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 f should be same as y, but still we can find the f closest to y in all the cases, if we take a new quantity, U = Σ(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 f or when y is greater, and we want their difference to be minimum, so if we would not square the term, then situations in which f

is greater and situation in which y is 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 :

Σy = nc + mΣx, and Σxy = cΣx + mΣx, which can be rearranged as - m = (n * Σxy - ΣxΣy) / (n * Σx - (Σx)), and c = (Σy - mΣx) / 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 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 function int 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++

// C++ Program to find m and c for a straight line given, // x and y #include<iostream> #include<cmath> 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 function int 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

// Java Program to find m and c for a straight line given, // x and y import 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); } }

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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