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Bug 101 - Investigate newer JacobiSVD techniques
Summary: Investigate newer JacobiSVD techniques
Status: NEW
Alias: None
Product: Eigen
Classification: Unclassified
Component: SVD (show other bugs)
Version: unspecified
Hardware: All All
: --- Unknown
Assignee: Nobody
Depends on:
Reported: 2010-10-29 15:40 UTC by Benoit Jacob
Modified: 2019-12-04 09:56 UTC (History)
2 users (show)


Description Benoit Jacob 2010-10-29 15:40:34 UTC
Reading e.g. this paper: "New fast and accurate Jacobi SVD algorithm: I. LAPACK working note" by Drmac and Veselic,

Stuff worth considering:
 * TFA claims that QR preconditioning makes JacobiSVD converge faster, and that a second LQ applied to R helps even more.
 * Parallelization (block Jacobi SVD).

However there is also lots of stuff that we DON'T want:
 * TFA claims that the second LQ does not need to be pivoting. But that will necessarily reduce the reliability of the algorithm. So if we do a LQ, either make it non-default or make it pivoting.
 * Large parts of TFA are about how working on the transposed matrix is more numerically stable: this is an artifact of them doing one-sided Jacobi SVD. Since our JacobiSVD is two-sided, we don't need to bother about this. The fact that they worry about this is quite a testament to how real the issues with one-sided Jacobi are!
Comment 1 Dirk Toewe 2019-07-19 20:23:25 UTC
One small change that I would suggest is to use a different stopping criterion. The Jacobi SVD can achieve higher relative accuracy than other Jacobi SVD methods, at least according to this paper:

In order to achieve this accuracy, however, I believe the stopping criterion of JacobiSVD would have to be changed. If I understand the code correctly, the current stopping criterion compares the off-diagonal values to the largest diagonal entry:

n*eps*max[i](S[i,i]) >= max[i != j]( S[i,j] )

The aforementioned paper suggests a stopping criterion along the lines of:

n*eps >= max[i != j]( |S[i,j]| / sqrt|S[i,i]*S[j,j]| )

The paper however only covers the symmetric case. Not sure how to generalize this to the non-symmetric case. My guess would be something like

n*eps >= max[i != j]( max{|S[i,j]|,|S[j,i]|} / sqrt|S[i,i]*S[j,j]| )
Comment 2 Nobody 2019-12-04 09:56:06 UTC
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