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. 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, … The types of graph occurring as interconnection topologies in parallel machines usually occur in recursively decomposable families where the individual members are of relatively small diameter (to aid message transit times) of relatively low degree (to speed local communications) highly symmetric (to aid programmers and communication) highly connected (so as to tolerate faults), …

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.

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.

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 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 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.

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 Qnk. 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 Qnk 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 Qnk, 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 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-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 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)

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 Q2k into 3 or 4 row-torus.

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).

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).

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).

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  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–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.

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?