LFD Book Forum  

Go Back   LFD Book Forum > Book Feedback - Learning From Data > Chapter 2 - Training versus Testing

Reply
 
Thread Tools Display Modes
  #1  
Old 09-21-2015, 09:33 AM
ilson ilson is offline
Member
 
Join Date: Sep 2015
Posts: 10
Default Exercise 2.1

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 \mathcal{H}. Thus, the break point does not exist for this \mathcal{H}. So then you can't really verify that m_{\mathcal{H}}(k)< 2^k 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?
Reply With Quote
  #2  
Old 09-21-2015, 03:50 PM
magdon's Avatar
magdon magdon is offline
RPI
 
Join Date: Aug 2009
Location: Troy, NY, USA.
Posts: 592
Default Re: Exercise 2.1

Yes, when there is no break point, the theorem says that m_H(N)=2^N for all N. So the theorem is trivially verified.
Quote:
Originally Posted by ilson View Post
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 \mathcal{H}. Thus, the break point does not exist for this \mathcal{H}. So then you can't really verify that m_{\mathcal{H}}(k)< 2^k 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?
__________________
Have faith in probability
Reply With Quote
  #3  
Old 09-19-2016, 12:03 AM
svend svend is offline
Junior Member
 
Join Date: Sep 2016
Posts: 2
Default Re: Exercise 2.1

I don't understand why the breaking point inequality holds for the positive rays or positive intervals .

For instance, it seems to me that no set of 3 real points can be shattered by a positive ray, since at least always the [cross, circle, cross] dichotomy cannot be achieved, no matter how large N is, so k=3 would be a breaking point and m_H(N) < 2^3, which is obviously not true for N>7 since the real growth function is m_H(N) = N + 1.

I understand that to be a breaking point, we need that no set of size k can be shattered, am I failing to imagine such set or did I misunderstand some of the definition?
Reply With Quote
  #4  
Old 09-21-2016, 06:11 PM
dubwub dubwub is offline
Junior Member
 
Join Date: Sep 2016
Posts: 2
Default Re: Exercise 2.1

For N > 7 you need your growth function to be less than 2^N, not 2^3.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -7. The time now is 10:41 PM.


Powered by vBulletin® Version 3.8.3
Copyright ©2000 - 2017, Jelsoft Enterprises Ltd.
The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. Abu-Mostafa, Malik Magdon-Ismail, and Hsuan-Tien Lin, and participants in the Learning From Data MOOC by Yaser S. Abu-Mostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity.