Data Science Asked on February 21, 2021
If we change the $ywx<0$ condition (for performing update) to $ywx<1$ like in SVM (but without adding regularization to maximize the margin), is there any difference from the basic perceptron (the one with the aforementioned $ywx<0$ condition)?
Old question, but in case anyone is still interested in an answer...
In the perceptron algorithm a point $x$ has a label $y$ equal to 1 or -1. The predicted label for a point is $wcdot x$. The goal is to separate the points with an hyperplane orthogonal to w in such a way that the points with label 1 are on one side and the points with label -1 are on the other side. Mathematically, the "sides" of the hyperplane are characterised by the equations $wcdot x>0$ and $wcdot x<0$, respectively. ($wcdot x=0$ is the equation of the hyperplane itself).
Therefore $x$ point is correctly labelled (it is on the right side of the hyperplane orthogonal to $w$) if
In both cases, a correctly labelled point corresponds to $ywcdot x > 0$, while an incorrectly labelled points corresponds to $ywcdot x leq 0$.
The whole algorithm and the proof of its convergence are based on this simple observation.
If we change the condition for update to $ywcdot x < 1$, the algorithm would not be able to find a separating hyperplane. For example, consider the simple dataset $x_1 = (2,0), y_1= 1$ , $x_2 = (-2,1), y_2= -1$.
TLDR; Update conditions different from $ywcdot x leq 0$ do not offer any guarantees for finding a separating hyperplane.
Answered by ValleyCrisps on February 21, 2021
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