A canonical tree decomposition for order types, and some applications

TL;DR

This work develops a canonical tree-structured decomposition for order types via chirotopes, introducing bowtie products and modules to assemble large point-set chirotopes from smaller pieces. It proves existence and uniqueness of a canonical chirotope tree, shows that almost all realizable chirotopes are indecomposable, and provides a framework to count triangulations of a chirotope from its tree using generating polynomials and a kernel-method-inspired path analysis. A key bijection and a set of counting tools reduce the triangulation problem to operations on node-level polynomials, enabling practical counting on highly decomposable instances and establishing asymptotic behavior for certain families. The approach offers new algorithmic avenues for counting geometric structures and raises open questions about extending to higher dimensions and other crossing-free configurations, with a concrete path to implementation and experimental validation.

Abstract

We introduce and study a notion of decomposition of planar point sets (or rather of their chirotopes) as trees decorated by smaller chirotopes. This decomposition is based on the concept of mutually avoiding sets (which we rephrase as \emph{modules}), and adapts in some sense the modular decomposition of graphs in the world of chirotopes. The associated tree always exists and is unique up to some appropriate constraints. We also show how to compute the number of triangulations of a chirotope efficiently, starting from its tree and the (weighted) numbers of triangulations of its parts.

Figures (13)

• Figure 1: A set of 9 points which can be decomposed into two modules of size $4$ and $5$ points.
• Figure 2: Illustration of the proof of \ref{['lem:caratheodory']} for a realizable chirotope. The points $\mathfrak p_{a_i}$ corresponding to the elements $a_i$ are seen from $\mathfrak p_t$ in clockwise order (since $\chi(t,a_i,a_{i+1})=-1$ for all $i$ by construction). Given $\mathfrak p_{a_i}$, either the line $\mathfrak p_{a_i} \mathfrak p_t$ splits the point set $\{\mathfrak p_{a_j}, j<i\}$ in two non trivial parts (which happens here for $i=4$), and we can find a triangle containing $\mathfrak p_t$ (namely $\mathfrak p_{a_0} \mathfrak p_{a_3} \mathfrak p_{a_4}$ in the picture), or all points $\{\mathfrak p_{a_j}, j<i\}$ are on the same side of $\mathfrak p_{a_i} \mathfrak p_t$ (which happens for $i=2$ and $i=3$ here), and we let $\mathfrak p_{a_{i+1}}$ be any point on the other side (there is necessarily one, since $\mathfrak p_t$ is not extreme). Since the set is finite, we cannot be always in the latter situation and we eventually find a triangle containing $\mathfrak p_t$.
• Figure 3: On the left, 10 points uniformly distributed on a circle; on the right, the same number of points uniformly distributed on a semicircle. Both configurations have the same chirotope. In the circle case, each point is in a bounded cell of the line arrangement defined by the other points (as illustrated for the left-most point). On the other hand in the semi-circle case, the left-most point is in an unbounded region of the line arrangement defined by the other points.
• Figure 4: Illustration of the proof of \ref{['prop:Realization_Substitution']}. Marked points are drawn in red.
• Figure 5: An example of chirotope tree.

Theorems & Definitions (92)

• Proposition 1.1
• Proposition 1.2
• Theorem 1.3
• Proposition 1.4
• Theorem 1.5
• Lemma 2.1
• proof
• Lemma 2.2
• proof
• Lemma 2.4