我怎样才能简化一个复杂的多边形？ [英] How could I simplify a complex polygon?

问题描述

最近我一直在想如何将一个复杂的多边形转化为一个非复杂的多边形。这是怎么做的？

这是我想要做的事情：

我将以JavaScript结尾，但是任何形式的解决方案都是好的（语言，算法或简单的英语）。

我会使用与手工绘制多边形时相同的启发式（这可能不是计算多边形最有效的数学算法，但可能是最容易理解/实现的）。

1. 从一个点开始


2. 查找我当前的点与我尝试访问的点之间的所有交点


3. 如果不存在，请到下一个点
   
4. 如果有，则绘制到此处，然后将下一个点设置到下一个点。 $ b
5. 如果您不回到开头，那么转到2。
   

这是jsfiddle上的一个示例实现。注意：它没有优化。

Recently I've been thinking about how to transform a complex polygon into a non-complex polygon. How is this done?

This is the sort of thing I want to do:

I'm going to end up with JavaScript when I'm done, but any form of a solution is fine (language, algorithm, or just plain English).

解决方案

I would use the same heuristic that I would use when drawing the polygon by hand (which is probably not the most mathematically efficient way to caluclaute that polygon, but probably the easiest to understand/implement).

1. Start at a point
2. Find all the intersections between my current point and the point I'm trying to get to
3. If none exist draw to the next point
4. If one does, then draw to there, and then set the next point to the next point from there
5. If you aren't back to the beginning then goto 2.

Here is an example implementation on jsfiddle. Note: it isn't optimized.

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