Abstract
Let F be a set of f <= 2n - 5 faulty nodes in an n-cube Q(n) such that every node of Q(n) still has at least two fault-free neighbors. Then we show that Q(n) - F contains a path of length at least 2(n) - 2f - 1 (respectively, 2(n) - 2f - 2) between any two nodes of odd (respectively, even) distance. Since the n-cube is bipartite, the path of length 2(n) - 2f - 1 (or 2(n) - 2f - 2) turns out to be the longest if all faulty nodes belong to the same partite set. As a contribution, our study improves upon the previous result presented by [J.-S. Fu, Longest fault-free paths in hypercubes with vertex faults, Information Sciences 176 (2006) 759-771] where only n - 2 faulty nodes are considered. (c) 2008 Elsevier Inc. All rights reserved.
Original language | English |
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Pages (from-to) | 667-681 |
Number of pages | 15 |
Journal | Information Sciences |
Volume | 179 |
Issue number | 5 |
DOIs | |
State | Published - 15 Feb 2009 |
Keywords
- Interconnection network; Hypercube; Fault tolerance; Conditional fault; Linear array; Path embedding