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 wolszhang 09-28-2016 08:29 PM

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.

 magdon 09-29-2016 12:01 PM

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.
Quote:
 Originally Posted by wolszhang (Post 12441) 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.

 EliLee 12-01-2017 05:31 PM

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.:clueless:

 pdsubraa 12-08-2017 11:11 PM

Re: Problem 2.10

@Magdon - Can you explain us in detail.

That would help!

 magdon 03-04-2018 05:48 AM

Re: Problem 2.10

For example, if you have 10 ways to dichotomize 4 points and 8 ways to dichotomize another 4 points , for each of the 10 ways for the first 4 points there are at most 8 ways to dichotomize the second 4 points (why?). Therefore, there are at most ways to dichotomize all 8 points.

Quote:
 Originally Posted by pdsubraa (Post 12876) @Magdon - Can you explain us in detail. That would help! Thanks for your time!

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