Perceptron VC dimension

Elroch · #1 04-24-2013, 04:21 AM

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:

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

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

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

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 $\sum\limits_{x\in X_A}a_x x$ 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

cannot be shattered, completing the proof.

yaser · #2 04-24-2013, 03:39 PM

Nice argument! (and thank you for the compliment)

Quote:

Originally Posted by Elroch

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)

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

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 · #3 07-23-2015, 01:09 AM

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 · #4 07-23-2015, 02:11 AM

Quote:

Originally Posted by yongxien

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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