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  #1  
Old 04-20-2013, 04:09 AM
jjepsuomi jjepsuomi is offline
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Post Homework set 3, problem 4 2013

Hi,

I'm puzzled with problem 4

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
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  #2  
Old 04-20-2013, 11:05 AM
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yaser yaser is offline
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Default Re: Homework set 3, problem 4 2013

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Originally Posted by jjepsuomi View Post
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.
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Old 04-20-2013, 12:39 PM
jjepsuomi jjepsuomi is offline
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Default Re: Homework set 3, problem 4 2013

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That seems like a good approach.
Thank you for the help professor
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Old 04-20-2013, 04:57 PM
marek marek is offline
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Default 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.
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Old 04-20-2013, 05:45 PM
Elroch Elroch is offline
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Default Re: Homework set 3, problem 4 2013

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Originally Posted by marek View Post
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.
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Old 04-20-2013, 05:51 PM
marek marek is offline
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Default Re: Homework set 3, problem 4 2013

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Originally Posted by Elroch View Post
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?"
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Old 04-20-2013, 05:55 PM
Elroch Elroch is offline
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Default Re: Homework set 3, problem 4 2013

Yes.
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