ISO module for binomial coefficients, etc.

Dave Angel davea at ieee.org
Sun Jan 24 22:11:03 CET 2010


Alf P. Steinbach wrote:
> <div class="moz-text-flowed" style="font-family: -moz-fixed">* Dave 
> Angel:
>> kj wrote:
>>> Before I go off to re-invent a thoroughly invented wheel, I thought
>>> I'd ask around for some existing module for computing binomial
>>> coefficient, hypergeometric coefficients, and other factorial-based
>>> combinatorial indices.  I'm looking for something that can handle
>>> fairly large factorials (on the order of 10000!), using floating-point
>>> approximations as needed, and is smart about optimizations,
>>> memoizations, etc.
>>>
>>> TIA!
>>>
>>> ~K
>>>
>>>   
>> You do realize that a standard. python floating point number cannot 
>> possibly approximate a number like 10000!
>
> I think what kj is looking for, provided she/he is knowledgable about 
> the subject, is code that does something like
>
>   >>> from math import *
>   >>> log_fac = 0
>   >>> for i in range( 1, 10000+1 ):
>   ...     log_fac += log( i, 10 )
>   ...
>   >>> print( "10000! = {}e{}".format( 10**(log_fac % 1), int( log_fac 
> ) ) )
>   10000! = 2.84625968062e35659
>   >>> _
>
> which turned out to be accurate to 10 digits.
>
>
>>  Better use longs.
>
> That would involve incredible overhead. E.g., how many bytes for the 
> number above? Those bytes translate into arithmetic overhead.
>
About 14k.
>
>> I'd check out the gamma function, which matches factorial for integer 
>> arguments (plus or minus 1).
>
> Or, e.g., logarithms... ;-)
>
>
> Cheers & hth.,
>
> - Alf
>
I didn't think of simply summing the logs.  I did have some 
optimizations in mind for the multiply of the longs.  If you do lots of 
partial products, you can do a good bit of the work with smaller 
numbers, and only get to longs when those partial products get big 
enough.   You could also use scaling when the numbers do start getting 
bigger.

But I still think there must be code for the gamma function that would 
be quicker.  But I haven't chased that lead.

DaveA



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