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#1
10-24-2012, 09:54 PM
 skyw2012 Junior Member Join Date: Oct 2012 Posts: 1
On Q9

Trying to solve this question (sadly I failed) I realized that I didn't quite understand the shattering correctly in the beginning. In the meantime I've clarified this but still I don't know how to answer this question.
I imagine that must be a particular way of either selecting the right sets of points from the options of the question to prove they can be shattered or to identify the break point for this case to ensure that any set of points equal or greater can't be shattered.
Any suggestions on how to approach the solution? Thank you
#2
10-24-2012, 10:56 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: On Q9

Quote:
 Originally Posted by skyw2012 Trying to solve this question (sadly I failed) I realized that I didn't quite understand the shattering correctly in the beginning. In the meantime I've clarified this but still I don't know how to answer this question. I imagine that must be a particular way of either selecting the right sets of points from the options of the question to prove they can be shattered or to identify the break point for this case to ensure that any set of points equal or greater can't be shattered. Any suggestions on how to approach the solution? Thank you
There is no systematic way of doing this, but here are some tips.

The choice of where to place the points should be done so as to maximize the chances that they will be shattered. Since the triangle is a convex region, having one of the points being 'internal' to other points will preclude the possibility of shattering (same situation as in the case of convex regions discussed in the lecture). Therefore, choosing the points at the perimeter of a circle (for example) would be a good starting point.

Now, imagine all kinds of triangles trying to split these points into different dichotomies, and look for how many points you can afford to have while being able to split them in every possible way using triangles. It's an interesting puzzle to work on.
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#3
01-26-2013, 11:05 AM
 melipone Senior Member Join Date: Jan 2013 Posts: 72
Re: On Q9

Can it be several overlapping triangles (like in picture) or just one triangle?

Last edited by melipone; 01-26-2013 at 11:06 AM. Reason: Check on picture
#4
01-26-2013, 11:30 AM
 ezreal Member Join Date: Jan 2013 Posts: 15
Re: On Q9

Since this thread doesn't say *Answer* in the title, I don't think we can discuss any hints further than the notion of clever point placement.

Quote:
 Originally Posted by melipone Can it be several overlapping triangles (like in picture) or just one triangle?
I believe it should only be one triangle.
#5
01-26-2013, 01:44 PM
 MartinT Junior Member Join Date: Jan 2013 Posts: 2
Re: On Q9

One word in addition - with triangles we can't afford the convenience of having arbitrarily many vertices (as with convex sets in general - just stating the obvious). It might therefore be useful to keep in mind that the triangle's vertices don't have to rest on the circle, though the data points better do. As the question states, for a point to be covered (h(x)=+1) it suffices if it lies within the triangle, not necessarily directly on its edges or vertices.
#6
04-17-2013, 06:14 PM
 Elroch Invited Guest Join Date: Mar 2013 Posts: 143
Re: On Q9

It's interesting to consider the relationship between the parametrisation of hypotheses and VC dimension. A triangle requires a fixed finite number of parameters (count them ), more general convex sets can require an unbounded number of parameters. I wouldn't expect a hypothesis set with a finite parametrisation to have an infinite VC dimension (although the two dimensions are only loosely related) but I haven't investigated the details much.

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