# Is the honeycomb lattice the double graph of another graph?

Associated to any planar graph $$G$$ there is another graph called its double graph, denoted by $$\mathcal{H}(G)$$, defined according to the following rules (see 'Trees and Matchings' section 2, page 3):

1. Embed $$G$$ and its dual $$G^{\ast}$$ simultaneously in the plane in such a way that any vertex $$v$$ of $$G$$ should lie inside its dual face $$F_{v}^{\ast}$$ in $$G^{\ast}$$ and any two dual edges of $$G$$ and $$G^{\ast}$$ intersect exactly once.
2. Declare all the new edge intersections from step 1 as new vertices and color them in white and all the remaining vertices coming from $$G$$ and $$G^{\ast}$$ as black vertices.
3. Denote the resulting graph $$\mathcal{H}(G)$$ and notice that by construction is bipartite.

Question: Does the bipartite honeycomb lattice arise as the double graph $$\mathcal{H}(G)$$ of any other graph $$G$$?

Note: It is indicated in 'Trees and Matchings' that the double graph construction can be generalized to directed graphs and in this case the bipartite honeycomb lattice appears as the double graph a directed triangular lattice, however, I'm interested in undirected graphs.

No: in the dual graph construction, all the white vertices created have degree $$4$$, but the honeycomb lattice is $$3$$-regular.