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