Minimize maximum manhattan distance of a point to a set of points

问题

For 3 points in 2D :

P1(x1,y1), 
P2(x2,y2), 
P3(x3,y3) 

I need to find a point P(x,y), such that the maximum of the manhattan distances

max(dist(P,P1), 
    dist(P,P2), 
    dist(P,P3))

will be minimal.

Any ideas about the algorithm?

I would really prefer an exact algorithm.

回答1:

There is an exact, noniterative algorithm for the problem; as Knoothe pointed out, the Manhattan distance is rotationally equivalent to the Chebyshev distance, and P is trivially computable for the Chebyshev distance as the mean of the extreme coordinates.

The points reachable from P within the Manhattan distance x form a diamond around P. Therefore, we need to find the minimum diamond that encloses all points, and its center will be P.

If we rotate the coordinate system by 45 degrees, the diamond is a square. Therefore, the problem can be reduced to finding the smallest enclosing square of the points.

The center of a smallest enclosing square can be found as the center of the smallest enclosing rectangle (which is trivially computed as the max and min of the coordinates). There is an infinite number of smallest enclosing squares, since you can shift the center along the shorter edge of the minimum rectangle and still have a minimal enclosing square. For our purposes, we can simply use the one whose center coincides with the enclosing rectangle.

So, in algorithmic form:

1. Rotate and scale the coordinate system by assigning x' = x/sqrt(2) - y/sqrt(2), y' = x/sqrt(2) + y/sqrt(2)
2. Compute x'_c = (max(x'_i) + min(x'_i))/2, y'_c = (max(y'_i) + min(y'_i))/2
3. Rotate back with x_c = x'_c/sqrt(2) + y'_c/sqrt(2), y_c = - x'_c/sqrt(2) + y'_c/sqrt(2)

Then x_c and y_c give the coordinates of P.

回答2:

If an approximate solution is okay, you could try a simple optimization algorithm. Here's an example, in Python

import random
def opt(*points):
    best, dist = (0, 0), 99999999
    for i in range(10000):
        new = best[0] + random.gauss(0, .5), best[1] + random.gauss(0, .5)
        dist_new = max(abs(new[0] - qx) + abs(new[1] - qy) for qx, qy in points)
        if dist_new < dist:
            best, dist = new, dist_new
            print new, dist_new
    return best, dist

Explanation: We start with the point (0, 0), or any other random point, and modify it a few thousand times, each time keeping the better of the new and the previously best point. Gradually, this will approximate the optimum.

Note that simply picking the mean or median of the three points, or solving for x and y independently does not work when minimizing the maximum manhattan distance. Counter-example: Consider the points (0,0), (0,20) and (10,10), or (0,0), (0,1) and (0,100). If we pick the mean of the most separated points, this would yield (10,5) for the first example, and if we take the median this would be (0,1) for the second example, which both have a higher maximum manhattan distance than the optimum.

Update: Looks like solving for x and y independently and taking the mean of the most distant points does in fact work, provided that one does some pre- and postprocessing, as pointed out by thiton.

来源：https://stackoverflow.com/questions/15346216/minimize-maximum-manhattan-distance-of-a-point-to-a-set-of-points