From aa374e3e9956acf3ab69a4a4f94f681e622b9d0c Mon Sep 17 00:00:00 2001 From: Remy Thomasse Date: Thu, 26 Jun 2014 10:43:21 +0200 Subject: [PATCH] reference biblio --- Documentation/biblio/geom.bib | 10 ++++++++++ .../doc/Generator/CGAL/random_convex_hull_in_disc_2.h | 2 +- 2 files changed, 11 insertions(+), 1 deletion(-) diff --git a/Documentation/biblio/geom.bib b/Documentation/biblio/geom.bib index cc6fbb1e01f..b8bd86a8613 100644 --- a/Documentation/biblio/geom.bib +++ b/Documentation/biblio/geom.bib @@ -151908,3 +151908,13 @@ pages = {179--189} year={2010} } +@TechReport{Devillers2014Generator +, author = "Olivier Devillers and Philippe Duchon and R{\'e}my Thomasse" +, title = "A generator of random convex polygon in a disc" +, institution = "INRIA" +, type = "Research Report" +, number = "RR-8467" +, month = "Feb" +, year = "2014" +, url = "http://hal.inria.fr/hal-00943409" +} diff --git a/Generator/doc/Generator/CGAL/random_convex_hull_in_disc_2.h b/Generator/doc/Generator/CGAL/random_convex_hull_in_disc_2.h index 2a84df06100..b8f22dcbf0c 100644 --- a/Generator/doc/Generator/CGAL/random_convex_hull_in_disc_2.h +++ b/Generator/doc/Generator/CGAL/random_convex_hull_in_disc_2.h @@ -27,7 +27,7 @@ The generated polygon will have an average number of vertices \f$ n^\frac{1}{3}( \cgalHeading{Implementation} The implementation is based on an incremental construction of a convex hull. At each step, we choose a number of points to pick uniformly at random in the disc. Then, a subset of these points, that won't change the convex hull, is evaluated using a Binomial law. -As these points won't be generated, the time and size complexities are reduced. +As these points won't be generated, the time and size complexities are reduced \cgalCite{Devillers2014Generator}. A tradeoff between time and memory is provided with the option `fast`, true by default. Using the `fast` option, both time and size expected complexities are \f$O\left(n^\frac{1}{3}\log^\frac{2}{3}n \right)\f$. If this option is disabled, the expected size complexity becomes \f$O\left(n^\frac{1}{3}\right)\f$ but the expected time complexity becomes \f$O\left(n^\frac{1}{3}\log^2 n \right)\f$.