On Coverings of Tori with Cubes

We obtain new bounds and exact values of the minimum number of cubes with side length \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varepsilon \in (0,\;1)$$\end{document} covering the torus \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{[\mathbb{R}{\text{/}}\mathbb{Z}]}^{3}}$$\end{document}.

For an arbitrary and , the problem is to find the minimum number of cubes with side length ε that cover the torus . As usual, by a cube of side length ε, we mean a set of the form , and, by a covering, we mean a collection of sets such that . It is known [1] that, for all d and , (1) where and . The case of growing dimension d has been rather well studied. Note that (1) implies . The general result of Erdős and Rogers on coverings [2] implies that 1/ε is the correct base of the exponential function, i.e., . More precisely, Erdős and Rogers proved that μ(d; ε) = . For the discrete torus covered with cubes of side length , this bound can easily be improved by removing the logarithmic factor (here, we mean that ) with the help of the probabilistic method (see, e.g., [3]). We were able to show that the discrete and continuous formulations are equivalent (see Lemmas 1 and 2). Thus, for both rational and irrational ε). Note that the lower bound (1) implies only .
For small values of d, we know only the work [1], where it is proved that, for d = 2, the lower bound (1) is sharp, i.e., .
Consider the case d = 3. Since and , it follows that (2) Note that, for d = 3, the lower bound is no longer sharp. For example, it was noted in [1] that μ(3; We were able to find the exact value of for all and for close to , . Finally, we note that, for , , the corresponding packing problem (i.e., the problem of finding the minimum number of disjoint cubes of side length in T d ) is related to finding the Shannon capacity of a simple cycle on r vertices [4]; specifically, (a similar relationship holds for all other rational ε, but the corresponding graphs are more difficult to describe). For even r, obviously, . If r is odd, then the Shannon capacity value is known only for r = 5, namely, [5].

NEW RESULTS
Assume first that .

Theorem 1. It is true that
Note that, for , the quantity is equal to its lower bound in (2) if and only if .
Since, for integer and , the lower and upper bounds in (2) coincide, we obviously have . Additionally, we found a left neighborhood of 1/r such that the lower bound is sharp for all ε from this neighborhood. Note that, for such ε, the difference between the upper and lower bounds is, on the contrary, large.
Additionally, the right neighborhood was expanded.

Theorem 3. Suppose that is an integer and
. Then .
In certain cases, we were able to strengthen the lower bound in (2).

Theorem 4. Suppose that is an integer and is such that
Additionally, assume that is coprime to s.
Finally, in certain cases, we managed to strengthen the upper bound in (2).
Theorem 5. Suppose that is an integer and . Additionally, assume that