Tricky Median Question

Question

Given n points, choose a point in the given list such that the sum of distances to this point is minimum ,compared to all others.

Distance is measured in the following manner.

For a point (x,y) all 8 adjacent points have distance 1.

(x+1,y)(x+1,y+1),(x+1,y-1),(x,y+1),(x,y-1),(x-1,y)(x-1,y+1),(x-1,y-1)

EDIT

More clearer explanation.

A function foo is defined as

foo(point_a,point_b) = max(abs(point_a.x - point_b.x),abs(point_a.y - point_b.y))

Find a point x such that sum([foo(x,y) for y in list_of_points]) is minimum.

Example

Input:

0 1
2 5
3 1
4 0

Output

3 1

Eg: Distance between (4,5) and 6,7) is 2.

This can be done in O(n^2) time, by checking the sum of each pair. Is there any better algorithm to do it?

Answer 1

I can imagine a scheme better than O(n^2), at least in the common case.

Build a quadtree out of your input points. For each node in the tree, compute the number and average position of the points within that node. Then for each point, you can use the quadtree to compute its distance to all other points in less than O(n) time. If you're computing the distance from a point p to a distant quadtree node v, and v doesn't overlap the 45 degree diagonals from p, then the total distance from p to all the points in v is easy to compute (for v which are more horizontally than vertically separated from p, it is just v.num_points * |p.x - v.average.x|, and similarly using y coordinates if v is predominately vertically seperated). If v overlaps one of the 45 degree diagonals, recurse on its components.

That should beat O(n^2), at least when you can find a balanced quadtree to represent your points.

Answer 2

Scala

object Median extends App
{
    var(xMin,yMin,xMax,yMax,a,x)=(100000000,100000000,-100000000,-100000000,"",Array.fill(2)(""))
    var points=Array((100000000,100000000))
    a=readLine
    while(a!=null)
    {
        x=a.split(" ")
        if(x(0).toInt<xMin)xMin=x(0).toInt
        if(x(1).toInt<yMin)yMin=x(1).toInt
        if(x(0).toInt>xMax)xMax=x(0).toInt
        if(x(1).toInt>yMax)yMax=x(1).toInt
        points=points:+((x(0).toInt,x(1).toInt))
        a=readLine
    }
    var middlex=(xMax-xMin)/2
    var middley=(yMax-yMin)/2

    var meanPoint=(100000000,100000000)
    var shortestDistance=100000000

    points.map(x=>
        {
            var s=(math.abs(x._1-middlex).max(math.abs(x._2-middley)))
            if(s<shortestDistance)
            {
                shortestDistance=s
                meanPoint=x
            }
        })

    println(meanPoint)
}

Takes in the points, keeping track of the highest and lowest values of x and y.
Uses those values to find the midpoint of the square defined by xMin,yMin,xMax,yMax.
Finds the point with the shortest distance to that midpoint.

Never was very good with that Big O business, so I couldn't tell you what it is for this algorithm.

Answer 3

Haskell, 136 characters

main=interact$show.n.q.map read.words
n x=snd$minimum$map(\p@(a,b)->(sum$map(\(x,y)->max(abs$a-x)$abs$b-y)x,p))x
q[]=[]
q(x:y:z)=(x,y):q z

Answer 4

Python

def median(lst):
    """ sort of median of a list """
    return sorted(lst)[len(lst) / 2]

def median_point(points):
    xs = [point[0] for point in points]
    ys = [point[1] for point in points]
    return (median(xs), median(ys))

points = [
    (0, 1),
    (2, 5),
    (3, 1),
    (4, 0),
]

print median_point(points) # (3, 1)

It calculates the median of the x and y coordinates separately. I'm not positive that that's correct in all cases, but it kind of seems like it might be. It's O(n log n).