#1




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.

#2




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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