# Mutually disjoint odd cycles in certain planar graph

Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.

Is it possible to construct an example of such G for which no two faces of odd length share a common vertex?

(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)

• This graph is not necesary simple? That is two vertices can be connected with more than 1 edge? – Aqua Dec 29 '18 at 16:55
• This would make a face of length 2, I think, which I ruled out. – Finallysignedup Dec 30 '18 at 5:04

## 1 Answer

Infinite counter examples of this form (just keep adding two more squares): The key is to have a face with even degree incident to only one face with odd degree. Here's another way to achieve this goal.

Consider any cubic polyhedron $$P$$, which has a face, $$f$$, of odd degree.

Chamfer $$P$$ once to get $$P'$$. $$P'$$ is still cubic, and any face originally in $$P$$ is now only incident to hexagons.

Now, chamfer $$P'$$ to get $$P''$$. Here, every face incident to $$f$$ will be a hexagon, which is incident to hexagons on every other edge. Thus you can subdivide any edge in $$f$$ to get your desired kind of graph.

• Thanks. I realize now I only wish to allow odd faces that are triangles (the rest even). I've posted this as a new question. – Finallysignedup Dec 30 '18 at 5:02