LFD Book Forum - Q10 higher bound

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Q10 higher bound

I think I have found the tighter lower bound.
But I cannot determine the tighter higher bound, among the two candidates.

From the tighter lower bound that I have found and the information in the question, it seems possible to determine the tighter higher bound.

But I would like to either properly admit or dismiss the K-1 factor.
Can anybody more advanced on the subject propose an indication, a clue, a line of reasoning ? (NOT the answer)

Thx

Re: Q10 higher bound

One helpful step is to check the problems in the book. :)

Re: Q10 higher bound

Quote:

Originally Posted by OlivierB (Post 10583)

Can anybody more advanced on the subject propose an indication, a clue, a line of reasoning?

Let me just comment that this problem is probably the hardest in the entire course. A discussion from the participants is most welcome.

Re: Q10 higher bound

In 2d, a positive "2d ray" in the x direction can shatter 1 point, and a positive "2d ray" in the y direction can shatter 1 point, whereas their union can shatter 2 points...

So at least I can feel comfortable that the union of hypothesis sets *can* achieve a VC dimension which is the sum of the VC dimensions of its parts, but can it exceed it??

Re: Q10 higher bound

Quote:

Originally Posted by alasdairj (Post 10614)

I am struggling with this question as well. Based on this discussion, I can intuit what the right answer should be, but I am struggling with the justification.

I figure if I can simply come up with an H1 and H2 such that d(H1 U H2) = d(H1) + d(H2) + 1, I would be done. I can come up 1 short of that (much like your example, even consider 1d positive and negative rays). But I cannot think of an example that would hit the higher bound.

That being said, I decided to just play around with the growth function. We know that $m_H(N) \leq \sum_{i=0}^{d} {N \choose i}$ . If all the terms in that expansion are present, we get 2^N and hence can shatter. The binomial expansion is symmetric, so if I had two growth functions that were "disjoint" one could get the terms from the bottom and the other could get terms from the top and these could together give the full expansion.

More formally: $\sum_{i=0}^{d_1} {N \choose i} + \sum_{i=0}^{d_2} {N \choose i} = \sum_{i=0}^{d_1} {N \choose i} + \sum_{i=N-d_2}^{N} {N \choose i}$ . This gives the full expansion if the second sum starts off one term after the first sum ends, specifically $N-d_2 \leq d_1 + 1 \Leftrightarrow N \leq d_1 + d_2 + 1$ .

This simple analysis gives me my desired

. But clearly something is missing. How can growth functions be "disjoint"? Any example I come up with, there is a significant overlap in dichotomies. I feel like I'm definitely on the right track here, but I also feel like something is clearly missing. Maybe B(N,k) is a better starting point than the growth function?

Hopefully someone can pick up the ball and carry it over the finish line...

Re: Q10 higher bound

@marek, Thanks for your contribution.
Let me try and rephrase, not adding much.

For a hypothesis set H with VD dimension d, we have
$m_{H}(N)\le B(N, d+1)=\sum_{i=0}^d \left(\begin{array}{c}N\\ i\end{array}\right)$

The second equality is the result of 2 inequalities in opposite direction. In class, we saw '≤' and '≥' is the subject of problem 2.4 in the book.

Let 2 hypothesis sets $H_{1}$ and $H_{2}$ with VC dimensions respectively $d_{1}$ and $d_{2}$ .

We can construct two disjoint $H'_{1}$ and $H'_{2}$ putting the intersection of both with one of them only.
$H'_{1}=H_{1}$
$H'_{2}=H_{2}-\bigcap(H_{1}, H_{2})$

The VC dimension of $H'_{1}$ is $d_{1}$
The VC dimension of $H'_{2}$ is $d'_{2}\le d_{2}$

Let H be the union of these hypothesis sets:
$H=\bigcup(H_{1}, H_{2})=\bigcup(H'_{1}, H'_{2})$
Let $N=d_{1}+d_{2}+1$

Now in order to try and determine the VC dimension of H, let us compute a higher bound of $m_{H}(N)$ :

$m_{H}(N)=m_{H'_1}(N)+m_{H'_2}(N)\le\sum_{i=0}^{d_1}\left(\begin{array}{c}N\\ i\end{array}\right)+\sum_{i=0}^{d'_2}\left(\begin{array}{c}N\\ i\end{array}\right)\le\sum_{i=0}^{d_1}\left(\begin{array}{c}N\\ i\end{array}\right)+\sum_{i=0}^{d_2}\left(\begin{array}{c}N\\ i\end{array}\right)$

which we can simplify:

$\sum_{i=0}^{d_1}\left(\begin{array}{c}N\\ i\end{array}\right)+\sum_{i=0}^{d_2}\left(\begin{array}{c}N\\ i\end{array}\right)=\sum_{i=0}^{d_1}\left(\begin{array}{c}N\\ i\end{array}\right)+\sum_{i=N-d_2}^{N}\left(\begin{array}{c}N\\ i\end{array}\right)=\sum_{i=0}^{d_1}\left(\begin{array}{c}N\\ i\end{array}\right)+\sum_{i=d_1+1}^{N}\left(\begin{array}{c}N\\ i\end{array}\right)$

For $m_{H}(N)=2^N$ to be true i.e. for H to shatter N, all inequalities must be equalities.
In other words, we must have:
1/ The growth functions of $H_{1}$ and $H_{2}$ must be exactly

and

2/ Removing the intersection of $H_{1}$ and $H_{2}$ from $H_{2}$ does not decrease the VC dimension of $H_{2}$ . If the intersection is empty then this condition holds.

If these 2 requirements are not contradictory (which seems plausible but I cannot prove - neither can I visualize with an example), then the VC dimension of

is at least

.

Now unless there is a mistake in the reasoning, the question is: Can these requirements be met ? Ideally via an example, because beyond the abstract equations, I would really like to 'visualize' a case where these inequalities are 'saturated'.

Re: Q10 higher bound

I am still confused on how to piece everything together. I like your much cleaner version, but I do have one comment. Having disjoint hypothesis sets does not necessarily mean that the set of dichotomies they create will also be disjoint.

For example, let H1 and H2 be the positive and negative 1d rays, respectively. These two hypothesis sets are disjoint. However, given any data set they can both create the dichotomies of all +1 or all -1. They won't have much overlap beyond that (and maybe THAT is the point), but they won't be entirely disjoint as far as our inequalities are concerned.

Re: Q10 higher bound

Your requirement 2 can be simply rendered from Q9 that the VC dimension of intersection of hypothesis could not exceed any one of them(hope I got Q9 right)

As for requirement 1, I come up with the case that
H1 := mapping all points to +1
H2 := mapping all points to -1
thus $d_1 = 0 , d_2 = 0 , d_{1+2} = 1 = d_1 + d_2 + 1$

Quote:

Originally Posted by OlivierB (Post 10620)

@marek, Thanks for your contribution.
Let me try and rephrase, not adding much.

.......

For $m_{H}(N)=2^N$ to be true i.e. for H to shatter N, all inequalities must be equalities.
In other words, we must have:
1/ The growth functions of $H_{1}$ and $H_{2}$ must be exactly

and

is at least

Re: Q10 higher bound

Quote:

Originally Posted by MindExodus (Post 10631)

As for requirement 1, I come up with the case that
H1 := mapping all points to +1
H2 := mapping all points to -1
thus $d_1 = 0 , d_2 = 0 , d_{1+2} = 1 = d_1 + d_2 + 1$

Fantastic! I can sleep well tonight =)

Re: Q10 higher bound

Quote:

Originally Posted by MindExodus (Post 10631)

Bad news: the problem requires that each hypothesis set has a finite positive VC dimension, so this particular example does not guarantee that the maximum is greater than the sum of the VC dimension for the case we're being asked about.

I'm still on the fence for this question - my first attempts at constructing an example similar to this where each hypothesis set had a VC dimension of 1 failed to shatter 3 points, but I'm not 100% sure yet that I can't get it to work.