Lagrange’s Interpolation  

源自於  https://www.geeksforgeeks.org/lagranges-interpolation/

// C++ program for implementation of Lagrange's Interpolation
#include<bits/stdc++.h>
using namespace std;

// To represent a data point corresponding to x and y = f(x)
struct Data
{
    int x, y;
};

// function to interpolate the given data points using Lagrange's formula
// xi corresponds to the new data point whose value is to be obtained
// n represents the number of known data points
double interpolate(Data f[], int xi, int n)
{
    double result = 0; // Initialize result

    for (int i=0; i<n; i++)
    {
        // Compute individual terms of above formula
        double term = f[i].y;
        for (int j=0;j<n;j++)
        {
            if (j!=i)
                term = term*(xi - f[j].x)/double(f[i].x - f[j].x);
        }

        // Add current term to result
        result += term;
    }

    return result;
}

// driver function to check the program
int main()
{
    // creating an array of 4 known data points
    Data f[] = {{0,2}, {1,3}, {2,12}, {5,147}};

    // Using the interpolate function to obtain a data point
    // corresponding to x=3
    cout << "Value of f(3) is : " << interpolate(f, 3, 4);

    return 0;
}

Lagrange’s Interpolation

What is Interpolation?
Interpolation is a method of finding new data points within the range of a discrete set of known data points (Source Wiki). In other words interpolation is the technique to estimate the value of a mathematical function, for any intermediate value of the independent variable.
For example, in the given table we’re given 4 set of discrete data points, for an unknown function f(x) :

How to find?
Here we can apply the Lagrange’s interpolation formula to get our solution.
The Lagrange’s Interpolation formula:
If, y = f(x) takes the values y0, y1, … , yn corresponding to x = x0, x1 , … , xn then,

This method is preferred over its counterparts like Newton’s method because it is applicable even for unequally spaced values of x.

We can use interpolation techniques to find an intermediate data point say at x = 3.

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// C++ program for implementation of Lagrange's Interpolation #include<bits/stdc++.h> using namespace std;    // To represent a data point corresponding to x and y = f(x) struct Data {     int x, y; };    // function to interpolate the given data points using Lagrange's formula // xi corresponds to the new data point whose value is to be obtained // n represents the number of known data points double interpolate(Data f[], int xi, int n) {     double result = 0; // Initialize result        for (int i=0; i<n; i++)     {         // Compute individual terms of above formula         double term = f[i].y;         for (int j=0;j<n;j++)         {             if (j!=i)                 term = term*(xi - f[j].x)/double(f[i].x - f[j].x);         }            // Add current term to result         result += term;     }        return result; }    // driver function to check the program int main() {     // creating an array of 4 known data points     Data f[] = {{0,2}, {1,3}, {2,12}, {5,147}};        // Using the interpolate function to obtain a data point     // corresponding to x=3     cout << "Value of f(3) is : " << interpolate(f, 3, 4);        return 0; }

Output:

Value of f(3) is : 35

Complexity:
The time complexity of the above solution is O(n2) and auxiliary space is O(1).