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 k1 points are classified 1 by any dichotomy.
Quote:
Originally Posted by cls2k
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.
