Represent a given set of points by the best possible straight line

22 Apr, 2021
Find the value of m and c such that a straight line y = mx + c, best represents the equation of a given set of points (x, y), (x, y), (x, y), ......., (x, y), given n >=2.
Examples:  
Input : n = 5
        x = 1, x = 2, x = 3, 
x = 4, x = 5
y = 14, y = 27, y = 40, 
y = 55, y = 68   
Output : m = 13.6
c = 0 
If we take any pair of number ( x, y ) 
from the given data, these value of m and c
should make it best fit into the equation 
for a straight line, y = mx + c. Take x = 1 
and y = 14, then using values
of m and c from the output, and putting it 
in the following equation,
y = mx + c,
L.H.S.: y = 14, R.H.S: mx + c = 13.6 x 1 + 0 = 13.6
So, they are approximately equal.
Now, take x = 3 and y = 40,
L.H.S.: y = 40, R.H.S: mx + c = 13.6 x 3 + 0 = 40.8
So, they are also approximately equal, and so on
for all other values.
Input : n = 6
x = 1, x = 2, x = 3, 
x = 4, x = 5, x = 6
y = 1200, y = 900, y = 600, 
y = 200, y = 110, y = 50
Output : m = -243.42
c = 1361.97
Approach
To best fit a set of points in an equation for a straight line, we need to find the value of two variables, m and c. Now, since there are 2 unknown variables and depending upon the value of n, two cases are possible – 
Case 1 – When n = 2 : There will be two equations and two unknown variables to find, so, there will be a unique solution . Case 2 – When n > 2 : In this case, there may or may not exist values of m and c, which satisfy all the n equations, but we can find the best possible values of m and c which can fit a straight line in the given points .
So, if we have n different pairs of x and y, then, we can form n no. of equations from them for a straight line, as follows 
f = mx + c,
f = mx + c,
f = mx + c,
......................................,
......................................,
f = mx + c,
where, f, is the value 
obtained by putting x in equation 
mx + c. 
Then, since ideally fshould be same as y, but still we can find the fclosest to yin all the cases, if we take a new quantity, U = ?(y– f), and make this quantity minimum for all value of i from 1 to n. 
Note:(y– f)is used in place of (y– f), as we want to consider both the cases when for when yis greater, and we want their difference to be minimum, so if we would not square the term, then situations in which fis greater and situation in which yis greater will ancel each other to an extent, and this is not what we want. So, we need to square the term.
Now, for U to be minimum, it must satisfy the following two equations –
 = 0 and  
 = 0. 
On solving the above two equations, we get two equations, as follows : 
?y = nc + m?x, and
?xy = c?x + m?x, which can be rearranged as - 
m = (n * ?xy - ?x?y) / (n * ?x - (?x)), and
c = (?y - m?x) / n, 
So, this is how values of m and c for both the cases are obtained, and we can represent a given set of points, by the best possible straight line. 
The following code implements the above given algorithm – 
C++
C
Java
Python3
C#
PHP
Javascript
// C++ Program to find m and c for a straight line given,// x and y#include <cmath>#include <iostream>using namespace std; // function to calculate m and c that best fit points// represented by x[] and y[]void bestApproximate(int x[], int y[], int n){    float m, c, sum_x = 0, sum_y = 0, sum_xy = 0, sum_x2 = 0;    for (int i = 0; i < n; i++) {        sum_x += x[i];        sum_y += y[i];        sum_xy += x[i] * y[i];        sum_x2 += pow(x[i], 2);    }     m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - pow(sum_x, 2));    c = (sum_y - m * sum_x) / n;     cout << "m =" << m;    cout << "\nc =" << c;} // Driver main functionint main(){    int x[] = { 1, 2, 3, 4, 5 };    int y[] = { 14, 27, 40, 55, 68 };    int n = sizeof(x) / sizeof(x[0]);    bestApproximate(x, y, n);    return 0;}
// C Program to find m and c for a straight line given,// x and y#include <stdio.h> // function to calculate m and c that best fit points// represented by x[] and y[]void bestApproximate(int x[], int y[], int n){    int i, j;    float m, c, sum_x = 0, sum_y = 0, sum_xy = 0, sum_x2 = 0;    for (i = 0; i < n; i++) {        sum_x += x[i];        sum_y += y[i];        sum_xy += x[i] * y[i];        sum_x2 += (x[i] * x[i]);    }     m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - (sum_x * sum_x));    c = (sum_y - m * sum_x) / n;     printf("m =% f", m);    printf("\nc =% f", c);} // Driver main functionint main(){    int x[] = { 1, 2, 3, 4, 5 };    int y[] = { 14, 27, 40, 55, 68 };    int n = sizeof(x) / sizeof(x[0]);    bestApproximate(x, y, n);    return 0;}
// Java Program to find m and c for a straight line given,// x and yimport java.io.*;import static java.lang.Math.pow; public class A {    // function to calculate m and c that best fit points    // represented by x[] and y[]    static void bestApproximate(int x[], int y[])    {        int n = x.length;        double m, c, sum_x = 0, sum_y = 0,                     sum_xy = 0, sum_x2 = 0;        for (int i = 0; i < n; i++) {            sum_x += x[i];            sum_y += y[i];            sum_xy += x[i] * y[i];            sum_x2 += pow(x[i], 2);        }         m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - pow(sum_x, 2));        c = (sum_y - m * sum_x) / n;         System.out.println("m = " + m);        System.out.println("c = " + c);    }     // Driver main function    public static void main(String args[])    {        int x[] = { 1, 2, 3, 4, 5 };        int y[] = { 14, 27, 40, 55, 68 };        bestApproximate(x, y);    }}
# python Program to find m and c for# a straight line given, x and y # function to calculate m and c that# best fit points represented by x[]# and y[]def bestApproximate(x, y, n):         sum_x = 0    sum_y = 0    sum_xy = 0    sum_x2 = 0         for i in range (0, n):        sum_x += x[i]        sum_y += y[i]        sum_xy += x[i] * y[i]        sum_x2 += pow(x[i], 2)     m = (float)((n * sum_xy - sum_x * sum_y)            / (n * sum_x2 - pow(sum_x, 2)));                 c = (float)(sum_y - m * sum_x) / n;         print("m = ", m);    print("c = ", c);          # Driver main functionx = [1, 2, 3, 4, 5 ]y = [ 14, 27, 40, 55, 68]n = len(x) bestApproximate(x, y, n)     # This code is contributed by Sam007.
// C# Program to find m and c for a// straight line given, x and yusing System; class GFG {     // function to calculate m and c that    // best fit points represented by x[] and y[]    static void bestApproximate(int[] x, int[] y)    {        int n = x.Length;        double m, c, sum_x = 0, sum_y = 0,                     sum_xy = 0, sum_x2 = 0;         for (int i = 0; i < n; i++) {            sum_x += x[i];            sum_y += y[i];            sum_xy += x[i] * y[i];            sum_x2 += Math.Pow(x[i], 2);        }         m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - Math.Pow(sum_x, 2));         c = (sum_y - m * sum_x) / n;         Console.WriteLine("m = " + m);        Console.WriteLine("c = " + c);    }     // Driver main function    public static void Main()    {        int[] x = { 1, 2, 3, 4, 5 };        int[] y = { 14, 27, 40, 55, 68 };         // Function calling        bestApproximate(x, y);    }} // This code is contributed by Sam007
<?php// PHP Program to find m and c// for a straight line given,// x and y // function to calculate m and// c that best fit points// represented by x[] and y[]function bestApproximate($x, $y, $n){    $i; $j;    $m; $c;    $sum_x = 0;    $sum_y = 0;    $sum_xy = 0;    $sum_x2 = 0;    for ($i = 0; $i < $n; $i++)    {        $sum_x += $x[$i];        $sum_y += $y[$i];        $sum_xy += $x[$i] * $y[$i];        $sum_x2 += ($x[$i] * $x[$i]);    }     $m = ($n * $sum_xy - $sum_x * $sum_y) /         ($n * $sum_x2 - ($sum_x * $sum_x));    $c = ($sum_y - $m * $sum_x) / $n;     echo "m =", $m;    echo "\nc =", $c;}     // Driver Code    $x =array(1, 2, 3, 4, 5);    $y =array (14, 27, 40, 55, 68);    $n = sizeof($x);    bestApproximate($x, $y, $n); // This code is contributed by ajit?>
<script> // Javascript Program to find m and c// for a straight line given, x and y // function to calculate m and c that// best fit points represented by x[] and y[]function bestApproximate(x, y, n){    let m, c, sum_x = 0, sum_y = 0,             sum_xy = 0, sum_x2 = 0;    for(let i = 0; i < n; i++)    {        sum_x += x[i];        sum_y += y[i];        sum_xy += x[i] * y[i];        sum_x2 += Math.pow(x[i], 2);    }     m = (n * sum_xy - sum_x * sum_y) /        (n * sum_x2 - Math.pow(sum_x, 2));    c = (sum_y - m * sum_x) / n;     document.write("m =" + m);    document.write("<br>c =" + c);} // Driver codelet x = [ 1, 2, 3, 4, 5 ];let y = [ 14, 27, 40, 55, 68 ];let n = x.length; bestApproximate(x, y, n); // This code is contributed by subham348 </script>
Output: 
m=13.6
c=0.0
Analysis of above code- Auxiliary Space : O(1) Time Complexity : O(n). We have one loop which iterates n times, and each time it performs constant no. of computations.
Reference- 1-Higher Engineering Mathematics by B.S. Grewal.
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