Is the honeycomb lattice the double graph of another graph?

Question

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.

Answer 1

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