Re: [Pysparse-users] reading the source code of pysparse

[Oz N. T. <na...@gm...>]

On Sat, Feb 11, 2012 at 6:57 PM, Dominique Orban
<dom...@gm...> wrote:

Hi Dominique,
Thanks for the answer. A few more questions.


> This is a macro that adjusts the call to various types of Fortran
> compilers (this particular call is a call to the BLAS library, which
> is written in Fortran). Some Fortran compilers add a trailing
> underscore to symbols, some add two, some add none, etc.
>

So this macro is found in BLAS? not in pysprase itself ?


> I wouldn't recommend coding bicgstab all over again. The only costly
> operations in Bi-CGSTAB (and other Krylov-type methods) are vector
> operations (most often, addition of vectors) and operator-vector
> products (e.g., A*x or A.T*x). I would say that to speed things up,
> you'll want to speed up your operator-vector operations; they are the
> dominant cost.
>
> You can take a look at PyKrylov (https://github.com/dpo/pykrylov)
> which contains a pure Python implementation of Bi-CGSTAB and allows
> you to input your operator in different ways (a Pysparse matrix being
> one of them). For instance, you could implement your operator in C or
> in Cython and that should speed things up. I believe that is the way
> to go.

Why is pure python Bi-CGSTAB should be faster than the one writen in C
contained in pysparse?


> Of course, vector operations in PyKrylov could also be speeded up with
> Cython. That's been on my list for a while.

You mean doing stuff on numpy vectores instead of for loops? I didn't see
so many of these in the code ...

And one last thing, in pysparse solvers, I pass a Matrix, initial x vector,
and b vector (representing my boundary conditions.) I was quite
confused to see the example here:
https://github.com/dpo/pykrylov/blob/master/examples/bmark.py

    for KSolver in [CGS, TFQMR, BiCGSTAB]:
        ks = KSolver( lambda v: A*v,
                      #precon = dp,
                      #verbose=False,
                      reltol = 1.0e-8
                      )
        ks.solve(rhs, guess = 1+np.arange(n, dtype=np.float), matvec_max=2*n)

you are not assigning the lambda v to anything? What does this do?

solveing is done only with rhs and guess ? is this because A is
already contained
in ks ? I'd be happy to see an example which does not confuse me so much :-)

Thanks again for the answers and the effort invested in this!

Oz