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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 f v denote the number of faulty vertices in Q n .

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cycles embedding in hypercubes with node failures

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
Outline
  • Introduction
  • Preliminaries
  • Results
  • Conclusions
introduction
Introduction
  • Let fvdenote 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 Qnlies on a fault-free cycle of every even length from 4 to 2n − 2fvinclusive even if fv≦n−2. Our results are optimal.
slide6
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 Q0n−1 and Q1n−1 such that u ∈ V (Q0n−1) and v ∈ V (Q1n−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.
slide7
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 Qn −F lies on a fault-free k-cycle where k = 4, 6, . . . , 8− 2|F|.
slide8
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)

slide9
Lemma 5. Let n≧3 be an integer and Qn 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.
slide10
Proof:

Since Qnis 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.

Qnis partitioned along dimension j into two (n − 1)-cubes, denoted by Q0n−1 and Q1n−1, respectively. Hence e is a crossing edge between Q0n−1 and Q1n−1.

slide11
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 Q0n−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 Q0n−1 which is of length containing from 4 to 2 n−1 − 2 inclusive.
slide12
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) ≦2n −2.
results
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.
slide14
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 (Qin−1)| for i = 0, 1, and thus, fv= f0+f1.Consequently, f0≦n−3 and f1≦n−3.
slide15
we establish every even k-cycle containing e where 4≦k≦2n − 2fv. We check two possible distributions of the fault-free edge e.
slide16
Case 1: e ∈ E(Q0n−1)∪E(Q1n−1), i.e., e lies on Q0n−1 or Q1n−1. We only consider that e ∈ E(Q0n−1) (the discussion of e ∈ E(Q1n−1) is the same).
  • Therefore, the cycle C is of length from 2n−1 −2f0 +2 to 2n−2f inclusive.
slide17
Case 2: e  E(Q0n−1)∪E(Q1n−1), i.e., e is a crossing edge between Q0n−1 and Q1n−1.
  • Therefore, l(C) = l(W0)+l(W1)+2. This implies that 8≦ l(C) ≦2n− 2f .
slide18
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
Conclusions
  • (1) For fe+ fv ≦n − 1, every fault-free edge of Qnlies on a fault-free cycle of every even length from 6 to 2n− 2fvinclusive 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− 2fvif fe≦n − 2 and fe+ fv≦ 2n− 4.