1 HISTORY AND FORMULATION OF THE PROBLEM

For an arbitrary \(d \in \mathbb{N}\) and \(\varepsilon \in (0,\;1)\), the problem is to find the minimum number \(\mu (d;\varepsilon )\) of cubes \({{A}_{1}},\; \ldots ,\;{{A}_{\mu }}\) with side length ε that cover the torus \({{T}^{d}}: = [\mathbb{R}{\text{/}}\mathbb{Z}{{]}^{d}}\). As usual, by a cube of side length ε, we mean a set of the form \(\{ ({{x}_{1}},\; \ldots ,\;{{x}_{d}}){\text{:}}\;{{x}_{i}} \in [x_{i}^{0},x_{i}^{0} + \varepsilon ]\} \), and, by a covering, we mean a collection of sets \({{A}_{1}},\; \ldots ,\;{{A}_{\mu }}\) such that \({{A}_{1}} \cup \; \ldots \; \cup {{A}_{\mu }} = {{T}^{d}}\). It is known [1] that, for all d and \(\varepsilon \),

$$\mu (d;\varepsilon )\ge \lceil{{1\text{/}\varepsilon }\rceil^{(d)}},$$
(1)

where \(\lceil{{x}\rceil^{(i)}}=\lceil x\lceil{{x}\rceil^{(i-1)}}\rceil \) and \(\lceil{{x}\rceil^{(1)}}=\lceil x \rceil\). The case of growing dimension d has been rather well studied. Note that (1) implies \(\mu (d;\varepsilon ) \geqslant {{(1{\text{/}}\varepsilon + o(1))}^{d}}\). The general result of Erdős and Rogers on coverings [2] implies that 1/ε is the correct base of the exponential function, i.e., \(\mu (d;\varepsilon ) = (1{\text{/}}\varepsilon + o{{(1))}^{d}}\). More precisely, Erdős and Rogers proved that μ(d; ε) = \(O(d{\text{log}}d{{(1{\text{/}}\varepsilon )}^{d}})\). For the discrete torus \({{T}^{d}} = [\mathbb{Z}{\text{/}}t\mathbb{Z}{{]}^{d}}\) covered with cubes of side length \(s \in \{ 1,\; \ldots ,\;t\} \), this bound can easily be improved by removing the logarithmic factor (here, we mean that \(\varepsilon = s{\text{/}}t\)) 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, \(\mu (d;\varepsilon ) = O(d{{(1{\text{/}}\varepsilon )}^{d}})\) (for both rational and irrational ε). Note that the lower bound (1) implies only \(\mu (d;\varepsilon ) = \Omega ({{(1{\text{/}}\varepsilon )}^{d}})\).

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., \(\mu (2;\varepsilon )=\lceil 1\text{/}\varepsilon \lceil 1\text{/}\varepsilon \rceil\rceil\).

Consider the case d = 3. Since \(\mu (1;\varepsilon )=\lceil 1\text{/}\varepsilon \rceil\) and \(\mu (2;\varepsilon )= \lceil 1\text{/}\varepsilon \lceil 1\text{/}\varepsilon \rceil\rceil\), it follows that

$$\lceil 1\text{/}\varepsilon \lceil 1\text{/}\varepsilon \lceil 1\text{/}\varepsilon \rceil\rceil\rceil \le \mu (3;\varepsilon )\le \lceil 1\text{/}\varepsilon \lceil 1\text{/}\varepsilon \rceil\rceil \lceil 1\text{/}\varepsilon \rceil .$$
(2)

Note that, for d = 3, the lower bound is no longer sharp. For example, it was noted in [1] that μ(3; \(3\text{/}7) > \lceil 7\text{/}3 \lceil 7\text{/}3 \lceil 7\text{/}3 \rceil\rceil\rceil .\) We were able to find the exact value of \(\mu (3;\varepsilon )\) for all \(\varepsilon \geqslant 7{\text{/}}15\) and for \(\varepsilon \) close to \(1{\text{/}}r\), \(r \in \mathbb{N}\).

Finally, we note that, for \(\varepsilon = 2{\text{/}}r\), \(r \in \mathbb{N}\), the corresponding packing problem (i.e., the problem of finding the minimum number \(\nu (d;\varepsilon )\) of disjoint cubes of side length \(\varepsilon \) in Td) is related to finding the Shannon capacity \(c({{C}_{r}})\) of a simple cycle on r vertices [4]; specifically, \(c({{C}_{r}}) = \mathop {\sup }\nolimits_{d \geqslant 1} {{(\nu (d;2{\text{/}}r))}^{{1/d}}}\) (a similar relationship holds for all other rational ε, but the corresponding graphs are more difficult to describe). For even r, obviously, \(c({{C}_{r}}) = r{\text{/}}2\). If r is odd, then the Shannon capacity value is known only for r = 5, namely, \(c({{C}_{5}}) = \sqrt 5 \) [5].

2 NEW RESULTS

Assume first that \(\varepsilon \geqslant 1{\text{/}}2\).

Theorem 1. It is true that

$$\begin{gathered} \mu (3;\varepsilon ) = 8,\quad \varepsilon \in [1{\text{/}}2,2{\text{/}}3),\quad \mu (3;\varepsilon ) = 5, \\ \varepsilon \in [2{\text{/}}3,3{\text{/}}4),\quad \mu (3;\varepsilon ) = 4,\quad \varepsilon \in [3{\text{/}}4,1). \\ \end{gathered} $$

Note that, for \(\varepsilon \in [1{\text{/}}2,1)\), the quantity \(\mu (3;\varepsilon )\) is equal to its lower bound in (2) if and only if \(\varepsilon \in [1{\text{/}}2,4{\text{/}}7) \cup [2{\text{/}}3,1)\).

Since, for integer \(r \geqslant 2\) and \(\varepsilon \in \left[ {\frac{1}{r},\frac{1}{{r - 1{\text{/}}{{r}^{2}}}}} \right)\), the lower and upper bounds in (2) coincide, we obviously have \(\mu (3;\varepsilon ) = {{r}^{3}}\). 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.

Theorem 2. Suppose that \(r \in \mathbb{N}\) and ε ∈ \(\left[ {\frac{1}{{r + 1{\text{/}}({{r}^{2}} + r + 1)}},\frac{1}{r}} \right)\). Then \(\mu (3;\varepsilon ) = {{r}^{3}} + {{r}^{2}} + r + 1\).

Additionally, the right neighborhood was expanded.

Theorem 3. Suppose that \(r \geqslant 2\) is an integer and \(\varepsilon \in \left[ {\frac{1}{r},\frac{{{{r}^{2}} - 1}}{{{{r}^{3}} - r - 1}}} \right)\). Then \(\mu (3;\varepsilon ) = {{r}^{3}}\).

In certain cases, we were able to strengthen the lower bound in (2).

Theorem 4. Suppose that \(r \geqslant 2\) is an integer and \(\xi \in \{ 1,\; \ldots ,\;r\} \) is such that

$${{\xi }^{2}} \leqslant \xi + (r + 1)\left\lfloor {\frac{{{{\xi }^{2}}}}{{r + 1}}} \right\rfloor .$$

Additionally, assume that

(i) \(s = {{r}^{2}} + r + \xi \),

(ii) \(t = {{r}^{3}} + {{r}^{2}} + 2\xi r + \left\lfloor {\frac{{{{\xi }^{2}}}}{{r + 1}}} \right\rfloor \) is coprime to s.

Then \(\mu \left( {3;\frac{s}{t}} \right) \geqslant t + 1\).

Since the condition \({{\xi }^{2}} \leqslant \xi + (r + 1)\left\lfloor {\frac{{{{\xi }^{2}}}}{{r + 1}}} \right\rfloor \) implies either \(\xi \geqslant \sqrt {r + 1} \) or ξ = 1, there are only two such values in the interval [1/3, 1/2): \(\frac{s}{t} \in \left\{ {\frac{7}{{16}},\frac{8}{{21}}} \right\}\).

Finally, in certain cases, we managed to strengthen the upper bound in (2).

Theorem 5. Suppose that \(r \geqslant 2\) is an integer and \(\xi \in \{ 1,\; \ldots ,\;r\} \). Additionally, assume that

(i) \(s = {{r}^{2}} + r + \xi \),

(ii) \(t = {{r}^{3}} + {{r}^{2}} + \xi (r + 1)\).

Then \(\mu \left( {3;\frac{s}{t}} \right) \leqslant t\).

To prove the above results, we established that the problem of finding \(\mu (d;\varepsilon )\) is reduced to its discrete analogue, namely, to the problem of finding the minimum number of cubes with side length s that cover \({{[\mathbb{Z}{\text{/}}t\mathbb{Z}]}^{d}}\) for some s, t such that s/t is close to \(\varepsilon \). Thus, these auxiliary assertions imply that the problem is reduced to considering a countable set of values of \(\varepsilon \), and the number of such values in each interval of the form \([1{\text{/}}r,1{\text{/}}(r - 1)]\) is finite.

3 TRANSITION TO THE DISCRETE CASE

Lemma 1. There exists an infinite sequence of rational numbers \(1 > \frac{{{{s}_{1}}}}{{{{t}_{1}}}} > \frac{{{{s}_{2}}}}{{{{t}_{2}}}} > \; \ldots \; > 0\) such that, for each \(i \in \mathbb{N}\), it is true that \({{t}_{i}} \leqslant \mu \left( {d;\frac{{{{s}_{i}}}}{{{{t}_{i}}}}} \right)\) and \(\mu (d;\varepsilon ) = \mu \left( {d;\frac{{{{s}_{i}}}}{{{{t}_{i}}}}} \right)\) for all \(\varepsilon \in \left[ {\frac{{{{s}_{i}}}}{{{{t}_{i}}}},\frac{{{{s}_{{i - 1}}}}}{{{{t}_{{i - 1}}}}}} \right)\), where \({{s}_{0}} = {{t}_{0}} = 1\).

It follows from (2) that, for each \(r \geqslant 2\), to solve the problem for all ε in the interval \(\left[ {\frac{1}{r},\frac{1}{{r - 1}}} \right)\), it suffices to consider at most \(\frac{{{{r}^{2}}\left( {{{r}^{3}} + 1} \right)}}{{2(r - 1)}} + {{r}^{3}}\) distinct values of ε.

Let s and t be positive integers such that \(s \leqslant t\). Let \({{\mu }_{0}}(d;s,t)\) be the smallest number of cubes with an side consisting of s points that cover the torus \({{[\mathbb{Z}{\text{/}}t\mathbb{Z}]}^{d}}\).

Lemma 2. Suppose that \(r \geqslant 2\) is an integer and \(\varepsilon \in \left[ {\frac{1}{r},\frac{1}{{r - 1}}} \right)\). Additionally, suppose that \(\frac{s}{t} \leqslant \varepsilon \) is the rational number nearest to \(\varepsilon \) such that \(t \leqslant {{r}^{d}}\). Then \(\mu (d;\varepsilon ) = {{\mu }_{0}}(d;s,t)\).

4 SOME SPECIAL CASES

For \(\varepsilon \in \left[ {\frac{1}{2},1} \right)\), the exact values of \(\mu (3;\varepsilon )\) are given in Theorem 2. Below are the exact results and some estimates following from Theorems 2–5 and Lemma 1 for \(\varepsilon \in \left[ {\frac{1}{3},\;\frac{1}{2}} \right)\):

• for \(\varepsilon \in \left[ {\frac{1}{3},\frac{8}{{23}}} \right)\), \(\mu (3;\varepsilon ) = 27\);

• for \(\varepsilon \in \left[ {\frac{8}{{21}},\frac{5}{{13}}} \right)\), \(\mu (3;\varepsilon ) \in [22,24]\);

• for \(\varepsilon \in \left[ {\frac{7}{{16}},\frac{4}{9}} \right)\), \(\mu (3;\varepsilon ) \in [17,21]\);

• for \(\varepsilon \in \left[ {\frac{4}{9},\frac{7}{{15}}} \right)\), \(\mu (3;\varepsilon ) \in [16,18]\);

• for \(\varepsilon \in \left[ {\frac{7}{{15}},\frac{1}{2}} \right)\), \(\mu (3;\varepsilon ) = 15\).