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Optimum location of point to minimize total distance

  • Difficulty Level : Hard
  • Last Updated : 11 Sep, 2021

Given a set of points as and a line as ax+by+c = 0. We need to find a point on given line for which sum of distances from given set of points is minimum. 

Example: 

In above figure optimum location of point of x - y - 3 = 0 line 
is (2, -1), whose total distance with other points is 20.77, 
which is minimum obtainable total distance.

If we take one point on given line at infinite distance then total distance cost will be infinite, now when we move this point on line towards given points the total distance cost starts decreasing and after some time, it again starts increasing which reached to infinite on the other infinite end of line so distance cost curve looks like a U-curve and we have to find the bottom value of this U-curve. 

As U-curve is not monotonically increasing or decreasing we can’t use binary search for finding bottom most point, here we will use ternary search for finding bottom most point, ternary search skips one third of search space at each iteration, you can read more about ternary search here



So solution proceeds as follows, we start with low and high initialized as some smallest and largest values respectively, then we start iteration, in each iteration we calculate two mids, mid1 and mid2, which represent 1/3rd and 2/3rd position in search space, we calculate total distance of all points with mid1 and mid2 and update low or high by comparing these distance cost, this iteration continues until low and high become approximately equal. 

C++




//  C/C++ program to find optimum location and total cost
#include <bits/stdc++.h>
using namespace std;
#define sq(x) ((x) * (x))
#define EPS 1e-6
#define N 5
 
//  structure defining a point
struct point {
    int x, y;
    point() {}
    point(int x, int y)
        : x(x)
        , y(y)
    {
    }
};
 
//  structure defining a line of ax + by + c = 0 form
struct line {
    int a, b, c;
    line(int a, int b, int c)
        : a(a)
        , b(b)
        , c(c)
    {
    }
};
 
//  method to get distance of point (x, y) from point p
double dist(double x, double y, point p)
{
    return sqrt(sq(x - p.x) + sq(y - p.y));
}
 
/*  Utility method to compute total distance all points
    when choose point on given line has x-coordinate
    value as X   */
double compute(point p[], int n, line l, double X)
{
    double res = 0;
 
    //  calculating Y of chosen point by line equation
    double Y = -1 * (l.c + l.a * X) / l.b;
    for (int i = 0; i < n; i++)
        res += dist(X, Y, p[i]);
 
    return res;
}
 
//  Utility method to find minimum total distance
double findOptimumCostUtil(point p[], int n, line l)
{
    double low = -1e6;
    double high = 1e6;
 
    // loop until difference between low and high
    // become less than EPS
    while ((high - low) > EPS) {
        // mid1 and mid2 are representative x co-ordiantes
        // of search space
        double mid1 = low + (high - low) / 3;
        double mid2 = high - (high - low) / 3;
 
        //
        double dist1 = compute(p, n, l, mid1);
        double dist2 = compute(p, n, l, mid2);
 
        // if mid2 point gives more total distance,
        // skip third part
        if (dist1 < dist2)
            high = mid2;
 
        // if mid1 point gives more total distance,
        // skip first part
        else
            low = mid1;
    }
 
    // compute optimum distance cost by sending average
    // of low and high as X
    return compute(p, n, l, (low + high) / 2);
}
 
//  method to find optimum cost
double findOptimumCost(int points[N][2], line l)
{
    point p[N];
 
    //  converting 2D array input to point array
    for (int i = 0; i < N; i++)
        p[i] = point(points[i][0], points[i][1]);
 
    return findOptimumCostUtil(p, N, l);
}
 
//  Driver code to test above method
int main()
{
    line l(1, -1, -3);
    int points[N][2] = {
        { -3, -2 }, { -1, 0 }, { -1, 2 }, { 1, 2 }, { 3, 4 }
    };
    cout << findOptimumCost(points, l) << endl;
    return 0;
}

Java




// A Java program to find optimum location
// and total cost
class GFG {
    static double sq(double x) { return ((x) * (x)); }
    static int EPS = (int)(1e-6) + 1;
    static int N = 5;
 
    // structure defining a point
    static class point {
        int x, y;
        point() {}
 
        public point(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
    };
 
    // structure defining a line of ax + by + c = 0 form
    static class line {
        int a, b, c;
 
        public line(int a, int b, int c)
        {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    };
 
    // method to get distance of point (x, y) from point p
    static double dist(double x, double y, point p)
    {
        return Math.sqrt(sq(x - p.x) + sq(y - p.y));
    }
 
    /* Utility method to compute total distance all points
        when choose point on given line has x-coordinate
        value as X */
    static double compute(point p[], int n, line l,
                          double X)
    {
        double res = 0;
 
        // calculating Y of chosen point by line equation
        double Y = -1 * (l.c + l.a * X) / l.b;
        for (int i = 0; i < n; i++)
            res += dist(X, Y, p[i]);
 
        return res;
    }
 
    // Utility method to find minimum total distance
    static double findOptimumCostUtil(point p[], int n,
                                      line l)
    {
        double low = -1e6;
        double high = 1e6;
 
        // loop until difference between low and high
        // become less than EPS
        while ((high - low) > EPS) {
            // mid1 and mid2 are representative x
            // co-ordiantes of search space
            double mid1 = low + (high - low) / 3;
            double mid2 = high - (high - low) / 3;
 
            double dist1 = compute(p, n, l, mid1);
            double dist2 = compute(p, n, l, mid2);
 
            // if mid2 point gives more total distance,
            // skip third part
            if (dist1 < dist2)
                high = mid2;
 
            // if mid1 point gives more total distance,
            // skip first part
            else
                low = mid1;
        }
 
        // compute optimum distance cost by sending average
        // of low and high as X
        return compute(p, n, l, (low + high) / 2);
    }
 
    // method to find optimum cost
    static double findOptimumCost(int points[][], line l)
    {
        point[] p = new point[N];
 
        // converting 2D array input to point array
        for (int i = 0; i < N; i++)
            p[i] = new point(points[i][0], points[i][1]);
 
        return findOptimumCostUtil(p, N, l);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        line l = new line(1, -1, -3);
        int points[][] = { { -3, -2 },
                           { -1, 0 },
                           { -1, 2 },
                           { 1, 2 },
                           { 3, 4 } };
        System.out.println(findOptimumCost(points, l));
    }
}
 
// This code is contributed by Rajput-Ji

Python3




# A Python3 program to find optimum location
# and total cost
import math
 
class Optimum_distance:
     
    # Class defining a point
    class Point:
         
        def __init__(self, x, y):
             
            self.x = x
            self.y =
         
    # Class defining a line of ax + by + c = 0 form
    class Line:
         
        def __init__(self, a, b, c):
             
            self.a = a
            self.b = b
            self.c = c
         
    # Method to get distance of point
    # (x, y) from point p
    def dist(self, x, y, p):
         
        return math.sqrt((x - p.x) ** 2 +
                         (y - p.y) ** 2)
       
    # Utility method to compute total distance
    # all points when choose point on given
    # line has x-coordinate value as X
    def compute(self, p, n, l, x):
         
        res = 0
         
        y = -1 * (l.a*x + l.c) / l.b
         
        # Calculating Y of chosen point
        # by line equation
        for i in range(n):
            res += self.dist(x, y, p[i])
             
        return res
     
    # Utility method to find minimum total distance
    def find_Optimum_cost_untill(self, p, n, l):
         
        low = -1e6
        high = 1e6
         
        eps = 1e-6 + 1
         
         
        # Loop until difference between low
        # and high become less than EPS
        while((high - low) > eps):
           
              # mid1 and mid2 are representative x
            # co-ordiantes of search space
            mid1 = low + (high - low) / 3
            mid2 = high - (high - low) / 3
             
            dist1 = self.compute(p, n, l, mid1)
            dist2 = self.compute(p, n, l, mid2)
             
            # If mid2 point gives more total
            # distance, skip third part
            if (dist1 < dist2):
                high = mid2
                 
            # If mid1 point gives more total
            # distance, skip first part
            else:
                low = mid1
                 
        # Compute optimum distance cost by
        # sending average of low and high as X
        return self.compute(p, n, l, (low + high) / 2)
     
    # Method to find optimum cost
    def find_Optimum_cost(self, p, l):
         
        n = len(p)
        p_arr = [None] * n
         
        # Converting 2D array input to point array
        for i in range(n):
            p_obj = self.Point(p[i][0], p[i][1])
            p_arr[i] =  p_obj
             
        return self.find_Optimum_cost_untill(p_arr, n, l)
       
 # Driver Code
if __name__ == "__main__":
     
    obj = Optimum_distance()
    l = obj.Line(1, -1, -3)
     
    p = [ [ -3, -2 ], [ -1, 0 ],
          [ -1, 2 ], [ 1, 2 ],
          [ 3, 4 ] ]
     
    print(obj.find_Optimum_cost(p, l))
     
# This code is contributed by Sulu_mufi

C#




// C# program to find optimum location
// and total cost
using System;
 
class GFG {
    static double sq(double x) { return ((x) * (x)); }
 
    static int EPS = (int)(1e-6) + 1;
    static int N = 5;
 
    // structure defining a point
    public class point {
        public int x, y;
        public point() {}
 
        public point(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
    };
 
    // structure defining a line
    // of ax + by + c = 0 form
    public class line {
        public int a, b, c;
 
        public line(int a, int b, int c)
        {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    };
 
    // method to get distance of
    // point (x, y) from point p
    static double dist(double x, double y, point p)
    {
        return Math.Sqrt(sq(x - p.x) + sq(y - p.y));
    }
 
    /* Utility method to compute total distance
    of all points when choose point on
    given line has x-coordinate value as X */
    static double compute(point[] p, int n, line l,
                          double X)
    {
        double res = 0;
 
        // calculating Y of chosen point
        // by line equation
        double Y = -1 * (l.c + l.a * X) / l.b;
        for (int i = 0; i < n; i++)
            res += dist(X, Y, p[i]);
 
        return res;
    }
 
    // Utility method to find minimum total distance
    static double findOptimumCostUtil(point[] p, int n,
                                      line l)
    {
        double low = -1e6;
        double high = 1e6;
 
        // loop until difference between
        // low and high become less than EPS
        while ((high - low) > EPS) {
            // mid1 and mid2 are representative
            // x co-ordiantes of search space
            double mid1 = low + (high - low) / 3;
            double mid2 = high - (high - low) / 3;
 
            double dist1 = compute(p, n, l, mid1);
            double dist2 = compute(p, n, l, mid2);
 
            // if mid2 point gives more total distance,
            // skip third part
            if (dist1 < dist2)
                high = mid2;
 
            // if mid1 point gives more total distance,
            // skip first part
            else
                low = mid1;
        }
 
        // compute optimum distance cost by
        // sending average of low and high as X
        return compute(p, n, l, (low + high) / 2);
    }
 
    // method to find optimum cost
    static double findOptimumCost(int[, ] points, line l)
    {
        point[] p = new point[N];
 
        // converting 2D array input to point array
        for (int i = 0; i < N; i++)
            p[i] = new point(points[i, 0], points[i, 1]);
 
        return findOptimumCostUtil(p, N, l);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        line l = new line(1, -1, -3);
        int[, ] points = { { -3, -2 },
                           { -1, 0 },
                           { -1, 2 },
                           { 1, 2 },
                           { 3, 4 } };
        Console.WriteLine(findOptimumCost(points, l));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// A JavaScript program to find optimum location
// and total cost
 
function sq(x)
{
    return x*x;
}
 
let EPS = (1e-6) + 1;
let N = 5;
 
// structure defining a point
class point
{
    constructor(x,y)
    {
        this.x=x;
        this.y=y;
    }
}
 
// structure defining a line of ax + by + c = 0 form
class line
{
    constructor(a,b,c)
    {
        this.a = a;
            this.b = b;
            this.c = c;
    }
     
}
 
// method to get distance of point (x, y) from point p
function dist(x,y,p)
{
    return Math.sqrt(sq(x - p.x) + sq(y - p.y));
}
 
/* Utility method to compute total distance all points
        when choose point on given line has x-coordinate
        value as X */
function compute(p,n,l,X)
{
    let res = 0;
  
        // calculating Y of chosen point by line equation
        let Y = -1 * (l.c + l.a * X) / l.b;
        for (let i = 0; i < n; i++)
            res += dist(X, Y, p[i]);
  
        return res;
}
// Utility method to find minimum total distance
function findOptimumCostUtil(p,n,l)
{
     let low = -1e6;
        let high = 1e6;
  
        // loop until difference between low and high
        // become less than EPS
        while ((high - low) > EPS) {
            // mid1 and mid2 are representative x
            // co-ordiantes of search space
            let mid1 = low + (high - low) / 3;
            let mid2 = high - (high - low) / 3;
  
            let dist1 = compute(p, n, l, mid1);
            let dist2 = compute(p, n, l, mid2);
  
            // if mid2 point gives more total distance,
            // skip third part
            if (dist1 < dist2)
                high = mid2;
  
            // if mid1 point gives more total distance,
            // skip first part
            else
                low = mid1;
        }
  
        // compute optimum distance cost by sending average
        // of low and high as X
        return compute(p, n, l, (low + high) / 2);
}
 
// method to find optimum cost
function findOptimumCost(points,l)
{
    let p = new Array(N);
  
        // converting 2D array input to point array
        for (let i = 0; i < N; i++)
            p[i] = new point(points[i][0], points[i][1]);
  
        return findOptimumCostUtil(p, N, l);
}
 
// Driver Code
let l = new line(1, -1, -3);
let points= [[ -3, -2 ],
             [ -1, 0 ],
             [ -1, 2 ],
             [ 1, 2 ],
             [ 3, 4 ]];
document.write(findOptimumCost(points, l));
 
 
// This code is contributed by rag2127
 
</script>
Output
20.7652

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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