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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