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Algorithm to find points that are furthest apart -- better than O(n^2)?

In my program, I have a set of points. For purposes of rescaling, I am searching for the two nodes that are furthest apart, and then computing a factor by which to multiply all coordinates so that the maximum distance is equal to some predefined one I define.

The algorithm I am using to find the two points furthest apart is, however, problematic for large sets of points, as it is O(n^2); pseudocode (distances that were already calculated are skipped):

 for each point in points:
     for each other point in points:
         if distance between point and other point > max
             max = distance between point and other point

Is there something faster?

like image 682
houbysoft Avatar asked Jun 29 '11 16:06

houbysoft


1 Answers

If you just need the scale and not the exact points you can do this in O(n) time with some error margin. Think about the simple case of making a bounding box. Calculate the minimum x value from all the points, the maximum x, the minimum y and the maximum y. These four numbers give you a maximum bounding box around your points with max error of 1 - (1/sqrt(2)) about 30%. You can reduce this by adding more sides to your square. Think about the case of an octagon. To calculate the min and max values for the other sides you have to rotate your coordinate system.

Error vs run time breaks down like this.

shape - run time - max error

  • square - 2N - 30%
  • octagon - 4N - 16%
  • 16 sides - 8N - 4%
  • 32 sides - 16N - 1%

Here's the equation for the max error I came up with.

angle = 180 / sides
max_error = (1 / cos angle) - cos angle

Let me know if I should add a diagram explaining this.

like image 159
gradbot Avatar answered Oct 12 '22 01:10

gradbot