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

Answer 1

I picked up the idea from Moron and andand and continued to form a deterministic O(n) algorithm.

I also assumed that the points are distinct and n is even (thought the algorithm can be changed so that uneven n with one point on the dividing line are also supported).

The algorithm tries to divide the points with a vertical line between them. This only fails if the points in the middle have the same x value. In that case the algorithm determines how many points with the same x value have to be on the left and lower site and and accordingly rotates the line.

I'll try to explain with an example. Let's asume we have 16 points on a plane. First we need to get the point with the 8th greatest x-value and the point with the 9th greatest x-value. With a selection algorithm this is possible in O(n), as pointed out in another answer. If the x-value of that points is different, we are done. We create a vertical line between that two points and that splits the points equal.

Problematically now is if the x-values are equal. So we have 3 sets of points. That on the left side (x < x_a), in the middle (x = x_a) and that on the right side (x > x_a). The idea now is to count the points on the left side and calculate how many points from the middle needs to go there, so that half of the points are on that side. We can ignore the right side here because if we have half of the points on the left side, the over half must be on the right side.

So let's asume we have we have 3 points (c=3) on the left side, 6 in the middle and 7 on the right side (the algorithm doesn't know the count from the middle or right side, because it doesn't need it, but we could also determine it in O(n)). We need 8-3=5 points from the middle to go on the left side. The points we already got in the first step are useless now, because they are only determined by the x-value and can be any of the points in the middle.

We want the 5 points from the middle with the lowest y-value on the left side and the point with the highest y-value on the right side. Again using the selection algorithm, we get the point with the 5th greatest y-value and the point with the 6th greatest y-value. Both points will have the x-value equal to x_a, else we wouldn't get to this step, because there would be a vertical line.

Now we can create the point Q in the middle of that two points. Thats one point from the resulting line. Another point is needed, so that no points from the left or right side are divided. To get that point we need the point from the left side, that has the lowest angle (b_h) between the the vertical line at x_a and the line determined by that point and Q. We also need that point from the right side (with angle a_g). The new point R is between the point with the lower angle and a point on the vertical line (if the lower angle is on the left side a point above Q and if the lower angle is on the right side a point below Q).

The line determined by Q and R divides the points in the middle so that there are a even number of points on both sides. It doesn't divide any points on the left or right side, because if it would that point would have a lower angle and would have been choosen to calculate R.

From the view of a mathematican that should work well in O(n). For computer programs it is fairly easy to find a case where precision becomes a problem. An example with 4 points would be A(0, 100000000), B(0, 100000001), C(0, 0), D(0.0000001, 0). In this example Q would be (0, 100000000.5) and R (0.00000005, 0). Which gives B and C on the left side and A and D on the right side. But it is possible that A and B are both on the dividing line, because of rounding errors. Or maybe only one of them. So it belongs to the input values if this algorithm suits to the requirements.

1. get that two points P_a(x_a, y_a) and P_b(x_b, y_b)
   which are the medians based on the x values > O(n)
2. if x_a != x_b you can stop here
   because a y-axis parallel line between that two points is the result > O(1)
3. get all points where the x value equals x_a > O(n)
4. count points with x value less than x_a as c > O(n)
5. get the lowest point P_c based on the y values from the points from 3. > O(n)
6. get the greatest point P_d based on the y values from the points from 3. > O(n)
7. get the (n/2-c)th greatest point P_e based on the y values from the points from 3. > O(n)
8. also get the next greatest point P_f based on the y values from the points from 3. > O(n)
9. create a new point Q (x_a, (y_e+y_f)/2) between P_e and P_f > O(1)
10. for all points P_i calculate
    the angle a_i between P_c, Q and P_i and
    the angle b_i between P_d, Q and P_i > O(n)
11. get the point P_g with the lowest angle a_g (with a_g>0° and a_g<180°) > O(n)
12. get the point P_h with the lowest angle b_h (with b_h>0° and b_h<180°) > O(n)
13. if there aren't any P_g or P_h (all points have same x value)
    create a new point R (x_a+1, 0) anywhere but with a different x value than x_a
    else if a_g is lower than b_h
    create a new point R ((x_c+x_g)/2, (y_c+y_g)/2) between P_c and P_g
    else
    create a new point R ((x_d+x_h)/2, (y_d+y_h)/2) between P_d and P_h > O(1)
14. the line determined by Q and R divides the points > O(1)