Consider a boolean cube $\{0,1\}^n$ and a collection of colored subcubes $\{(C_a, \ell_a)\}_{a \in [N]}$, where $C_a \in \{0,1,*\}^n$ is a subcube and $\ell_a \in \mathbb{N}$ is a color. I will identify $C_a$ with a set $\{x \in \{0,1\}^n \mid \{x_j, *\} \ni (C_a)_j\}$.
Call a collection almost unique cover if $\Pr[\exists a\colon {\bf x} \in C_a] = 1$ and $\Pr[\exists a, b\in [N]\colon {\bf x} \in C_a \cap C_b \land \ell_a \neq \ell_b] \le 1/100$, i.e. the cubes cover the entire boolean cube, but the volume where different colors mix is small.
Question: suppose that codimension of each cube is bounded by $m$ (equivalently $|C_a| \ge 2^{n-m}$). Can we bound the number of distinct colors the cubes have by some $2^{m^{O(1)}}$? (I am interested in the larger and even ramsey-type bounds as well, but would prefer an exponent).
If we lift the requirement that cubes must form a cover, then the number of colors is unbounded in $m$: let $(C_a)_j = 0$ if $j \in [m-1]$ and if $j - (m-1) = a$, otherwise let $(C_a)_j = *$. Let all colors be different: $\ell_a = a$. Then the cubes intersect in the volume $2^{-(m-1)}$, but the number of colors is $N$ which does not depend on $m$.
Here is my naive attempt to bound the number of colors: is there is no point that is covered by more than $2^{m^3}$ colors, then since each color is presented in at least $2^{-m}$-fraction of points, the total number of colors is bounded by $2^{m^4}$. If on the other hand there is a point $y$ covered by $\ge 2^{m^3}$ points, then by the sunflower lemma we get that there is a cubcube $D_y \ni y$ of dimension $<m$ such that $\Pr[{\bf x} \text{ is covered by two different colors} \mid {\bf x} \in D_y] = 1 - \exp(2^{m^2 - o(m^2)})$. We can try to continue applying the sunflower lemma until we have covered all the points with $>2^{m^3}$ color coverage by subcubes where points are very likely to be covered by at least two subcubes, so in total we should cover at most $1/100$ of the volume. If we have, say $2^{m^3 - m^2}$ distinct cubes $D_y$, we can apply induction on $m$, but I don't see how to show that there must be a lot of different cubes.
