Yes, when there is no break point, the theorem says that
for all N. So the theorem is trivially verified.
Quote:
Originally Posted by ilson
Hi,
For the convex set case, it seems to me that since N points on a circle can always be shattered, there's always at least one data set of size k that can be shattered by  . Thus, the break point does not exist for this . So then you can't really verify that  for this case - or can you say that it's trivially true since break point k doesn't even exist? Is this the correct interpretation of this exercise? |