Re: ddimensional Perceptrons and break points (related to Q4 of homework)
Thanks for intervening Professor.
The 4point set I have in mind is the set of 4 corners of a square. The concept of "shattering" I am working under corresponds to being able to propose a hypothesis that conforms to the classification given for each of the points in the dataset. So in the case of a 2d perceptron, It is easy to "shatter" (if my def. of shattering is correct) the 4 corners in a square if 2 of the corners in one side of the square are red and the other 2 corners are blue  just pass a line in the middle of the square such that the two red points are on one side of the line and the other 2 are on the opposite side. Got your colors inverted? No problem, just multiply the w vector by 1.
Now if the red points are on opposite corners (and the blue as well), then we couldn't shatter them because no matter what boundary line we choose, we always get either both sides with the same color or two colors on each side. That's how I understand that the break point for the 2d perceptron is 4 because there exists a 4point set that is not shatterable.
Obviously there is a mistake in my concepts somewhere because if you choose a 3 point set in which all points are collinear and you set the middle point to blue and the outer points to red, this is not shatterable by a 2d perceptron, or a 3d perceptron or any higher dimensional perceptron.
I hope I made clear what my doubts are and where is my confusion.
