Comparing square root methods - Brute force vs Babylonian method. An understanding of mathematical convergence, and time complexity

square_root.c

/*
The MIT License (MIT)

Copyright (c) 2014 Quoc Le <[email protected]>

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*/

#include <stdio.h>
#include <math.h> 
#include <stdlib.h>     /* atof */


/*  sqrt(n) = x * x 

We can brute force calculate the square root by picking a number x as N, and decrement
until the accuracy converges to what we what.  However, the time complexity is 
directly proportional to the acccuracy.

O(complexity)  = n - actual_result  / accuracy

So time increases as accuracy gets better and/or number gets larger. What if there is a method
to quickly converge where we can achieve both lower time & higher accuracy & support for large numbers.
 
*/
float sqrt_brute_force(float n) {

	float result;
	int count = 0;
	if (n > 1) {
		result = n;		
		while(true) {

			float predicted = result * result;			
			float diff =  n - predicted;
			printf("value = %f predicted = %f diff = %f\n ", result, predicted, diff );
			if (diff < 0 ) {
				result = result - 0.00001;
			} else {
				break;
			}

			count++;			
		}

		printf("count=%d\n", count);
	}
	else {

	}

	return result;
}


/* 
 * Babylonian Method, found in 60 AD.
 * very quick convergence, and works  when S < 1.0
 *
 * 1. Begin with an arbitrary positive starting value x0 (the closer to the actual square root of S, the better).
 * 2. Let xn+1 be the average of xn and S / xn (using the arithmetic mean to approximate the geometric mean).
 * 3. Repeat step 2 until the desired accuracy is achieved.
 *
 * http://en.wikipedia.org/wiki/Methods_of_computing_square_roots
 */
double sqrt(float S) {
	
	printf("***");

	double x0 = S / 2.0f;
	double x1;
	double accuracy;
	do {
		x1 = 0.5 * (x0 + (S/x0));
		accuracy = fabs((x1* x1) - S);
		printf("X1 = %f Accuracy %f\n", x1, accuracy);
		x0 = x1;
	}
	while(accuracy > 0.000001);

	return x1;
}

int main(int argc, char* argv[]) {

	float x = atof(argv[1]);
  	printf("**** input = %f result = %f \n", x, sqrt(x));

}