We consider the problem of counting straight-edge triangulations of a given set $P$ of $n$ points in the plane. Until very recently it was not known whether the exact number of triangulations of $P$ can be computed asymptotically faster than by enumerating all triangulations. We now know that the number of triangulations of $P$ can be computed in $O^{*}(2^{n})$ time, which is less than the lower bound of $\Omega(2.43^{n})$ on the number of triangulations of any point set. In this paper we address the question of whether one can approximately count triangulations in sub-exponential time. We present an algorithm with sub-exponential running time and sub-exponential approximation ratio, that is, denoting by $\Lambda$ the output of our algorithm, and by $c^{n}$ the exact number of triangulations of $P$, for some positive constant $c$, we prove that $c^{n}\leq\Lambda\leq c^{n}\cdot 2^{o(n)}$. This is the first algorithm that in sub-exponential time computes a $(1+o(1))$-approximation of the base of the number of triangulations, more precisely, $c\leq\Lambda^{\frac{1}{n}}\leq(1 + o(1))c$. Our algorithm can be adapted to approximately count other crossing-free structures on $P$, keeping the quality of approximation and running time intact. In this paper we show how to do this for matchings and spanning trees.

Thanks. We have received your report. If we find this content to be in
violation of our guidelines,
we will remove it.

Ok