optimal if n 4. In particular, the fact that Pn− F is hamiltonian when F consists of only a single vertex implies the existence of a cycle of length n! − 1. As a simple consequence, we improve the result in [10].

To discuss the fault hamiltonicity of the pancake graphs, we need the concept of fault hamiltonian connectivity. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two vertices of G. All hamiltonian connected graphs except K1and

K2are hamiltonian. A graph G is k-fault hamiltonian

connected if G− F remains hamiltonian connected

for every F ⊂ V (G) ∪ E(G) with |F | k. The

fault hamiltonian connectivity, Hκ_f(G), is defined to be the maximum integer k such that G is k-fault hamiltonian connected if G is hamiltonian connected and undefined if otherwise. It can be checked that Hκ

f(G) δ(G)−3 if Hfκ(G) is defined and|V (G)|

4. In this paper, we prove that Hκ_f(Pn)= n − 4 if

n 4. Again, the fault hamiltonian connectivity of the pancake graph Pnis optimal if n 4.

2. Some properties of the pancake graph Pn

Let u= u1u2. . . un be any vertex of the pancake

graph Pn. We say that ui is the ith coordinate of u,

denoted by (u)i, for 1 i n. By the definition of

Pn, there is exactly one neighbor v of u such that u

and v are adjacent through an i-dimensional edge with 2 i n. For this reason, we use i(u) to denote the unique i-neighbor of u. Obviously, i(i(u))= u. For 1 i n, let Pn(i) denote the subgraph of Pninduced

by those vertices u with (u)n= i. Obviously, Pncan

be decomposed into n subgraph Pn(i), 1 i n,

such that each Pn(i) is isomorphic to Pn−1. Thus,

the pancake graph can be constructed recursively. Let I ⊆ n. We use Pn(I ) to denote the subgraph of Pn

induced by _i_∈IV (Pn(i)). For 1 i = j n, we

use Ei,j to denote the set of edges between Pn(i) and

Pn(j ). Obviously, we have the following lemmas.

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Lemma 2. Let u and v be two distinct vertices of

Pn with (u)n = (v)n such that d(u, v) 2. Then

(n(u))n = (n(v))n.

Let F⊂ V (Pn)∪E(Pn) be any faulty set of Pn. An

edge (u, v) is F -fault if (u, v)∈ F , u ∈ F , or v ∈ F ; and (u, v) is F -fault free if (u, v) is not F -fault. Let H= (V, E) be a subgraph of Pn. We use F (H ) to

denote the set (V∪ E)∩ F .

Lemma 3. Assume that n 5 and I = {i1, i2, . . . , im}

is a subset of n such that |I| = m 2. Let F ⊂ V (Pn)∪ E(Pn) be any faulty set such that Pn(i)− F

is hamiltonian connected for any i∈ I and there are at least three F -fault free edges in Eij,ij+1 _{for any}

1 j < m. Then there exists a hamiltonian path of Pn(I )− F joining any two vertices u and v with u∈ V (Pn(i1))− F and v ∈ V (Pn(im))− F .

Proof. Let u1= u and vm= v. Since there are at least

three F -fault free edges in Eij,ij+1_{for any 1}_{j < m,} we can easily choose two different vertices uij _and vij+1 _{in P}_n_(i_j_{) such that (v}ij_{, u}ij+1_{) is F -fault free.} Obviously, uij = vij_{. Since P}

n(ij)− F is hamiltonian

connected for all ij ∈ I, there is a hamiltonian path

Qj of Pn(ij) joining uij and vij. Thus,ui1, Q1, vi1, ui2_{, Q}

2, . . . , vim−1, uim, Qm, vim forms a hamiltonian

path of Pn(I )− F joining u and v. The lemma is

proved. ✷

3. Fault hamiltonicity and fault hamiltonian connectivity of the pancake graphs

Lemma 4. P4is 1-fault hamiltonian and hamiltonian

connected.

Proof. To prove P4is 1-fault hamiltonian we need to

prove P4− F is hamiltonian for any F = {f } with

f ∈ V (P4)∪ E(P4). Without loss of generality, we may assume that f = 1234 if f is a vertex, or f ∈

{(1234, 2134), (1234, 3214), (1234, 4321)} if f is an

edge. The corresponding hamiltonian cycles of P4−F

are listed in Table 1.

To prove P4 is hamiltonian connected, we have

to find the hamiltonian path joining any two vertices

u and v. By the symmetric property of P4, we may

assume that u= 1234 and v is any vertex in V (P4)−

{u}. The corresponding hamiltonian paths are listed in

Table 2. Thus, the lemma is proved. ✷

Lemma 5. Suppose that n 5. If Pn₋₁is (n− 4)-fault hamiltonian and (n− 5)-fault hamiltonian connected, then Pnis (n− 3)-fault hamiltonian.

Table 1

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Proof. Assume that F is any faulty set of Pn with |F | n − 3. Since n 5, |Ei,j _{− F | (n − 2)! −}

(n− 3) 4 for any 1 i, j n. Thus, there are at least four F -fault free edges between Pn(i) and Pn(j )

for any 1 i = j n. We may assume that

FPn(i1) F Pn(i2) ···F Pn(in). Case 1: |F (Pn(i1))| = n − 3. Thus, F ⊂ Pn(i1).

Choose any element f in F (Pn(i1)). By the

assump-tion of this lemma, there exists a hamiltonian cy-cle Q of Pn(i1)− F + {f }. We may write Q as u, Q1, v, f, u where f= f if f is incident with

Q or f is any edge of Q if otherwise. Obviously,

d(u, v) 2. By Lemma 2, (n(u))n = (n(v))n. Then

by Lemma 3, there exists a hamiltonian path Q2

of Pn(n − {i1}) joining n(u) and n(v). Then u,

n(u), Q2, n(v), v, Q1, u forms a hamiltonian cycle of

Pn− F .

Case 2:|F (Pn(i1))| = n−4. Thus, |F −F (Pn(i1))| 1. Hence, there exists an index i2such that

FPn(n − {i1, i2}) =0.

Since|Ei1,i2 − F | (n − 2)! − (n − 3), there exists

an F -fault free edge (u, v) in Ei1,i2 _{such that u}∈

V (Pn(i1)) and v∈ V (Pn(i2)). By the assumption of

this lemma, there exists a hamiltonian cycle C1 of

Pn(i1)− F and there exists a hamiltonian cycle C2

of Pn(i2)− F . We may write C1 as u, w, Q1, z, u

and C2 as v, y, Q2, x, v. Since d(x, y) 2 and

d(w, z) 2, by Lemma 2

n(x)_n =n(y)_n and n(w)_n =n(z)_n. Thus, we can choose a vertex from x and y, say x, and we can choose a vertex from w and z, say w, such that

(n(w))n = (n(x))n and (w, n(w)) and (x, n(x)) are

F -fault free. By Lemma 3, there exists a hamiltonian path Q3 of Pn(n − {i1, i2}) − F joining n(w) and

n(x). Hence, u, v, y, Q2, x, n(x), Q3, n(w), w, Q1, u forms a hamiltonian cycle of Pn− F .

Case 3:|F (Pn(i1))| n−5. We can choose any F

-fault free edge (u, v) in Ei1,i2 _{such that u}∈ V (Pn_(i

1))

and v∈ V (Pn(i2)). By the assumption of this lemma,

any Pn(i)− F is hamiltonian connected for i ∈ n.

Then by Lemma 3, there exists a hamiltonian path Q1

of Pn− F joining u and v. Then u, Q1, v, u forms a

hamiltonian cycle of Pn− F . ✷

Lemma 6. Suppose that n 5. If Pn₋₁is (n− 4)-fault hamiltonian and (n− 5)-fault hamiltonian connected, then Pnis (n− 4)-fault hamiltonian connected.

Proof. Assume that F is any faulty set of Pn with |F | n − 4. Let u and v be any two arbitrary vertices

of Pn− F . We want to construct a hamiltonian path

of Pn− F joining u and v. Obviously, |Ei,j − F |

(n− 2)! − (n − 4) 5 for any 1 i, j n with n 5. Thus, there are at least five F -fault free edges between Pn(i) and Pn(j ) for any 1 i = j n. We assume

Case 1:|F (Pn(i1))| = n − 4. Hence, F ⊂ Pn(i1).

Subcase 1.1: (u)n = (v)n = i1. Choose any

el-ement f in F (Pn(i1)). By the assumption of this

lemma, there exists a hamiltonian path Q of Pn(i1)−

F + {f } joining u and v. We may write Q as

u, Q1, x, f, y, Q2, v where f = f if f is

inci-dent with Q or f is any edge of Q if otherwise.

Obviously, d(x, y) 2. By Lemma 2, (n(x))n =

(n(y))n. By Lemma 3, there exists a hamiltonian path

Q3 of Pn(n − {i1}) joining n(x) and n(y). Then u, Q1, x, n(x), Q3, n(y), y, Q2, v forms a

hamil-tonian path of Pn− F joining u to v.

Subcase 1.2: (u)n = i1 and (v)n = ij with j =

1. By the assumption of this lemma, there exists a hamiltonian cycle C1 of Pn(i1) − F . We may

write C1 as u, y, Q1, x, u. Since d(x, y) 2, by

Lemma 2 (n(x))n = (n(y))n. Thus, we can choose

a vertex from x and y, say x, such that (n(x))n =

(v)n. By Lemma 3, there exists a hamiltonian path

Q2 of Pn(n − {i1}) joining v and n(x). Then u, y, Q1, x, n(x), Q2, v forms a hamiltonian path of

Pn− F joining u to v.

Subcase 1.3: (u)n= (v)n= ij with j = 1. Since

there are at least five F -fault free edges in Ei1,ij_{, there}

exists an F -fault free edge (w, x) in Ei1,ij _{such that}

(w)n= i1, (x)n= ij, and x = v. By the assumption

of this lemma, there exists a hamiltonian cycle C1 of

Pn(i1)− F and a hamiltonian path Q1of Pn(ij)

join-ing u and v. We may write Q1asu, Q2, x, y, Q3, v

and C1 as w, z, Q4, z, w. Since d(z, z) 2, by

Lemma 2 (n(z))n = (n(z))n. Thus, we can choose

a vertex from z and z, say z, such that (n(z))n =

(n(y))n. By Lemma 3, there exists a hamiltonian

path Q5 of Pn(n − {i1, ij}) joining n(y) and n(z).

Thenu, Q2, x, w, z, Q4, z, n(z), Q5, n(y), y, Q3,v

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Subcase 1.4: (u)n = ij and (v)n= ik with ij, ik

and i1 are all distinct. Since there are at least five

F -fault free edges in Ei1,ij_{, there exists an F -fault free}

edge (w, x) in Ei1,ij _{such that (w)}

n= i1, (x)n= ij,

and x = u. By the assumption of this lemma, there exists a hamiltonian cycle C1 of Pn(i1)− F and a

hamiltonian path Q1 of Pn(ij) joining u and x. We

may write C1 as w, z, Q2, y, w. Since d(y, z)

2, by Lemma 2 (n(y))n = (n(z))n. Thus, we can

choose a vertex from y and z, say y, such that (n(y))n = (v)n. By Lemma 3, there exists a

hamil-tonian path Q3of Pn(n − {i1, ij} joining n(y) and v. Thus, u, Q1, x, w, z, Q2, y, n(y), Q3, v forms a

hamiltonian path of Pn− F joining u and v.

Case 2:|F (Pn(i1))| n − 5. By the assumption of

this lemma, Pn(i) is hamiltonian connected for every

1 i n.

Subcase 2.1: (u)n= (v)n= ij. By the assumption

of this lemma, there exists a hamiltonian path Q1 of

Pn(ij)− F joining u to v. We claim that there exists

an F -fault free edge (x, y) in Q1such that (x, n(x))

V (F (Pn(ij)))|/2 (n − 1)!/2 > n − 3 for n 5.

However,|F | n − 3. We get a contradiction. Hence, such edge exists.

Write Q1asu, Q2, x, y, Q3, v. Since d(x, y) =

1, by Lemma 2 (n(x))n = (n(y))n. By Lemma 3, there

exists a hamiltonian path Q4of Pn(n − {ij}) joining

n(x) and n(y). Then u, Q2, x, n(x), Q4, n(y), y,

Q3, v forms a hamiltonian path of Pn− F joining u

and v.

Subcase 2.2: (u)n = (v)n. By Lemma 3, there exists

a hamiltonian path of Pn− F joining u and v. ✷

Theorem 1. Let n be a positive integers with n 4.

Then Pnis (n− 3)-fault hamiltonian and (n − 4)-fault

hamiltonian connected.

Proof. We prove this theorem by induction. The

induction base, n= 4, is proved in Lemma 4. With Lemmas 5 and 6, we prove the induction step. ✷

Since δ(Pn)= n − 1, we have the following

corol-lary.

Corollary 1. Hf(Pn)= n − 3 and Hκ_f(Pn)= n − 4

for any positive integer n with n 4.

Acknowledgements

The authors would like to thank the anonymous referees for their comments and suggestions. These comments and suggestions are helpful for improve the quality of this paper.

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