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  #1  
Old 09-28-2016, 08:29 PM
wolszhang wolszhang is offline
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Default Problem 2.10

I understand that for the case m(N) = 2^N, we can show that this is true. But how do we prove it when m(N) < 2^N? It seems like every single theorem is giving me an upper bound. Any hints would be super appreciated.
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Old 09-29-2016, 12:01 PM
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magdon magdon is offline
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Default Re: Problem 2.10

Hint: Any dichotomy 2N points can be viewed as a dichotomy on the first N points plus a dichotomy on the second N points.
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Originally Posted by wolszhang View Post
I understand that for the case m(N) = 2^N, we can show that this is true. But how do we prove it when m(N) < 2^N? It seems like every single theorem is giving me an upper bound. Any hints would be super appreciated.
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Old 12-01-2017, 05:31 PM
EliLee EliLee is offline
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Default Re: Problem 2.10

I am still not quite clear about this problem. To prove this problem is true for all N values, I think we should discuss different relationship between break point k, N and 2N. I can prove that when N < 2N < k or k is infinite, the inequality holds. Also when N < k < 2N, the in equality also holds. When k < N < 2N, do you mean mH(2N) = 2mH(N)? However I did not figure it out.
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Old 12-08-2017, 11:11 PM
pdsubraa pdsubraa is offline
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Default Re: Problem 2.10

@Magdon - Can you explain us in detail.

That would help!

Thanks for your time!
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