#include<stdio.h>
#include<conio.h>
#include<math.h>
#define e 0.01
void main() {
int i, j, k, n;
float a[10][10], x[10];
float sum, temp, error, big;
printf("Enter the number of equations: ");
scanf("%d", & n);
printf("Enter the co-efficients of the equations: \n");
for (i = 1; i <= n; i++) {
for (j = 1; j <= n + 1; j++) {
printf("a[%d][%d]= ", i, j);
scanf("%f", & a[i][j]);
}
}
for (i = 1; i <= n; i++) {
x[i] = 0;
}
do {
big = 0;
for (i = 1; i <= n; i++) {
sum = 0;
for (j = 1; j <= n; j++) {
if (j != i) {
sum = sum + a[i][j] * x[j];
}
}
temp = (a[i][n + 1] - sum) / a[i][i];
error = fabs(x[i] - temp);
if (error > big) {
big = error;
}
x[i] = temp;
printf("\nx[%d] =%f", i, x[i]);
}
printf("\n");
}
while (big >= e);
printf("\n\nconverges to solution");
for (i = 1; i <= n; i++) {
printf("\nx[%d]=%f", i, x[i]);
}
getch();
}
#include<conio.h>
#include<math.h>
#define e 0.01
void main() {
int i, j, k, n;
float a[10][10], x[10];
float sum, temp, error, big;
printf("Enter the number of equations: ");
scanf("%d", & n);
printf("Enter the co-efficients of the equations: \n");
for (i = 1; i <= n; i++) {
for (j = 1; j <= n + 1; j++) {
printf("a[%d][%d]= ", i, j);
scanf("%f", & a[i][j]);
}
}
for (i = 1; i <= n; i++) {
x[i] = 0;
}
do {
big = 0;
for (i = 1; i <= n; i++) {
sum = 0;
for (j = 1; j <= n; j++) {
if (j != i) {
sum = sum + a[i][j] * x[j];
}
}
temp = (a[i][n + 1] - sum) / a[i][i];
error = fabs(x[i] - temp);
if (error > big) {
big = error;
}
x[i] = temp;
printf("\nx[%d] =%f", i, x[i]);
}
printf("\n");
}
while (big >= e);
printf("\n\nconverges to solution");
for (i = 1; i <= n; i++) {
printf("\nx[%d]=%f", i, x[i]);
}
getch();
}
Enter the number of equations: 3
Enter the co-efficients of the equations:
a[1][1]= 2
a[1][2]= 1
a[1][3]= 1
a[1][4]= 5
a[2][1]= 3
a[2][2]= 5
a[2][3]= 2
a[2][4]= 15
a[3][1]= 2
a[3][2]= 1
a[3][3]= 4
a[3][4]= 8
x[1] =2.500000
x[2] =1.500000
x[3] =0.375000
x[1] =1.562500
x[2] =1.912500
x[3] =0.740625
x[1] =1.173437
x[2] =1.999688
x[3] =0.913359
x[1] =1.043477
x[2] =2.008570
x[3] =0.976119
x[1] =1.007655
x[2] =2.004959
x[3] =0.994933
x[1] =1.000054
x[2] =2.001995
x[3] =0.999474
converges to solution
x[1]=1.000054
x[2]=2.001995
x[3]=0.999474
Hello friend, thank you very much for sharing your code.
ReplyDeleteI wanted to ask you a question, which means when you define "# define and 0.01"?
Tea appreciate your nswer, we are in contact, greetings.
its the Alowed Error
Deleteits show 5 error
Deletecan you please convert this coding in c++ also as while doing so its rhs values are not able to be assigned properly and its showing abnormal termination
ReplyDeleteSorry, never mind!!!
ReplyDeleteThank you very much.
ReplyDeletethanks....very much....!!!!!!!
ReplyDeletethanks....very much....!!!!!!!
ReplyDeletehow can you print the percent's values of errors?
ReplyDeleteThanks
ReplyDeleteThanks dude
ReplyDeleteRun to ho hi nhi raha program sir
ReplyDeletewrong program
ReplyDelete