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  1. #include<iostream>
  2. #include<iomanip>
  3. #include<cmath>
  4. using namespace std;
  5. //jacobi
  6. // Functions to calculate next iteration values
  7. inline double f1(double x, double y, double z) { return (17 - y + 2 * z) / 20; }
  8. inline double f2(double x, double y, double z) { return (-18 - 3 * x + z) / 20; }
  9. inline double f3(double x, double y, double z) { return (25 - 2 * x + 3 * y) / 20; }
  10.  
  11. bool hasConverged(double e1, double e2, double e3, double tolerance) {
  12.     return (e1 <= tolerance && e2 <= tolerance && e3 <= tolerance);
  13. }
  14.  
  15. int main() {
  16.     double x = 0, y = 0, z = 0, newX, newY, newZ, tol;
  17.     int step = 1;
  18.  
  19.     cout << "Enter tolerable error: ";
  20.     cin >> tol;
  21.     cout << fixed << setprecision(6);
  22.     cout << "\nIteration\tx\t\ty\t\tz" << endl;
  23.  
  24.     do {
  25.         newX = f1(x, y, z);
  26.         newY = f2(x, y, z);
  27.         newZ = f3(x, y, z);
  28.  
  29.         cout << step++ << "\t" << newX << "\t" << newY << "\t" << newZ << endl;
  30.  
  31.         double e1 = fabs(newX - x);
  32.         double e2 = fabs(newY - y);
  33.         double e3 = fabs(newZ - z);
  34.  
  35.         x = newX;
  36.         y = newY;
  37.         z = newZ;
  38.  
  39.         if (hasConverged(e1, e2, e3, tol)) {
  40.             break;
  41.         }
  42.     } while (true);
  43.  
  44.     cout << "\nSolution: x = " << x << ", y = " << y << ", z = " << z << endl;
  45.     return 0;
  46. }
Success #stdin #stdout 0.01s 5284KB
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0.0001
stdout
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Enter tolerable error: 
Iteration	x		y		z
1	0.850000	-0.900000	1.250000
2	1.020000	-0.965000	1.030000
3	1.001250	-1.001500	1.003250
4	1.000400	-1.000025	0.999650
5	0.999966	-1.000078	0.999956
6	1.000000	-0.999997	0.999992

Solution: x = 1.000000, y = -0.999997, z = 0.999992
https://ideone.com/b8JRQ3
Jacobi
language:
C++14 (gcc 8.3)
created:
8 days ago
visibility:
public

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