#1




Perceptron VC dimension
First I'd like to say that Professor AbuMostafa'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 coordinate 1. There is a nontrivial linear relation on these points: Rearrange so all the coefficients are positive (changing labels for convenience) and must be nonempty subsets of because the relation is nontrivial, and the first coordinate 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. 
#2




Re: Perceptron VC dimension
Nice argument! (and thank you for the compliment)
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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#3




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?

#4




Re: Perceptron VC dimension
Quote:
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