TITLE: Tessellating  $\mathbb{R}^n$ by bricks.
QUESTION [10 upvotes]: Let us call the $\ell_1$-product of intervals $[0,k_1]\times...\times [0,k_n]$ a  brick of size  $k_1+...+k_n$. Consider a tessellation $T$ of $\mathbb{R^n}$ by (shifted) bricks so that every point belongs to at most $n+1$ bricks and the $\ell_1$-distance between any two disjoint bricks is at least 1 (thus every two bricks can share only the boundary points). Let $s(T)$ be the maximal size of a brick in that tessellation. Let $s(n)$ be the minimum of all $s(T)$. For example, $s(1)=1$ (we tessellate $\mathbb{R}$ by intervals $[i,i+1]$), $s(2)\le 3$ (we can tessellate ${\mathbb R}^2$ by $2\times 1$-bricks so that each point belongs to at most 3 bricks). It is easy to have a tessellation with $s(T) \sim 2^n$, so $s(n)$ is at most exponential.
 Question.  What is $s(n)$?  Does $s(n)$ grow exponentially with $n$?
 Update  It looks like the question is completely answered when we assume that all bricks are isometric by Will (upper bound) and Eric (lower bound). What if we tile by different bricks? How about arbitrary convex regions (same properties: every point belongs to at most $n+1$ regions, every two disjoint rejions are at distance at least 1, and, of course, different regions may share only boundary points)? Can we achive livear upper bound on the diameter of a tile? Can there be a constant upper bound?

REPLY [5 votes]: Since Ycor took the liberty to edit this 5 year old question and thus to bring it to the front page again, I'll throw in my one cent (not two yet; that may come later).
If you do a lattice tiling by translates, then, as Eric showed, we have no chance to get below the quadratic rate. However, not all tilings are like that. We cannot gain too much, but I think we still can gain somewhat (though I'm not ready to present an example yet).
First, the foolproof part: why not too much. By our conditions, any shift of the ball $B$ or radius $1/2$ in $\ell_1$ cannot intersect two disjoint bricks simultaneously, so we have all bricks intersecting $B$ also intersecting pairwise, which, by the special feature of bricks, means having a common point (bricks are like intervals on the line in this respect). Thus, the covering number for $(K_j+B)$ is at most $d+2$ but the volume of each brick increased $\sum_{\ell\ge 0}\sigma_\ell(\frac{1}{k_1},\dots,\frac 1{k_d})\frac 1{\ell!}$ times where $\sigma_\ell$ is the usual symmetric sum. By the AM-GM used twice, this is at least (here $q=d/s$)
$$
\sum_{\ell=0}^d \frac{d!}{(\ell!)^2{(d-\ell)!}}q^\ell\ge\max_{\ell\le d/2}\left(\frac{dq}{2\ell^2}\right)^\ell\approx{e^{c\sqrt{dq}}} 
$$
if $q\le d$ and, clearly, increasing beyond that point. Thus $dq\le C\log^2 (d+2)$ and $s\ge cd^2\log^{-2}(d+2)$ regardless of whether we use same size bricks or not and whether we follow any regular pattern or not, or even whether we do a tesselation or just covering, killing all Gerhard's hopes for a linear bound.
To be continued...