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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.

A note on using the resistance-distance matrix to solve Hamiltonian cycle problem