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-   -   HW4, Question 1 (http://book.caltech.edu/bookforum/showthread.php?t=412)

stordoff 04-30-2012 09:29 AM

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.

sakumar 04-30-2012 09:51 AM

Re: HW4, Question 1
 
Quote:

Originally Posted by stordoff (Post 1680)
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).

rodrigo 04-30-2012 02:01 PM

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

yaser 04-30-2012 02:38 PM

Re: HW4, Question 1
 
Quote:

Originally Posted by rodrigo (Post 1685)
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 {2N \choose i} since you are evaluating the growth function at 2N.

rodrigo 04-30-2012 04:45 PM

Re: HW4, Question 1
 
Quote:

Originally Posted by yaser (Post 1686)
I take it you used {2N \choose i} since you are evaluating the growth function at 2N.

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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