conditional fault binpancyclicity of the hypercube n.
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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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definition
Definition
  • Bipancyclic
  • Conditional fault
result
Result
  • The hypercube Qn is (2n-5)-conditional fault bipancyclic (at least 2 good neighbors).
proof
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
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.

slide6
會有兩種情況:
  • Case 2.1. 只切到一條faulty edge.
  • Case 2.2. 切到兩條以上faulty edges.
case 2 1
Case 2.1

2n-6

<=n-3

case 2 2
Case 2.2

<=n-3

<=2n-7

abstract
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
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 q k n
k-ary n-cube Qkn
  • Qkn has kn nodes
  • Vertex set:
  • Edge set:
k ary n cube q k n cont
k-ary n-cube Qkn (cont.)
  • Qkn can be partitioned into k disjoint k-ary (n-1)-cubes Qk,in-1.
proof1
Proof
  • Lemma 2: CFD(Qn) >= n +2, for k>=4, n>=2.
proof cont1
Proof (cont.)
  • Lemma 3: CFD(Qn) <= n +2, for k >= 4, n >= 2.
  • Proof: By induction on n (induction basis n=2).
proof cont2
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 cont3
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
跳到 Case 2
  • Case 2. xm != ym for all m, 0 <= m < n.
  • Proof: 4n-2 nod disjoint paths.
counter example
Counter example
  • X=000, Y=111