#1




on the right track?
Since probably most of us are writing our own regular RBF implementations by hand, I thought it would be helpful to compare the results of a couple simple test cases to make sure our implmentations are correct.
data set = (0, 0) (0, 1) (1, 0) (1, 1) labels = 1 1 1 1 centers = (0, 0.2) (1, 0.7) Case 1: =1 weights = (7.848, 7.397, 7.823) (first weight is the bias) Case 2: =100 weights = (1.0, 109.196, 16206.168) 
#2




Re: on the right track?
I concur with those results.

#3




Re: on the right track?
I agree with those results as well.

#4




Re: on the right track?
My results concur with these figures too

#5




Re: on the right track?
Same results here.

#6




Re: on the right track?
Quote:
I believe that Lloyd's algorithm would produce unstable results with this configuration of points and number of centers. Indeed, these are the 3 different cases that I get from running my algorithm (the results depend on the random starting point of course): centres = [[ 0., 0.], [ 0.66666667, 0.66666667]] clusters = [[[0, 0]], [[0, 1], [1, 0], [1, 1]]] centres = [[ 0.66666667, 0.33333333], [ 0., 1.]] clusters = [[[0, 0], [1, 0], [1, 1]], [[0, 1]]] centres = [[ 1. , 0.5], [ 0. , 0.5]] clusters = [[[1, 0], [1, 1]], [[0, 0], [0, 1]]] These seem pretty reasonable to me. Do other people get the same results? 
#7




Re: on the right track?
I have run my algo on the last dataset that you have provided and I get the same results

#8




Re: on the right track?
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? 
#9




Re: on the right track?
Quote:
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 nonseparable data by the RBF kernel (i.e. SVM hard margin with RBF kernel). In about 1000 runs that I have tried I never encountered nonseparable 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). 
#10




Re: on the right track?
Thanks for your reply, reviewed the lecture again and spotted my thinking error.
I changed my code and it now gives Ein=0 for the examples. I still have doubts however whether my code is correct because when I use Lloyds with these examples my Ein is never zero. As a matter of fact my Ein is rather large all the time; I am trying to find the flaw in my code. What Ein values do you get in these examples when using Lloyds? I agree that the centers are not calculated according to Lloyds and also agree with your centers. However I think that there are more centers than you have, for example: Code:
[1 , 1] [1/3, 1/3] [1/2, 1] [1/2, 0] [1 , 0] [1/3, 2/3] 
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