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  1. #include <stdio.h>
  2. #include <math.h>
  3. #include <stdlib.h>
  4.  
  5. #define EPSILON 1e-6   // Convergence tolerance
  6. #define MAX_ITER 100   // Maximum iterations
  7.  
  8. // Define the nonlinear equations
  9. double f1(double x, double y) {
  10.     return x * x + y * y - 4; 
  11. }
  12.  
  13. double f2(double x, double y) {
  14.     return x * x * x * x + x - y*y + 1; // Hyperbola: xy = 1
  15. }
  16.  
  17. // Partial derivatives for Jacobian matrix
  18. double df1_dx(double x, double y) { return 2 * x; }
  19. double df1_dy(double x, double y) { return 2 * y; }
  20. double df2_dx(double x, double y) { return 4 * x * x*x + 1; }
  21. double df2_dy(double x, double y) { return -2* y; }
  22.  
  23. int main() {
  24.     double x = 1, y = 3; // Initial guess
  25.     int iter = 0;
  26.  
  27.     printf("Initial guess: x = %.6f, y = %.6f\n", x, y);
  28.  
  29.     while (iter < MAX_ITER) {
  30.         // Function values
  31.         double F1 = f1(x, y);
  32.         double F2 = f2(x, y);
  33.  
  34.         // Check convergence
  35.         if (fabs(F1) < EPSILON && fabs(F2) < EPSILON) {
  36.             printf("Converged after %d iterations.\n", iter);
  37.             printf("Solution: x = %.6f, y = %.6f\n", x, y);
  38.             return 0;
  39.         }
  40.  
  41.         // Jacobian matrix
  42.         double J11 = df1_dx(x, y);
  43.         double J12 = df1_dy(x, y);
  44.         double J21 = df2_dx(x, y);
  45.         double J22 = df2_dy(x, y);
  46.  
  47.         // Determinant of Jacobian
  48.         double det = J11 * J22 - J12 * J21;
  49.         if (fabs(det) < 1e-12) {
  50.             printf("Jacobian determinant too small. Method fails.\n");
  51.             return 1;
  52.         }
  53.  
  54.         // Inverse Jacobian * F
  55.         double dx = (-F1 * J22 + F2 * J12) / det;
  56.         double dy = (-J11 * F2 + J21 * F1) / det;
  57.  
  58.         // Update variables
  59.         x += dx;
  60.         y += dy;
  61.  
  62.         iter++;
  63.     }
  64.  
  65.     printf("Did not converge within %d iterations.\n", MAX_ITER);
  66.     return 1;
  67. }
Success #stdin #stdout 0s 5300KB
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Standard input is empty
stdout
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Initial guess: x = 1.000000, y = 3.000000
Converged after 4 iterations.
Solution: x = 1.000000, y = 1.732051
https://ideone.com/91XJwQ
language:
C (gcc 8.3)
created:
4 days ago
visibility:
public

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