优雅的“左手”测试Polyline [英] Elegant "Left-of" test for Polyline

问题描述

给定：

• （X，Y）坐标，它是车辆的位置。 / li>
  
• （X，Y）的数组，它们是多段线中的顶点。请注意，折线仅包含直线段，不包含弧线。

我想要的：

• 计算车辆是在折线的左边还是右边（或者位于酒店顶层）。
  

我的方法： b
$ b

• 迭代所有线段，并计算到每个线段的距离。然后，对于最接近的部分，您可以做一个简单的剩余测试（如）。
  
  
  

  可能的问题：

  
  
  
  
  
  • 当三个点形成一个小于90度的角度时（如图像所示）复杂的情况出现。当车辆处于如下所示的红色区段中时，最近的区段可以是两者中的一个。但是，如果第一个段被选为最接近的段，那么 left-of 测试将产生右，否则离开。我们可以很容易地看到（至少，我希望），正确的结果应该是车辆是折线的。
  
  
  
  
  
  
  
  

  我的问题：

  
  
  
  
  
  • 我如何优雅地 处理这种特定情况？
  
  
  
  
  

  到目前为止我的解决方法：

  
  
  
  
  
  • 计算这两个线段上该点上的一个点，从顶点开始计算。
  
  
  • 计算距离使用欧几里得距离来保持计算点最接近的分段
  
  
  
  
  

  我对这个修补程序并不满意，因为我觉得我缺少一个更优雅的解决方案，我的修补程序感觉相当黑客。效率是关键，因为它被用于实时嵌入式系统中。

  
  
  

  现有的代码库是用C ++编写的，所以如果你想用特定的语言编写，C ++有我的偏爱。
  Thanks！

  
  
  

  
  我更改了我的修补程序指向一个平行点，因为我认为跟踪线段比计算外向法线更容易。 解决方案

一直没有活动，我相信它已经死了。不过，我有一个解决方案。

但是，左边的测试会产生 em>如果第一个细分受众群被选为最接近的细分受众群，那么离开，否则。 >你已经使用了一些含糊不清的语言。我将使用段来说明折线和象限中的线段，以说明由它们分隔的区域。因此，对于您的情况，您会看到一个红色象限，它似乎位于一个段的右侧并位于另一个的左侧。

如果左边的测试对不同的分段产生不同的答案，则应该对分段本身重做测试。在你的情况下，你会有：

• 象限在第一段的右边


• 象限位于第二段的左侧

两个线段都不同意象限所在的位置，因此您需要进一步消除歧义测试：

• 第二个分段位于第一个分段的右侧


• 第一个分段是在第二段的右侧

这可以让我们得出结论：第二段 之间第一部分和象限 - 因为这两部分中的每一部分位于第二部分的不同侧。因此，第二部分比第一部分接近象限，它的答案是左右测试应该用作正确的测试。

（I'我几乎可以肯定，你可以只使用两个消歧测试中的一个，为了清楚起见，我已经把它们都放进去了）。为了完整起见，我相信这个解决方案也是可行的考虑到您对效率和优雅的要求，因为它使用了您从一开始就使用的相同方法（左侧测试），因此它符合所有指定的条件：优雅，高效并且照顾问题。

Given:

• (X,Y) coordinate, which is the position of a vehicle.
• Array of (X,Y)'s, which are vertices in a polyline. Note that the polyline consists of straight segments only, no arcs.

What I want:

• To calculate whether the vehicle is to the left or to the right of the polyline (or on top, ofcourse).

My approach:

• Iterate over all line-segments, and compute the distance to each segment. Then for the closest segment you do a simple left-of test (as explained here for instance).

Possible issues:

• When three points form an angle smaller than 90 degrees (such as shown in the image blow), a more complicated scenario arises. When the vehicle is in the red segment as shown below, the closest segment can be either one of the two. However, the left-of test will yield right if the first segment is chosen as the closest segment, and left otherwise. We can easily see (at least, I hope), that the correct result should be that the vehicle is left of the polyline.

My question:

• How can I elegantly, but mostly efficiently take care of this specific situation?

My fix so far:

• Compute for both segments a point on that segment, starting from the vertex point.
• Compute the distance from the vehicle to both of the points, using Euclidian distance
• Keep the segment for which the computed point is the closest.

I am not very happy with this fix, because I feel like I am missing a far more elegant solution, my fix feels rather "hacky". Efficiency is key though, because it is used on a realtime embedded system.

Existing codebase is in C++, so if you want to write in a specific language, C++ has my preference. Thanks!

[edit] I changed my fix, from a perpendicular point to a parallel point, as I think it is easier to follow the line segment than compute the outward normal.

解决方案

This topic has been inactive for so long that I believe it's dead. I have a solution though.

However, the left-of test will yield right if the first segment is chosen as the closest segment, and left otherwise.

You've used slightly ambiguous language. I'm gonna use segments to speak of the line segments in the polyline and quadrants to speak of the areas delimited by them. So in your case, you'd have a red quadrant which seems to be on the right of one segment and on the left of the other.

If the left-of test yields different answers for different segments, you should redo the test on the segments themselves. In your case, you'd have:

• The quadrant is on the RIGHT of the first segment
• The quadrant is on the LEFT of the second segment

Both segments disagree on where the quadrant lies, so you do two further disambiguation tests:

• The second segment is on the RIGHT of the first segment
• The first segment is on the RIGHT of the second segment

This allows us to conclude that the second segment is in between the first segment and the quadrant—since each of those two lies on a different side of the second segment. Therefore, the second segment is "closer" to the quadrant than the first and it's answer to the left-right test should be used as the correct one.

(I'm almost sure you can do with only one of the two disambiguation tests, I've put both in for clarity)

For the sake of completeness: I believe this solution also accounts for your demands of efficiency and elegance, since it uses the same method you've been using from the start (the left-of test), so it meets all the conditions specified: it's elegant, it's efficient and it takes care of the problem.

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