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. 
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!

Re: Problem 2.2
Quote:

Re: Problem 2.2
It must be no less than 4.
run 3 
All times are GMT 7. The time now is 10:17 AM. 
Powered by vBulletin® Version 3.8.3
Copyright ©2000  2021, Jelsoft Enterprises Ltd.
The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. AbuMostafa, Malik MagdonIsmail, and HsuanTien Lin, and participants in the Learning From Data MOOC by Yaser S. AbuMostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity.