LFD Book Forum Concentric circles in Q10

#1
07-28-2012, 09:11 AM
 ESRogs Member Join Date: Jul 2012 Posts: 12
Concentric circles in Q10

The inequality a^2 <= x1^2 + x2^2 <= b^2 in question 10 implies

1) that the points have to be outside of the smaller circle and inside the larger circle, and
2) that the circles have to be centered at the origin.

Is that correct?
#2
07-28-2012, 12:45 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: Concentric circles in Q10

Quote:
 Originally Posted by ESRogs The inequality a^2 <= x1^2 + x2^2 <= b^2 in question 10 implies 1) that the points have to be outside of the smaller circle and inside the larger circle, and 2) that the circles have to be centered at the origin. Is that correct?
Correct. To be exact about inside and and outside, the perimeters of the inner circle and the outer circle are included in the region.
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#3
07-28-2012, 04:01 PM
 ESRogs Member Join Date: Jul 2012 Posts: 12
Re: Concentric circles in Q10

Thanks!
#4
07-31-2012, 05:54 AM
 ilya239 Senior Member Join Date: Jul 2012 Posts: 58
Re: Concentric circles in Q10

Quote:
 Originally Posted by yaser Correct. To be exact about inside and and outside, the perimeters of the inner circle and the outer circle are included in the region.
So once the set of N points is fixed, and we pick different subsets of the N points and try to cover them with "donuts" -- all donuts must be centered at the same point? We can't use a donut centered at p1 to cover {x1,x2} but a donut centered at a different point p2 to cover {x3,x4}? In other words, after picking N points, we pick ONE origin, and _then_ can pick donuts centered at that origin only?
#5
07-31-2012, 06:08 AM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: Concentric circles in Q10

Quote:
 Originally Posted by ilya239 So once the set of N points is fixed, and we pick different subsets of the N points and try to cover them with "donuts" -- all donuts must be centered at the same point? We can't use a donut centered at p1 to cover {x1,x2} but a donut centered at a different point p2 to cover {x3,x4}? In other words, after picking N points, we pick ONE origin, and _then_ can pick donuts centered at that origin only?
Correct. You can choose any points you want to work with (with a view to maximizing the number of dichotomies), so the origin is effectively arbitrary, but fixed, for a given set of points.
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#6
09-21-2012, 06:14 PM
 doris Junior Member Join Date: Sep 2012 Posts: 3
Re: Concentric circles in Q10

In order to test the effective number of hypotheses in H, how can we fix the center of the circles? by fixing the center we restrict ourselves to one hypothesis. So, I am more confused by the last comment. My thought process is find a set of N points and look through all possible concentric circles, so all radii and centers, that will give me each dichotomy possible on N. And this is how H shatters N, not each single hypothesis. Is this correct?

What is a correct strategy to approach this problem? can we reduce it to 1D with an interior interval an exterior interval (to infinity) =-1 and the 2 in-between regions (between the 2 circles) =+1

then it becomes a more complex version of the 2.3 c, the positive, negative intervals?
#7
09-22-2012, 01:57 PM
 doris Junior Member Join Date: Sep 2012 Posts: 3
Re: Concentric circles in Q10

the last comment confused me a little bit.
For a given set of N points, we should change the center of the sphere to get as many dichotomies as we can, thus measuring the effective number of hypotheses (spheres) in this hypothesis set.

Does it make sense to move project the spheres from 3D to 1D and look at the problem as intervals of +1 for a<=x<=b and -a>=x>=b?
#8
01-25-2013, 02:13 PM
 geekoftheweek Member Join Date: Jun 2012 Posts: 26
Re: Concentric circles in Q10

Quote:
 Originally Posted by ilya239 So once the set of N points is fixed, and we pick different subsets of the N points and try to cover them with "donuts" -- all donuts must be centered at the same point? We can't use a donut centered at p1 to cover {x1,x2} but a donut centered at a different point p2 to cover {x3,x4}? In other words, after picking N points, we pick ONE origin, and _then_ can pick donuts centered at that origin only?
This approach confuses my understanding of the problem. Why doesn't the symmetry of the problem reduce it to 1-D (r instead of x) with two parameters, i.e. the "positive intervals" case in example 2.2.2?
Thanks
#9
01-25-2013, 11:14 PM
 Suhas Patil Senior Member Join Date: Dec 2012 Posts: 57
Re: Concentric circles in Q10

Quote:
 Originally Posted by geekoftheweek This approach confuses my understanding of the problem. Why doesn't the symmetry of the problem reduce it to 1-D (r instead of x) with two parameters, i.e. the "positive intervals" case in example 2.2.2? Thanks
Sounds logical...the linear scale representing radius can range from 0 (center of the concentric circles) to infinity with positive interval a-b contained within.
#10
04-17-2013, 05:18 PM
 Elroch Invited Guest Join Date: Mar 2013 Posts: 143
Re: Concentric circles in Q10

When I skimmed this question first of all, I started thinking about annuli with arbitrary centres, which is an interesting more powerful hypothesis set.

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