LFD Book Forum HW4, Question 1

#1
04-30-2012, 10:29 AM
 stordoff Junior Member Join Date: Apr 2012 Posts: 2
HW4, Question 1

For the VC bound, we have a m_H(2N) term. Should this be replaced with (2N)^dvc, or merely N^dvc?

I presume the former, but both give answers around 400,000-500,000, so confirmation would be useful.
#2
04-30-2012, 10:51 AM
 sakumar Member Join Date: Apr 2012 Posts: 40
Re: HW4, Question 1

Quote:
 Originally Posted by stordoff For the VC bound, we have a m_H(2N) term. Should this be replaced with (2N)^dvc, or merely N^dvc? I presume the former, but both give answers around 400,000-500,000, so confirmation would be useful.
The VC Generalization bound is defined in terms of m_H(2N), so that's what I used. Specifically m_H(2N) <= (2N)^dvc + 1, so I substituted the Right hand term for m_H(2N).
#3
04-30-2012, 03:01 PM
 rodrigo Member Join Date: Apr 2012 Location: London Posts: 20
Re: HW4, Question 1

I used the bound we derived in lecture 6: sum(i=0 to dvc, of N choose i ). This should yield a tighter bound on e as sum(i=0 to dvc of, N choose i ) <= (2N)^dvc
#4
04-30-2012, 03:38 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: HW4, Question 1

Quote:
 Originally Posted by rodrigo I used the bound we derived in lecture 6: sum(i=0 to dvc, of N choose i ). This should yield a tighter bound on e as sum(i=0 to dvc of, N choose i ) <= (2N)^dvc
I take it you used since you are evaluating the growth function at .
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#5
04-30-2012, 05:45 PM
 rodrigo Member Join Date: Apr 2012 Location: London Posts: 20
Re: HW4, Question 1

Quote:
 Originally Posted by yaser I take it you used since you are evaluating the growth function at .
Yes indeed! On my previous post I was referring to how one can substitute mH(N), as shown on lecture 6. But reading through the thread again I can see how that may have been misleading given the context.

Thanks professor.

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