Efficient computation of minimum-area rectilinear convex hull under rotation and generalizations

TL;DR

The rectilinear convex hull of P, denoted by RHP(θ), has minimum (or maximum) area in optimal O(nlogn) space, improving the previous O(n2) space.

Abstract

Let $P$ be a set of $n$ points in the plane. We compute the value of $θ\in [0,2π)$ for which the rectilinear convex hull of $P$, denoted by $\mathcal{RH}_θ(P)$, has minimum (or maximum) area in optimal $O(n\log n)$ time and $O(n)$ space, improving the previous $O(n^2)$ bound. Let $\mathcal{O}$ be a set of $k$ lines through the origin sorted by slope and let $α_i$ be the sizes of the $2k$ angles defined by pairs of two consecutive lines, $i=1, \ldots , 2k$. Let $Θ_{i}=π-α_i$ and $Θ=\min\{Θ_i \colon i=1,\ldots,2k\}$. We obtain: (1) Given a set $\mathcal{O}$ such that $Θ\ge\fracπ{2}$, we provide an algorithm to compute the $\mathcal{O}$-convex hull of $P$ in optimal $O(n\log n)$ time and $O(n)$ space; If $Θ< \fracπ{2}$, the time and space complexities are $O(\frac{n}Θ\log n)$ and $O(\frac{n}Θ)$ respectively. (2) Given a set $\mathcal{O}$ such that $Θ\ge\fracπ{2}$, we compute and maintain the boundary of the ${\mathcal{O}}_θ$-convex hull of $P$ for $θ\in [0,2π)$ in $O(kn\log n)$ time and $O(kn)$ space, or if $Θ< \fracπ{2}$, in $O(k\frac{n}Θ\log n)$ time and $O(k\frac{n}Θ)$ space. (3) Finally, given a set $\mathcal{O}$ such that $Θ\ge\fracπ{2}$, we compute, in $O(kn\log n)$ time and $O(kn)$ space, the angle $θ\in [0,2π)$ such that the $\mathcal{O}_θ$-convex hull of $P$ has minimum (or maximum) area over all $θ\in [0,2π)$.