Problem 2.2

[Pallen]

I've been having trouble proving that the break point is 5 for problem 2.2 I don't know what special properties of pentagons might come into play.

For one thing, I know that in the case where at least one point is in the convex hull of the other four points, then the dichotomy with -1 for that one point and +1 for all others won't work. Therefore, we must only consider the case where no point is in the convex hull of the four other points (so that, I think, we have a convex pentagon). But from here I'm completely stuck.

[molab]

Pag 13 of (http://www.cs.rpi.edu/~magdon/course...idesLect05.pdf) can help you to visualize. It is interesting to note that in the case of 4 colinear points, this learning model cannot shatter de 4 points (e.g the dichotomy 1010). So the breakpoint must be less than 4. This is correct? Thanks!

[sharov_am]

Originally Posted by molab

Pag 13 of (http://www.cs.rpi.edu/~magdon/course...idesLect05.pdf) can help you to visualize. It is interesting to note that in the case of 4 colinear points, this learning model cannot shatter de 4 points (e.g the dichotomy 1010). So the breakpoint must be less than 4. This is correct? Thanks!

Hi. Probably correct to say must be no less than 4.

[alanhealey]

It must be no less than 4.
run 3