[petsc-users] SLEPc EPS Tolerance

[Eda Oktay]

Okay thanks again!

On Fri, Jul 10, 2020, 7:52 PM Jose E. Roman <jroman at dsic.upv.es> wrote:

> I don't know. This question belongs to the application part, not the
> solver part.
>
> > El 10 jul 2020, a las 18:45, Eda Oktay <eda.oktay at metu.edu.tr> escribió:
> >
> > I looked at a2965*2965 sized matrix. I want to ask one last question.
> >
> > For e-8:
> >
> > Linear eigensolve converged (2 eigenpairs) due to CONVERGED_TOL;
> iterations 64
> > ---------------------- --------------------
> >            k             ||Ax-kx||/||kx||
> > ---------------------- --------------------
> >        0.002486            8.60766e-09
> >        0.004466            1.68813e-09
> > ---------------------- --------------------
> >
> > For e-10:
> > Linear eigensolve converged (2 eigenpairs) due to CONVERGED_TOL;
> iterations 74
> > ---------------------- --------------------
> >            k             ||Ax-kx||/||kx||
> > ---------------------- --------------------
> >        0.002486             9.5257e-11
> >        0.004466            2.12622e-11
> > ---------------------- --------------------
> >
> > But my partition is more qualified when tol=e-8, not e-10.
> >
> > My last question is: Is this because of the iteration number that
> > eigensolver tries to find suitable eigenvalues up to this tolerance?
> >
> > Again, thank you so much!
> >
> > Jose E. Roman <jroman at dsic.upv.es>, 10 Tem 2020 Cum, 19:36 tarihinde
> şunu yazdı:
> >>
> >> For such small matrix, the solver is essentially solving with a direct
> method via LAPACK. If you understand how projection methods work, iteration
> gets in action when the matrix size is larger than the subspace size (ncv).
> The tolerance is relevant when the solver iterates, not when it computes
> the solution with LAPACK.
> >>
> >> Jose
> >>
> >>
> >>> El 10 jul 2020, a las 18:32, Eda Oktay <eda.oktay at metu.edu.tr>
> escribió:
> >>>
> >>> No, it is of size 9
> >>>
> >>> Jose E. Roman <jroman at dsic.upv.es>, 10 Tem 2020 Cum, 19:31 tarihinde
> şunu yazdı:
> >>>>
> >>>> [Please respond to the list.]
> >>>>
> >>>> Is your matrix of size 8? This would explain the residuals.
> >>>>
> >>>> Jose
> >>>>
> >>>>> El 10 jul 2020, a las 17:10, Eda Oktay <eda.oktay at metu.edu.tr>
> escribió:
> >>>>>
> >>>>> I computed residual norm via -eps_error_relative::ascii_info_detail
> >>>>> for different tolerance numbers (e-4, e-6, e-8, e-10). In each
> >>>>> tolerance, I got the same table below:
> >>>>>
> >>>>> ---------------------- --------------------
> >>>>>          k             ||Ax-kx||/||kx||
> >>>>> ---------------------- --------------------
> >>>>>      3.000000            6.25528e-16
> >>>>>      3.000000            7.13774e-16
> >>>>>      3.438447            2.64362e-16
> >>>>>      5.000000            4.39333e-16
> >>>>>      6.000000            1.63943e-16
> >>>>>      6.000000            2.93737e-16
> >>>>>      6.000000            3.95997e-16
> >>>>>      7.561553            3.48664e-16
> >>>>> ---------------------- --------------------
> >>>>>
> >>>>> I understood that since relative error is E-16 and this table shows
> >>>>> eigenvalues whose relative error are below the tolerance, I am
> getting
> >>>>> the same table but I still couldn't understand although relative
> >>>>> errors are so small, how am I getting the most qualified partition in
> >>>>> e-4 tolerance, not e-8. I am not computing zero eigenvalue I believe,
> >>>>> since I am using EPSSetWhichEigenpairs(eps,EPS_SMALLEST_MAGNITUDE);
> >>>>> and I am not getting zero eigenvalue.
> >>>>>
> >>>>> Thanks so much for answering!
> >>>>>
> >>>>> Jose E. Roman <jroman at dsic.upv.es>, 10 Tem 2020 Cum, 14:09
> tarihinde şunu yazdı:
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>>> El 10 jul 2020, a las 12:54, Eda Oktay <eda.oktay at metu.edu.tr>
> escribió:
> >>>>>>>
> >>>>>>>> How do you measure accuracy?
> >>>>>>> Using the word accuracy may be not true actually, I am sorry. I am
> >>>>>>> using eigenvectors corresponding to these eigenvalues in k-means
> >>>>>>> algorithm, then do spectral graph partitioning. I must look at the
> >>>>>>> partition quality. By quality, I mean, the resulting edge cut of my
> >>>>>>> partitioned graph. I thought that the less tolerance results in
> more
> >>>>>>> accuracy, hence more qualified partition.
> >>>>>>>
> >>>>>>>> What do you mean "the result was still the same"?
> >>>>>>> I mean I am still not getting the most qualified solution in E-10,
> >>>>>>> still E-2 or E-6 gives more qualified partitions, i.e. they give
> less
> >>>>>>> edge cut.
> >>>>>>>
> >>>>>>>> What is the eigenvalue you are computing?
> >>>>>>> I am computing the smallest eigenvalue of a Laplacian matrix.
> >>>>>>
> >>>>>> You should compute the residual norm, for instance with
> -eps_error_relative ::ascii_info_detail (see section 2.5.4 of the manual).
> >>>>>> The relative residual error should be in the order of the tolerance
> (or smaller) if using the default convergence test, but if you are
> computing a zero eigenvalue then you may want to use an absolute
> convergence criterion (see table 2.6 or the manual). A graph Laplacian has
> at least one zero eigenvalue, unless you deflate it as explained in section
> 2.6.2 of the manual, see also ex11.c
> https://slepc.upv.es/documentation/current/src/eps/tutorials/ex11.c.html
> >>>>>>
> >>>>>> Jose
> >>>>>>
> >>>>
> >>
>
>
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