Embedding long paths in k-ary n-cubes with faulty nodes and links Yonghong Xiang Durham University

Embedding long paths in k-ary n-cubes with faulty nodes and links

Yonghong Xiang

Durham University

Yonghong Xiang: Embedding long paths in k-ary n-cubes with faulty nodes and links: Slide 2 of 20

Interconnection networks

• Parallel machines can be constructed in essentially two ways: – shared-memory machines

– distributed-memory machines.

• Technology restricts the incorporation of a large number of processors into shared-memory machines; but not so in distributed-memory machines.

• There are many design decisions to make when working with distributed-memory machines, one of which is the interconnection topology of the processors: and there are many trade-offs in the choice of topology– diameter

– degree

– symmetry

– connectivity, …


Some topologies: hypercubes

• The hypercube family of interconnection networks is probably the most popular.

• For n dimensional Hypercube Qn, nodes are named

{x : x {0, 1}n}and there is a link (x, y) if, and only if, x and y differ in exactly one bit position.

• Hypercubes can also be defined recursively.

• Some properties:– 2n nodes, n2n-1 links– diameter n– degree n– (n-1)-connected.

01

00 1 0

1 1

0 0 1

0 0 0

0 1 1

0 1 0

1 1 1

1 1 0

1 0 0

1 0 1

0 0 0 1

0 0 0 0

0 0 1 1

0 01 0

0 1 1 1

0 1 1 0

0 1 00

0 1 01 1 0 0 1

1 0 0 0

1 0 1 1

1 0 1 0

1 1 1 1

1 1 1 0

11 0 0

11 0 1


Some topologies: k-ary n-cubes

• The k-ary n-cube family of interconnection networks is also very popular.

• Nodes are named{x : x {0, 1, …, k-1}n}

and there is a link (x, y) if, and only if, x and y differ in exactly one bit position, i say, and in that position

|xi – yi| = 1 (mod k).• k-ary n-cubes can also be defined

recursively.• Some properties:

– kn nodes, nkn links– diameter nk/2– degree 2n– (2n-1)-connected.

• Advantage over hypercube? 2n nodes can be connected as a low-degree k-ary n-cube.


Definitions

• A graph is hamiltonian if it has a hamiltonian cycle. A graph G is hamiltonian connected if, for any two arbitrary vertices x and y in G, there is a hamiltonian path connecting x and y.

• Conditional fault assumption (CFA): each fault-free node is adjacent to at least two fault-free nodes.

• A graph G = (V0 V1, E) is bipartite if V0 V1 = and E {(a,b) | a V0, and b V1}

A bipartite graph G=(V0V1, E) is hamiltonian laceable if there is a hamiltonian path between any two vertices x and y which are in different partite sets.

• Let f=|F|, fe=|Fe|, fn=|Fn|.


Existing results – Hypercube

• The hypercube is bipartite graph [1].• Under CFA, for n 3 and fn 2n – 5, the authors find a fault-free

longest path between two arbitrary distinct fault-free nodes in different partite sets (resp. the same partite set) of length at least 2n – 2fn – 1 (resp. 2n – 2fn – 2) [2].

• Under CFA, n 3 and fe 2n – 5, the wounded Qn is hamiltonian laceable [3].

• In [4], the authors proved that the wounded Qn is hamiltonian laceable if fe n – 2. …

-------------[1] F. Leighton, Introduction to parallel algorithms and architecture: arrays, trees,

hypercubes, Morgan Kaufmann, San Mateo, 1992.[2] S. Hsieh, N. Chang, Optimal node-to-node path embedding in hypercubes with

conditional faults, to appear.[3] C. Tsai, Linear array and ring embeddings in conditional faulty hypercubes,

Theoretical Computer Science 314 (2004) 431-443.[4] C. Tsai, J. Tan, T. Liang, L. Hsu, Fault-tolerant hamiltonian laceability of

hypercubes, Information Processing Letters 83 (2002) 301-306.


Existing results – k-ary n-cube

• [1] shows that under CFA, if fe 4n – 5, then the wounded Qnk is

hamiltonian.• [2] shows that for odd integer k 3, if f 2n – 2, then the wounded

Qnk is hamiltonian, and if f 2n – 3, then the wounded Qn

k is hamiltonian connected.

-------------[1] Yaagoub A. Ashir, Iain A. Stewart, Fault-tolerant embeddings of Hamiltonian

circuits in k-ary n-cubes, SIAM Journal on Discrete Mathematics, Volume 15(2002), Number 3, pp. 317-328.

[2] M. Yang, J. Tan, L. Hsu, Hamiltonian circuit and linear array embeddings in faulty k-ary n-cubes, to appear in Journal of Parallel and Distributed Computing.


Our Result - 1

• Qnk is bipartite if and only if k is even. When k is even and there is a

faulty node, there exists neither a hamiltonian cycle nor a hamiltonian path between two vertices in different partite sets in a wounded Qn

k. Therefore, in paper [1], the authors supposed that k is an odd integer with k ≥ 3, and f 2n – 2. They constructed a hamiltonian cycle.

• The parity of a node in Qnk is the sum modulo 2 of the elements in the

n-tuple over {0, 1, . . . , k −1} representing the node.• Theorem 1 Let k ≥ 4 be even and let n ≥ 2. In a faulty k-ary n-cube Qn

k in which the number of node faults fn and the number of link faults fe are such that fn + fe ≤ 2n − 2, given any two healthy nodes s and e of Qn

k, there is a path from s to e of length at least kn − 2fn − 1 (resp. kn − 2fn − 2) if the nodes s and e have different (resp. the same) parities.

----------[1] M. Yang, J. Tan, L. Hsu, Hamiltonian circuit and linear array embeddings in

faulty k-ary n-cubes, to appear in Journal of Parallel and Distributed Computing.


Our Result - 2

• We first prove the base case for the proposed problem, that is we can find such a path in Q2

k with 2 faults. Then, by induction, we prove it for Qn

k with k 3. The most hard is to prove the base case.

• We consider Q2k as a k × k grid with wrap-around and we think of a

node vi,j as indexed by its row i and column j. Given two row indices i, j {0, 1, . . . , ∈ k − 1}, where j i, we define the row-torus rt(i, j) to be

the subgraph of Q2k induced by the nodes on rows i, i + 1, . . . , j, if i < j,

or rows i, i + 1, . . . , k − 1, 0, . . . , j, if j < i, but with all column links between nodes on row j and nodes on row i removed if i = j +1 or (i = 0 and j = k − 1).

• Lemma 2 Let k ≥ 4 be even and consider the row-torus rt(0, 1) in Q2k

where 1 node of the row-torus is faulty. If the pair of distinct, healthy nodes {s, e} of the row-torus is odd (resp. even) then there is a path in the row-torus joining s and e of length at least 2k − 3 (resp. 2k − 4).




Our result - 3

• Lemma 3 Let k ≥ 4 be even and consider the row-torus rt(0, p − 1) in Q2

k where 2 ≤ p ≤ k. If the pair of distinct nodes {s, e} of the row-torus is odd (resp. even) then there is a path in the row-torus joining s and e of length pk − 1 (resp. pk − 2). (No Faults)


Our result - 4

• Proposition 4 Consider the k-ary 2-cube Q2k where k ≥ 6 is even and

where 2 of the nodes are faulty. Let s and e be any two distinct, non-faulty nodes. There is a path of length at least k2 − 5 (resp. k2 − 6) from s to e if {s, e} is odd (resp. even).


• W.l.o.g. suppose that the two faulty nodes are f0 = v0,0 and f1 = vp,p′ with p 0. We begin by partitioning Q2

k into 3 or 4 row-torus.


Our result - 5


where 1 of the nodes is faulty. Let s and e be any two distinct, non-faulty nodes. There is a path of length at least k2 − 3 (resp. k2 − 4) from s to e if {s, e} is odd (resp. even).


Our result - 6


where there is 1 faulty link. Let s and e be any two distinct nodes in the row-torus rt(0, p − 1), where 2 ≤ p ≤ k. There is a path in rt(0, p − 1) from s to e of length pk − 1 (resp. pk − 2) if {s, e} is odd (resp. even).


Our result - 7


where 2 of the links are faulty. Let s and e be any two distinct nodes. There is a path of length k2 − 1 (resp. k2 − 2) from s to e if {s, e} is odd (resp. even).


Our result - 8


where there is a faulty node and a faulty link. Let s and e be any two distinct, non-faulty nodes. There is a path of length at least k2 − 3 (resp. k2 − 4) from s to e if {s, e} is odd (resp. even).

• Now, we have got the result for base case:

• For even k 4 and n = 2, in the faulty k-ary 2-cube Q2k in which fn + fe

2, given any two healthy nodes s and e of Q2k, there is a path from s

to e of length at least k2 − 2fn − 1 (resp. k2 − 2fn − 2) if the nodes s and e have different (resp. the same) parities.


Our result - Inductive step

• Suppose we can find a long path of length at least kn-1 − 2fn − 1 (resp. kn-1 − 2fn − 2) if the nodes s and e have different (resp. the same) parities while fn + fe ≤ 2(n – 1)− 2, given any two healthy nodes s and e of Qn–1

k.

• Partition the k-ary n-cube on the dimension which has the most faulty links.

• We look at different cases of how the faults are distributed in the k-ary (n-1)-cubes and use our inductive assumption to build a path of the required length.


Open questions and further work

1. Can we design distributed algorithm to implement finding a longest path in wounded k-ary n-cube?

2. Under CFA, can the k-ary n-cube tolerate more faults?

3. Can we find a Hamiltonian cycle or long path for some other interconnection networks with faults in: star graph, arrangement graph, and so on?

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Embedding long paths in k-ary n-cubes with faulty nodes and links Yonghong Xiang Durham University