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Old 03-17-2013, 02:16 AM
boulis boulis is offline
Join Date: Feb 2013
Location: Sydney, Australia
Posts: 29
Default Re: on the right track?

Originally Posted by heer2351 View Post
I get the same weights. But do not understand how you get to an Ein of 0 with these points and clusters.

I would say point 1 and 2 belong to center 1, these points have opposite labels. Same for point 3 and 4 which belong to center 2. So how can Ein be zero?
I get the same weights too if I take the centres given as input. And I also get Ein = 0

It is not strange to have Ein = 0, I am not sure why you are confused about it. You are correct that the clusters are as you name them, but the sum of the weighted RBFs is such that we achieve the right sign in the right place (and in fact almost the right value, not just the sign). The key in this case I believe is the bias (W0 or b). This is negative so it gives the field a negative start(sign). Then the two RBFs work to make it positive. The points that are close to the centres are getting affected most and they become positive. The other two are staying negative.

Can someone please revisit my comment earlier? (That these are not the centres when Lloyd's algorithm is applied.)

One more question: Q14 asks the % of times that we get non-separable data by the RBF kernel (i.e. SVM hard margin with RBF kernel). In about 1000 runs that I have tried I never encountered non-separable data. Is this normal? I am using libsvm so it's harder to make a mistake about it... but who knows. I have checked that it identifies non separable data correctly (the 4 point example given in this thread is one such case).
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