Cycles embedding in hypercubes with node failures

Information Processing Letters

作者； Chang-Hsiung Tsai

老師：洪春男 學生 : ：林雨淳

Outline

Introduction Preliminaries Results Conclusions

Introduction

Let fv denote the number of faulty vertices in Qn.

For n 3, in this paper, we prove that every ≧fault-free edge and fault-free vertex of Qn lies on a fault-free cycle of every even length from 4 to 2n − 2fv inclusive even if fv≦n−2. Our results are optimal.

Preliminaries

Lemma 1. Let u and v be two distinct vertices of an n-cube, denoted by Qn. Then, there is a partition which can be partition Qn into two (n − 1)-cubes, denoted by Q0

n−1 and Q1n−1 such

that u V (Q∈ 0n−1) and v V (Q∈ 1

n−1).

Lemma 2. Let e be any edge of Qn for n 2. ≧Then, there are n − 1 cycles with length four contained e in common.

Lemma 3. (See [6].) For n 2, every edge of ≧Qn lies on a cycle of every even length from 4 to 2n.

Lemma 4. For any subset F of V (Q3) with |F| 1, every edge of Q≦ n −F lies on a fault-free

k-cycle where k = 4, 6, . . . , 8− 2|F|.

Q3 壞掉點 000, there are nine fault-free edges such as (001, 011), (001,101), (010, 011), (010, 110), (011, 111), (100, 110),(100, 101), (101, 111), and (111, 110).

有 three 4-cycles （ 100, 110, 111, 101, 100 ） , （ 010,110, 111, 011, 010 ） , and （ 101, 111, 011, 001,101 ）

有 Two 6-cycles （ 100, 110,111, 011, 001, 101, 100 ） and （ 100, 110, 010, 011, 111,101, 100 ）

Lemma 5. Let n 3 be an integer and Q≧ n has exactly one faulty vertex. Then, every fault-free edge of Qn lies on a fault-free cycle of every even length from 4 to 2n−2 inclusive.

Proof ：Since Qn is vertex-symmetric , we may assume that the faulty vertex is w = 00 . . . 0. Let e, denoted by (u, v), be a fault-free edge of dimension j , i.e., u = v(j ) for 0 ≦ j≦n−1.

Qn is partitioned along dimension j into two (n − 1)-cubes, denoted by Q0

n−1 and Q1n−1,

respectively. Hence e is a crossing edge between Q0

n−1 and Q1n−1.

Let （ u, u(i), v(i), v,u ） be a fault-free 4-cycle including (u, v) for some 0≦i ≠j≦n−1. Obviously, u and u(i) (respectively, v and v(i)) are adjacent in Q0

n−1 (respectively, Q1n−1).

Since Q0n−1 contains a faulty vertex and n 4, ≧

by induction hypothesis, the edge (u, u (i)) lies on a fault-free cycle in Q0

n−1 which is of length containing from 4 to 2 n−1 − 2 inclusive.

there exists a path P[u, u(i) ] in Q0n−1 joining u

and u(i) where l(P) = 1, 3, 5, . . . , 2n−1 − 3. By Lemma 3, there exists a path S[v(i) , v] in

Q1n−1 joining v(i) and v where l(S) = 1, 3,

5, . . . , 2n−1 −1. C = （ v ,u ,P[u, u(i) ] ,u(i) , v(i) ,S[v(i), v],

v ） l(C) = l(P) + l(S) + 2 Therefore, the edge (u, v) lies on the cycle C and 4 l(C) 2≦ ≦ n −2.

Results

Theorem 1. Assume that n 3. For any ≧subset F of V (Qn) with |F| = fv n − 2, every ≦edge of Qn − F lies on a cycle of every even length from 4 to 2n − 2fv inclusive.

Proof ：By Lemmas 3 and 5, the theorem holds for fv 1. Thus, we only consider the case of 2 ≦ ≦fv n− 2.≦

Let w and z are two distinct faulty vertices. 在不同的 (n − 1)-cubes 上

Let fi = |F ∩V (Qi n−1)| for i = 0, 1, and thus, fv = f0+f1.Consequently, f0≦n−3 and f1≦n−3.

we establish every even k-cycle containing e where 4 k 2≦ ≦ n − 2fv. We check two possible distributions of the fault-free edge e.

Case 1: e E(Q∈ 0n−1) E(Q∪ 1

n−1), i.e., e lies on Q0

n−1 or Q1n−1. We only consider that e ∈

E(Q0n−1) (the discussion of e E(Q∈ 1

n−1) is the same).

Therefore, the cycle C is of length from 2n−1 −2f0 +2 to 2n −2f inclusive.

Case 2: e E(Q0n−1) E(Q∪ 1

n−1), i.e., e is a crossing edge between Q0

n−1 and Q1n−1.

Therefore, l(C) = l(W0)+l(W1)+2. This implies that 8 ≦ l(C) 2≦ n − 2f .

Corollary 1. Assume that n 3. For any ≧subset F of V (Qn) with |F| n − 2, every ≦vertex of Qn − F lies on a fault-free cycle of every even length from 4 to 2n −2|F| inclusive.

Conclusions

(1) For fe + fv ≦n − 1, every fault-free edge of Qn lies on a fault-free cycle of every even length from 6 to 2n − 2fv inclusive provided n 4 and every fault-free vertex is incident ≧with at least two fault-free edges.

(2) For n 5, ≧ Qn exists a fault-free cycle of every even length from 4 to 2n − 2fv if fe≦n − 2 and fe + fv ≦ 2n− 4.