At 47:00, an upper bound beta is defined. There is this problem though that beta might not be finite, as H1 might be unbounded. I don't see how the proof can actually work given this.
I think that a finite training set H silently is assumed. In that case, the solution for you problem is the following. If |H| = {h0, h1, ..., hA}, then x0 = h0, ..., xA = hA, x(A + 1) = h0, x(A + 2) = h1 ... i.e. you cycle the training examples. However, he did not prove the theorem, because he ignored the wrongly classified negative examples ...
Very large value of n means that the number of iterations are large enough for convergence to occur. Thus, a large value of N will ensure that we can actually reach convergence at which both the extremes will start to come closer (lower bound will increase and upper bound will decrease) and the bounds become equal.
What an amazing lecture !!! Flawless !! Thank you professor
best lecture on pct
At 47:00, an upper bound beta is defined. There is this problem though that beta might not be finite, as H1 might be unbounded. I don't see how the proof can actually work given this.
I think that a finite training set H silently is assumed. In that case, the solution for you problem is the following. If |H| = {h0, h1, ..., hA}, then x0 = h0, ..., xA = hA, x(A + 1) = h0, x(A + 2) = h1 ... i.e. you cycle the training examples. However, he did not prove the theorem, because he ignored the wrongly classified negative examples ...
but how do you actually know that a very large value of n, n times beta is going to decrease? and the other term is going to increase
Very large value of n means that the number of iterations are large enough for convergence to occur. Thus, a large value of N will ensure that we can actually reach convergence at which both the extremes will start to come closer (lower bound will increase and upper bound will decrease) and the bounds become equal.
i think last formula he derived is wrong
n0>=nmax
then only w(n0)=w(n0+1)=w(n0+2)=------
am i correct?
If n0>=nmax the perceptron shall diverge. I think derivation shown here is correct i.e n0
Lecture is good but very less intuitive just covering Haykin book word by word.