发现（近）最小覆盖集的光盘上的2-D平面 [英] find (near-)minimal covering set of discs on a 2-D plane

问题描述

好了，说我有一堆光盘中已知的固定位置上坐的飞机。每盘1个单位的半径。平面被完全覆盖的一组片中的，事实上，它是广泛过度覆盖的一组片中的，数量级或两个的一些地区的订单。我想找到仍然完全覆盖面盘的一个子集。最佳的是好的，但不是必需的。

OK, say I've got a bunch of discs sitting on a plane in fixed known locations. Each disc is 1 unit in radius. The plane is fully covered by the set of discs, in fact, it is extensively over-covered by the set of discs, by an order of magnitude or two in some areas. I'd like to find a subset of the discs that still fully cover the plane. Optimal is nice, but not necessary.

下面是前图所示：

和这里的说明后：

在我看来，这有不得不做Delauney三角双重问题，但我不能肯定，可以帮助我。我也知道，这是类似的，但不一样的，在计算几何光盘覆盖问题。这是一个标准的问题，他的名字我不知道吗？

It seems to me that there's a dual problem having to do with Delauney triangulation, but I'm not quite sure that helps me. I also know that this is similar, but not the same as, the disc covering problem in computational geometry. Is this a standard problem whose name I don't know?

可能的方法在我看来，包括不断增长的使用本地贪婪搜索一个覆盖集，并反复使用的最近对查询取出光盘一次。我不知道如果任是保证正常工作，我也没有通过细节工作。

Possible approaches seem to me to include growing a covering set using a local greedy search, and iteratively using a nearest-pair query to remove discs one at a time. I'm not sure if either is guaranteed to work well, and I haven't worked through the details.

呵呵，且应用程序，如果你没有猜对了，找了ZIP code形心的子样本进行查询时，包括地图，所以的 N 的是约50,000。

Oh, and the application is, if you haven't guessed it, finding a subsample of ZIP code centroids to cover a map when making queries, so n is about 50,000.

推荐答案

下面基本上是你的问题的一个更precise重述，但它可能会帮助：

Game Plan

The following is basically just a more precise restatement of your problem, but it might help:

1. 枚举所有的连接区域，导致当所有磁盘的边界绘制的平面。由假设，每个区域被水覆盖着1个或多个磁盘。
2. 在每个地区是一个东西被覆盖，并且每个磁盘是一个涵盖的东西。找到最小集合覆盖这一组区域。这是一个NP难不幸。

1. Enumerate every connected region in the plane that results when the boundaries of all disks are drawn. By assumption, each of these regions is covered by 1 or more disks.
2. Each region is a "thing to be covered", and each disk is a "covering thing". Find the minimum set cover on this set of regions. This is NP-hard unfortunately.

这可能无法利用所有可用的结构问题，但它肯定会给你一个最佳答案。

This might not be exploiting all the structure available in the problem, but it will definitely give you an optimal answer.

枚举的区域和记录的磁盘覆盖每个步骤1中是棘手的部分。区域不是一般凸这使得相交测试棘手，并且每圈添加可能会加倍区域的数量。下面是我会接近了：

Enumerating the regions and recording which disks cover each in step 1 is the tricky part. Regions are not in general convex which makes intersection tests tricky, and every circle you add potentially doubles the number of regions. Here is how I would approach that:

忘掉每个区域的实际位置，并且仅在其中术语磁盘它是里面并且它是外侧限定的区域。即一个区域是由0/1值的长度 - 正向量定义，每一个指示内部或外部磁盘的区域是否是被包括在交叉 - 有问题的区域是由相交的所有这些n个区域形成。所以，原则上你可以有多达2 ^ n个区域，但在实践中的一些（大部分）载体产生空地区，因为他们带来相交的两个磁盘有没有交集 - 这个很容易测试，谢天谢地。它应该是简单的递归生成所有非空的区域，不同的是...

Forget about the actual location of each region, and define a region only in terms of which disks it is inside and which it is outside. I.e. a region is defined by a length-n vector of 0/1 values, each indicating whether the region inside or outside that disk is to be included in the intersection -- the region in question is formed by intersecting all these n regions. So in principle you could have up to 2^n regions, but in practice some (most) vectors produce empty regions because they entail intersecting two disks that have no intersection -- this is easy to test for, thankfully. It should be straightforward to recursively generate all non-empty regions, except that...

不幸的是我现在看到它的是的需要进行全面相交测试，因为它并不总是能够知道什么时候一个区域将是空的。关键的反例是，由于两个磁盘A和B有一个小条子重叠，另外C盘取决于所有3个磁盘的位置重叠每个A和B，以及所有3的交集可以有或者没有为非空。 （看到这一点，得出3盘以不同的颜色与在绘图程序中50％不透明度，并移动它们。）

Unfortunately I now see that it is necessary to perform full intersection testing, because it's not always possible to tell when a region will be empty. The critical counterexample is that, given two disks A and B that have a small sliver of overlap and another disk C that overlaps each of A and B, depending on the positions of all 3 disks, the intersection of all 3 either may or may not be non-empty. (To see this, draw 3 disks in different colours with 50% opacity in a drawing program, and move them around.)

由于生成非空区的确切名单看起来这将是一个大量的工作，需要很长的时间，由于相交测试，并且你声称你不需要最优的解决方案，您可以使用网格尽量只样本点为一组的事物被覆盖，而不是非空区域的确切列表。这是简单的，以确定哪些磁盘覆盖给定样品点。那么解决的最大集合覆盖和以前一样。

Since generating the exact list of non-empty regions looks like it will be a lot of work and take a long time due to intersection testing, and you claim you don't need optimal solutions, you could try just using a grid of sample points as the set of "things to be covered" instead of the exact list of non-empty regions. It's straightforward to determine which disks cover a given sample point. Then solve maximum set cover as before.

要获得信任，有没有间隙，重新运行几次，每次随机引起的颤动样本点的坐标。增加的采样点的密度，直到有在最终结果未发生变化。

To get confidence that there are no gaps, rerun several times, randomly jittering the sample points' co-ordinates each time. Increase the density of sample points until there is no change in the final result.

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