Exercises and Problems - Page 2

yaser · #11 08-29-2012, 12:44 PM

Quote:

Originally Posted by vsthakur

But all we know about the growth functions (in general) is their bound

Actually, we also know the definition of growth functions, and this may be the key to answering the question.

htlin · #12 08-29-2012, 08:09 PM

Quote:

Originally Posted by vsthakur

Sorry for the delayed response here, but i still find that

$m_{H}(N) = 2 \sum_{i=0}^d \left({N-1}\atop i\right)$

is not the case for a perceptron in d-dimensional space.

When N=6 and d=2, this equation says $m_{H}(n) = 32$ , while i was able to get 38 dichotomies (by picking 6 equidistant points on the circumference of a circle).

If i am missing something, then can you please point me to the proof.

Thank you.

I checked the case you are describing, and the number of dichotomies in the case is

. (Hint: did you double-calculate the case of 3-positive and 3-negative?) Hope this helps.

vsthakur · #13 08-30-2012, 04:45 AM

Quote:

Originally Posted by htlin

I checked the case you are describing, and the number of dichotomies in the case is

. (Hint: did you double-calculate the case of 3-positive and 3-negative?) Hope this helps.

You are right, that was the mistake. My apologies.
Thank you.

vsthakur · #14 09-01-2012, 10:34 AM

Quote:

Originally Posted by yaser

Actually, we also know the definition of growth functions, and this may be the key to answering the question.

I think i get it now. Let

. Now, if we partition any set of

points into two sets of

points each, each of these two partitions will produce

dichotomies at best. If we now combine these two sets, then the maximum no. of dichotomies possible will be the cross product of the two sets of dichotomies (with N points each), i.e.,
$m_H(2N) \le k^2 = m_h(N)^2$

Thank you.

vsthakur · #15 09-01-2012, 10:46 AM

Quote:

Originally Posted by magdon

The problem, though an over-estimate seems correct.

Hint: If you have $\ell$ points, then ${\cal H}_1$ can implement at most $\ell^{d_{VC}}+1\le\ell^{d_{VC}+1}$ dichotomies on those points. Now try to upper bound the number of dichotomies that all

hypothesis sets can implement on these $\ell$ points and proceed from there.

Got it, thanks. The inequality should be strict i think ( $d_{vc}(H) < l, when \; 2^l > Kl^{d_{vc}+1}$ ).

magdon · #16 09-03-2012, 06:54 AM

Yes, well done.

Quote:

Originally Posted by vsthakur

I think i get it now. Let

. Now, if we partition any set of

points into two sets of

points each, each of these two partitions will produce

dichotomies at best. If we now combine these two sets, then the maximum no. of dichotomies possible will be the cross product of the two sets of dichotomies (with N points each), i.e.,
$m_H(2N) \le k^2 = m_h(N)^2$

Thank you.

doris · #17 09-22-2012, 03:00 PM

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?

magdon · #18 09-23-2012, 10:42 AM

You will note from the definition of the hypothesis set: $\cal H$ contains functions which are +1 for

$a\le\sqrt{x_1^2+\cdots+x_d^2}\le b$

You only get to vary

, and so the two spheres are restricted to be centered on the origin.

And yes, the $m_{\cal H}(N)$ for this hypothesis set is very related to the growth function for positive intervals.

Quote:

Originally Posted by doris

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?

mileschen · #19 09-23-2012, 06:19 PM

In (a), it said that we should provide a monotonic classifier. Then, why there are +1 and -1 regions?

Also, as it said in (b) that generating the next point by increasing the first component and decreasing the second component. Then, how can we determine which point is larger? Because X1>=X2 if and only if the inequality is satisfied for every component. However, the next point is just with one component larger than that of the first one, while another component is less than that of the first one. So, it's a little confusing.

magdon · #20 09-23-2012, 08:23 PM

The function

is +1 in some region and is -1 in the complement - i.e. it takes on two values. Any function can be monotonic, even one that takes on just 2 values.

In (b), you are asked to compute m(N). To compute m(N) you need to count the maximum number of implementable dichotomys on some N points. The problem suggest a set of N points which might be helpful. The fact that no point is larger than another is crucial [hint: because if a point were larger than another, there is a dichotomy that you cannot implement].

Quote:

Originally Posted by mileschen

In (a), it said that we should provide a monotonic classifier. Then, why there are +1 and -1 regions?

Also, as it said in (b) that generating the next point by increasing the first component and decreasing the second component. Then, how can we determine which point is larger? Because X1>=X2 if and only if the inequality is satisfied for every component. However, the next point is just with one component larger than that of the first one, while another component is less than that of the first one. So, it's a little confusing.

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