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 ilson 09-21-2015 09:33 AM

Exercise 2.1

Hi,

For the convex set case, it seems to me that since N points on a circle can always be shattered, there's always at least one data set of size k that can be shattered by . Thus, the break point does not exist for this . So then you can't really verify that for this case - or can you say that it's trivially true since break point k doesn't even exist? Is this the correct interpretation of this exercise?

 magdon 09-21-2015 03:50 PM

Re: Exercise 2.1

Yes, when there is no break point, the theorem says that for all N. So the theorem is trivially verified.
Quote:
 Originally Posted by ilson (Post 12050) Hi, For the convex set case, it seems to me that since N points on a circle can always be shattered, there's always at least one data set of size k that can be shattered by . Thus, the break point does not exist for this . So then you can't really verify that for this case - or can you say that it's trivially true since break point k doesn't even exist? Is this the correct interpretation of this exercise?

 svend 09-19-2016 12:03 AM

Re: Exercise 2.1

I don't understand why the breaking point inequality holds for the positive rays or positive intervals .

For instance, it seems to me that no set of 3 real points can be shattered by a positive ray, since at least always the [cross, circle, cross] dichotomy cannot be achieved, no matter how large is, so would be a breaking point and , which is obviously not true for since the real growth function is .

I understand that to be a breaking point, we need that no set of size k can be shattered, am I failing to imagine such set or did I misunderstand some of the definition?

 dubwub 09-21-2016 06:11 PM

Re: Exercise 2.1

For N > 7 you need your growth function to be less than 2^N, not 2^3.

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