English

The minimum number of peeling sequences of a point set

Combinatorics 2024-11-15 v2

Abstract

Let PP be a set of nn points in Rd\mathbb{R}^d, in general position. We remove all of them one by one, in each step erasing one vertex of the convex hull of the current remaining set. Let gd(P)g_d(P) denote the number of different removal orders we can attain while erasing all points of PP this way, and let gd(n)g_d(n) be the \emph{minimum} of gd(P)g_d(P) over all nn-element point sets PRdP\subset \mathbb{R}^d. Dumitrescu and T\'oth showed that gd(n)=(d+1)(d+1)2ng_d(n)=(d+1)^{(d+1)^2n}. We substantially improve their bound, by proving that gd(n)=O((d+dln(d))(2+(d1)dlnd)n)g_d(n)= O((d+d\ln{(d)})^{(2+\frac{(d-1)}{\lfloor d\ln{d}\rfloor})n}). It follows that, for any ϵ>0\epsilon>0, there exist sufficiently high dimensional point sets PRdP\subset \mathbb{R}^d with gd(P)O(d(2+ϵ)n)g_d(P)\leq O(d^{(2+\epsilon)n}). This almost closes the gap between the upper bound and the best-known lower bound (d+1)n(d+1)^n for large values of dd.

Keywords

Cite

@article{arxiv.2312.00244,
  title  = {The minimum number of peeling sequences of a point set},
  author = {Dániel Gábor Simon},
  journal= {arXiv preprint arXiv:2312.00244},
  year   = {2024}
}