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Given a bicubic planar graph consisting of faces with degree $4$ and $6$, so called Barnette graphs. We can show that there are exactly six squares. Kundor and I found six types of arrangements of the six squares:

  1. three pairs of squares $(2+2+2)$
  2. two triples arranged in row $(\bar3+\bar3)$
  3. two triples arranged like a triangle $(3^\triangle+3^\triangle)$
  4. six isolated squares $(1+1+1+1+1+1)$
  5. two pairs and two isolated squares $(2+2+1+1)$
  6. one pair and four isolated squares $(2+1+1+1+1)$

Is there anything known if any of these arrangements can be proven to be Hamiltonian?

For example, "3. two triples arranged like a triangle" gives much less structural degrees of freedom compared to "4. six isolated squares"...

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Yes! They are all Hamiltonian. Paul Goodey proved in his 1975 paper Hamiltonian circuits in polytopes with even sided faces (MR410565) that any 3-regular planar 3-connected graph with only 4-sided or 6-sided faces is Hamiltonian.

These are exactly the graphs you are talking about, except that they are required to be 3-connected. However, any bicubic planar graph which is only 2-connected has some faces with at least 8 sides; any bicubic planar graph with only quadrilateral and hexagonal faces is 3-connected.

This is because being 2-connected but not 3-connected means there must be two faces which share more than one (non-adjacent) side, and you can quickly check that this cannot be compatible with the other requirements. (With two squares, you get digons, and I am assuming we are dealing with simple graphs. With a square and a hexagon, you get forced cut-edges, which are impossible in a regular bipartite graph. With two hexagons, you are forced to make faces with more than six sides in order to close the graph.)

Nick Matteo
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