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I started drawing planar, cubic, bipartite graphs consisting of faces with $4$ or $6$ vertices only. I found that six $4$-faces are sufficient to do that. The smallest graph is a planar drawing of the cube.

How many vertices or $6$-faces are necessary to get two non-isomorphic graphs?

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  • $\begingroup$ Your question is a little confusing. Are you asking for the smallest (in terms of vertices) pair of planar cubic graphs with 6-sided faces? $\endgroup$ – gilleain Jul 20 '15 at 12:49
  • $\begingroup$ @gilleain yes; 6 4-sided and x 6-sided faces... $\endgroup$ – draks ... Jul 20 '15 at 14:17
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Since the outer face must have length 4 or 6, it should be possible to prove your graphs must be at least 3-connected. But then your graphs are examples of Barnette graphs (cubic bipartite 3-connected planar graphs) which are conjectured to all have Hamiltonian cycles.

These have been studied fairly extensively, and you can find a drawings of all the Barnette graphs on up to 20 vertices in Appendix A here. Browsing through that appendix, it seems there is exactly 1 of your graphs on 8, 12, 14, 16, and 18 vertices (0, 2, 3, 4, and 5 6-faces), no such graph on 10 vertices, and there are 3 non-isomorphic such graphs on 20 vertices (20-3, 20-5, and 20-8 in the Appendix).

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