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Efficient computation of minimum-area rectilinear convex hull under rotation and generalizations
Journal of Global Optimization ( IF 1.3 ) Pub Date : 2020-10-07 , DOI: 10.1007/s10898-020-00953-5
Carlos Alegría , David Orden , Carlos Seara , Jorge Urrutia

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 \(\mathcal {RH}_{P}({\theta })\), 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 \(\alpha _i\) 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 \(\Theta =\min \{\Theta _i :i=1,\ldots ,2k\}\). 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 \(\theta \in [0,2\pi )\).



中文翻译:

旋转条件下最小面积直线凸包的高效计算与归纳

P为平面中n个点的集合。我们计算\(\ theta \ in [0,2 \ pi)\)的值,其中P的直线凸包由\(\ mathcal {RH} _ {P}({\ theta})\)表示在最佳\(O(n \ log n)\)时间和On)空间中具有最小(或最大)面积,从而改善了先前的\(O(n ^ 2)\)界限。让\(\ mathcal {ö} \)是一组 ķ线过原点排序由斜率和让\(\阿尔法_i \)是2的尺寸ķ角度通过对接连的两行所定义,\(ⅰ = 1,\ ldots,2k \)。设\(\ Theta _ {i} = \ pi-\ alpha _i \)\(\ Theta = \ min \ {\ Theta _i:i = 1,\ ldots,2k \} \)。我们得到:(1)给定一个\(\ mathcal {O} \)使得\(\ Theta \ ge \ frac {\ pi} {2} \),我们提供了一种计算\(\ mathcal {在最佳\(O(n \ log n)\)时间和On)空间中P的O} \)-凸包 ;如果\(\ Theta <\ frac {\ pi} {2} \),则时间和空间复杂度为\(O(\ frac {n} {\ Theta} \ log n)\)\(O(\ frac {n} {\ Theta})\)。(2)给定一个\(\ mathcal {O} \)使得\(\西塔\ GE \压裂{\ PI} {2} \) ,我们计算和保持的边界\({\ mathcal {ö}} _ {\ THETA} \)的船体-凸 P\( \ theta \ in [0,2 \ pi} \)\(O(kn \ log n)\)时间和Okn)空间中,或者如果\(\ Theta <\ frac {\ pi} {2} \ ),在\(O(k \ frac {n} {\ Theta} \ log n)\)时间和\(O(k \ frac {n} {\ Theta})\)空间中。(3)最后,给定一个\(\ mathcal {O} \)使得\(\ Theta \ ge \ frac {\ pi} {2} \),我们以\(O(kn \ log n) \)时间和Okn)空间,角度\(\ theta \ in [0,2 \ pi } \),使得P\({\ mathcal {O}} _ {\ theta} \)-凸包在所有\( \ theta \ in [0,2 \ pi)\)

更新日期:2020-10-07
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