find all multiplicands and multipliers for a number

Dave Angel davea at
Mon Apr 13 07:43:07 CEST 2015

On 04/13/2015 01:25 AM, Paul Rubin wrote:
> Dave Angel <davea at> 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.

Well that value x has some 3300 digits, and I seem to recall that float 
only can handle 10**320 or so.  But if the point of all this is to 
decide when to stop dividing, I think our numbers here are somewhere 
beyond the heat death of the universe.

   Thus (not completely tested):
>      def isqrt(x):
>          def log2(x): return math.log(x,2)  # python 2 compatibility
>          if x < 1e9:

Now 10**9 is way below either limit of floating point.  So i still don't 
know which way you were figuring it.  Just off the top of my head, I 
think 10**18 is approx when integers don't get exact representation, and 
10**320 is where you can't represent numbers as floats at all.

>             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.

If you're willing to use the 10**320 or whatever it is for the limit, I 
don't see what's wrong with just doing floating point sqrt.  Who cares 
if it can be an exact int, since we're just using it to get an upper limit.

And I can't figure out your code at this hour of night, but it's much 
more complicated than Newton's method would be anyway.


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