On Barnette's conjecture

Barnette’s conjecture is the statement that every 3-connected cubic planar bipartite graph is Hamiltonian. Goodey showed that the conjecture holds when all faces of the graph have either 4 or 6 sides. We generalize Goodey’s result by showing that when the faces of such a graph are 3-colored, with adjacent faces having different colors, if two of the three color classes contain only faces with either 4 or 6 sides, then the conjecture holds. More generally, we consider 3-connected cubic planar graphs that are not necessarily bipartite, and show that if the faces of such a graph are 2-colored, with every vertex incident to one blue face and two red faces, and all red faces have either 4 or 6 sides, while the blue faces are arbitrary, provided that blue faces with either 3 or 5 sides are adjacent to a red face with 4 sides (but without any assumption on blue faces with 4, 6, 7, 8, 9, . . . sides), then the graph is Hamiltonian. The approach is to consider the reduced graph obtained by contracting each blue face to a single vertex, so that the reduced graph has faces corresponding to the original red faces and with either 2 or 3 sides, and to show that such a reduced graph always contains a proper quasi spanning tree of faces. In general, for a reduced graph with arbitrary faces, we give a polynomial time algorithm based on spanning tree parity to decide if the reduced graph has a spanning tree of faces having 2 or 3 sides, while to decide if the reduced graph has a spanning tree of faces with 4 sides or of arbitrary faces is NP-complete for reduced graphs of even degree. As a corollary, we show that whether a reduced graph has a noncrossing Euler tour has a polynomial time algorithm if all vertices have degree 4 or 6, but is NP-complete if all vertices have degree 8. Finally, we show that if Barnette’s conjecture is false, then the question of whether a graph in the class of the conjecture is Hamiltonian is NP-complete.