Exercises and Problems
Please comment on the chapter problems in terms of difficulty, clarity, and time demands. This information will help us and other instructors in choosing problems to assign in our classes.
Also, please comment on the exercises in terms of how useful they are in understanding the material. 
ERRATA: Small mistake in description of Exercise 2.1
I won't "mathify" this correction, as I don't know how in this forum (my LaTex has escaped me after years of neglect!). It's a minor point, but I feel like it should be corrected. In the wording for Exercise 2.1 on page 45:
ERROR: "Verify that m_H(n) < 2^k" CORRECTION: "Verify that m_H(k) < 2^k" Thanks! 
Re: ERRATA: Small mistake in description of Exercise 2.1
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Problem 2.9 : Growth function of perceptron, seems incorrect
The problem says that in case of perceptron in ddimensional space, growth function is equal to B(N,k). Consider the following case :
d = 2, implies dvc = 2+1 = 3 N = 4 B(N,k) = 15 but the maximum no. of dichotomies possible in this case is only 14 Can someone please comment if i am missing something. Thanks. Vishwajeet. 
Re: Problem 2.9 : Growth function of perceptron, seems incorrect
Thank you

Re: Problem 2.9 : Growth function of perceptron, seems incorrect
Sorry for the delayed response here, but i still find that
is not the case for a perceptron in ddimensional space. When N=6 and d=2, this equation says , while i was able to get 38 dichotomies (by picking 6 equidistant points on the circumference of a circle). If i am missing something, then can you please point me to the proof. Thank you. 
Problem 2.10
To prove :
As this is a generic statement, it has to apply to every growth function. But all we know about the growth functions (in general) is their bound, in terms of and . Also, we know that if then is an increasing function whose value is . But, if , then we can only say that is nondecreasing and is bounded by . I guess my question is that how can we prove the generic statement above. Kindly shed some light on the proof strategy. Thank you, Vishwajeet. 
Possible correction to Problem 2.14 (b)

Re: Possible correction to Problem 2.14 (b)

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

Exercise 2.4b
The final part of the hint in this question says:
"Now, if you choose the class of these other vectors carefully, then the classification of the dependent vector will be dictated." The other vectors refers to the set of linearly independent vectors that make up the d+2th vector. What do you mean by class? Do you mean class of vector, (e.g., unit vector), or class based on the PLA algorithm (i.e., +1 or 1)? 
Re: Exercise 2.4b
Class means . (Note: there is no PLA or algorithm here; the VC dimension has only to do with the hypothesis set.)
At this point you have established that some input vector is linearly dependent on the others. If you assign the class () of the other vectors appropriately, you should be able to show that the linear dependence dictates that the class of must be (say) +1. This means you cannot implement 1 with the other points having those appropriately chosen classifications, and hence this data set cannot be shattered. This argument will apply to any data set of d+2 points, and so you cannot shatter any set of d+2 points. Quote:

Problem 2.24
When we design the numerical experiment, shall we randomly generate more datasets to determine g_bar(x), E_out, bias, and var?

Re: Problem 2.24
Yes, when computing bias and var numerically you need to generate many data sets. For example, is the average function that results from learning on each of these data sets.
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Re: Exercises and Problems
Professor  in my version of the text, for 2.14.b, the inequality is:
. It looks like others are using a version of the inequality that is: , which I think makes the problem a little more transparent. I'm wondering if there's a typo in my version? If not, any additional hints on how to treat this version of the inequality would be helpful. 
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