Counting triangulations and other crossing-free structures approximately

Victor Alvarez, Karl Bringmann, Saurabh Ray, Raimund Seidel

Research output: Contribution to journalArticlepeer-review

Abstract

We consider the problem of counting straight-edge triangulations of a given set P of npoints 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∗(2n) time [9], which is less than the lower bound of Ω(2.43n) on the number of triangulations of any point set [30]. 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 Λ the output of our algorithm and by cn the exact number of triangulations of P, for some positive constant c, we prove that cn ≤ Λ ≤ cn· 2°(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 ≤ Λn-1≤ (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.

Original languageEnglish (US)
Pages (from-to)386-397
Number of pages12
JournalComputational Geometry: Theory and Applications
Volume48
Issue number5
DOIs
StatePublished - 2015

Keywords

  • Algorithmic geometry
  • Approximation algorithms
  • Counting algorithms
  • Crossing-free structures
  • Triangulations

ASJC Scopus subject areas

  • Computer Science Applications
  • Geometry and Topology
  • Control and Optimization
  • Computational Theory and Mathematics
  • Computational Mathematics

Fingerprint

Dive into the research topics of 'Counting triangulations and other crossing-free structures approximately'. Together they form a unique fingerprint.

Cite this