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 Elroch 04-24-2013 04:21 AM

Perceptron VC dimension

First I'd like to say that Professor Abu-Mostafa's lectures are unsurpassed in clarity and effectiveness in communicating understanding of the key elements of this fascinating subject. So it is unusual (actually, unique so far) for me to think I can see a way in which clarity of a point can be improved. Maybe I'm wrong: please judge!

The proof I am referring to is the second side of the proof of the VC dimension of a perceptron, where it is necessary to show that no set of points can be shattered. Here's my slightly different version.

Consider a set of points in -dimensional space, all of which have first co-ordinate 1. There is a non-trivial linear relation on these points:

Rearrange so all the coefficients are positive (changing labels for convenience)

and must be non-empty subsets of because the relation is non-trivial, and the first co-ordinate of all the points is 1.

If some perceptron is positive on and negative on then the value of the perceptron on is positive and its value on is negative.

But from the above these are the same point. Hence such a perceptron does not exist, so cannot be shattered, completing the proof.

 yaser 04-24-2013 03:39 PM

Re: Perceptron VC dimension

Nice argument! (and thank you for the compliment)

Quote:
 Originally Posted by Elroch (Post 10579) There is a non-trivial linear relation on these points: Rearrange so all the coefficients are positive (changing labels for convenience) and must be non-empty subsets of because the relation is non-trivial, and the first co-ordinate of all the points is 1.
Just to elaborate, the coefficients are not all zeros so because of the constant 1 coordinate, there has to be at least one positive and one negative coefficient, hence the rest of the argument even if all other coefficients are zeros.

 yongxien 07-23-2015 01:09 AM

Re: Perceptron VC dimension

Isn't d+1 greater than the dimension too? Such that the proof works on d+1? why then is it not d_vc <= d?

 yaser 07-23-2015 02:11 AM

Re: Perceptron VC dimension

Quote:
 Originally Posted by yongxien (Post 11989) Isn't d+1 greater than the dimension too? Such that the proof works on d+1? why then is it not d_vc <= d?
is greater than or equal to the VC dimension. Together with the other fact that is also less than or equal to that dimension gets you the equality.

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