#1




A little puzzle
Pondering the loose relationship between parametrisation of a hypothesis set and VC dimension (or the minimum break point, ) led me to the following example and puzzle.
Let the set of points be the natural numbers Let elements of the hypothesis set be made up of alternating intervals of the same size, like a 1dimensional checkerboard with varying scale What is the VC dimension of this hypothesis set? [there is also a continuous version on the real line, but all the structure is in this simplified version] 
#2




Re: A little puzzle
Don't have the rules too clear. Can your alternating segments contain the +1s, _or_ all the 1s? If not, I imagine the VC dimension is 0.
If you allow both, I guess the first pattern you can't achieve is +1, 1, 1 so the VC dimension would be 2. Or do you allow them shifted as well, then you'd get that one by starting at 2. But I don't think you could do +1, 1, +1, +1. 
#3




Re: A little puzzle
Forgive me if it wasn't clear. There is exactly one hypothesis for each positive integer , as described in the first post.
Intuitively, each hypothesis (i.e. permitted subset of )is an alternating sequence of "black" and "white" intervals of equal size starting at 1, and continuing indefinitely. "Black" = +1 = inclusion in the set, and "white" = 1 = exclusion from the set. ok? Do remember you have great freedom as to how to choose a set of N points. Use it well. A good attack might be to try to shatter sets for increasing from upward. 
#4




Re: A little puzzle
Ah  not the first time today I was tripped up by the rule that we only need to find a single set of xns with the maximum shattering. I keep forgetting and looking for some set of xns which doesn't.
So here's a method that will work for any N, and I don't even need all of your hypotheses; it's enough just to use the set of H's where n=2^k. H0=+1 for 1,3,5,7,... H1=+1 for 1,2,5,6,9,10,... H2=+1 for 1,2,3,4,9,10,11,12,17,18,19,20,... H3=+1 for 1,2,3,4,5,6,7,8,17,18,19,... The easy way to see it is to use binary notation. H0=+1 for any n with the last digit=1 H1=+1 for any n with the secondtolast digit=1 H2=+1 for any n with the thirdtolast digit=1 etc. Now gH(N)=2^N. Here's how we build our set of x1,...,xN that we can shatter. For N, take the set Z of every combination of N 1s and 0s and use them to build our x1,...,xN. They will be very big numbers indeed, with 2^N digits or so. The first digit of xi, call it xi1, is the ith digit of the first element of Z. If that is 000000...0 (N 0s), the first digits of all the xis will all be zero as well, so all of the numbers x1,...,xN will start with 0. If the second element of Z is 000000..01  might as well keep them in order  the second digit of x1,...,xN will be 0, 0, ..., 0, and 1 respectively. And so forth, for all 2^N digits. These N very special binary numbers are shattered by H0 through H(2^N1), as H(i) picks out the ith digit of each of our numbers, and the digits cover every possibility in Z. Way to go  this is a neat problem. 
#5




Re: A little puzzle
Thank you and well done!
Exactly the solution I found. 
#6




Re: A little puzzle
Cool. I'll copy the thread into the Create New Homework Problems forum.
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#7




Re: A little puzzle
This problem makes me wonder if the VC dimension formula for the growth function isn't (sometimes?) far too restrictive. It's true that the growth function here is 2^N, and therefore our problem is maybe unlearnable. But is that really true? Because we could find one wacko example with numbers 2^50 digits long, does that mean that our hypotheses are really much too general? Actually, this set of hypotheses is awfully restricted, and you (probably) can't hardly do anything with them for almost all sets of x1,...xn. Maybe we should have a modified version of the VC formula, where the bound works for almost all sets of xn. Is it still true that the growth will be polynomial, almost always?

#8




Re: A little puzzle
Quote:
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#9




Re: A little puzzle
Ah  I see that in your book (footnote p. 51), the case of convex regions is mentioned as an example where an "estimated" growth bound works. I'm guessing that's because even though points on the rim of a circle and such can be shattered, "almost every" set of N points is going to have some on the interior, which cannot be shattered.

#10




Re: A little puzzle
In this example, Michael, it's not really correct to think of the shattered sets of points as being untypical. Almost all integers are very large! For example, given N if you pick a number M and choose a set of N points randomly in [M, 2M], the probability of the points not being shattered by this hypothesis set will tend to zero as M tends to infinity. [exercise for reader ]

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