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  1. #include <iostream>
  2. #include <iomanip>
  3. #include <vector>
  4. #include <cmath>
  5. #include <string>
  6.  
  7. using namespace std;
  8.  
  9. void newton_backward_interpolation() {
  10.     int n = 4;
  11.     double x_vals[] = {0.0, 1.0, 2.0, 3.0};
  12.     double y_vals[] = {1.0, 0.0, 1.0, 10.0};
  13.     double a = 0.5;
  14.  
  15.     vector<vector<double>> diff_table(n, vector<double>(n));
  16.  
  17.     for (int i = 0; i < n; ++i) {
  18.         diff_table[i][0] = y_vals[i];
  19.     }
  20.  
  21.     for (int j = 1; j < n; ++j) {
  22.         for (int i = 0; i < n - j; ++i) {
  23.             diff_table[i][j] = diff_table[i + 1][j - 1] - diff_table[i][j - 1];
  24.         }
  25.     }
  26.  
  27.     cout << setprecision(6) << fixed;
  28.     cout << "--- Newton's Backward Interpolation ---" << endl;
  29.     cout << "Input table (x, y):" << endl;
  30.     for (int i = 0; i < n; ++i) {
  31.         cout << "(" << x_vals[i] << ", " << y_vals[i] << ") ";
  32.     }
  33.     cout << "\nPoint of Interpolation (a): " << a << endl;
  34.     cout << string(60, '-') << endl;
  35.  
  36.     cout << "Output: The backward difference table is (same as forward, but values used start at the bottom):" << endl;
  37.     cout << setw(10) << "x" << setw(10) << "y";
  38.     for (int j = 1; j < n; ++j) {
  39.         cout << setw(10) << "Nabla^" << j << "y";
  40.     }
  41.     cout << endl;
  42.     cout << string(10 + 10 * n, '-') << endl;
  43.  
  44.     for (int i = 0; i < n; ++i) {
  45.         cout << setw(10) << x_vals[i] << setw(10) << diff_table[i][0];
  46.         for (int j = 1; j < n - i; ++j) {
  47.             cout << setw(10) << diff_table[i][j];
  48.         }
  49.         cout << endl;
  50.     }
  51.     cout << string(10 + 10 * n, '-') << endl;
  52.  
  53.     double h = x_vals[1] - x_vals[0];
  54.     double u = (a - x_vals[n - 1]) / h; 
  55.     double sum = diff_table[n - 1][0]; 
  56.     double p = 1.0;
  57.  
  58.     for (int j = 1; j < n; ++j) {
  59.         double backward_diff = diff_table[n - 1 - j][j];
  60.         p = p * (u + j - 1) / j;
  61.         sum = sum + p * backward_diff;
  62.     }
  63.  
  64.     cout << "\nThe value of y at x=" << a << " is " << round(sum * 1000.0) / 1000.0 << endl;
  65.     cout << "Calculated value: " << sum << endl;
  66. }
  67.  
  68. int main() {
  69.     newton_backward_interpolation();
  70.     return 0;
  71. }
Success #stdin #stdout 0s 5316KB
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stdin
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Standard input is empty
stdout
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--- Newton's Backward Interpolation ---
Input table (x, y):
(0.000000, 1.000000) (1.000000, 0.000000) (2.000000, 1.000000) (3.000000, 10.000000) 
Point of Interpolation (a): 0.500000
------------------------------------------------------------
Output: The backward difference table is (same as forward, but values used start at the bottom):
         x         y    Nabla^1y    Nabla^2y    Nabla^3y
--------------------------------------------------
  0.000000  1.000000 -1.000000  2.000000  6.000000
  1.000000  0.000000  1.000000  8.000000
  2.000000  1.000000  9.000000
  3.000000 10.000000
--------------------------------------------------

The value of y at x=0.500000 is 0.625000
Calculated value: 0.625000
https://ideone.com/3utOve
language:
C++ (gcc 8.3)
created:
4 hours ago
visibility:
public

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