Infinite loop in babylonian method for square roots - loops

Question

In some specific numbers my algorithm gets stuck. It never reaches the minimum approximation so we never get out of the while. I think I can either low my approximation requisites or use double for my numbers, but I'm trying to figure out other solutions.
I'm programming a babylonian algorithm to calculate roots. First I'm doing this in C and later I will do this in Assembly(University homework). When I try to find the root of numbers like 99999 the program iterates to infinity. I have already tried two different stop conditions, one of them I did exactly like this tutorial from geeks4geeks(the first one inside the site).
https://www.geeksforgeeks.org/square-root-of-a-perfect-square/
The second stop condition I tested was this:
while ((x*x - n) > e) {}
I tried something like this because it is more "relatable" to the method enunciation. The full code is showed below:
#include <stdio.h>
#include <math.h>
/*Returns the square root of n. Note that the function */
float squareRoot(float n)
{
/*We are using n itself as initial approximation
This can definitely be improved */
float x = n;
float y = 1;
float e = 0.000001; /* e decides the accuracy level*/
while ((x*x - n) > e) {
x = (x + y) / 2;
y = n / x;
// if(prev_err == x-y){
// printf("A aproximação por ponto flutuante alcançou o máximo possível para o caso\n");
// return x;
// }
// prev_err = x-y;
}
return x;
}
/* Driver program to test above function*/
int main()
{
int n;
printf("Insira o número cuja raiz deseja calcular\n");
scanf("%d", &n);
printf("Square root of %d is %.8f\n", n, squareRoot(n));
return 0;
}

Answer 1

An absolute tolerance will never work. If n is large, x.x - n can remain large and the loop will never stop. If n is tiny, x.x - n can too quickly become small and the result will be quite inaccurate.
Your test has another big flaw: if x.x - n < e, the iterations will stop immediately if the LHS is negative, whatever its value.
The cure is to take the absolute value and a relative tolerance.
A better cure is to
adapt the starting approximation to the magnitude of n (such as the nearest power of 4),
use a fixed number of iterations (with a good starting approximation, 7 iterations are enough).

Answer 2

float only can take 4 Size (bytes),
so If you want to calculate the square root for number greater than 4 bytes
You need to replace all float to double
like this:
#include <stdio.h>
#include <math.h>
/*Returns the square root of n. Note that the function */
double squareRoot(double n)
{
/*We are using n itself as initial approximation
This can definitely be improved */
double x = n;
double y = 1;
double e = 0.000001; /* e decides the accuracy level*/
while ((x*x - n) > e) {
x = (x + y) / 2;
y = n / x;
// if(prev_err == x-y){
// printf("A aproximação por ponto flutuante alcançou o máximo possível para o caso\n");
// return x;
// }
// prev_err = x-y;
}
return x;
}
/* Driver program to test above function*/
int main()
{
int n;
printf("Insira o número cuja raiz deseja calcular\n");
scanf("%d", &n);
printf("Square root of %d is %.8f\n", n, squareRoot(n));
return 0;
}
if u want learn more about Primitive Data Types size in c:
Primitive Data Types sizes
https://www.programiz.com/c-programming/c-data-types#:~:text=The%20size%20of%20float%20(single,data%20type)%20is%208%20bytes.