I was reading some post about interview questions of 2011 and came across one that stated “find the square root of a number”.

Assuming we can’t use the sqrtf function of the standard C math library, let’s see how we can calculate the square root of a number **x**.

Given **n**, we know that its square root is a number **x** that holds:

Let’s work on this equation a little. Raise both sides to the second power:

Move to the left of the equality:

If we found the roots of this last equation somehow, we would have found the square root of **n**. We can do this by using the Newton-Raphson iteration.

The Newton-Raphson iteration states that we can find the root of an equation using the following formula iteratively:

Where **f'(x)** is the derivative of function **f(x)**. We will approximate the derivative using the definition of derivative at a point (we could also note that the derivative could be trivially calculated; this method is more general).

The error of the Newton-Raphson iteration is given by:

Starting with a hardcoded seed value, we will perform this iteration in a loop until the error is less than a given value. I have chosen to iterate until the error is less than 1×10^(-10): 0.00000000001.

Let us see what a tentative “pythonesque” pseudocode for this loop could be:

def sqrt(n): f = function(x*x - n) x = 1 # seed xant = 0 do: f1 = (f.eval(x+h) - f.eval(x)) / h xant = x x = x - f.eval(x)/f1 while abs(x - xant) > err; return x

Assuming we have a symbolic function type, that loop does not seem too difficult. In order to code this in C, since the equation is always the same, I will hardcode it as a plain function.

typedef double real; // change to float for single precision real f(real x, real n) { return x*x - n; } real sqrt(real n) { real err = 0.00000000001f; real h = 0.01f; real x = 1.0f; // seed real xant = 0.0f; do { xant = x; real df = (f(x+h, n) - f(x, n))/h; x = x - f(x, n)/df; } while (abs(x - xant) > err); return x; }

Here are the results of running our custom square root function, compared to the standard version provided with the C programming language:

[ale@syaoran sqrt]$ ./sqrt 1.0 Custom sqrt of: 1 = 1 libm sqrt of: 1 = 1 [ale@syaoran sqrt]$ ./sqrt 2.0 Custom sqrt of: 2 = 1.41421 libm sqrt of: 2 = 1.41421 [ale@syaoran sqrt]$ ./sqrt 4.0 Custom sqrt of: 4 = 2 libm sqrt of: 4 = 2 [ale@syaoran sqrt]$ ./sqrt 16.0 Custom sqrt of: 16 = 4 libm sqrt of: 16 = 4 [ale@syaoran sqrt]$ ./sqrt 32.0 Custom sqrt of: 32 = 5.65685 libm sqrt of: 32 = 5.65685 [ale@syaoran sqrt]$ ./sqrt 100.0 Custom sqrt of: 100 = 10 libm sqrt of: 100 = 10 [ale@syaoran sqrt]$ ./sqrt 1000000.0 Custom sqrt of: 1e+06 = 1000 libm sqrt of: 1e+06 = 1000