给定一组点，找出三个点是否共线

Question

What is the best algorithm to find if any three points are collinear in a set of points say n. Please also explain the complexity if it is not trivial.

Thanks
Bala

Answer 1

If you can come up with a better than O(N^2) algorithm, you can publish it!

This problem is 3-SUM Hard , and whether there is a sub-quadratic algorithm (ie better than O(N^2)) for it is an open problem. Many common computational geometry problems (including yours) have been shown to be 3SUM hard and this class of problems is growing. Like NP-Hardness, the concept of 3SUM-Hardness has proven useful in proving 'toughness' of some problems.

For a proof that your problem is 3SUM hard, refer to the excellent surver paper here: http://www.cs.mcgill.ca/~jking/papers/3sumhard.pdf

Your problem appears on page 3 (conveniently called 3-POINTS-ON-LINE) in the above mentioned paper.

So, the currently best known algorithm is O(N^2) and you already have it :-)

Answer 2

A simple O(d*N^2) time and space algorithm, where d is the dimensionality and N is the number of points (probably not optimal):

• Create a bounding box around the set of points (make it big enough so there are no points on the boundary)
• For each pair of points, compute the line passing through them.
• For each line, compute its two collision points with the bounding box.
• The two collision points define the original line, so if there any matching lines they will also produce the same two collision points.
• Use a hash set to determine if there are any duplicate collision point pairs.
• There are 3 collinear points if and only if there were duplicates.

Answer 3

Another simple (maybe even trivial) solution which doesn't use a hash table, runs in O(n ² log n) time, and uses O(n) space:

Let S be a set of points, we will describe an algorithm which finds out whether or not S contains some three collinear points.

1. For each point o in S do:
   1. Pass a line L parallel to the x -axis through o .
   2. Replace every point in S below L , with its reflection. (For example if L is the x axis, (a,-x) for x>0 will become (a,x) after the reflection). Let the new set of points be S'
   3. The angle of each point p in S' , is the right angle of the segment po with the line L . Let us sort the points S' by their angles.
   4. Walk through the sorted points in S' . If there are two consecutive points which are collinear with o - return true.
2. If no collinear points were found in the loop - return false.

The loop runs n times, and each iteration performs nlog n steps. It is not hard to prove that if there're three points on a line they'll be found, and we'll find nothing otherwise.