## Enumerating triangulations of quadrangulations in cycles

Consider a cycle of length $(2n+2)$. Now we quadrangulate this cycle into $n$ quadrants. We want to enumerate the number of quadrangulations, and we denote this number by $q_n$. Now we triangulate this quadrangulation by triangulating each quadrant. We denote the number of triangulations $t_n$. It is clear that $t_n = 2^nq_n$.

Do I count some triangulations multiple times? That is, will some triangulation be the result of multiple different quadrangulations?

-

No triangulation occurs multiple times. At least one of the triangles $T$ of the triangulation has two edges on the boundary. There is a unique quadrant (I would say quadrilateral) $Q$ made up of two triangles of the triangulation, one of which is $T$. Remove the three edges of $T$, obtaining two quadrangulated cycles (possibly degenerate), and induct.

Note that not every triangulation of a $(2n+2)$-cycle is obtained by bisecting the quadrilaterals of a quadrangulation. Also, it is well-known that the number of quadrangulations of a $(2n+2)$-gon is $\frac{1}{2n+1}\binom{3n}{n}$.

-

Another way to think about this is that each triangulation gives you a binary tree, while each quadrangulation gives you a ternary tree (consider the dual graphs). If a triangulation comes from a quadrangulation then the correspondence gives you a perfect matching on the binary tree. Now show that every tree has at most one perfect matching. This translates to: every triangulation comes from at most one quadrangulation.

-
 This approach seems easiest to prove. – utdiscant Sep 18 2010 at 23:11