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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.)

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

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  • $\begingroup$ 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. $\endgroup$ – Finallysignedup Dec 30 '18 at 5:02

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