Have a look here. For instance, at 3(a) there is this method, which is trivially adaptable to do a 64->32 bit square root, and also trivially transcribable to assembler: /* by Jim Ulery */ static unsigned julery_isqrt(unsigned long val) { unsigned long temp, g=0, b = 0x8000, bshft = 15; do { if (val >= … Read more
Yes, it is possible even without trickery: sacrifice accuracy for speed: the sqrt algorithm is iterative, re-implement with fewer iterations. lookup tables: either just for the start point of the iteration, or combined with interpolation to get you all the way there. caching: are you always sqrting the same limited set of values? if so, … Read more
Perhaps you want the result to be an Int as well? isqrt :: Int -> Int isqrt = floor . sqrt . fromIntegral You may want to replace floor with ceiling or round. (BTW, this function has a more general type than the one I gave.)
Here’s a version without sqrt, though I’m not sure whether it is faster than a version which has only one sqrt (it may depend on the distribution of values). Here’s the math (how to remove both sqrts): ad = a2-a1 bd = b2-b1 a1+sqrt(b1) < a2+sqrt(b2) // subtract a1 sqrt(b1) < ad+sqrt(b2) // square it … Read more
Option 1: math.sqrt() The math module from the standard library has a sqrt function to calculate the square root of a number. It takes any type that can be converted to float (which includes int) as an argument and returns a float. >>> import math >>> math.sqrt(9) 3.0 Option 2: Fractional exponent The power operator … Read more
Note: There is now math.isqrt in stdlib, available since Python 3.8. Newton’s method works perfectly well on integers: def isqrt(n): x = n y = (x + 1) // 2 while y < x: x = y y = (x + n // x) // 2 return x This returns the largest integer x for … Read more
I don’t know how to encode the float number using integer format. There is a function for that: f32::to_bits which returns an u32. There is also the function for the other direction: f32::from_bits which takes an u32 as argument. These functions are preferred over mem::transmute as the latter is unsafe and tricky to use. With … Read more
In Python 2, sqrt=x**(1/2) does integer division. 1/2 == 0. So x(1/2) equals x(0), which is 1. It’s not wrong, it’s the right answer to a different question. If you want to calculate the square root without an import of the math module, you’ll need to use x**(1.0/2) or x**(1/2.). One of the integers needs … Read more