将椭圆扫掠到消失点
Sweeping an oval to a vanishing point
摘要：
给定平面上的一个凸区域，以扫描线为工具，通过一系列扫描将该区域缩小为一个点的最佳方法是什么？文献[2]首次研究了正交扫描点的扫描问题。在这里，我们考虑最近在[1]中引入的扫描的以下\emph{slanted}变体：在单次扫描中，扫描线被放置在平面中某处的起始位置，然后根据扫描矢量$\vec v$（不一定与扫描线正交）连续移动到另一个平行的结束位置，然后从平面提起。扫描序列的成本是扫描向量的长度之和。区域的（最优）扫描成本是该区域的所有有限扫描序列上的成本的下确界。一个区域的最佳扫描序列是一个最小的总成本，如果它存在。另一个感兴趣的参数是扫描次数。我们发现，存在凸区域的最佳扫描成本不能达到两次扫描。这反驳了Bousany，Karker，O 'Rourke和Sparaco的猜想，即两次扫描（沿着最小周长包围的两个相邻边的向量）总是足够的[1]。此外，我们推测，对于某些凸区域，没有有限的扫描序列是最佳的。另一方面，我们证明了基于最小周长外接矩形的2-sweep算法和基于最小周长外接矩形的2-sweep算法在该扫描模型下都达到了$4/π\approx 1.27$近似。
英文摘要：
Given a convex region in the plane, and a sweep-line as a tool, what is best way to reduce the region to a single point by a sequence of sweeps? The problem of sweeping points by orthogonal sweeps was first studied in [2]. Here we consider the following \emph{slanted} variant of sweeping recently introduced in [1]: In a single sweep, the sweep-line is placed at a start position somewhere in the plane, then moved continuously according to a sweep vector $\vec v$ (not necessarily orthogonal to the sweep-line) to another parallel end position, and then lifted from the plane. The cost of a sequence of sweeps is the sum of the lengths of the sweep vectors. The (optimal) sweeping cost of a region is the infimum of the costs over all finite sweeping sequences for that region. An optimal sweeping sequence for a region is one with a minimum total cost, if it exists. Another parameter of interest is the number of sweeps. We show that there exist convex regions for which the optimal sweeping cost cannot be attained by two sweeps. This disproves a conjecture of Bousany, Karker, O'Rourke, and Sparaco stating that two sweeps (with vectors along the two adjacent sides of a minimum-perimeter enclosing parallelogram) always suffice [1]. Moreover, we conjecture that for some convex regions, no finite sweeping sequence is optimal. On the other hand, we show that both the 2-sweep algorithm based on minimum-perimeter enclosing rectangle and the 2-sweep algorithm based on minimum-perimeter enclosing parallelogram achieve a $4/π\approx 1.27$ approximation in this sweeping model.
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