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-   -   Homework set 3, problem 4 2013 (http://book.caltech.edu/bookforum/showthread.php?t=4224)

jjepsuomi 04-20-2013 03:09 AM

Homework set 3, problem 4 2013
 
Hi,

I'm puzzled with problem 4 :clueless:

Where should I start...should I try to visualize the problem by drawing a picture and simply trying what seems intuitive or what?... :)

Thank you for any advice :bow:

yaser 04-20-2013 10:05 AM

Re: Homework set 3, problem 4 2013
 
Quote:

Originally Posted by jjepsuomi (Post 10500)
Where should I start...should I try to visualize the problem by drawing a picture and simply trying what seems intuitive

That seems like a good approach.

jjepsuomi 04-20-2013 11:39 AM

Re: Homework set 3, problem 4 2013
 
Quote:

Originally Posted by yaser (Post 10502)
That seems like a good approach.

Thank you for the help professor :)

marek 04-20-2013 03:57 PM

Re: Homework set 3, problem 4 2013
 
I think I can answer my own question here, but it would add peace of mind to get a confirmation. In the perceptron model, once you fix a dividing line/plane, you still have the choice of which region gets assigned the +1, correct?

I believe this to be the case, because the vector w and the vector -w both give the same geometry, but sign of the two vectors give opposite values. I ask because if this is not the case, and the geometry does fix the region (say "above" the plane is +1), then that would change my answer for #4.

Elroch 04-20-2013 04:45 PM

Re: Homework set 3, problem 4 2013
 
Quote:

Originally Posted by marek (Post 10511)
I think I can answer my own question here, but it would add peace of mind to get a confirmation. In the perceptron model, once you fix a dividing line/plane, you still have the choice of which region gets assigned the +1, correct?

I believe this to be the case, because the vector w and the vector -w both give the same geometry, but sign of the two vectors give opposite values. I ask because if this is not the case, and the geometry does fix the region (say "above" the plane is +1), then that would change my answer for #4.

A little more completely, a vector w gives a real value w.x to every point x. The points at which w.x=0 form a hyperplane. The points at which w.x>0 are on one side of the hyperplane, the points at which w.x<0 are on the other side. Changing from w to -w, all the values of w.x are multiplied by -1. As a result, the hyperplane where w.x=0 stays the same, but the two regions defined by the sign of w.x swap over.

marek 04-20-2013 04:51 PM

Re: Homework set 3, problem 4 2013
 
Quote:

Originally Posted by Elroch (Post 10512)
A little more completely, a vector w gives a real value \phi(x) = w.x to every point x. The points at which \phi(x) is zero form a hyperplane. The points at which \phi(x) is positive are on one side of the hyperplane, the points at which \phi(x) is negative are on the other side. Changing from w to -w, all the values of \phi(x) are multiplied by -1. As a result, the hyperplane stays the same, but the two regions defined by the sign of the value of the scalar product swap over.

So then the answer to my question was "yes?"

Elroch 04-20-2013 04:55 PM

Re: Homework set 3, problem 4 2013
 
Yes.


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