Determining Floating Point Square Root

Question

How do I determine the square root of a floating point number? Is the Newton-Raphson method a good way? I have no hardware square root either. I also have no hardware divide

Answer 1

Easiest to implement (you can even implement this in a calculator):

def sqrt(x, TOL=0.000001):
    y=1.0
    while( abs(x/y -y) > TOL ):
        y= (y+x/y)/2.0
    return y

This is exactly equal to newton raphson:

y(new) = y - f(y)/f'(y)

f(y) = y^2-x and f'(y) = 2y

Substituting these values:

y(new) = y - (y^2-x)/2y = (y^2+x)/2y = (y+x/y)/2

If division is expensive you should consider: http://en.wikipedia.org/wiki/Shifting_nth-root_algorithm .

Shifting algorithms:

Let us assume you have two numbers a and b such that least significant digit (equal to 1) is larger than b and b has only one bit equal to (eg. a=1000 and b=10). Let s(b) = log_2(b) (which is just the location of bit valued 1 in b).

Assume we already know the value of a^2. Now (a+b)^2 = a^2 + 2ab + b^2. a^2 is already known, 2ab: shift a by s(b)+1, b^2: shift b by s(b).

Algorithm:

Initialize a such that a has only one bit equal to one and a^2<= n < (2*a)^2. 
Let q=s(a).    
b=a
sqra = a*a

For i = q-1 to -10 (or whatever significance you want):
    b=b/2
    sqrab = sqra + 2ab + b^2
    if sqrab > n:
        continue
    sqra = sqrab
    a=a+b

n=612
a=10000 (16)

sqra = 256

Iteration 1:
    b=01000 (8) 
    sqrab = (a+b)^2 = 24^2 = 576
    sqrab < n => a=a+b = 24

Iteration 2:
    b = 4
    sqrab = (a+b)^2 = 28^2 = 784
    sqrab > n => a=a

Iteration 3:
    b = 2
    sqrab = (a+b)^2 = 26^2 = 676
    sqrab > n => a=a

Iteration 4:
    b = 1
    sqrab = (a+b)^2 = 25^2 = 625
    sqrab > n => a=a

Iteration 5:
    b = 0.5
    sqrab = (a+b)^2 = 24.5^2 = 600.25
    sqrab < n => a=a+b = 24.5

Iteration 6:
    b = 0.25
    sqrab = (a+b)^2 = 24.75^2 = 612.5625
    sqrab < n => a=a


Iteration 7:
    b = 0.125
    sqrab = (a+b)^2 = 24.625^2 = 606.390625
    sqrab < n => a=a+b = 24.625

and so on.