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

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

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.