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Old 04-24-2013, 03:21 AM
Elroch Elroch is offline
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Default 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 (n+2) points can be shattered. Here's my slightly different version.

Consider a set X of (n+2) points in (n+1)-dimensional space, all of which have first co-ordinate 1. There is a non-trivial linear relation on these points:

\sum\limits_{x\in X}a_x x = 0

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

\sum\limits_{x\in X_A}a_x x =\sum\limits_{x\in X_B}b_x x

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

If some perceptron is positive on X_A and negative on X_B then the value of the perceptron on \sum\limits_{x\in X_A}a_x x is positive and its value on \sum\limits_{x\in X_B}b_x x is negative.

But from the above these are the same point. Hence such a perceptron does not exist, so X cannot be shattered, completing the proof.
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Old 04-24-2013, 02:39 PM
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yaser yaser is offline
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Default Re: Perceptron VC dimension

Nice argument! (and thank you for the compliment)

Quote:
Originally Posted by Elroch View Post
There is a non-trivial linear relation on these points:

\sum\limits_{x\in X}a_x x = 0

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

\sum\limits_{x\in X_A}a_x x =\sum\limits_{x\in X_B}b_x x

X_A and X_B must be non-empty subsets of X 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.
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Old 07-23-2015, 12:09 AM
yongxien yongxien is offline
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Default 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?
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Old 07-23-2015, 01:11 AM
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yaser yaser is offline
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Default Re: Perceptron VC dimension

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Originally Posted by yongxien View Post
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?
d+1 is greater than or equal to the VC dimension. Together with the other fact that d+1 is also less than or equal to that dimension gets you the equality.
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