Optimum location of point to minimize total distance

25 Jun, 2022
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++
Java
Python3
C#
Javascript
//  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 pointstruct point {    int x, y;    point() {}    point(int x, int y)        : x(x)        , y(y)    {    }}; //  structure defining a line of ax + by + c = 0 formstruct 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 pdouble 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 distancedouble 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 costdouble 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 methodint 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;}
// A Java program to find optimum location// and total costclass 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
# A Python3 program to find optimum location# and total costimport math class Optimum_distance:         # Class defining a point    class Point:                 def __init__(self, x, y):                         self.x = x            self.y = 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 Codeif __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# program to find optimum location// and total costusing 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
<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 pointclass point{    constructor(x,y)    {        this.x=x;        this.y=y;    }} // structure defining a line of ax + by + c = 0 formclass line{    constructor(a,b,c)    {        this.a = a;            this.b = b;            this.c = c;    }     } // method to get distance of point (x, y) from point pfunction 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 distancefunction 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 costfunction 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 Codelet 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>
20.7652
Time Complexity: O(n2) Auxiliary Space: O(n)
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Rajput-Ji
29AjayKumar
suluksm
sumitgumber28
rag2127
ruhelaa48
akshaysingh98088
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