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 Pallen 06-26-2018 08:15 PM

Problem 2.2

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 03-12-2019 01:31 PM

Re: Problem 2.2

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 12-22-2020 02:01 AM

Re: Problem 2.2

Quote:
 Originally Posted by molab (Post 13245) 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 01-21-2021 07:21 PM

Re: Problem 2.2

It must be no less than 4.
run 3

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