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Old 04-22-2012, 04:56 PM
TomKetola TomKetola is offline
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Default Clarification on finding a break point

I watched the lectures and went through the book, but I had some questions with my understanding after working on the homework exercises. I seem to be lost in somewhat of a circular argument. My understanding is that if k is a breakpoint, then m_H(n) < 2^k. I can find B(n, k), which if k is a breakpoint for H, m_H(n) < B(n, k). I think I follow the logic of the proofs and bounding, where if we can prove that B(n, k) is bounded by a polynomial, then clearly values smaller than B(n, k) are also bounded by polynomials. What I thought I had seen/heard when I was reading the text book and watching the lectures was a formula for discovering m_H(n) without using pure inspection, but now I can't seem to find it. Is there a formula for discovering m_H(n), or is it found only by inspection? If so, can someone point me to where this is discussed?
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Old 04-22-2012, 07:47 PM
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yaser yaser is offline
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Default Re: Clarification on finding a break point

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Originally Posted by TomKetola View Post
I watched the lectures and went through the book, but I had some questions with my understanding after working on the homework exercises. I seem to be lost in somewhat of a circular argument. My understanding is that if k is a breakpoint, then m_H(n) < 2^k. I can find B(n, k), which if k is a breakpoint for H, m_H(n) < B(n, k). I think I follow the logic of the proofs and bounding, where if we can prove that B(n, k) is bounded by a polynomial, then clearly values smaller than B(n, k) are also bounded by polynomials. What I thought I had seen/heard when I was reading the text book and watching the lectures was a formula for discovering m_H(n) without using pure inspection, but now I can't seem to find it. Is there a formula for discovering m_H(n), or is it found only by inspection? If so, can someone point me to where this is discussed?
There is a formula for an upper bound on the growth function, based on a break point k. The exact evaluation of a growth function is a case-by-case exercise, and a difficult one in many cases. Fortunately, it's not needed since the upper bound achieves the probability bound that we need in the VC inequality.

The bound comes from the bound on B(N,k) (which in turn bounds the growth function), and it is the formula given in class:

\sum_{i=0}^{k-1} {N \choose i}
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