C-code for Lab 2

Gauss Elimination with pivoting

Note: the code without pivoting is the same as the one given below except that

/* This is a gaussian elimination program with pivoting.
The matrix is directly programmed into this
code so that the emphasis is on the mathematics. An operational
code would more than likely read in the matrix from a file
so that the program would not have to be re-compiled each
time a new matrix is used. The matrix given is the one
for the circuit problem stated at the beginning of
chapter 6.
The matrix is in an array called a; which is an n by n+1 array.
The choice of pivots is kept in an array called
row[].
*/
#include 
#include 
main()
{
int n, n1, k, j, j1, l, i, i1, p, trow, row[10];
float a[5][6]={5.,5,0,0,0,5.5,0,0,1,-1,-1,0,0,0,0,2,-3,0,1,-1,-1,0,0,0,0,5,-7,-2,0,0};
float x[10], temp, temp1, mult, sum;


n=5;
n1=n-1;
/*print the matrix */
for(k=0;k<=n1;k++)
{for(l=0;l<=n;l++) printf("%e \t",a[k][l]);
printf("\n");
}
for(i=0;i<=n1;i++) row[i]=i;
/*initialize the array row*/

/*This outermost of 3 loops that performs the elimination step */
for(i=0; i<=n1;i++) 
	/* First, find the choice of pivot */
	{
	p=i;
	n1=n-1;
	i1=i+1;
	temp=fabs(a[row[i]][i]);
	for(j=i1;jtemp)
      			{p=j; temp=temp1;}
  	 	}
	if(temp<1.E-9) {printf("no solution is likely \n"); return;}
	/* The largest pivot value is in temp and p is the
	corresponding index of the array row.
	If the largest pivot is small (< 10^(-9)), then
	a message is printed that no solution is likely, or can be trusted*/

	/*This step performs a simulated row switch by switching
	the values of row[i] with row[p] */
	trow=row[i];
	row[i]=row[p];
	row[p]=trow;
	/* This is the elimination step */
	for(j=i1; j<=n1;j++)
   		{j1=j+1;
   		mult=a[row[j]][i]/a[row[i]][i];
   		for(l=i1;l<=n;l++)
      			{a[row[j]][l]=a[row[j]][l]-mult*a[row[i]][l];}
  	 	}
	}

	 	
	
 
/* For a debug check, the gaussian reduced matrix together
with the row[] array are printed */

printf("\n");
printf("the gauss reduced matrix is \n");

for(k=0;k<=n1;k++) 
		{for(j=0;j<=n;j++)
		{if(k>j) printf("%f \t",0.);
		else printf("%f \t",a[row[k]][j]);}
		printf("\n");
		}
for(k=0;k<=n1;k++) printf("%d \t",row[k]);
printf("\n");

/* Now initialize the solution array, x[] */

for(k=0;k<=n1;k++) x[k]=0;

/* Now perform back substitution
to calculate the solutions, x[k]*/

n1=n-1;
x[n1]=a[row[n1]][n]/a[row[n1]][n1];
for(i=2;i<=n;i++)
   {sum=0;
   for(j=n-i+1;j

 The answer to the circuit problem is the following current
values (in amps) i1=.678505, i2=.421495, i3=.257009, i4=.154206, i5=.102804.


 Last updated 10/25/95