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#1
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Two of the questions (7 & 9) ask how many iterations it takes for the PLA to converge. I would expect this to be a function of both the size of N that was mandated as well as the alpha (learning rate) that is selected. Is this not correct?
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#2
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#3
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Thanks! I didn't realize that the learning rate wasn't present in the model you had discussed during the first lecture. I had previously run all my data at .5 so it will be interesting to see what the difference is when I set it to 1.
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#4
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If you take a deeper look at the steps of the PLA algorithm, you'll find that setting the learning rate to any positive value gives you equivalent results (subject to the same random sequence and equivalent starting weights, of course). For instance, if you start with the zero vector, the final weights that you get for learning rate 1 are simply twice the final weights that you get for learning rate 0.5.
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#5
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I would expect that if your learning rate is too large it would be possible to "overshoot" the convergence values and therefore require some back and forth before they settle. Depending upon the extent of that oscillation it may or may not require more iterations than a smaller value. I guess you could also say a similar thing about too small a learning value. It could slowly inch up to one possible set of convergence weight values and get stuck in a "local minima" of sorts without truly finding the "global minima". |
#6
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__________________
When one teaches, two learn. |
#7
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After thinking about why this is the case I can almost understand it ![]() Thanks for following up on my original question! Last edited by tcristo; 04-08-2012 at 12:07 PM. Reason: spelling correction |
#8
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What is the range that we should expect the weighting factors to be around? Should it be less than 1, less than 10, or greater?
Any help appreciated - thanks. |
#9
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__________________
Where everyone thinks alike, no one thinks very much |
#10
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In my simulation, I plotted the random data points and the "true" f(x) line, and it helped me intuitively see that with many points (e.g. N=100), there is much less "wiggle room" for two lines to fit between the same set of "boundary" points (i.e. the set of points that are "closest" to the f(x) line.) With less points (say, N=10), there could be a huge variation in the slope and intercept of two lines that both "fit" the data.
I would think that you could get a better starting point than w = 0 by examining the data at the "boundary" points, where the result y changes from -1 to +1 and somehow use that information? |
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