# Optimum location of point to minimize total distance

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 canâ€™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 ` `=` `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

**Time Complexity:** O(n^{2}) **Auxiliary Space:** O(n)

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