d-dimensional Perceptrons and break points (related to Q4 of homework)

[yaser]

Originally Posted by jlaurentum

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 2-d perceptron is 4- because there exists a 4-point set that is not shatterable.

Not mere existence. The key is that no other arrangement of the 4 points would be shatterable either.

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 2-d perceptron, or a 3-d perceptron or any higher dimensional perceptron.

Your analysis is right, but the problem is with the conclusion. Although this set of 3 points cannot be shattered, another set of 3 points can. Since we are allowed to choose the points with a view to maximizing the number of possible dichotomies, the existence of any 3 points that can be shattered means 3 is a not a break point, notwithstanding the fact that some set of 3 points cannot be shattered.

[jlaurentum]

Thank you for your kind patience, Professor.

After reading the feedback on this thread and rewatching the videos, I realized that my mistake was in the concept of "shatters". From minute 51:20 of lecture 5, we can say that "a data set of size k can be shattered by script H (the hypothesis set) if there exists a choice of points in the data set for which all dichotomies are possible".
I realize now the 3 colinear points don't imply that a data set of size 3 can't be shattered by 2-d perceptrons because there are other choices of 3 points for which all dichotomies are possible.

Slowly the fog is begining to clear...

[kafar]

Originally Posted by yaser

There is no systematic way that applies to all cases, but it is usually not too difficult to get some upper bound of when the growth function breaks. Fortunately, only an upper bound is needed to carry through the theory, as you will see in Lecture 7.

thanks a lot. i feel relieved