pythonize this!

[Stefan Behnel]

Peter Otten, 18.06.2010 12:14:
> Stefan Behnel wrote:
>
>> Andre Alexander Bell, 18.06.2010 11:23:
>>> On 06/16/2010 12:47 PM, Lie Ryan wrote:
>>>> Probably bending the rules a little bit:
>>>>
>>>>>>> sum(x**2 - 8*x - 20 for x in range(1, 2010, 5))
>>>> 536926141
>>>
>>> Bending them even further, the sum of the squares from 1 to N is given by
>>>
>>> (1) N*(N+1)*(2*N+1)/6.
>>>
>>> The given problem can be divided into five sums of every fifth square
>>> starting from 1,2,3,4,5 respectively. This is given by
>>>
>>> (2) S(a,k) = sum_{i=1}^k (5*i-4+a)^2
>>>
>>> for a in range(5) and we are finally interested in
>>>
>>> (3) S(k) = S(0,k) + S(1,k) + S(2,k) - S(3,k) - S(4,k)
>>>
>>> Substituting (2) and (1) in (3) gives (in python code)
>>>
>>>>>> S = lambda k: (50*k**3 - 165*k**2 - 47*k) / 6
>>>>>> S(2010/5)
>>> 536926141
>>>
>>> However, this only works for full cycles of (1,1,1,-1,-1) and you would
>>> have to add/subtract the modulus parts yourself. (e.g. if you are
>>> interested in your sum from 1..2011...
>>
>> The thing is, if you can't do the math in the time that your processor
>> needs to run the brute force loop, it's often not worth doing the math at
>> all.
>
> By that standard using Cython for the problem doesn't pay either;)

True. Running the C compiler certainly takes a lot longer than starting up 
Python and running the loop.

Stefan