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

Embedding long paths in k-ary n-cubes with faulty nodes and links. Yonghong Xiang Durham University. Interconnection networks. Parallel machines can be constructed in essentially two ways: shared-memory machines distributed-memory machines.

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

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

  2. 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, …

  3. 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.

  4. 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 nk/2 • degree 2n • (2n-1)-connected. • Advantage over hypercube? 2n nodes can be connected as a low-degree k-ary n-cube.

  5. 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=(V0V1, E) is hamiltonian laceableif 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|.

  6. 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.

  7. 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 Qnk 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.

  8. Our Result - 1 • Qnkis 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 Qnk. Therefore, in paper [1], the authors supposed that k is an odd integer with k ≥ 3, andf  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 Qnk inwhich the number of node faults fn and the number of link faults fe are such thatfn + fe ≤ 2n − 2, given any two healthy nodes s and e of Qnk, there is a path froms to e of length at least kn − 2fn − 1 (resp. kn − 2fn − 2) if the nodes s and e havedifferent (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.

  9. Our Result - 2 • We first prove the base case for the proposed problem, that is we can find such a path in Q2k with 2 faults. Then, by induction, we prove it for Qnk 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-torusrt(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).

  10. Our result - 3 • Lemma 3 Let k ≥ 4 be even and consider the row-torusrt(0, p − 1) in Q2k 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)

  11. 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).

  12. 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 Q2k into 3 or 4 row-torus.

  13. Our result - 5 • Proposition 5 Consider the k-ary 2-cube Q2k where k ≥ 6 is even and 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).

  14. Our result - 6 • Proposition 6 Consider the k-ary 2-cube Q2k where k ≥ 6 is even and 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).

  15. Our result - 7 • Proposition 7 Consider the k-ary 2-cube Q2k where k ≥ 6 is even and 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).

  16. Our result - 8 • Proposition 8 Consider the k-ary 2-cube Q2k where k ≥ 6 is even and 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  4andn= 2,in the faulty k-ary 2-cube Q2k inwhich fn + fe 2, given any two healthy nodes s and e of Q2k, there is a path froms to e of length at least k2 − 2fn − 1 (resp. k2 − 2fn − 2) if the nodes s and e havedifferent (resp. the same) parities.

  17. 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 havedifferent (resp. the same) parities while fn + fe ≤ 2(n– 1)− 2, given any two healthy nodes s and e of Qn–1k. • 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.

  18. Open questions and further work • Can we design distributed algorithm to implement finding a longest path in wounded k-ary n-cube? • Under CFA, can the k-ary n-cube tolerate more faults? • 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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