Let P be a set of n points in the plane and _{\theta}(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 } O be a set of k lines passing through the origin . We show=\pi-\alpha_i and \Theta=\min\{\Theta_i:i=1,\ldots,2k\}. We further obtain} : (1) How such that \Theta}\ge to compute the O -hull of P in \Theta (n\log n) time and O(n) space, ({2} the complexities are O(\frac{n}{\Theta}\log n) time and O(\frac{n}{\Theta}) space. (} 2) how to such that \Theta}\ge compute and maintain the rotated hull \mathcal{OH_{\theta}(P ) }  }} for \theta\in [0,2\pi) in O(kn\log n) time and O(kn) space, and (3) how to compute in \Theta(n{\Theta}} \log n) time and O( n) space a value of \theta for which the rectilinear convex hull, \mathcal{RH_{\theta}(P ), has minimum area, thus improving the previously best O(n^2) algorithm presented by Bae et al.in 2009.} {\Theta}) space if \Theta<\frac{\pi}{2}. (3) Finally, given a set \mathcal{O} such that \Theta}\ge }}-convex hull of P of minimum (or maximum) area over all \theta\in }[
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(n^2) bound. Let } O be a set of k lines through the origin sorted by slope and let \alpha_i be the aperture angles of the 2k sectors defined by every pair of two consecutive lines. Let \Theta_{i=\pi-\alpha_i and \Theta=\min\{\Theta_i:i=1,\ldots,2k\}. We further obtain} : (1) Given a set \mathcal{O such that \Theta}\ge\frac{\pi to compute the O -convex hull of P in optimal O (n\log n) time and O(n) space, while if \Theta<\frac{\pi{2} the complexities are O(\frac{n}{\Theta}\log n) time and O(\frac{n}{\Theta}) space. (} 2) Given a set \mathcal{O such that \Theta}\ge\frac{\pi compute and maintain the _{\theta}(P ) } boundary of the \mathcal{O}}_{\theta for \theta\in [0,2\pi) in O(kn\log n) time and O(kn) space, or in O(k\frac{n{\Theta}} \log n) time and O( _{\theta}(P ), has minimum area, thus improving the previously best O(n^2) algorithm presented by Bae et al.in 2009.} k\frac{n{\Theta}) space if \Theta<\frac{\pi}{2}. (3) Finally, given a set \mathcal{O} such that \Theta}\ge\frac{\pi \mathcal{O}}_{\theta-convex hull of P of minimum (or maximum) area over all \theta\in }[0,2\pi) in O(kn\log n) time and O(kn) space.
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