Fault-free Hamiltonian cycles in faulty arrangement graphs

The arrangement graph An,k, which is a generalization of the star graph (n驴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 |Fe| and |Fv| denote the numbers of edge faults and vertex faults, respectively. We show that An,k is Hamiltonian when 1) (k = 2 and n驴k驴 4, or k驴 3 and $n-k\ge 4+\left\lceil {{\textstyle{k \over 2}}} \right\rceil$), and |Fe| 驴k(n驴k) 驴 2, or 2) k驴 2, $n-k\ge 2+\left\lceil {{\textstyle{k \over 2}}} \right\rceil,$ and |Fe| 驴k(n驴k驴 3) 驴 1, or 3) k驴 2, n驴k驴 3, and |Fe| 驴k, or 4) n驴k驴 3 and |Fv| 驴n驴 3, or 5) n驴k驴 3 and |Fv| + |Fe| 驴k. Besides, for An,k with n驴k = 2, we construct a cycle of length at least 1) ${\textstyle{{n!} \over {\left( {n-k} \right)!}}}-2$ if |Fe| 驴k驴 1, or 2) ${\textstyle{{n!} \over {\left( {n-k} \right)! }}}-\left| {F_v} \right|-2\left( {k-1} \right)$ if |Fv| 驴k驴 1, or 3) ${\textstyle{{n!} \over {\left( {n-k} \right)! }}}-\left| {F_v} \right|-2\left( {k-1} \right)$ if |Fe| 驴 + |Fv| 驴k驴 1, where ${\textstyle{{n!} \over {\left( {n-k} \right)!}}}$ is the number of nodes in An,k.