#1




Doubt on weights and pla
suppose i have a line f(x) = x , (for 2 dim) one side of which is +ve 1 and other side ve 1..
Does this mean that target function is : [ 1 , 1] ? (since: x + f(x) > 0 implies +ve 1 and x + f(x) < 0 implies ve 1) and on running the PLA is the set of weights [2 , 2] or any [n, n] a valid solution for f(x) = x ? (since its just a constant to divide throughout?) and when selecting random points, the PLA seems to run till infinity at times with the same value of w repeating at regular intervals am i getting the idea right here btw? 
#2




Re: Doubt on weights and pla
If it is already known that the target function is linear and passes through (0,0) (such as f(x) = x), then it is known that bias term will be zero. [ 1 , 1] makes sense, which is indeed equivalnet to [n,n]. However, if it's not the case, you do need the bias term (weight vector will have length of 3), otherwise the algorithm may not converge.

#3




Re: Doubt on weights and pla
Thanks for the idea about the bias it was really helpful! ... i tried a simple case for understanding better...
The PLA ran indefinitely because of the absence of bias (bias = 0 because of the target function i was having in my mind) The problem was the random training points i generated based on the expected target function had a bit of a flaw that some points were present on the target function and not on either of its sides...further bias of 0 caused the PLA to generate lines that rotated about the origin alone..this at times prevented the algorithm to get to a solution and ran indefinitely! 
#4




Re: Doubt on weights and pla
This is what happened without bias ..this time it terminated...
(many at times it did not as data points increased) (Green = expected , yellow = final result , red = trials) 
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