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Greatest linear dimension 2d set of points可能重复:最大线性尺寸2d点集
I could compute the distance between each point and take the largest but that doesn't sound like a very efficient way to do it when there are a large (> 1000) number of points.
我可以计算每个点之间的距离并取最大值,但是当有大(> 1000)点数时,这听起来不是一种非常有效的方法。
Note: This is for iPhone so I don't have a ton of processing power.
注意:这适用于iPhone,因此我没有大量的处理能力。
4 个解决方案
#1
8
You're asking to compute the diameter of the set. The standard technique is to first compute the convex hull, which reduces the problem to finding the diameter of a convex polygon. Even in the case where you don't eliminate any points, this added information is exactly what's needed to solve the problem efficiently. However, finding the diameter of a convex polygon is not entirely trivial; several published papers with algorithms for this task turn out to be incorrect.
你要求计算集合的直径。标准技术是首先计算凸包,这减少了找到凸多边形直径的问题。即使在您没有消除任何积分的情况下,这些增加的信息正是有效解决问题所需要的。然而,找到凸多边形的直径并非完全无关紧要;一些已发表的论文与此任务的算法结果是不正确的。
Here's a fairly readable discussion of a correct O(n) algorithm for the task (where n is the number of points in the convex hull).
这是对任务的正确O(n)算法的相当可读的讨论(其中n是凸包中的点数)。
Also, note the the iphone isn't that limited; a carefully written implementation of even the completely naive algorithm can handle 1000 points in less than a tenth of a second. Of course, using the correct algorithm will let you go much faster =)
另外,请注意iphone不是那么有限;即使是完全天真的算法,精心编写的实现也可以在不到十分之一秒的时间内处理1000个点。当然,使用正确的算法会让你走得更快=)
#2
9
Why not just compute the convex hull of the points? Depending on the algorithm you use, it takes either O(n) or O(n log n) time and eliminates all the inner points from consideration. Then, only check these outermost points to find the two that are the farthest away.
为什么不计算点的凸包呢?根据您使用的算法,它需要O(n)或O(n log n)时间并消除考虑的所有内部点。然后,只检查这些最外面的点,找到距离最远的两个点。
#3
0
Start with point with lowest x-coord. (Call it Point X) Construct set of "boundary points" starting with point x, and vertical line through the point, There should be no other points to left of PointX) find the next point in boundary by slowly rotating the line clockwise (Or counterclockwise) until the line touches some other point, (see below). Add that point to the set and repeat with that next point to get the next one, until you eventually get back to the original point x. You npw have a set of points forming the boundary of the complete set. Compare distance between each pair in this reduced set to find the pair that are furthest apart.
从具有最低x-coord的点开始。 (称之为点X)构造从点x开始的“边界点”集合,以及通过点的垂直线,PointX左边应该没有其他点)通过顺时针缓慢旋转线来找到边界中的下一个点(或者逆时针方向)直到线接触其他一些点,(见下文)。将该点添加到该集合并重复该下一个点以获取下一个点,直到最终返回到原始点x。你npw有一组点形成整套的边界。比较该减少的集合中每对之间的距离,以找到最远的对。
To "rotate the line" (to find each sequential boundary point), take the point which is "furthest" in the direction perpindicular to the line used for the last boundary point, and construct a new line between the last boundary point and that "next" point. Then verify that there are no other points furthur in the new perpindicular direction formed by that new line. If there are no other points "furthur out" in the dirtection perpindicular to this line or to the last line, then this is the right cjhoice for the next boundary point, if there is such a point, switch to that one and retest.
要“旋转直线”(找到每个连续的边界点),将在垂直方向上“最远”的点取向用于最后一个边界点的直线,并在最后一个边界点与“之间”构造一条新线。下一个“点。然后验证在新线形成的新的perpindicular方向上没有其他点。如果在这条线或最后一条线路上没有其他点“向外”,那么这就是下一个边界点的正确选择,如果有这样一个点,切换到那个并重新测试。
#4
0
See these pages (the one linked to and the pages reachable by clicking on the "next" links) on computing the diameter of the convex hull of a set of points.
在计算一组点的凸包的直径时,请参阅这些页面(链接到的页面和通过单击“下一个”链接可到达的页面)。
My quick summary:
我的快速摘要:
- compute set of points in convex hull (= O(n log n), the only time you get O(n) is if you sort the list first which takes O(n log n) anyway)
- order along boundary (you get this for free if you use a Graham scan for #1)
- use one of the O(n) diameter algorithms to scan for antipodal points with greatest diameter. Shamos algorithm looks good to me as it's one of the rotating calipers algorithms.
计算凸包中的点集(= O(n log n),唯一得到O(n)的是你首先对列表进行排序,无论如何都需要O(n log n)
沿着边界排序(如果你使用格雷厄姆扫描#1,你可以免费获得这个)
使用O(n)直径算法之一扫描具有最大直径的对映点。 Shamos算法对我来说很好,因为它是旋转卡尺算法之一。
#1
8
You're asking to compute the diameter of the set. The standard technique is to first compute the convex hull, which reduces the problem to finding the diameter of a convex polygon. Even in the case where you don't eliminate any points, this added information is exactly what's needed to solve the problem efficiently. However, finding the diameter of a convex polygon is not entirely trivial; several published papers with algorithms for this task turn out to be incorrect.
你要求计算集合的直径。标准技术是首先计算凸包,这减少了找到凸多边形直径的问题。即使在您没有消除任何积分的情况下,这些增加的信息正是有效解决问题所需要的。然而,找到凸多边形的直径并非完全无关紧要;一些已发表的论文与此任务的算法结果是不正确的。
Here's a fairly readable discussion of a correct O(n) algorithm for the task (where n is the number of points in the convex hull).
这是对任务的正确O(n)算法的相当可读的讨论(其中n是凸包中的点数)。
Also, note the the iphone isn't that limited; a carefully written implementation of even the completely naive algorithm can handle 1000 points in less than a tenth of a second. Of course, using the correct algorithm will let you go much faster =)
另外,请注意iphone不是那么有限;即使是完全天真的算法,精心编写的实现也可以在不到十分之一秒的时间内处理1000个点。当然,使用正确的算法会让你走得更快=)
#2
9
Why not just compute the convex hull of the points? Depending on the algorithm you use, it takes either O(n) or O(n log n) time and eliminates all the inner points from consideration. Then, only check these outermost points to find the two that are the farthest away.
为什么不计算点的凸包呢?根据您使用的算法,它需要O(n)或O(n log n)时间并消除考虑的所有内部点。然后,只检查这些最外面的点,找到距离最远的两个点。
#3
0
Start with point with lowest x-coord. (Call it Point X) Construct set of "boundary points" starting with point x, and vertical line through the point, There should be no other points to left of PointX) find the next point in boundary by slowly rotating the line clockwise (Or counterclockwise) until the line touches some other point, (see below). Add that point to the set and repeat with that next point to get the next one, until you eventually get back to the original point x. You npw have a set of points forming the boundary of the complete set. Compare distance between each pair in this reduced set to find the pair that are furthest apart.
从具有最低x-coord的点开始。 (称之为点X)构造从点x开始的“边界点”集合,以及通过点的垂直线,PointX左边应该没有其他点)通过顺时针缓慢旋转线来找到边界中的下一个点(或者逆时针方向)直到线接触其他一些点,(见下文)。将该点添加到该集合并重复该下一个点以获取下一个点,直到最终返回到原始点x。你npw有一组点形成整套的边界。比较该减少的集合中每对之间的距离,以找到最远的对。
To "rotate the line" (to find each sequential boundary point), take the point which is "furthest" in the direction perpindicular to the line used for the last boundary point, and construct a new line between the last boundary point and that "next" point. Then verify that there are no other points furthur in the new perpindicular direction formed by that new line. If there are no other points "furthur out" in the dirtection perpindicular to this line or to the last line, then this is the right cjhoice for the next boundary point, if there is such a point, switch to that one and retest.
要“旋转直线”(找到每个连续的边界点),将在垂直方向上“最远”的点取向用于最后一个边界点的直线,并在最后一个边界点与“之间”构造一条新线。下一个“点。然后验证在新线形成的新的perpindicular方向上没有其他点。如果在这条线或最后一条线路上没有其他点“向外”,那么这就是下一个边界点的正确选择,如果有这样一个点,切换到那个并重新测试。
#4
0
See these pages (the one linked to and the pages reachable by clicking on the "next" links) on computing the diameter of the convex hull of a set of points.
在计算一组点的凸包的直径时,请参阅这些页面(链接到的页面和通过单击“下一个”链接可到达的页面)。
My quick summary:
我的快速摘要:
- compute set of points in convex hull (= O(n log n), the only time you get O(n) is if you sort the list first which takes O(n log n) anyway)
- order along boundary (you get this for free if you use a Graham scan for #1)
- use one of the O(n) diameter algorithms to scan for antipodal points with greatest diameter. Shamos algorithm looks good to me as it's one of the rotating calipers algorithms.
计算凸包中的点集(= O(n log n),唯一得到O(n)的是你首先对列表进行排序,无论如何都需要O(n log n)
沿着边界排序(如果你使用格雷厄姆扫描#1,你可以免费获得这个)
使用O(n)直径算法之一扫描具有最大直径的对映点。 Shamos算法对我来说很好,因为它是旋转卡尺算法之一。