  An instance of Hamiltonian cycle problem can be solved by converting it to an
instance of Travelling salesman problem, assigning any choice of weights to
edges of the underlying graph. In this note we demonstrate that, for difficult
instances, choosing the edge weights to be the resistance distance between its
two incident vertices is often a good choice. We also demonstrate that arguably
stronger performance arises from using the inverse of the resistance distance.
Examples are provided demonstrating benefits gained from these choices.
