find all multiplicands and multipliers for a number
Steven D'Aprano
steve+comp.lang.python at pearwood.info
Mon Apr 13 09:21:27 CEST 2015
On Monday 13 April 2015 15:25, Paul Rubin wrote:
> Dave Angel <davea at davea.name> writes:
>> But doesn't math.pow return a float?...
>> Or were you saying bignums bigger than a float can represent at all?
>> Like:
>>>>> x = 2**11111 -1 ...
>>>>> math.log2(x)
>> 11111.0
>
> Yes, exactly that. Thus (not completely tested):
>
> def isqrt(x):
> def log2(x): return math.log(x,2) # python 2 compatibility
> if x < 1e9:
> return int(math.ceil(math.sqrt(x)))
> a,b = divmod(log2(x), 1.0)
> c = int(a/2) - 10
> d = (b/2 + a/2 - c + 0.001)
> # now c+d = log2(x)+0.001, c is an integer, and
> # d is a float between 10 and 11
> s = 2**c * int(math.ceil(2**d))
> return s
>
> should return slightly above the integer square root of x. This is just
> off the top of my head and maybe it can be tweaked a bit. Or maybe it's
> stupid and there's an obvious better way to do it that I'm missing.
Check the archives: I started a thread last November titled "Challenge:
optimizing isqrt" which is relevant. Also:
http://code.activestate.com/recipes/577821-integer-square-root-function/
--
Steve
More information about the Python-list
mailing list