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-   -   Q4 and Q5 (http://book.caltech.edu/bookforum/showthread.php?t=3903)

tathagata 01-23-2013 04:18 AM

Q4 and Q5
 
Regarding Q4, which asks us to determine the break point of a 3D Perceptron, just a clarification: I am thinking that the 2D case will also be a pathological case in 3D, as it is just a special case for the 3D plane, but if there exists any setting of 4 points in 3D that can be shattered by the 3D Perceptron then break point is greater than four, since we take the maximum? (like the 2D case with collinear points for N = 3)

Regarding Q5, as I understand it, any monotonically increasing function <= 2^N for all N, can be a possible growth function, is that correct or are there more restrictions?
Also we have a N choose 2 term in one of the options that is not defined for N = 1, but that can be logically regarded as zero, right?

yaser 01-23-2013 10:27 AM

Re: Q4 and Q5
 
Quote:

Originally Posted by tathagata (Post 8946)
Regarding Q4, which asks us to determine the break point of a 3D Perceptron, just a clarification: I am thinking that the 2D case will also be a pathological case in 3D, as it is just a special case for the 3D plane, but if there exists any setting of 4 points in 3D that can be shattered by the 3D Perceptron then break point is greater than four, since we take the maximum? (like the 2D case with collinear points for N = 3)

You are right.

Quote:

Regarding Q5, as I understand it, any monotonically increasing function <= 2^N for all N, can be a possible growth function, is that correct or are there more restrictions?
There are more restrictions that were discovered in Lecture 6.

Quote:

Also we have a N choose 2 term in one of the options that is not defined for N = 1, but that can be logically regarded as zero, right?
Correct. For k>N, we have {N \choose k}=0.

tathagata 01-24-2013 05:08 AM

Re: Q4 and Q5
 
Quote:

Originally Posted by yaser (Post 8947)
There are more restrictions that were discovered in Lecture 6.

But doesn't lecture 6 discuss a more strict bound only if we have a break point? Whereas Q5 asks for any possible growth function, so being less than 2^N is sufficient I would have thought. What am i missing?

geekoftheweek 01-24-2013 11:30 AM

Re: Q4 and Q5
 
Quote:

Originally Posted by yaser (Post 8947)
You are right.

I don't follow. In 2D, all four points must be coplanar so no line can shatter the points. In 3D the two x's and two o's can be separated along the third axis and this can easily be shattered by a plane. You need k > 4 as a breakpoint. Am I missing something? Perhaps I did not understand the OP's question.

yaser 01-24-2013 01:11 PM

Re: Q4 and Q5
 
Quote:

Originally Posted by tathagata (Post 8965)
But doesn't lecture 6 discuss a more strict bound only if we have a break point? Whereas Q5 asks for any possible growth function, so being less than 2^N is sufficient I would have thought. What am i missing?

We either have a break point, or else we don't. In the latter case, the growth function is identically 2^N. In the former case, the growth function is constrained such that many formulas cannot possibly be valid growth functions.

yaser 01-24-2013 01:13 PM

Re: Q4 and Q5
 
Quote:

Originally Posted by geekoftheweek (Post 8969)
I don't follow. In 2D, all four points must be coplanar so no line can shatter the points. In 3D the two x's and two o's can be separated along the third axis and this can easily be shattered by a plane. You need k > 4 as a breakpoint. Am I missing something? Perhaps I did not understand the OP's question.

You are not missing anything. You have just articulated why a break point in 2D may not be a break point in 3D.

Anne Paulson 01-24-2013 01:29 PM

Re: Q4 and Q5
 
The terminology can be a bit confusing, because a break point exists if there is no pathological case that can be shattered. If you can shatter, you don't break.

yaser 01-24-2013 01:59 PM

Re: Q4 and Q5
 
Quote:

Originally Posted by Anne Paulson (Post 8978)
If you can shatter, you don't break.

:D

tathagata 01-24-2013 02:38 PM

Re: Q4 and Q5
 
Quote:

Originally Posted by yaser (Post 8974)
We either have a break point, or else we don't. In the latter case, the growth function is identically 2^N. In the former case, the growth function is constrained such that many formulas cannot possibly be valid growth functions.

Ah.. I got it thanks! :)

Elroch 04-17-2013 06:25 PM

Re: Q4 and Q5
 
One tip I would give for studying is this: if you are shattered, take a break. :D


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