Fault-free Hamiltonian cycles in faulty arrangement graphs

The arrangement graph A_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_e| and |F_v| denote the numbers of edge faults and vertex faults, respectively. We show that A_n,k is Hamiltonian when 1) (k=2 and n-k⩾4, or k⩾3 and n-k⩾4+[k/2]), and |F_e|⩽k(n-k)-2, or 2) k⩾2, n-k⩾2+[k/2], and |F_e|⩽k(n-k-3)-1, or 3) k⩾2, n-k⩾3, and |F_e |⩽k, or 4) n-k⩾3 and |F_v|⩽n-3, or 5) n-k⩾3 and |F_v|+|F_e|⩽k. Besides, for A_n,k with n-k=2, we construct a cycle of length at least 1) [n!/(n-k!)]-2 if |F_e|⩽k-1, or 2) [n!/(n-k)!]-|F_v |-2(k-1) if |F_v|⩽k-1, or 3) [n!/(n-k)!]-|F_v |-2(k-1) if |F_e|+|F_v|⩽k-1, where [n!/(n-k)!] is the number of nodes in A_n,k