Vertex fault tolerance for multiple spanning paths in hypercube

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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

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
• Conclusion
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

X

X

Q4:

|Fw|=0,|Fb|=2,|Kw|=4 ,|Kb|= 0

|Fw|=0,|Fb|=2,|Kw|=2 ,|Kb|= 0？

Q4

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

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

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

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

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

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

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 )