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Old 05-05-2016, 07:55 AM
MaciekLeks MaciekLeks is offline
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Join Date: Jan 2016
Location: Katowice, Upper Silesia, Poland
Posts: 17
Default Re: Help in understanding proof for VC-dimension of perceptron.

Context:
To show that d_{vc}\leq d+1 we need to prove we cannot shatter any set od d+2 points. In this case we have d+2 vectors of \mathbf{x}_{i}, where 1\leq i\leq d+2, hence we have +1 vector than the total number of dimensions (our space is d+1 -dimensional). According to the theory of vector spaces when we have more vectors than dimensions of the space then they are linearly dependent (one of the vectors is a linear combination of the others). Which means that for instance our extra point \mathbf{x}_{d+2}, can be computed by the equation (in d+1-dimensional space): \mathbf{x}_{d+2}=\sum_{i=1}^{d+1}a_{i}\mathbf{x}_{i}, (where not all a_{i} coefficients are zeros).
We need to show that we can't get \mathbf{y}=\begin{bmatrix}\pm1\\
\pm1\\
\pm1\\
\pm1\\
\pm1
\end{bmatrix} ( 2^{d+2}) dichotomies, where \mathbf{y} is of d+2 size.

Problem:

And now let me define my problem I have with understanding the video proof. Why do we correlate a_{i} coefficient with the y_{i}=sign(a_{i}) while they don't correlate? On what basis we can draw that sign(a_{i})=sign(\mathbf{w}^{T}\mathbf{x}_{i}), and that's why \mathbf{w}^{T}\mathbf{x}_{d+2}=\sum_{i=1}^{d+1}a_{i}\mathbf{w}^{T}\mathbf{x}_{i}>0, hence we can't generate \mathbf{y}=\begin{bmatrix}\pm1\\
\pm1\\
...\\
\pm1\\
-1
\end{bmatrix}?
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