Empty Squares in Arbitrary Orientation Among Points

TL;DR

This paper investigates the combinatorial properties of empty squares in arbitrary orientations among planar points, establishing bounds, and introduces algorithms for detecting such squares and related geometric structures efficiently.

Contribution

It provides tight bounds on the number of empty squares with contact pairs and develops algorithms for their detection and related geometric optimization problems.

Findings

Number of empty squares with four contact pairs is between Ω(n) and O(n^2).

An algorithm reports all such squares in O(s log n) time.

Algorithms for largest empty square and minimum width/area square annulus are proposed.

Abstract

This paper studies empty squares in arbitrary orientation among a set $P$ of $n$ points in the plane. We prove that the number of empty squares with four contact pairs is between $\Omega(n)$ and $O(n^2)$, and that these bounds are tight, provided $P$ is in a certain general position. A contact pair of a square is a pair of a point $p\in P$ and a side $\ell$ of the square with $p\in \ell$. The upper bound $O(n^2)$ also applies to the number of empty squares with four contact points, while we construct a point set among which there is no square of four contact points. These combinatorial results are based on new observations on the $L_\infty$ Voronoi diagram with the axes rotated and its close connection to empty squares in arbitrary orientation. We then present an algorithm that maintains a combinatorial structure of the $L_\infty$ Voronoi diagram of $P$, while the axes of the plane continuously rotates by $90$ degrees, and simultaneously reports all empty squares with four contact pairs among $P$ in an output-sensitive way within $O(s\log n)$ time and $O(n)$ space, where $s$ denotes the number of reported squares. Several new algorithmic results are also obtained: a largest empty square among $P$ and a square annulus of minimum width or minimum area that encloses $P$ over all orientations can be computed in worst-case $O(n^2 \log n)$ time.