$\epsilon$-nets with respect to the cut norm

Question

The cut norm $||A||_C$ of a real matrix $A = (a_{i,j}) \in \mathcal{R}^{n\times n}$ is the maximum over all $I \subseteq [n], J \subseteq [n]$ of the quantity $\left|\sum_{i \in I, j \in J}a_{i,j}\right|$.

Define the distance between two matrices $A$ and $B$ to be $d_C(A,B) = ||A-B||_C$

What is the cardinality of the smallest $\epsilon$-net of the metric space $([0,1]^{n\times n}, d_C)$?

i.e. the size of the smallest subset $S \subset [0,1]^{n\times n}$ such that for all $A \in [0,1]^{n\times n}$, there exists an $A' \in S$ such that $d_C(A, A') \leq \epsilon$.

(EDIT: I forgot to mention, but I am also interested in "non-proper" $\epsilon$-nets, with $S \subset \mathcal{R}_+^{n\times n}$ -- i.e. if the elements of the $\epsilon$-net have entries outside of [0,1], that is also interesting.)

I am interested in both upper bounds and lower bounds.

Note that cut sparsifier techniques imply $\epsilon$-nets for cut metrics, but give something stronger than I need -- they give an $\epsilon$-net for which you can efficiently find an $\epsilon$-close point to any matrix simply by sampling from that matrix. One might imagine that there exist much smaller $\epsilon$-nets for which you cannot simply sample do find an $\epsilon$-close point to an arbitrary matrix.

I initially asked this question here on mathoverflow.

Answer 1

Here is an easy estimate. Here we call a set S⊆X an ε-net of a metric space X when for every point x∈X, there exists a point s∈S such that the distance between x and s is at most ε. If you want a strict inequality in the definition of ε-net, you can tweak the value of ε slightly.

It holds that ||A||_∞ ≤ ||A||_C ≤ n²||A||_∞, where ||A||_∞ denotes the entrywise max-norm of an n×n matrix A.

It is easy to construct an ε-net of the metric space ([0,1]^N, d_∞) with size ⌈1/(2ε)⌉^N, and it is not hard to show that this size is the minimum. (To show the minimality, consider the ⌈1/(2ε)⌉^N points whose coordinates are multiples of 1/⌈1/(2ε)−1⌉ and show that the distance between any two of these points is greater than 2ε.) By setting N=n² and combining this with the aforementioned comparison between the cut norm and the max-norm, the minimum cardinality of an ε-net with respect to the cut norm is at least ⌈1/(2ε)⌉ⁿ² and at most ⌈n²/(2ε)⌉ⁿ².

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Update: If my calculation is correct, a better lower bound Ω(n/ε)ⁿ² can be obtained by the volume argument. To do this, we need an upper bound on the volume of an ε-ball with respect to the cut norm.

First we consider the “cut norm” of a single vector, which is the maximum between the sum of positive elements and the negated sum of negative elements. It is not hard to show that the volume of an ε-ball in ℝⁿ with respect to this “cut norm” is equal to

$$ \varepsilon^n \sum_{I \subseteq \{1,\dotsc,n\}} \frac{1}{|I|!} \cdot \frac{1}{(n-|I|)!} = \varepsilon^n \sum_{r=0}^n \binom{n}{r} \frac{1}{r!(n-r)!} $$

$$ = \frac{\varepsilon^n}{n!} \sum_{r=0}^n \binom{n}{r}^2 = \frac{\varepsilon^n}{n!} \binom{2n}{n} = \frac{(2n)!\varepsilon^n}{(n!)^3}. $$

Next, since the cut norm of an n×n matrix A is greater than or equal to the cut norm of each row, the volume of an ε-ball in ℝ^n×n is at most the nth power of the volume of an ε-ball in ℝⁿ. Therefore the size of an ε-net of [0,1]^n×n must be at least

$$ \frac{(n!)^{3n}}{(2n)!^n \varepsilon^{n^2}} = \left(\Omega\left(\frac{n}{\varepsilon}\right)\right)^{n^2}, $$

where the last equality is a boring calculation in which we use Stirling’s formula: ln n! = n ln n − n + O(log n).