Abstract We analyse a polyhedron which contains the convex hull of all Hamiltonian cycles of a given undirected connected cubic graph. Our constructed polyhedron is defined by polynomially-many linear constraints in polynomially-many continuous (relaxed) variables. Clearly, the emptiness of the constructed polyhedron implies that the graph is non-Hamiltonian. However, whenever a constructed polyhedron is non-empty, the result is inconclusive. Hence, the following natural question arises: if we assume that a non-empty polyhedron implies Hamiltonicity, how frequently is this diagnosis incorrect? We prove that, in the case of bridge graphs, the constructed polyhedron is always empty. We also demonstrate that some non-bridge non-Hamiltonian cubic graphs induce empty polyhedra as well. We compare our approach to the famous Dantzig-Fulkerson-Johnson relaxation of a TSP, and give empirical evidence which suggests that the latter is infeasible if and only if our constructed polyhedron is also empty. By considering special edge cut sets which are present in most cubic graphs, we describe a heuristic approach, built on our constructed polyhedron, for which incorrect diagnoses of non-Hamiltonian graphs as Hamiltonian appear to be very rare. In particular, for cubic graphs containing up to 18 vertices, only four out of 45,982 undirected connected cubic graphs were so misdiagnosed. By constrast, we demonstrate that an equivalent heuristic, when built on the Dantzig-Fulkerson-Johnson relaxation of a TSP, is mostly unsuccessful in identifying additional non-Hamiltonian graphs. These empirical results suggest that polynomial algorithms based on our constructed polyhedron may be able to correctly identify Hamiltonicity of a cubic graph in all but rare cases.
|Number of pages||10|
|Journal||COMPUTERS & OPERATIONS RESEARCH|
|Issue number||Article: 3804|
|Publication status||Published - 15 Jul 2015|
- Hamiltonian cycles
- Linear feasibility
- Traveling salesman problem