Theoretical Computer Science
graph-theory co.combinatorics metrics max-cut epsilon-nets
Updated Wed, 22 Jun 2022 08:47:50 GMT

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

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.


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

It holds that ||A|| ||A||C n2||A||, where ||A|| denotes the entrywise max-norm of an nn matrix A.

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

Update: If my calculation is correct, a better lower bound (n/)n2 can be obtained by the volume argument. To do this, we need an upper bound on the volume of an -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 -ball in n 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 nn matrix A is greater than or equal to the cut norm of each row, the volume of an -ball in nn is at most the nth power of the volume of an -ball in n. Therefore the size of an -net of [0,1]nn 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 Stirlings formula: ln n! = n ln n n + O(log n).

Comments (4)

  • +0 – In response to the edit (revision 4) of the question, the lower bound stated in this answer is also applicable to non-proper -nets. — Jan 30, 2011 at 13:59  
  • +0 – Looks correct, nicely done! — Jan 31, 2011 at 04:15  
  • +0 – @Hsien-Chih: Thanks. The part I like the most is the use of binomial coefficients in the calculation of the volume of an -ball in ^n. — Jan 31, 2011 at 05:15  
  • +0 – I suspect that the lower bound on the size of the net (equivalently, the upper bound on the volume) can be improved. I asked a related question on MathOverflow. — Feb 21, 2011 at 21:22