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

Published 30 Oct 2017 in cs.CG and math.CO | (1710.10888v3)

Abstract: Let $P$ be a set of $n$ points in the plane. We compute the value of $\theta\in [0,2\pi)$ for which the rectilinear convex hull of $P$, denoted by $\mathcal{RH}\theta(P)$, has minimum (or maximum) area in optimal $O(n\log n)$ time and $O(n)$ space, improving the previous $O(n2)$ bound. Let $\mathcal{O}$ be a set of $k$ lines through the origin sorted by slope and let $\alpha_i$ be the sizes of the $2k$ angles defined by pairs of two consecutive lines, $i=1, \ldots , 2k$. Let $\Theta{i}=\pi-\alpha_i$ and $\Theta=\min{\Theta_i \colon i=1,\ldots,2k}$. We obtain: (1) Given a set $\mathcal{O}$ such that $\Theta\ge\frac{\pi}{2}$, 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 $\Theta < \frac{\pi}{2}$, the time and space complexities are $O(\frac{n}{\Theta}\log n)$ and $O(\frac{n}{\Theta})$ respectively. (2) Given a set $\mathcal{O}$ such that $\Theta\ge\frac{\pi}{2}$, we compute and maintain the boundary of the ${\mathcal{O}}{\theta}$-convex hull of $P$ for $\theta\in [0,2\pi)$ in $O(kn\log n)$ time and $O(kn)$ space, or if $\Theta < \frac{\pi}{2}$, in $O(k\frac{n}{\Theta}\log n)$ time and $O(k\frac{n}{\Theta})$ space. (3) Finally, given a set $\mathcal{O}$ such that $\Theta\ge\frac{\pi}{2}$, we compute, in $O(kn\log n)$ time and $O(kn)$ space, the angle $\theta\in [0,2\pi)$ such that the $\mathcal{O}{\theta}$-convex hull of $P$ has minimum (or maximum) area over all $\theta\in [0,2\pi)$.

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