# Challenge: optimizing isqrt

Stephen Tucker stephen_tucker at sil.org
Fri Nov 21 09:15:57 CET 2014

```Serhiy,

This looks very good indeed. As a matter of interest, is there any
particular reason you have used 2*b instead of b+b? Might b+b be faster
than b*2?

Also, in various lines, you use //2. Would >>1 be quicker? On reflection,
perhaps you have had to use //2 because >>1 cannot be used in those
situations.

Stephen Tucker.

On Thu, Nov 20, 2014 at 6:00 PM, Serhiy Storchaka <storchaka at gmail.com>
wrote:

> On 01.11.14 03:29, Steven D'Aprano wrote:
>
>> There is an algorithm for calculating the integer square root of any
>> positive integer using only integer operations:
>>
>
> Here is my optimized implementation inspired by Christian's ideas.
>
> import math, sys
>
> C1 = sys.float_info.radix ** sys.float_info.mant_dig
> C2 = int(sys.float_info.max)
> C3 = C2.bit_length() - 2
>
> def isqrt(n):
>     if n < 0:
>         raise ValueError
>     if n == 0:
>         return 0
>     if n <= C1:
>         return int(math.sqrt(n))
>     if n <= C2:
>         x = int(math.sqrt(n))
>     else:
>         b = (n.bit_length() - C3) // 2
>         x = int(math.sqrt(n >> (2 * b))) << b
>     y = (x + n // x) // 2
>     if y == x:
>         return x
>     while True:
>         x = y
>         y = (x + n // x) // 2
>         if y >= x:
>             return x
>
>
> --
> https://mail.python.org/mailman/listinfo/python-list
>
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