Let k 4 be even and let n 2. Consider a faulty k-ary n-cube Qk n in which the number of node faults fv and the number of link faults fe are such that fv +fe 2n-2. We prove that given any two healthy nodes s and e of Qk n, there is a path from s to e of length at least kn - 2fv - 1 (resp. kn - 2fv - 2) if the nodes s and e have different (resp. the same) parities (the parity of a node in Qk n is the sum modulo 2 of the elements in the n-tuple over {0, 1, . . . , k - 1} representing the node). Our result is optimal in the sense that there are pairs of nodes and fault configurations for which these bounds cannot be improved, and it answers questions recently posed by Yang, Tan and Hsu, and by Fu. Furthermore, we extend known results, obtained by Kim and Park, for the case when n = 2.
Iain A. Stewart, Yonghong Xiang