That's perfect. Thanks!
On Mar 26, 5:41 pm, Jason Grout <[email protected]> wrote:
> Burcin Erocal wrote:
> > Hi,
>
> > On Thu, 26 Mar 2009 09:31:09 -0700 (PDT)
> > Patrick <[email protected]> wrote:
>
> >> I'd like to count the iterations of a loop nest by evaluating a sum.
> >> Consider the following loop nest:
>
> >> for (k = 0; k < N; k++)
> >> for (i = k+1; i < N; i++)
> >> for (j = k+1; j <= i; j++)
> >> ...
>
> >> I'd like to count the total number of iterations of that loop nest.
> >> I've used mathematica a bit, and you could solve this problem with
> >> something like this:
>
> >> Sum[1, {k, 0, N}, {i, k+1, N}, {j, k+1, i-1}]
>
> >> The result should be in terms of N. For this loop nest the closed
> >> form is:
>
> >> 1 / 6 * (1 + N) * (N^2 - N)
>
> >> I've been trying to figure out how to do this sort of problem in Sage,
> >> but I have been unsuccessful. Any help would be greatly appreciated.
>
> > Unfortunately Sage doesn't have native code to find closed forms for
> > sums at the moment. You will have to use maxima directly to solve this
> > problem.
>
> You can use maxima as illustrated in William's message here:
>
> http://groups.google.com/group/sage-support/browse_thread/thread/4584...
>
> sage: a=maxima('sum(sum(sum(1,j,k+1,i-1), i, k+1, N), k, 0, N),
> simpsum').sage()
> sage: a
> (2*N^3 + 3*N^2 + N)/12 - (N^3 + N^2)/2 + (N^2 + N)/4 + N^2*(N + 1)/2 -
> N*(N + 1)/2
> sage: a.simplify_full()
> (N^3 - N)/6
>
> Jason
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