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Vertex fault tolerance for multiple spanning paths in hypercube

Vertex fault tolerance for multiple spanning paths in hypercube. Department of Computer Science and Information Engineering Dayeh University Professor: Chun-Nan Hung 洪春男 教授 Report by Guan-Yu Shi 施冠宇. Outline. Introduction Vertex fault tolerance for multiple spanning paths in hypercube

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Vertex fault tolerance for multiple spanning paths in hypercube

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  1. Vertex fault tolerance for multiple spanning paths in hypercube Department of Computer Science and Information Engineering Dayeh University Professor: Chun-Nan Hung 洪春男 教授 Report by Guan-Yu Shi 施冠宇

  2. Outline • Introduction • Vertex fault tolerance for multiple spanning paths in hypercube • Conclusion

  3. Introduction • A graph: Hypercube • Q1=K2 , Qn=Qn-1 X K2 • 2nvertices 100 101 01 000 00 001 0 1 111 110 10 11 010 011 Q1 Q2 Q3

  4. Q3 Q3 Bipartite graph

  5. Hamiltonian laceable and hyper-Hamiltonian laceable Q4

  6. Hamiltonian laceable and hyper-Hamiltonian laceable Q4

  7. Introduction(cont.)

  8. X X Q4: |Fw|=0,|Fb|=2,|Kw|=4 ,|Kb|= 0 |Fw|=0,|Fb|=2,|Kw|=2 ,|Kb|= 0？ Q4

  9. Introduction(cont.)

  10. Q8: |Fw|=|Fb|=0,|Kw|=|Kb|=4 Q7: |Fw|=|Fb|=0,|Kw|=|Kb|=3 |Fw|=1,|Fb|=0,|Kw|=2,|Kb|=4 t1 S2 S1 S4 t2 t4 t3 S3 t4能否直接連到Q12n-1? Q07 Q17

  11. Vertex fault tolerance for multiple spanning paths in hypercube

  12. Lemma 3 Case 1: |Fw| = 0,|F1| = 0 case 1.1: |K1b| = 0 case1.1.1 :|K1w| = 0 case1.1.2 :|K1w|  1 case 1.2: |K1b|  1, |K1w| = 0 case 1.3: |K1b|  1, |K1w|  1 case 1.3.1: |K11| = 0, |K1w|  |K1b| case 1.3.2: |K11| = 0, |K1b| > |K1w| and |K01w|  1 case 1.3.3: |K11| = 0, |K1b| > |K1w| and |K01w| = 0 case 1.3.4: |K11|  2, |K0w|  1 case 1.3.5: |K11|  2, |K0w| = 0 Case 2: |Fw| = 0, |F0b|  1, |F1b|  1 case 2.1: Kw  V0w or Kw  V1w case 2.2: |K0w|  1, |K1w|  1 Case 3: |Fb|  1, ,|Fw|  1 and (|F0| = 0 or |F1| = 0) case 3.1: |K1| = 0 case 3.2: |K1w| = 0 or |K1b| = 0 case 3.3: |K1w|  1, |K1b|  1 case 3.3.1: |K11| = 0 case 3,3.2: |K11|  2 Case 4: |Fb|  1, ,|Fw|  1, |F0|  1 ,|F1|  1 case 4.1: |K0b|+|F0b| = 0 or |K0w|+|F0w| = 0 or |K1b|+|F1b| = 0 or |K1w|+|F1w| = 0 case 4.2: |K0b|+|F0b|  1 or |K0w|+|F0w|  1 or |K1b|+|F1b|  1 or |K1w|+|F1w|  1

  13. Case 1: |Fw| = 0,|F1| = 0 case1.1: |K1b| = 0 case1.1.1 :|K1w| = 0 Q17: |Fw|=0,|Fb|=3,|Kw|=9 ,|Kb|=3 Q16: |Fw|=0,|Fb|=3,|Kw|=8 ,|Kb|=2 |Fw|=0,|Fb|=2,|Kw|=8 ,|Kb|=4 S4 S3 S2 S5 w S1 (w) t4 (t1) t1 S6 (z) z t5 t3 t2 y t6 (y) x x x x Q02n-2 Q12n-2

  14. Case 2: |Fw| = 0,|F0b|  1, ,|F1b|  1 case2.1: Kw V0wor Kw V1w Q17: |Fw|=0,|Fb|=3,|Kw|=9 ,|Kb|=3 Q16: |Fw|=0,|Fb|=3,|Kw|=8 ,|Kb|=2 |Fw|=0,|Fb|=2,|Kw|=8 ,|Kb|=4 |Fw|=0,|Fb|=1,|Kw|=5 ,|Kb|=3 t1 S2 S6 S5 (y1) y1 (S1) S1 y2 S3 t3 (y2) t6 t2 t5 t4 S4 x x x Q02n-2 Q12n-2

  15. Case 3: |Fb|  1, ,|Fw|  1 and (|F0| = 0 or |F1| = 0) case3.1: |K1| = 0 S2 Q17: |Fw|=1,|Fb|=2,|Kw|=7 ,|Kb|=5 Q16: |Fw|=1,|Fb|=2,|Kw|=6 ,|Kb|=4 |Fw|=0,|Fb|=1,|Kw|=8 ,|Kb|=6 t2 S5 S3 S4 S1 y1 (y1) t5 t4 t3 x1 t1 S6 y2 (y2) x t6 x2 x x Q02n-2 Q12n-2

  16. Case 4: |Fb|  1, ,|Fw|  1, |F0|  1 ,|F1|  1 case4.1: |K0b|+|F0b| = 0 or |K0w|+|F0w| = 0 or |K1b|+|F1b| = 0 or |K1w|+|F1w| = 0 t2 Q17: |Fw|=1,|Fb|=2,|Kw|=7 ,|Kb|=5 Q16: |Fw|=|Fb|=0,|Kw|=|Kb|=8 |Fw|=1,|Fb|=2,|Kw|=6 ,|Kb|=4 |Fw|=1,|Fb|=0,|Kw|=6 ,|Kb|=8 |Fw|=0,|Fb|=2,|Kw|=8 ,|Kb|=4 |Fw|=0,|Fb|=2,|Kw|=7 ,|Kb|=3 S2 t4 S3 t5 t3 S4 S5 S6 t1 t6 x (y) y S1 x x x Q02n-2 Q12n-2

  17. Lemma 4 Case 1: |K0b| = n or |K0w| = n case 1.1: |K0w| = n case1.1.1 :|K00bb|  1 case1.1.2 :|K00bb| = 0 case 1.2: |K01ww|  1 or |K11ww|  1 case 1.3: |K01ww| = 0 and |K1w| = 1 case 1.4: |K01ww| = 0 and |K11ww|  1 and |K1w|  2 Case 2: |K0b|  (n -1) and |K0w|  (n-1) case 2.1: |K11| > 0 case 2.2: |K11| = 0 and |K0|  (n+1) and |Kbb| > 0 case 2.3: |K11| = 0 and ( |K0| = n or |Kbb| = 0 )

  18. Conclusion

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