Department of Computer Science and Information Engineering

Dayeh UniversityProfessor: Chun-Nan Hung 洪春男 教

授Report by Guan-Yu Shi 施冠宇

OutlineIntroductionVertex fault tolerance for multiple

spanning paths in hypercubeConclusion

2

IntroductionA graph: Hypercube Q1=K2 , Qn=Qn-1 X K2

2n vertices

0 1

00 01

10 11Q1 Q2

000001

010 011

Q3

110

100 101

111

3

Bipartite graph

Q3Q3

4

Hamiltonian laceable and hyper-Hamiltonian laceable

Q45

Hamiltonian laceable and hyper-Hamiltonian laceable

Q46

Introduction(cont.)

7

Q4

X XQ4:|Fw|=0,|Fb|=2,|Kw|=4 ,|Kb|= 0

|Fw|=0,|Fb|=2,|Kw|=2 ,|Kb|= 0 ？

8

Introduction(cont.)

9

S1

t1

Q8:|Fw|=|Fb|=0,|Kw|=|Kb|=4

Q7:|Fw|=|Fb|=0,|Kw|=|Kb|=3|Fw|=1,|Fb|=0,|Kw|=2,|Kb|=4

Q07 Q1

7

t2t4

S3

S4

t3

S2

t4 能否直接連到 Q12n-1?

10

Vertex fault tolerance for multiple spanning paths in hypercube

11

12

Lemma 3Case 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, |K1

w| = 0case 1.3: |K1

b| 1, |K1w| 1

case 1.3.1: |K11| = 0, |K1w| |K1

b|case 1.3.2: |K11| = 0, |K1

b| > |K1w| and |K01

w| 1case 1.3.3: |K11| = 0, |K1

b| > |K1w| and |K01

w| = 0case 1.3.4: |K11| 2, |K0

w| 1case 1.3.5: |K11| 2, |K0

w| = 0Case 2: |Fw| = 0, |F0

b| 1, |F1b| 1

case 2.1: Kw V0w or Kw V1

w

case 2.2: |K0w| 1, |K1

w| 1Case 3: |Fb| 1, ,|Fw| 1 and (|F0| = 0 or |F1| = 0)

case 3.1: |K1| = 0case 3.2: |K1

w| = 0 or |K1b| = 0

case 3.3: |K1w| 1, |K1

b| 1case 3.3.1: |K11| = 0case 3,3.2: |K11| 2

Case 4: |Fb| 1, ,|Fw| 1, |F0| 1 ,|F1| 1 case 4.1: |K0

b|+|F0b| = 0 or |K0

w|+|F0w| = 0 or |K1

b|+|F1b| = 0 or |K1

w|+|F1w| = 0

case 4.2: |K0b|+|F0

b| 1 or |K0w|+|F0

w| 1 or |K1b|+|F1

b| 1 or |K1w|+|F1

w| 1

13

Case 1: |Fw| = 0,|F1| = 0case1.1: |K1

b| = 0case1.1.1 :|K1

w| = 0

xx x

S1

S3

S4

S2S5

t1

t4

t5t3

t6

t2

S6

x

y

w

z

(w)

(z)

(t1)

(y)

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

Q02n-2 Q1

2n-214

Case 2: |Fw| = 0,|F0b| 1, ,|F1

b| 1case2.1: Kw V0

w or Kw V1w

xx x

S1

S3

S4

S2

S5

t1

t4

t5

t3

t6 t2

S6

y1

(y1)

(y2)

(S1

)

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

Q02n-2 Q1

2n-2

y2

15

Case 3: |Fb| 1, ,|Fw| 1 and (|F0| = 0 or |F1| = 0)case3.1: |K1| = 0

x

S1

S3

S6

S2

S5

t2

t3t4

S4

y1(y1)

(y2)

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

Q02n-2 Q1

2n-2

y2

x

x

t5

t6

t1

x1

x2

16

Case 4: |Fb| 1, ,|Fw| 1, |F0| 1 ,|F1| 1 case4.1: |K0

b|+|F0b| = 0 or |K0

w|+|F0w| = 0 or

|K1b|+|F1

b| = 0 or |K1w|+|F1

w| = 0

x

S2

t2Q17:|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

Q02n-2 Q1

2n-2

xx x

S3

S4

S5

S1

t3

t6

t4

t5

t1S6

y (y)

17

18

Lemma 4

Case 1: |K0b| = n or |K0

w| = ncase 1.1: |K0

w| = n case1.1.1 :|K00

bb| 1case1.1.2 :|K00

bb| = 0case 1.2: |K01

ww| 1 or |K11ww| 1

case 1.3: |K01ww| = 0 and |K1

w| = 1case 1.4: |K01

ww| = 0 and |K11ww| 1 and |K1

w| 2

Case 2: |K0b| (n -1) and |K0

w| (n-1)case 2.1: |K11| > 0case 2.2: |K11| = 0 and |K0| (n+1) and |Kbb| > 0case 2.3: |K11| = 0 and ( |K0| = n or |Kbb| = 0 )

19

20

21

22

23

Conclusion

24