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Old 08-24-2012, 08:10 AM
patrickjtierney patrickjtierney is offline
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Join Date: Jul 2012
Location: Toronto, Canada
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Default Re: Quadratic programming

This was discussed extensively in the previous class. A user elkka found a solution to the non-terminating problem for Octave. The trick is to run qp on an altered version of H (using all the same other parameters) and then to use the output as the initial value alpha0 for qp on the original H. The second run seems to always terminate after one iteration.

The altered H is just to add a small amount to the diagonal of H (say 10^-15). This makes its determinant non-zero, which is helpful for qp. So HH=H+eye(n)*10^-15 should work. [ n is just the length of Y ] Also, the alpha0 I refer to goes first in the arg list for the second qp call.

The results are much better, both in terms of termination and in producing a noticeably higher prediction accuracy (ie 10+% rise in better than PLA results).

I posted about this here: http://book.caltech.edu/bookforum/showthread.php?t=1133 , but also there is the original thread: http://book.caltech.edu/bookforum/sh...p?t=513&page=5
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