Consider a connected graph whose cycles have length at least 4, the maximum valence is 4, there are 2 vertices of valence 2 and 4 of valence 3. I would like to show graphs of this type are planar . I am not sure whether the result is true or not. I am essentially trying to show that any connected subgraph whose vertices have at least valence 2 satisfy $2V -4 \geq E$. You may also note that the two vertices of valence 2 are not connected by a single edge, in fact the two vertices are antipodal points of a 4 cycle.



Determining whether a certain graph is planar
With the information given, you can not guarantee planarity. One counter example would be a graph consisting of the complete bipartite graph $K\_{4,4}$ (not planar) and any bipartite graph with two degree-2-vertices, four degree-three-vertices, and the remaining vertices degree 4, such that the degree-2-vertices are opposite in a $C\_4$. If you want it connected, reroute a couple edges between the two parts, keeping all degrees the same, and keeping $K\_{3,3}$ as a subgraph.


