如何找到包围所有给定点的最小半径圆？
假设平面上有大约 1000 个奇数点。
然后，我认为可以做的是丢弃不以任何方式影响圆半径的点 - 凸包 未通过 [使用 几种算法]。这给我们留下了重要的要点。
现在从现在开始，可以做什么来找到最小半径圆？
一旦我了解了如何对圆形进行此操作，我就希望将其推广到椭圆。
任何指向某些“公共源代码”的链接都会有所帮助，以便我可以将其修改为省略号。
Suppose I have some 1000 odd points on a plane.
Then, what I think could be done is to discard the points that do not affect the radius of the circle in any way - the points through which the convex hull does not pass [using one of the several algorithms]. This leaves us with points that do matter.
Now from here on, what can be done to find that minimum radius circle?
I am looking to generalize this for ellipses once I understand how it can be done for circles.
Any link to some "public source code" would be helpful, so that I can modify it for ellipses.
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这被称为最小封闭圆问题（我很困惑为什么你的谷歌搜索没有显示任何内容），并讨论了 此处，此处、这里以及许多其他地方。
This is known as the Minimal Enclosing Circle problem (I'm puzzled why your google search didn't show up anything), and discussed here, here, here, and in many other places.
一种选择是 CGAL 计算几何算法库。它是开源的，但它也很大——我怀疑你遇到的最大问题是大海捞针。
当然（这在一定程度上是为了向马丁道歉），您可以使用谷歌轻松找到更有针对性的选项。如果您不介意 Prolog，当我尝试时，列出的第二项看起来不错，并且结果的第一页上至少有一个 C 示例和一个 Javascript。你几乎不能再向谷歌声称不知道这些词了。
One option is the CGAL Computational Geometry Algorithms Library. It is open source, but it is also large - the biggest problem you'll have, I suspect, is finding the needle in the haystack.
Of course (and this is partly in apology to Martin), you can easily find more focussed options using Google. The second item listed looked OK when I tried, if you don't mind Prolog, and there was at least one C example and one Javascript on the first page of results. And you can hardly claim not to know the words to Google for any more.