Gauss Jacobi Method

Algorithm : Gauss-Jacobi Method

GaussJacobi(a,b,n)
{
This is an algorithm to find the roots of the equation using Gauss Jacobi method.
Where a is the coefficient matrix and b is the constant matrix of size n.
The limitation of this method is that the matrix has to be strictly
diagonally dominant
i.e a(1,i)>≠|a(i,j)| for j=1 to n and j≠i.


for(i=0 to i<n)
{
  prex[i]=0                                    // It initializes the previous and
                                                present values to 0
  newx[i]=0
}
do
{
 for(i=0 to i<n)
 {
     sum=0
     for(j=0 to j<n)
      {
         if(i!=j)
         sum=sum+a[i][j]*prex[j]
       }
  newx[i]=(b[i]-sum)/a[i][i]
}
 error=0
 for(i=0 to i<n)
 {
    error=error+fabs(prex[i]-newx[i])
    prex[i]=newx[i]
 }
}
while(error>e)			                       //To check the precision 
for(i=0 to i<n)
print("%2f  ",newx[i])	                                // Prints the root
print("\n\n")
}





C Program For Gauss Jacobi Method

Source Code:

#include<stdio.h>
#include<conio.h>
#include<math.h>

void show(float A[20][20],int n)
{
   int i,j;
   for(i=1;i<=n;i++)
   {
      for(j=1;j<=n;j++)
      {
  printf("\t%g",A[i][j]);
      }
      printf("\n");
   }
}

void input(float A[20][20],int n)
{
  int i,j;
  for(i=1;i<=n;i++)
   {
      for(j=1;j<=n;j++)
      {
   printf("\nEnter term %d of equation %d:",j,i);
   scanf("%f",&A[i][j]);
      }
   }
}


void build(float A[20][20],float B[20],int n)
{
   float max,sum,C[20],D[20],y;
   int i,j,pos,k,h;
   for(i=1;i<=n;i++)
   {
      max=-9999;
      for(j=1;j<=n;j++)
      {
  if(A[i][j]>max)
  {
     max=A[i][j];
     pos=j;
  }
      }
      y=B[i];
      B[i]=B[pos];
      B[pos]=y;
      for(k=1;k<=n;k++)
      {
  C[k]=A[i][k];
  D[k]=A[pos][k];
      }
      for(k=1;k<=n;k++)
      {
  A[i][k]=D[k];
  A[pos][k]=C[k];
      }
   }
   for(h=1;h<=n;h++)
   {
      printf("\n%g",B[h]);
   }
   for(i=1;i<=n;i++)
   {
     sum=0;
     for(j=1;j<=n;j++)
     {
 if(i!=j)
 {
   sum=sum+fabs(A[i][j]);
 }
     }
     if(fabs(A[i][i])<=sum)
     {
 printf("\nNot a strictly diagonally matrix");
 getch();
 exit(0);
     }
   }
}


void gauss_jacobi(float A[20][20],float B[20],int n)
{
   float E=0.000001,sum,PREVX[20],NEWX[20],error;
   int i,j;
   printf("\nThe co-efficient matrix(A):\n");
   show(A,n);
   printf("\nThe const. matrix(B):\n");
   for(i=1;i<=n;i++)
   {
     printf("\n%g",B[i]);
   }
   for(i=1;i<=n;i++)
   {
      sum=0;
      for(j=1;j<=n;j++)
      {
  sum=sum+fabs(A[i][j]);
      }
      if(2*fabs(A[i][i])<=sum)
      {
  printf("\nNot a strictly diagonally dominant matrix");
  return;
      }
   }
   for(i=1;i<=n;i++)
   {
      PREVX[i]=0;
      NEWX[i]=0;
   }
   do
   {
      for(i=1;i<=n;i++)
      {
  sum=0;
  for(j=1;j<=n;j++)
  {
    if(i!=j)
    {
      sum=sum+A[i][j]*PREVX[j];
    }
  }
  NEWX[i]=(B[i]-sum)/A[i][i];
      }
      error=0;
      for(i=1;i<=n;i++)
      {
  error=error+fabs(PREVX[i]-NEWX[i]);
  PREVX[i]=NEWX[i];
      }
   }while(error>E);
   printf("\n");
   for(i=1;i<=n;i++)
   {
     printf("\nX%d=%g",i,NEWX[i]);
   }
}



void main()
{
   float A[20][20],B[20];
   int i,j,n;
   clrscr();
   printf("\nEnter number of variables:");
   scanf("%d",&n);
   printf("\nEnter the co-efficient matrix:\n");
   input(A,n);
   printf("\nEnter the const. values:\n");
   for(i=1;i<=n;i++)
   {
     scanf("%f",&B[i]);
   }
   build(A,B,n);
   gauss_jacobi(A,B,n);
   getch();
}


OUTPUT:

Enter number of variables:3                                                     
                                                                                
Enter the co-efficient matrix:                                                  
                                                                                
Enter term 1 of equation 1:1                                                    
                                                                                
Enter term 2 of equation 1:1                                                    
                                                                                
Enter term 3 of equation 1:4                                                    
                                                                                
Enter term 1 of equation 2:8

Enter term 2 of equation 2:-3

Enter term 3 of equation 2:2

Enter term 1 of equation 3:4

Enter term 2 of equation 3:11

Enter term 3 of equation 3:-1

Enter the const. values:
9
20
33

20
33
9
The co-efficient matrix(A):
 8       -3      2
 4       11      -1
 1       1       4

The const. matrix(B):

20
33
9

X1=3
X2=2
X3=1