用最好的直线表示一组给定的点

求m和c的值，使直线y = mx + c最好地表示给定点集(x $_1$ ，Y $_1$ )， (X $_2$ ，Y $_2$ )， (X $_3$ ，Y $_3$ )， ……。， (X $_n$ ，Y $_n$ )，给定n> = 2。

例子：

输入：n = 5 x $_1$ = 1，x $_2$ = 2，x $_3$ = 3，x $_4$ = 4，x $_5$ = 5年$_1$ = 14，y $_2$ = 27，y $_3$ = 40，y $_4$ = 55，y $_5$ = 68输出：m = 13.6 c = 0如果我们取任意一对数字(x $_i$ ，Y $_i$ )从给定的数据来看，m和c的这些值应使其最适合直线方程y = mx + c。取x $_1$ = 1和y $_1$ = 14，然后使用输出中的m和c值，并将其放在以下等式中，y = mx + c，LHS：y = 14，RHS：mx + c = 13.6 x 1 + 0 = 13.6大致相等。现在，取x $_3$ = 3和y $_3$ = 40，LHS：y = 40，RHS：mx + c = 13.6 x 3 + 0 = 40.8因此，它们也近似相等，对于所有其他值，依此类推。输入：n = 6 x $_1$ = 1，x $_2$ = 2，x $_3$ = 3，x $_4$ = 4，x $_5$ = 5，x $_6$ = 6年$_1$ = 1200，y $_2$ = 900，y $_3$ = 600，y $_4$ = 200，y $_5$ = 110，y $_6$ = 50输出：m = -243.42 c = 1361.97

方法

为了使方程中的一组点最适合一条直线，我们需要找到两个变量m和c的值。现在，由于存在2个未知变量，并且取决于n的值，两种情况是可能的–

情况1 –当n = 2时：将找到两个方程和两个未知变量，因此，将有一个唯一的解。
情况2 –当n> 2时：在这种情况下，可能存在或不存在满足所有n个方程式的m和c的值，但是我们可以找到m和c的最佳可能值，它们可以拟合一条直线给定的点。

因此，如果我们有n对不同的x和y对，那么我们可以形成n个不。来自它们的方程组的直线，如下

F $_1$ = mx $_1$ + c，f $_2$ = mx $_2$ + c，f $_3$ = mx $_3$ + c，………………………………..，…….. …………………………， F $_n$ = mx $_n$ + c，其中，f $_i$ ，是将x放入的值$_i$在等式mx + c中。

那么，由于理想上f $_i$应该与y相同$_i$ ，但是我们仍然可以找到f $_i$最接近y $_i$在所有情况下，如果我们取一个新的量，U =?(y $_i$ – F $_i$ ) $^2$ ，并使该数量对于i的所有值(从1到n)最小。

注意： (y $_i$ – F $_i$ ) $^2$用于代替(y $_i$ – F $_i$ )，因为我们要考虑f时的两种情况$_i$或当y $_i$更大，并且我们希望它们的差最小，因此，如果我们不对项求平方，则在这种情况下，f $_i$
更大，且其中y的情况$_i$更大的程度会互相抵消，这不是我们想要的。因此，我们需要对术语进行平方。

现在，要使U最小，它必须满足以下两个方程式–

$\frac{\partial U}{\partial m}$ = 0且$\frac{\partial U}{\partial c}$ = 0。

在求解上述两个方程式时，我们得到两个方程式，如下所示：

?y = nc + m?x，?xy = c?x + m?x $^2$ ，可以重新排列为-m =(n *?xy-?x?y)/(n *?x $^2$ – (?X) $^2$ )，而c =(?y-m?x)/ n，

因此，这就是两种情况下m和c的值的获取方式，并且我们可以通过最佳可能的直线表示一组给定的点。

以下代码实现了上述算法-

C++

// C++ Program to find m and c for a straight line given,
// x and y
#include 
#include 
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;
}

C

// C Program to find m and c for a straight line given,
// x and y
#include 
 
// 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;
}

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);
    }
}

Python3

# 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 function
x = [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#

// C# Program to find m and c for a
// straight line given, x and y
using 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

Javascript

 
// 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 code
let 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
 

输出：

m=13.6
c=0.0

分析以上代码
辅助空间：O(1)
时间复杂度：O(n)。我们有一个循环迭代n次，每次循环执行常数no。计算。

参考-
BS Grewal的1-高等工程数学。