# Does every Maximal Planar Graph have this Representation?

Can every Maximal Planar Graph be represented by 'inner' and 'outer' triangulations of a regular polygon? For example for n vertices, draw a regular polygon with n vertices and n edges. Triangulate the inside with n-3 edges and triangulate again on the outside without repeating connections, another n-3 edges. 3n-6 in total. A proof seems to be obtained by taking a 'starting triangle' in the Maximal Planar Graph, adding a new vertex which joins two of the existing vertices, but doesn't enclose a vertex and repeating until all vertices are included, none should be 'inside'. Perhaps this representation would be useful in some situations. Is the 'proof' valid?

The proof is not valid. In particular, to add a new vertex, it doesn't just need to join two existing vertices, but ones that are adjacent around the polygon built so far; there is no guarantee that such a vertex exists.

The statement is not always true, either. The smallest counterexample is the Goldner–Harary graph shown below: In this maximal planar graph, there simply is no cycle of length $$n$$ that visits every vertex once. A quick way to see that is to look at what happens when we delete the $$5$$ vertices marked with an $$X$$ below: The result is $$6$$ isolated vertices. However, if you have an $$11$$-gon and delete any $$5$$ of its vertices, you'll leave the remaining $$6$$ vertices in $$5$$ or fewer pieces; there will be some edges between them left. So this graph cannot have an $$11$$-gon inside it. (In other words, the Goldner-Harary graph is not Hamiltonian because it's not $$1$$-tough.)

• thankyou, good answer. Feb 3 at 9:23