Dividing a plane of points into two equal halves [closed]

Answer 1

The median equally divides a set of numbers in the manner similar to what you're trying to accomplish, and it can be computed in O(n) time using a selection algorithm (the writeup in Cormen et al is better, so you may want to look there instead). So, find the median of your x values M_x (or your y values M_y if you prefer) and set x = M_x (or y = M_y) and that line will be axially aligned and split your points equally.

If the nature of your problem requires that no more than one point lies on the line (if you have an odd number of points in your set, at least one of them will be on the line) and you discover that's what's happened (or you just want to guard against the possibility), rotate all of your points by some random angle, θ, and compute the median of the rotated points. You then rotate the median line you computed by -θ and it will evenly divide points.

The likelihood of randomly choosing θ such that the problem manifests itself again is very small with a finite number of points, but if it does, try again with a different θ.

Answer 2

Here is how I approach this problem (with the assumption that n is even and NO three points are collinear):

1) Pick up the point with smallest Y value. Let's call it point P.

2) Take this point as the new origin point, so that all other points will have positive Y values after this transformation.

3) For every other point (there are n - 1 points remaining), think it under the polar coordinate system. Each other point can be represented with a radius and angle. You could ignore the radius and just focus on the angle.

4) How can you find a line that split the points evenly? Find the median of (n - 1) angles. The line from point P to the point with that median angle will split the points evenly.

Time complexity for this algorithm is O(n).

Answer 3

1. Create an arbitrary line in that plane. Project each point onto that line a.k.a for each point, get the closest point on that line to that point.

2. Order those points along the line in either direction, and choose a point on that line such that there is an equal number of points on the line in either direction.

3. Get the line perpendicular to the first line which passes through that point. This line will have half the original points on either side.

There are some cases to avoid when doing this. Most importantly, if all the point are themselves on a single line, don't choose a perpendicular line which passes through it. In fact, choose that line itself so you don't have to worry about projecting the points. In terms of the actual mathematics behind this, vector projections will be very useful.

Answer 4

This is a modification of Dividing a plane of points into two equal halves which allows for the case with overlapping points (in which case, it will say whether or not the answer exists).

If number of points is odd, return "impossible".

Pick a random line (two random points)
Project all points onto this line (`O(N)` operation)
    (i.e. we pretend this line is our new X'-axis, and write down the
     X'-coordinate of each point)
Perform any median-finding algorithm on the X'-coordinates
    (`O(N)` or faster-if-desired operation)
    (returns 2 medians if no overlapping points)
Return the line perpendicular to original random line that splits the medians

In rare case of overlapping points, repeat a few times (it would take
a pathological case to prevent a solution from existing).

This is O(N) unlike other proposed solutions.

Assuming a solution exists, the above method will probably terminate, though I don't have a proof.

Try the above algorithm a few times unless you detect overlapping points. If you detect a high number of overlapping points, you may be in for a rough ride, but there is a terribly inefficient brute-force solution that involves checking all possible angles:

For every "critical slope range", perform the above algorithm 
  by choosing a line with a slope within the range.
If all critical slope ranges fail, the solution is impossible.

A critical angle is defined as the angle which could possibly change the result (imagine the solution to a previous answer, rotate the entire set of points until one or more points swaps position with one or more other points, crossing the dividing line. There are only finitely many of these, and I think they are bounded by the number of points, so you're probably looking at something in the range O(N^2)-O(N^2 log(N)) if you have overlapping points, for a brute-force approach.