On Mon, 11 Aug 2008 02:37:40 -0700 "William Stein" <wstein@gmail.com> wrote:

On Mon, Aug 11, 2008 at 1:30 AM, Burcin Erocal <burcin@erocal.org> wrote:

<snip>

...

At the ACA'08 conference, I've met some physicists using GiNaC (http://www.ginac.de/), a C++ library that provides symbolic manipulation and pattern matching facilities. Its lastest release was on Apr 04 2008, it also has extensive documentation, and active mailing lists. I wrote a basic Cython wrapper to benchmark its performance. Here are the numbers:

sage: var("x y z",ns=1) (x, y, z) sage: a = (x+y)^2 sage: %time t = sum( [a^i for i in xrange(10000)] ) CPU times: user 1.83 s, sys: 0.00 s, total: 1.83 s Wall time: 1.83 s

This benchmark might be lame. Even the current "insanely slow" maxima-based symbolics in Sage seem to do better than what you just wrote above:

sage: var('x,y') (x, y) sage: a = (x+y)^2 sage: time t = sum(a^i for i in xrange(10000)) CPU times: user 0.51 s, sys: 0.03 s, total: 0.54 s Wall time: 0.55 s

and

sage: var('y,z'); b = (y+z)^2 sage: time t = sum([a^i*b^j for i in xrange(1,200,5) for j in xrange(1, 300, 3)] ) CPU times: user 0.30 s, sys: 0.01 s, total: 0.31 s Wall time: 0.32 s

That's because the benchmark is maybe testing nothing but memory allocation.

Your benchmark doesn't call maxima at all. When you quit Sage after these lines, there is no Exiting spawned Maxima process. notice printed.

I have seen a lot of *real* benchmarks involving symbolic calculus when sad users and students come to me and ask why their code is 1000 times slower than Mathematica. This happened a *few* times today at Sage Days 9 today. It also comes up in sage-support all the time. I really need to come up with a large list of specific examples like this, since I'm getting way too confused by benchmarks.

I couldn't find any standard benchmarks, and had to come up with the one above. I think it is reflects some aspect of performance, as the power operator is very similar to a symbolic function application, and the system has to do a basic simplification like ((x+y)^2)^i = (x+y)^(2*i) for each step. Cheers, Burcin