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Old 02-25-2013, 02:57 PM
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magdon magdon is offline
Join Date: Aug 2009
Location: Troy, NY, USA.
Posts: 597
Default Re: Exercise problem 2.4

Here is a hint. Lets consider showing B(5,2)>=1+5; It suffices to list 6 dichotomies on 5 points such that no subset of size 2 is shattered.

Consider the following 6 dichotomies on 5 points:

[1 1 1 1 1] (zero -1s)

[-1 1 1 1 1] (5 ways of having one -1)
[1 -1 1 1 1]
[1 1 -1 1 1]
[1 1 1 -1 1]
[1 1 1 1 -1]

Can you show that no subset of 2 points is shattered? Hint: are there any two points that are classified -1,-1?

So, in general, you can guarantee that no subset of size k can be shattered if no subset of size k is classified all -1. This means that at most k-1 points are classified -1 by any dichotomy.

Originally Posted by cls2k View Post
I'm stuck at the exercise problem 2.4 behind the book despite the hint. My approach is to characterize the B(N+1,K) >= recursion as an upper bound on the lower order (in N) terms and then follow the approach of proving the Sauer lemma. However I'm stuck on constructing the "specific set" of dichotomies. I fail to see how the special property to this set (limiting the number of -1 as hinted) can make this proof go easier.

I'm not very good at mathematical proofs so any additional hints will be greatly appreciated.
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