Basic C Program, Babylonian Algorithm - c

Question

I'm new to the C language and am trying to do a lab tutorial that we were given at uni.
We've been asked to do the following:
Task 1.
The Babylonian algorithm to compute the square root of a number n is as follows:
1. Make a guess at the answer (you can pick n/2 as your initial guess).
Compute r = n / guess
Set guess = (guess +r) / 2
Go back to step 2 for as many iterations as necessary. The more that steps 2 and 3 are
repeated, the closer guess will become to the square root of n.
Write a program that inputs an integer for n, iterates through the Babylonian algorithm
five times, and outputs the answer as a double to two decimal places. Your answer will
be most accurate for small values of n.
Here is what I have written:
#include <stdio.h>
#include <math.h>
int n;
main(void){
printf("Enter a value for n: ");
scanf("%d",&n);
double guess = n / 2;
for (int i = 0; i < 5; i++) {
double r = n / guess;
double guess = (guess + r) / 2;
}
printf("%d",guess);
}
Where have I gone wrong? It spits out ridiculous results; for example if I input "4" as n, the answer should be around "2", but it gives different huge results each time.

Answer 1

Another solution would be:
guess = guess / 2.0;
This would "force" a floating-point operation.
And the variable guess is already in the scope. You can´t redeclare it (as you did inside the loop). You can only set it a new value.
And you also need to change the printf to :
printf("%f",guess);
Check this link for more info about the printf formatters:
http://www.cplusplus.com/reference/cstdio/printf/

Answer 2

A few things wrong here.
First, you have scoped a second instance of guess inside the loop. Take away the double declaration on that line. So it should become:
guess = (guess + r) / 2;
Second, because guess is a double you need to use %f instead of %d in the printf call.
printf( "%f", guess );
Once you get it working, consider running the algorithm until a certain accuracy is achieved.
const double epsilon = 0.0001;
double guess = (double)n / 2.0;
double r = 0.0;
while( fabs(guess * guess - (double)n) > epsilon )
{
r = (double)n / guess;
guess = (guess + r) / 2.0;
}

Answer 3

The Babylonian Algorithm seems incorrect to me, it should be like this,
int i;
float n,guess=1;
printf("\nEnter the Number: ");
scanf("%f",&n);
for(i=0;i<PRECISION;i++)
{
guess=(guess+n/guess)/2;
}
printf("\nThe Square root of %f is %f",n,guess);
There are other possible errors also in your program,
There might be the problem of integer division,
The line double guess = n / 2;
should be double guess = (double) n / 2;
Also the printf() should be printf("%lf",guess);