Fault-free Hamiltonian cycles in faulty arrangement graphs

Abstract

The arrangement graph A/sub n,k/, which is a generalization of the star graph (n-k=1), presents more flexibility than the star graph in adjusting the major design parameters: number of nodes, degree, and diameter. Previously, the arrangement graph has proved Hamiltonian. In this paper, we further show that the arrangement graph remains Hamiltonian even if it is faulty. Let |F/sub e/| and |F/sub v/| denote the numbers of edge faults and vertex faults, respectively. We show that A/sub n,k/ is Hamiltonian when 1) (k=2 and n-k/spl ges/4, or k/spl ges/3 and n-k/spl ges/4+[k/2]), and |F/sub e/|/spl les/k(n-k)-2, or 2) k/spl ges/2, n-k/spl ges/2+[k/2], and |F/sub e/|/spl les/k(n-k-3)-1, or 3) k/spl ges/2, n-k/spl ges/3, and |F/sub e/|/spl les/k, or 4) n-k/spl ges/3 and |F/sub v/|/spl les/n-3, or 5) n-k/spl ges/3 and |F/sub v/|+|F/sub e/|/spl les/k. Besides, for A/sub n,k/ with n-k=2, we construct a cycle of length at least 1) [n!/(n-k!)]-2 if |F/sub e/|/spl les/k-1, or 2) [n!/(n-k)!]-|F/sub v/|-2(k-1) if |F/sub v/|/spl les/k-1, or 3) [n!/(n-k)!]-|F/sub v/|-2(k-1) if |F/sub e/|+|F/sub v/|/spl les/k-1, where [n!/(n-k)!] is the number of nodes in A/sub n,k/.