C code using Gauss Seidel Method

Problem: Solve the following systems using gauss seidel method
5×1-x2-x3-x4=-4
-x1+10×2-x3-x4=12
-x1-x2+5×3-x4=8
-x1-x2-x3+10×4=34

Code:

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define acc 0.0001
#define X1(x2,x3,x4) ((x2 + x3 + x4 -4)/5)
#define X2(x1,x3,x4) ((x1 + x3 + x4 +12)/10)
#define X3(x1,x2,x4) ((x1 + x2 + x4 +8)/5)
#define X4(x1,x2,x3) ((x1 + x2 + x3 +34)/10)

void main()
{
  double x1=0,x2=0,x3=0,x4=0,y1,y2,y3,y4;
  int i=0;
  system("cls");
  printf("\n______________________________________________________________\n");
  printf("\n   x1\t\t   x2\t\t   x3\t\t    x4\n");
  printf("\n______________________________________________________________\n");
  printf("\n%f\t%f\t%f\t%f",x1,x2,x3,x4);
  do
  {
   y1=X1(x2,x3,x4);
   y2=X2(x1,x3,x4);
   y3=X3(x1,x2,x4);
   y4=X4(x1,x2,x3);
   if(fabs(y1-x1)<acc && fabs(y2-x2)<acc && fabs(y3-x3)<acc &&fabs(y4-x4) )
   {
     printf("\n_____________________________________________________________\n");
     printf("\n\nx1 = %.3lf",y1);
     printf("\n\nx2 = %.3lf",y2);
     printf("\n\nx3 = %.3lf",y3);
     printf("\n\nx4= %.3lf",y4);
     i = 1;
   }
   else
   {
     x1 = y1;
     x2 = y2;
     x3 = y3;
     x4 = y4;
     printf("\n%f\t%f\t%f\t%f",x1,x2,x3,x4);
   }
  }while(i != 1);
getch();
}

 Output:

gauss seidel method

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Author: zakilive

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