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First the basic definitions: Let $H$ be a family of sets, and let $P$ be a set of points. Then $H$ is said to shatter $P$ if $\{ h \cap P:~h \in H\}=2^P$, that is, if every subset of $P$ can be obtained by intersecting $P$ with an element of $H$. The Vapnik-Chervonenkis dimension of $H$ is the maximal cardinality of a point set $P$ that is shattered by $H$. See also https://en.wikipedia.org/wiki/VC_dimension.

Let $A$ be the family of axis-parallel boxes in the $d$-dimensional unit cube $[0,1]^d$ having one vertex at the origin. It is known that the VC dimension of $A$ is $d$. Let $B$ be the family of all axis-parallel boxes in $[0,1]^d$ (not necessarily anchored at the origin). The VC dimension of $B$ is known, it is $2d$.

Now the question: Let $C$ be the class of all axis-parallel boxes on the $d$-dimensional unit torus. You could also thing of $C$ as the class of all sets in $[0,1]^d$ which are the $d$-dimensional Cartesian product of elements of $D$, where $D$ is the collection of all subintervals and all complements of subintervals of $[0,1]^d$. Now, what is the VC dimension of $C$?

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  • $\begingroup$ Do you have some bounds in general or for $d=2$? $\endgroup$ – domotorp Jan 29 '17 at 21:27
  • $\begingroup$ Well, obviously a lower bound is the dimension for "non-periodic" boxes (boxes contained in the proper cube, not in the torus), which is 2d. For an upper bound, I don't have any idea. Probably you could use the fact that a "periodic" box on the torus splits into $2^d$ "non-periodic" boxes, but I do not see how this should lead to a reasonable result. My guess would be that the solution could be something linear in $d$, maybe $4d$ or something similar. $\endgroup$ – Kurisuto Asutora Jan 31 '17 at 10:03
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This answer doesn't provide the exact VC-dimension of $C$, but does provide an almost linear upper bound.

The observation that a periodic box is a union of at most $2^d$ boxes could be used get a bound of $(2d^2+o(d^2))2^d$ for the VC-dimension of non-periodic boxes (using the fact that boxes have VC-dimension $2d$). I will use a similar idea to obtain a much better bound (but still not linear in $d$, unfortunately).

Lemma. Suppose $H_1,\ldots,H_n$ are set systems on a set $X$, each of VC-dimension at most $k$. Suppose $H$ is one of the following two set systems:

  1. $H_1\vee\ldots\vee H_n:=\{S_1\cup\ldots\cup S_n:S_i\in H_i\}$, or
  2. $H_1\wedge\ldots\wedge H_n:=\{S_1\cap\ldots\cap S_n:S_i\in H_i\}$.

Then $H$ has VC-dimension at most $k(1+o(1))n\log_2(n)$ (where $o(1)$ depends only on $n$).

Proof. Let $\pi_i$ be the shatter function for $H_i$, i.e., $$ \pi_i(m)=\max\{|\{Y\cap S:S\in H_i\}|:Y\subseteq X,~|Y|=m\}. $$ Let $\pi$ be the shatter function for $H$. One can show $\pi(m)\leq\pi_1(m)\cdot\ldots\cdot\pi_n(m)$ for any $m$. By the Sauer-Shelah lemma, we have $\pi_i(m)\leq (em/k)^k$ for all $i$ and $m\geq k$. So $\pi(m)\leq (em/k)^{kn}$ for any $m\geq k$. In particular, if $m\geq k$ and $(em/k)^{kn}<2^m$, then $\pi(m)<2^m$, and so the VC-dimension of $H$ is less than $m$. So we just need to optimize $m$ satisfying these inequalities. The following works and is of the form stated in the lemma: $$ m:=kn\log_2(cn\log_2(cn)) $$ where $c=e+\log_2(e)$.

So if we use case $(1)$ and write a periodic box as a union of at most $2^d$ boxes, then we get $(2d^2+o(d^2))2^d$.

Instead, we can use case $(2)$ to get a better bound. In particular, for $i\leq d$, let $C_i$ be the set of periodic boxes of the form $I_1\times\ldots\times I_d$ such that $I_j=[0,1]$ for all $j\neq i$. Then the collection $C$ of all periodic boxes is precisely $C_1\wedge\ldots\wedge C_d$. Moreover, each $C_i$ has the same VC-dimension as the set of $1$-dimensional periodic boxes, which is $3$. So altogether, this yields:

Corollary. The VC-dimension of $C$ is at most $(3+o(1))d\log_2(d)$.

The precise VC-dimension of $C$ appears to still be an open problem. I only found one paper discussing it, which only conjectures that the VC-dimension is linear in $d$ (but gives no known bounds). One major defect in my argument as that geometry is only being used in dimension $1$, and then the rest is just abstract combinatorics. One would expect intersections of elements from the $C_i$'s to be much better behaved than intersections of arbitrary sets. Probably the bound in the lemma can be improved a little bit, but not to something linear.

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