Dave Angel <da...@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.
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