Fault- -free Hamiltonian Cycles in Pancake free Hamiltonian Cycles - - PowerPoint PPT Presentation

Fault Fault-

• free Hamiltonian Cycles in Pancake

free Hamiltonian Cycles in Pancake Graphs with Conditional Edge Faults Graphs with Conditional Edge Faults

Ping Ping-

• Ying Tsai(

Ying Tsai(蔡秉穎 蔡秉穎), ), Jung Jung-

• Sheng

Sheng Fu( Fu(傅榮勝 傅榮勝), ), Gen Gen-

• Huey Chen(

Huey Chen(陳健輝 陳健輝) )

Outline Outline

• Hamiltonian problems
• Fault tolerant problems
• Pancake graphs
• Problem and previous results
• Main result and proof idea

Hamiltonian Problems Hamiltonian Problems

• Ring (or linear array) embedding in networks
• Cycle (or path) embedding in graphs
• A cycle (or path) in a graph G is called a

Hamiltonian cycle (or Hamiltonian path ) if it contains every vertex of G exactly once.

• G is called Hamiltonian-connected if every

two vertices of G are connected by a Hamiltonian path.

Fault Tolerant Problems Fault Tolerant Problems

• Vertex faults (node faults).
• Edge faults (link faults).
• After some faults occur,

does the network still work?

Fault Models Fault Models

• Random fault model: assumed that the faults

might

• ccur

everywhere without any restriction.

• Conditional fault model: assumed that the

distribution of faults must satisfy some properties.

• It is more difficult to solve problems under the

conditional fault model than the random fault model.

Random Fault Random Fault – – Cycle Cycle Embedding

Embedding

No cycle can pass this vertex

• No Hamiltonian cycle.

Conditional Fault Conditional Fault – – Cycle Cycle Embedding

Embedding

Each vertex has two healthy edges

• No Hamiltonian cycle.

Pancake Graphs Pancake Graphs

• A n-dimensional pancake graph, denoted by

℘n , has the vertex set V(℘n) = {a1a2…an | a1a2…an is a permutation of 1, 2, …, n}, and edge set E(℘n) = (a1a2⋅⋅⋅an, b1b2⋅⋅⋅bn) | a1a2⋅⋅⋅ak = bkbk−1⋅⋅⋅b1 and ak+1ak+2⋅⋅⋅an = bk+1bk+2⋅⋅⋅bn for some 2 ≤ k ≤ n }.

12 21 123 321 231 132 213 312

℘2 ℘3

℘4

1234 3214 4321 2314 4132 1432 1423 2341 3241 4123 2143 3412 4312 1243 2413 3142 4231 1324 2134 3421 1342 3124 2431 4213

Some Properties of Some Properties of ℘ ℘n

n

• ℘n is regular of degree n−1.
• ℘n has n! vertices and (n−1)n!/2 edges.
• The girth of ℘n is 6, where n ≥ 3.
• ℘n belongs to the class of Cayley graphs.
• ℘n is vertex symmetric, but not edge symmetric.
• ℘n is Hamiltonian-connected.
• ℘n is a recursive structure.

℘4

1234 3214 4321 2314 4132 1432 1423 2341 3241 4123 2143 3412 4312 1243 2413 3142 4231 1324 2134 3421 1342 3124 2431 4213 ) 1 ( 4

℘

) 3 ( 4

℘

) 2 ( 4

℘

) 4 ( 4

℘

A famous problem of A famous problem of ℘ ℘n

n

• The pancake graph is named from the famous

“pancake problem” whose answer is exactly the diameter of the corresponding pancake graph 1.

• The diameter of ℘n is bounded above by 3(n + 1)/2.

It is still an open problem to compute the exact diameter of ℘n 2.

• 1 W. H. Gates and C. H. Papadimitriou, “Bounds for sorting by prefix reversal,”

Discrete Mathematics, vol. 27, pp. 47-57, 1979.

• 2 M. H. Heydari and I. H. Sudborough, “On the diameter of the pancake

network,” Journal of Algorithms, vol. 25, pp. 67-94, 1997.

Our Problem Our Problem

• Fault component: edge faults only.
• Fault model: conditional fault model.
• Assumption: each vertex has at least two non-

faulty edges.

• How many edge faults can ℘n tolerate while

retaining a fault-free Hamiltonian cycle? (n ≥ 4)

• This is the first result on the fault tolerance of

the pancake graph under the conditional fault model.

Previous results on Previous results on ℘ ℘n

n

§

• C. N. Hung, H. C. Hsu, K. Y. Liang and L. H. Hsu, “Ring embedding in faulty

pancake graphs,” Information Processing Letters, vol. 86, pp. 271-275, 2003. ∗

• P. Y. Tsai, “Edge-fault-tolerant path/cycle embedding on some Cayley graphs,”

Ph.D. Thesis, National Taiwan University, Taipei, Taiwan, to appear (2008).

k ≤ n – 4 * k ≤ n – 4 §

k-edge-fault-tolerant Hamiltonian-connected

k ≤ n – 3 §

k-edge-fault-tolerant Hamiltonian Conditional fault model Random fault model model property

k ≤ 2n − 7

Lemmas Lemmas

• Lemma 1: | (℘n)| = (n − 2)! for all p, q ∈{1, 2, ... ,

n} and p ≠ q, where n ≥ 3.

• Lemma 2: ℘n − F is Hamiltonian if |F| ≤ n − 3, and

Hamiltonian-connected if |F| ≤ n − 4, where n ≥ 4 (F denotes a set of edge faults in ℘n) .

,

%

, p q

E %

Lemmas Lemmas

• Lemma 3: Suppose that u, v ∈ V(℘n) and 〈u〉n ≠ 〈v〉n,

where n ≥ 5. For any I ⊆ {1, 2, …, n} and |I| ≥ 2, there exists a Hamiltonian path from u to v in − F provided the following two conditions hold: (C1) | (℘n) − F| ≥ 3 for all i, j ∈ I and i ≠ j; (C2) − F is Hamiltonian-connected for all r ∈ I.

• Lemma 4: Suppose that u, v ∈ V( ) and u ≠ v, where

r ∈ {1, 2, ⋅⋅⋅, n} and n ≥ 4. If du,v ≤ 2, then 〈N(n)(u)〉n ≠ 〈N(n)(v)〉n, where du,v is the distance between u and v.

( ) r n

℘

, i j

%

, i j

E %

I n

℘

( ) r n

℘

Lemmas Lemmas

• Lemma 5: Suppose that e1, e2 ∈ E(℘4) and e1 ≠ e2.

There exists a Hamiltonian cycle in ℘4 − {e2} that contains e1.

• Lemma 6: Suppose that s, t ∈ V(℘n), s ≠ t, and 〈s〉1 =

〈t〉1, where n ≥ 4. For every (x, y) ∈ E(℘n) with {x, y} ∩ {s, t} = ∅, there exists a Hamiltonian path from s to t in ℘n that contains (x, y).

Main Result Main Result

• Theorem: ℘n − F is Hamiltonian provided

|F| ≤ 2n − 7 and δ (℘n − F) ≥ 2 , where n ≥ 4 (F denotes a set of edge faults in ℘n) .

Proof idea Proof idea

• The theorem holds for ℘4, which is assured by

Lemma 2 (2n − 7 = n − 3 as n = 4).

• Prove by induction on n.
• Suppose the theorem holds for ℘k, now we

construct a Hamiltonian cycle in ℘k+1 − F, where k ≥ 4 and |F| ≤ 2k − 5.

Proof idea Proof idea

• Assume that |E(

) ∩ F| ≥ |E( ) ∩ F| ≥⋅⋅⋅ ≥ |E( ) ∩ F| .

• By Lemma 1, we have | (℘k+1)| = (k − 1)! ≥

2k − 2 ≥ |F| + 3, i.e., | (℘k+1) − F| ≥ 3 for all p, q ∈ {1, 2, ⋅⋅⋅, k + 1} and p ≠ q.

• Four cases are discussed.

1 + +

℘

( 1) 1 k k + +

℘

1 +

℘

1 +

℘

( ) 1 k k+

℘

(1) 1 k+

℘

, p q

E %

, p q

E %

Case 1 Case 1

• |E( ) ∩ F| ≤ k − 4.
• The induction hypothesis assures a Hamiltonian cycle

C in − F.

• An edge (u1, v1) can be determined from C so that

there exist (v1, u2), (u1, vk+1) ∈ E(k+1)(℘k+1) − F with u2, vk+1 ∈ V( ), where I = {1, 2, ⋅⋅⋅, k}.

• Lemma 2 assures that − F is Hamiltonian-

connected for all 1 ≤ j ≤ k.

• By Lemma 3, a Hamiltonian path in − F exists.

( 1) 1 k k + +

℘

1

( ) 1

+

+

℘

k

k ( ) 1 +

℘ j

k 1 I k+

℘

1 I k+

℘

2

u

1 k

v +

1

v

1

u

( 1) 1 k k + +

℘

1 I k+

℘

Case 2 Case 2

• k − 3 ≤ |E( ) ∩ F| ≤ 2k − 7.
• Subcase 1: |E( ) ∩ F| ≤ k − 4.
• If δ( − F) ≥ 2, the Hamiltonian cycle can be
• btained by the construction method of Case 1.
• If δ( − F) = 1, we can construct the Hamiltonian

cycle by slightly modifying the construction method

• f Case 1 (use the same figure).

( 1) 1 k k + +

℘

( ) 1 k k+

℘

( 1) 1 k k + +

℘

( 1) 1 k k + +

℘

Case 2(continue) Case 2(continue)

• Subcase 2: |E( ) ∩ F| ≥ k − 3.
• We have |E( ) ∩ F| = k − 3 or k − 2, |E( ) ∩ F| =

k − 3 (hence, δ( − F) ≥ 2), and |E(k+1)(℘k+1) ∩ F| ≤ 1.

• First we consider the situation of |E( ) ∩ F| = k − 3,

which occurs only when k = 4.

• We have |E( ) ∩ F| = |E( ) ∩ F| = |E( ) ∩ F|

= 1, |E( ) ∩ F| = |E( ) ∩ F| = 0, and |E(5)(℘5) ∩ F| = 0.

• Assisted by Lemma 5.

( 1) 1 k k + +

℘

( ) 1 k k+

℘

( ) 1 k k+

℘

( ) 1 k k+

℘

( 1) 1 k k − +

℘

(5) 5

℘

(4) 5

℘

(3) 5

℘

(2) 5

℘

(1) 5

℘

5

u

5

v

4

u

4

v

3

u

3

v

1

v

1

u

2

u

2

v

(5) 5

℘

(4) 5

℘

(2) 5

℘

(3) 5

℘

(1) 5

℘

Case 2(continue) Case 2(continue)

• Now we consider the situation of |E( ) ∩ F| ≤ k − 4.
• When δ( − F) ≥ 2.
• When δ( − F) = 1.

( 1) 1 k k + +

℘

( 1) 1 k k − +

℘

( 1) 1 k k + +

℘

3

u

1 k

v +

1

v

1

u

2

u

2

v

( 1) 1 k k + +

℘

1 I k+

℘

( ) 1 k k+

℘

3

u

3

v

...

1 k

u +

1 k

v +

1

v

1

u

2

u

2

v

4

u

4

v

5

u

5

v

( ) 1 k α +

℘

( ) 1 k τ +

℘

( ) 1 k β +

℘

( ) 1 k γ +

℘

( 1) 1 k k + +

℘

( ) 1 k k+

℘

3

u

3

v

...

1 k

u +

1 k

v +

1

v

1

u

2

u

2

v

4

u

4

v

( ) ( ) 1 1 k k α τ + +

=

℘ ℘

( ) 1 k γ +

℘

( 1) 1 k k + +

℘

( ) 1 k k+

℘

( ) 1 k β +

℘

Case 3 Case 3

• |E( ) ∩ F| = 2k − 6.
• We have |E(k+1)(℘k+1) ∩ F| ≤ 1 and |E( ) ∩ F| ≤ 1

for all 1 ≤ j ≤ k.

• When k ≥ 5, the Hamiltonian cycle can be obtained by

slightly modifying the construction method of Case 1.

• When k = 4, if |E( ) ∩ F| = 0 for all 1 ≤ j ≤ 4, the

Hamiltonian cycle can be obtained all the same as the situation of k ≥ 5. Otherwise, the Hamiltonian cycle can be obtained similar to the construction method of Case 2.

( 1) 1 k k + +

℘

( ) 1 +

℘ j

k ) ( 5 j

℘

Case 4 Case 4

• |E( ) ∩ F| = 2k − 5.
• First, two edges (x, x'), (y, y') ∈ E( ) ∩ F are

determined so that {x, x'} ∩ {y, y'} = ∅ and δ( − (F − {(x, x'), (y, y')})) ≥ 2.

• The induction hypothesis assures a Hamiltonian cycle

C in − (F − {(x, x'), (y, y')}).

• If (x, x') or (y, y') is not contained in C, the Hamil-

tonian cycle can be obtained by the construction method of Case 1.

• Otherwise, five figures are considered.

( 1) 1 k k + +

℘

( 1) 1 k k + +

℘

( 1) 1 k k + +

℘

( 1) 1 k k + +

℘

...

4

v

4

u

1 k

v +

3

u

3

v

2

u

2

v

1 k

u + x x' y' y

( 1) 1 k k + +

℘

( ) 1 k α +

℘

( ) 1 k β +

℘

( ) 1 k γ +

℘

( ) 1 k τ +

℘

. . .

1 k

v +

3

u

3

v

2

u

2

v

1 k

u + x x' y' y

( 1) 1 k k + +

℘

( ) 1 k α +

℘

( ) 1 k τ +

℘

( ) 1 k β +

℘

x x' y' y

...

2

u

2

v

3

u

3

v

4

v

4

u

5

u

5

v w w'

( 1) 1 k k + +

℘

( ) 1 k α +

℘

( ) 1 k β +

℘

( ) 1 k σ +

℘

( ) 1 k γ +

℘

5

u

5

v

3

u

3

v

2

u

2

v x x' y' y w w'

( 1) 1 k k + +

℘

( ) 1 k α +

℘

( ) 1 k β +

℘

1 I k+

℘

x x' y' y

2

u

2

v

5

u

5

v

6

v

6

u z z' w w'

4

u

4

v

( 1) 1 k k + +

℘

( ) 1 k β +

℘

( ) 1 k α +

℘

( ) 1 k σ +

℘

1 I k+

℘

A distribution of 3n − 9 edge faults over an n-dimensional pancake graph. No fault-free Hamiltonian cycle can be found for this situation.

... ...

edge faults

3 n −

edge faults

3 n −

edge faults

3 n −

u v w x y z

...

Open problem Open problem

• There is an upper bound of 3n − 10 on the greatest

number of tolerable edge faults for the problem.

• It is an open problem to narrow down the gap

between 2n − 7 and 3n − 10.