Homework set 3, problem 4 2013

[jjepsuomi]

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

[yaser]

Originally Posted by jjepsuomi

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]

Originally Posted by yaser

That seems like a good approach.

Thank you for the help professor

[marek]

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]

Originally Posted by marek

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 gives a real value to every point . The points at which form a hyperplane. The points at which are on one side of the hyperplane, the points at which are on the other side. Changing from to , all the values of are multiplied by . As a result, the hyperplane where stays the same, but the two regions defined by the sign of swap over.

[marek]

Originally Posted by Elroch

A little more completely, a vector gives a real value to every point . The points at which is zero form a hyperplane. The points at which is positive are on one side of the hyperplane, the points at which is negative are on the other side. Changing from to , all the values of are multiplied by . 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?"