# Are 3-valent, balanced bipartite, proper-edge coloured spherical graphs Hamiltonian?

I have not too much experience with graph theory, so I am basically asking if someone knows if the answer to the following question follows from some theorem of graph theory (which I'm probably unaware of):

Let $$\mathcal{G}=(\mathcal{V},\mathcal{E})$$ be a finite and simple graph with the following properties:

1. It is bipartite.
2. It is $$3$$-valent and admits a proper $$3$$-edge colouring, i.e. we can colour all edges by $$3$$ colours such that the $$3$$ edges adjacent to some vertex have different colours.
3. It is spherical, i.e. $$\vert\mathcal{F}\vert-\vert\mathcal{E}\vert+\vert\mathcal{V}\vert=2$$, where $$\mathcal{F}$$ is the set of "faces", i.e. the set of all the bi-coloured cycles (=maximal connected components of two colours). It is actually a well-known fact from "crystallization theory" that this is equivalent to saying that the graph can be drawn in such a way that is planar and such that all the faces of the graph (=closed paths enclosing some area) are actually also "faces" in the sense defined above, i.e. the only consist of $$2$$ colours.

Under these assumptions, does there exists a closed path in the graph which is maximal in the sense that it touches every vertex of $$\mathcal{G}$$ exactly once?

I think graphs with these properties are called Hamiltonian in the graph theory literature. If this has any relevance, note that the graph $$\mathcal{G}$$ is actually balanced bipartite, which follows by combining bipartitness and the fact that it admits a proper edge colouring.

EDIT: I just found on the internet that what I am trying to prove for a few days now is very similar to what is called "Barnette's conjecture". So, I guess the question might be much harder then I thought^^ • not sure which vertices you are suggesting to delete, but I don't think 8 vertices are enough - the only balanced bipartite, 3-regular graph on 8 vertices is $K_{4,4}$ minus a matching (which is isomorphic to the 3-cube), and this graph has a Hamiltonian cycle Aug 2, 2022 at 19:58