Re: Problem 2.10
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Re: Problem 2.9 : Growth function of perceptron, seems incorrect

Re: Problem 2.10
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Thank you. 
Re: Possible correction to Problem 2.14 (b)
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Re: Problem 2.10
Yes, well done.
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Problem 2.3 c
the last comment confused me a little bit.
For a given set of N points, we should change the center of the sphere to get as many dichotomies as we can, thus measuring the effective number of hypotheses (spheres) in this hypothesis set. Does it make sense to move project the spheres from 3D to 1D and look at the problem as intervals of +1 for a<=x<=b and a>=x>=b? 
Re: Problem 2.3 c
You will note from the definition of the hypothesis set: contains functions which are +1 for
You only get to vary , and so the two spheres are restricted to be centered on the origin. And yes, the for this hypothesis set is very related to the growth function for positive intervals. Quote:

Problem 2.15
In (a), it said that we should provide a monotonic classifier. Then, why there are +1 and 1 regions?
Also, as it said in (b) that generating the next point by increasing the first component and decreasing the second component. Then, how can we determine which point is larger? Because X1>=X2 if and only if the inequality is satisfied for every component. However, the next point is just with one component larger than that of the first one, while another component is less than that of the first one. So, it's a little confusing. 
Re: Problem 2.15
The function is +1 in some region and is 1 in the complement  i.e. it takes on two values. Any function can be monotonic, even one that takes on just 2 values.
In (b), you are asked to compute m(N). To compute m(N) you need to count the maximum number of implementable dichotomys on some N points. The problem suggest a set of N points which might be helpful. The fact that no point is larger than another is crucial [hint: because if a point were larger than another, there is a dichotomy that you cannot implement]. Quote:

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