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#11
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Using x = [1 0;0 1;0 -1.00002; -1 0; 0 2.0001; 0 -2; -2 0]; I still get one less s.v. for qp than libsvm (ie same values I get without perturbing). This remains the case when only perturbing one s.v. The most notable change is that the second weight entry grows, although the first and b also change.
Also, thanks to fgpancorbo for the code for getting w & b from libsvm. Useful for the future. |
#12
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In z space X6 and X7 map to the same point in z space, i.e, X6 (0, -2) and X7 (-2,0) map to (3,5). I wonder if this has any effect on the computation. |
#13
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I noticed that in Q12, but I believe that the z-space is defined by the polynomial kernel in Q13, and not the mapping from Q12. This can be seen in slide 10 of week 15 where g(x) is specified.
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#14
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Using Matlab libsvm I perturbed [0,-1] to [0,-0.94] and it reduced the number of support vectors by one. W and b agree with what others have seen for libsvm.
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#15
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I used libsvm and got this question right.
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#16
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Predictions of qp and libsvm are identical (for 10000 uniformly distributed samples on x1,x2 = [-5,5]).
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#17
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As a quick (read minimal effort) check of model equivalence, I compared the predicted results of 1,000,000 randomly selected points within [-3,3][-3,3] using the support vectors from both Octave/QP and Python/libsvm. Only three points were classified differently despite the difference in the number of support vectors returned by the two approaches. I'm certain that an analytical comparison of the support vectors would prove their equivalence; however, it hardly seems necessary given the empirical results.
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#18
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I had quite some problems with solving this problem and it forced me to play around with different approaches (qp, libsvm). One more approach to consider is this: Lecture 15, slide 5:
In the case of the kernel used in exercice 13, there is a corresponding transformation, given explicitly on that slide. So why not giving it a try? I got some confidence in the result after reading the slides title ![]() |
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