Conditional Fault Binpancyclicity of the Hypercube

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# Conditional Fault Binpancyclicity of the Hypercube - PowerPoint PPT Presentation

Conditional Fault Binpancyclicity of the Hypercube. Definition. Bipancyclic Conditional fault. Result. The hypercube Q n is (2n-5)-conditional fault bipancyclic (at least 2 good neighbors). Proof. Case 1. 4~2 n-1 . Proof: 找一條 faulty edge 切下去 , 使得兩個 subcube 都有至少兩個 good neighbors.

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## Conditional Fault Binpancyclicity of the Hypercube

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### Conditional Fault Binpancyclicity of the Hypercube

Definition
• Bipancyclic
• Conditional fault
Result
• The hypercube Qn is (2n-5)-conditional fault bipancyclic (at least 2 good neighbors).
Proof
• Case 1. 4~2n-1.

Proof: 找一條faulty edge切下去, 使得兩個subcube都有至少兩個good neighbors.

Proof (cont.)
• Case 2. 2n+1~2n.

Proof: 如果有點的周圍有(n-2)個faults, 則在其某個faulty edge切下去, 只有一種情況會是兩個點周圍都有(n-2)個faults, 不然就隨便找個faulty edge 切下去, 切完後兩個subcubes的每個點一定都有兩個good neighbors.

• Case 2.1. 只切到一條faulty edge.
• Case 2.2. 切到兩條以上faulty edges.
Case 2.1

2n-6

<=n-3

Case 2.2

<=n-3

<=2n-7

### The Conditional Fault-Diameter of the k-ary n-Cube

Khaled Day and Abderezak Touzene

Abstract
• Node connectivity C(Qkn)=2n.
• Conditional node connectivity CC(Qkn)=4n-2.
• In this paer, conditional fault diameter CFD(Qkn)= diameter D(Qkn)+2.
Previous results
• CC(Qn)=2n-2 (Esfahanian, 1989).
• CFD(Qn)=D(Qn)+2 (Latifi, 1993).
• CFD(Sn)=D(Sn)+2 (Y. Rouskov, S. Latifi, and P. K. Srimani, 1996).
k-ary n-cube Qkn
• Qkn has kn nodes
• Vertex set:
• Edge set:
k-ary n-cube Qkn (cont.)
• Qkn can be partitioned into k disjoint k-ary (n-1)-cubes Qk,in-1.
Proof
• Lemma 2: CFD(Qn) >= n +2, for k>=4, n>=2.
Proof (cont.)
• Lemma 3: CFD(Qn) <= n +2, for k >= 4, n >= 2.
• Proof: By induction on n (induction basis n=2).
Proof (cont.)
• Case 1. 兩點至少有一digit一樣
• Case1.1 If Qk,in-1 has at most 4(n-1)-3=4n-7 faults.
• Proof: By induction
• 滿足1g?
Proof (cont.)
• Case 1.2 If Qk,in-1 has more than 4n-7 faults.
• Case 1.2.1. If each of X and Y has all its 2n-2 neighbors inside Qk,in-1 faulty.
• Proof: There exist at most one faulty node outside Qk,in-1.

• Case 2. xm != ym for all m, 0 <= m < n.
• Proof: 4n-2 nod disjoint paths.
Counter example
• X=000, Y=111