Hamiltonian Tetrahedralizations with Steiner Points

TL;DR

This paper demonstrates how to add a limited number of Steiner points to a 3D point set to ensure a Hamiltonian tetrahedralization exists, providing an efficient algorithm and exploring special cases.

Contribution

It introduces a method to achieve Hamiltonian tetrahedralizations with Steiner points and presents an efficient algorithm for constructing such configurations.

Findings

Adding at most           Steiner points suffices.

All sets with up to 20 convex hull points have Hamiltonian tetrahedralizations without Steiner points.

Some point sets (e.g., 84 points) cannot have a Hamiltonian tetrahedralization with all tetrahedra sharing a vertex.

Abstract

Let $S$ be a set of $n$ points in 3-dimensional space. A tetrahedralization $\mathcal{T}$ of $S$ is a set of interior disjoint tetrahedra with vertices on $S$, not containing points of $S$ in their interior, and such that their union is the convex hull of $S$. Given $\mathcal{T}$, $D_\mathcal{T}$ is defined as the graph having as vertex set the tetrahedra of $\mathcal{T}$, two of which are adjacent if they share a face. We say that $\mathcal{T}$ is Hamiltonian if $D_\mathcal{T}$ has a Hamiltonian path. Let $m$ be the number of convex hull vertices of $S$. We prove that by adding at most $\lfloor \frac{m-2}{2} \rfloor$ Steiner points to interior of the convex hull of $S$, we can obtain a point set that admits a Hamiltonian tetrahedralization. An $O(m^{3/2}) + O(n \log n)$ time algorithm to obtain these points is given. We also show that all point sets with at most 20 convex hull points admit a Hamiltonian tetrahedralization without the addition of any Steiner points. Finally we exhibit a set of 84 points that does not admit a Hamiltonian tetrahedralization in which all tetrahedra share a vertex.