For How Many Edges is a Digraph Almost Certainly Hamiltonian?

Is the number of those Hamiltonian cycles in the complete digraph on nodes, each of which has just edges in common with a particular Hamiltonian cycle, and <!-- MATH $B(n,r) = n!/\{ r!(n - r)!\}$ --> . We show that <!-- MATH ${K_n}(r) = B(n,r){K_{n - r}}(0)$ --> and deduce that <!-- MATH ${K_n}(0)\sim(n - 1)!{e^{ - 1}}$ --> for large and that <!-- MATH ${K_n}(r)\sim(n - 1)!{e^{ - 1}}/r!$ --> if . An digraph is one with labelled nodes and edges. From the result for , we deduce that, if <!-- MATH $q{n^{ - 3/2}} \to \infty$ --> as <!-- MATH $n \to \infty$ --> , then almost all digraphs are Hamiltonian. If <!-- MATH $q{n^{ - 3/2}} \to c$ --> as <!-- MATH $n \to \infty$ --> , then the proportion of digraphs which are non-Hamiltonian is at most <!-- MATH $1 - \exp ( - {c^{ - 2}})$ --> as <!-- MATH $n \to \infty$ --> .