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Conditional Fault Binpancyclicity of the Hypercube

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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  1. Conditional Fault Binpancyclicity of the Hypercube

  2. Definition • Bipancyclic • Conditional fault

  3. Result • The hypercube Qn is (2n-5)-conditional fault bipancyclic (at least 2 good neighbors).

  4. 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.

  5. 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.

  6. 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.

  7. Case 2.1 2n-6 <=n-3

  8. Case 2.2 <=n-3 <=2n-7

  9. The Conditional Fault-Diameter of the k-ary n-Cube Khaled Day and Abderezak Touzene

  10. 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.

  11. 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).

  12. k-ary n-cube Qkn • Qkn has kn nodes • Vertex set: • Edge set:

  13. k-ary n-cube Qkn (cont.) • Qkn can be partitioned into k disjoint k-ary (n-1)-cubes Qk,in-1.

  14. Proof • Lemma 2: CFD(Qn) >= n +2, for k>=4, n>=2.

  15. Proof (cont.) • Lemma 3: CFD(Qn) <= n +2, for k >= 4, n >= 2. • Proof: By induction on n (induction basis n=2).

  16. 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?

  17. 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.

  18. 跳到 Case 2 • Case 2. xm != ym for all m, 0 <= m < n. • Proof: 4n-2 nod disjoint paths.

  19. Counter example • X=000, Y=111

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