Heron square root implementations in a few languages.

heron.c

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>

double heron_sqrt(long n, double percision);
void compare_sqrt(long n);

int main(int argc, char **argv)
{
    if (argc < 2)
    {
        fprintf(stderr, "Usage: %s n\nn - number to find the square root of\n", argv[0]);
        return(EXIT_FAILURE);
    }

    long n = strtol(argv[1], NULL, 10);
    compare_sqrt(n);
    return(EXIT_SUCCESS);
}

double heron_sqrt(long n, double percision)
{
    if (n == 0)
    {
        return 0.0;
    }
    double guess = n / 3.0;
    while (true)
    {
        double lastGuess = guess;
        guess = (guess + (n / guess)) / 2.0;
        if (abs(lastGuess - guess) <= percision) break;
    }
    return guess;
}

void compare_sqrt(long n)
{
    double percision = 0.000000000001;
    double heron = heron_sqrt(n, percision);
    double real_sqrt = sqrt(n);
    if (heron == real_sqrt)
    {
        printf("The square root of %ld is %f\n", n, heron);
    } else
    {
        printf(
            "The square root of %ld using heron (%f) differs by %f\n",
            n,
            heron,
            heron - real_sqrt
       );
    }
}

heron.py

#!/usr/bin/python2
import math
import sys
def heron_sqrt(n, percision=0.000000000001):
    if n == 0:
        return 0
    guess = n / 3.0
    while True:
        lastguess = guess
        guess = (guess + (n / guess)) / 2.0
        if abs(lastguess - guess) <= percision:
            break
    return guess

def compare_sqrt(n, percision=0.000000000001):
    heron = heron_sqrt(n, percision)
    py_sqrt = math.sqrt(n)
    if heron == py_sqrt:
        print "The sqrt of {0} is {1}".format(n, heron)
    else:
        print "The sqrt of {0} using heron ({1}) differs by {2}".format(
                n,heron, heron - py_sqrt
               )

if __name__ == "__main__":
    if len(sys.argv) > 1:
        compare_sqrt(int(sys.argv[1]))