Is it possible to draw a planar graph on 11 vertices in which each face (country) has 3 neighbours?
And is there some method after to draw it to confirm that it is in fact indeed planar? I drew some figures, but then I'm not entirely sure if it is actually planar...
Thanks.
First, the sum of the degrees must be an even number, since $\Sigma Deg(v_i)= 2|E(G)|$. If your graph is "simple", i.e., it is not bipartite, then it is planar iff it does not contain a $K_5$ as a subgraph. Since in your problem, $Deg(v_i)\leq 3$ , it cannot have a $K_5$ as a subgraph, since every vertex of a $K_5$ has degree $4$. If your graph is bipartite, then you have to avoid a $K_{3,3}$ as a subgraph.