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Old 01-13-2013, 04:12 PM
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magdon magdon is offline
Join Date: Aug 2009
Location: Troy, NY, USA.
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Default Re: How is linearity of PLA obvious?

It is not obvious at all, and for this reason I would recommend Problem 1.2 in the text in case anyone is having similar doubts.

The classification function: \mathbf{w}^T\mathbf{x}>0\implies +1 is called linear because the weights appear 'linearly' in this formula. Now to see that this corresponds to what in high-school we learned as a line requires a little algebra.

First, the classification function is not a line. The classification function is given by the formula above which assigns +1 to some regions of the space and -1 to others. What corresponds to the line is the boundary between the region where the hypothesis is +1 and the region where the hypothesis is -1. It is the separator that is a line.

Fix \mathbf{w}. The boundary corresponds exactly to those points \mathbf{x} for which


Lets consider the case in 2-dim. Then \mathbf{w}=[w_0,w_1,w_2] and \mathbf{w}^T\mathbf{x}=0 translates to


Rearrange a little and you have the following equation:


that must be satisfied by any point \mathbf{x}=[x_1,x_2] that lies on the decision boundary. Thus, the decision boundary is indeed a line, and one can identify m=-w_1/w_2, b=-w_0/w_2.

Originally Posted by scottedwards2000 View Post
The perceptron learning algo is very cool in its simplicity, but, although it was introduced in the course as sort of obviously linear, that was immediately apparen to me. Just wondering what made it obvious to others if it was. Is it the formula on page 6 (w-sub1*x-sub1 + w-sub2*x-sub2 + b = 0)? That seems similar to the old line equation of y=mx+b but not exactly same. However, after playing on paper with different choices of w's and b, I think I see that regardless of what values I choose for the weights and threshold that a line is formed. But how is that obvious theoretically that the shape is a line, without trying various values like i did?
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