Neighborly boxes and strings with jokers; constructions and asymptotics

TL;DR

This paper investigates the maximum size of families of axis-aligned boxes in high-dimensional space with intersection dimension constraints, providing new constructions and asymptotic bounds especially when the intersection parameter is close to the dimension.

Contribution

The authors introduce a novel construction of $k$-neighborly boxes that improves lower bounds for large $k$, and establish asymptotic formulas for the maximum number of such boxes.

Findings

New construction improves lower bounds for $n(k,d)$ when $k$ is close to $d$.

Asymptotic formula for $n(d-s,d)$ as $d$ grows large.

Use of Hamming cube interpretation with joker symbols in constructions.

Abstract

We study families of axis-aligned boxes in a $d$-dimensional Euclidean space $\mathbb{R}^d$ whose placement is restricted by bounds on the dimension of their pairwise intersections. More specifically, two such boxes in $\mathbb{R}^d$ are said to be \emph{$k$-neighborly} if their intersection has dimension at least $d-k$ and at most $d-1$. The maximum number of pairwise $k$-neighborly boxes in $\mathbb{R}^d$ is denoted by $n(k,d)$. It is known that $n(k,d)=\Theta(d^k)$, for fixed $1\leqslant k\leqslant d$, however, exact formulas are known only in three cases: $k=1$, $k=d-1$, and $k=d$. In particular, the equality $n(1,d)=d+1$ is equivalent to the famous theorem of Graham and Pollak concerning partitions of complete graphs into complete bipartite graphs. In our main result we give a new construction of families of $k$-neighborly boxes which improves the lower bound for $n(k,d)$ when $k$ is close to $d$. Together with some recent upper bounds on $n(k,d)$, it gives the asymptotic equality $n(d-s,d)\thicksim\frac{2^s+1}{2^{s+1}}\cdot2^d$, for every fixed $s\leqslant d/2$. In our constructions we use a familiar interpretation of the problem in the language of Hamming cubes represented by binary strings with a special blank symbol, called \emph{joker}.

Equations

$$\frac{1}{k^k}\cdot d^k⩽n(k,d)⩽\frac{2\cdot (2e{)}^k}{k^k}\cdot d^k.$$

$$n(k,d)⩾(1-o(1))\frac{d^k}{k!}.$$

$$\underset{d\to \infty }{\lim }\frac{n(k,d)}{d^k}={\gamma }_k.$$

$$n(k,d)⩽n(k-1,d-1)+n(k,d-1),$$

$$n(d-s,d)⩾\sum _{i=0}^{\frac{d}{2}-s}\left(\genfrac{}{}{0ex}{}{d-s}{i}\right)+\sum _{k=1}^{s-1}{2}^k\left(\genfrac{}{}{0ex}{}{d-s}{\frac{d}{2}-s+k}\right)+2^s\sum _{i=\frac{d}{2}}^{d-s}\left(\genfrac{}{}{0ex}{}{d-s}{i}\right).$$

$$\left(\frac{1}{2}+\frac{1}{2^{s+1}}\right)2^d-o(2^d)⩽n(d-s,d)⩽\left(\frac{1}{2}+\frac{1}{2^{s+1}}\right)2^d.$$

$$\underset{d\to \infty }{\lim }\frac{n(d-s,d)}{2^d}=\frac{2^s+1}{2^{s+1}}.$$

$$\mathrm{dist}(u,v)=|\{1⩽i⩽d:u_i\ne v_i\text{ and }{u}_i,v_i\in \{0,1\}\}|.$$

$$\mathrm{diam}(A,B)=\max \{\mathrm{dist}(a,b):a\in A,b\in B\}.$$

$$A_{\epsilon }=\{{\epsilon }_1\}\times \cdots \times \{{\epsilon }_i\}\times I^{s-1-i}\subseteq I^{s-1}.$$

$$k=|\epsilon |=\sum _{n=1}^{i-1}{\epsilon }_n+1+\sum _{n=i+1}^{s}1>\sum _{n=1}^{i-1}{\epsilon }_n+0+\sum _{n=i+1}^s{\gamma }_n=|\gamma |,$$

$$X_0$$

$$X_k$$

$$X_s$$

$$B_{\epsilon }=\{x=(x^{\prime },x^{\prime \prime })\in I^{d-2s+1}\times I^{s-1}:x\in X_{|\epsilon |},x^{\prime \prime }\in A_{\epsilon }\}$$

$$B_{\epsilon }=X_{|\epsilon |}\cap (I^{d-2s+1}\times A_{\epsilon })$$

$$⟨\epsilon ,\gamma ⟩⩽⟨x,y⟩.$$

$$⟨\epsilon ,\gamma ⟩=⟨\epsilon |i,\gamma |i⟩.$$

$$⟨x,y⟩=⟨x^{\prime },y^{\prime }⟩+⟨x^{\prime \prime },y^{\prime \prime }⟩⩾⟨x^{\prime \prime }|i,y^{\prime \prime }|i⟩,$$

$$\overline{B_{\alpha }}=B_{\omega }=B_{\bar{\alpha }},\text{and}\overline{B_{\omega }}=B_{\alpha }=B_{\bar{\omega }}.$$

$$\overline{B_{\epsilon }}=B_{\bar{\epsilon }}.$$

$$A_{\epsilon }=\{{\epsilon }_1\}\times \cdots \times \{{\epsilon }_i\}\times I^{s-1-i}\text{and}{A}_{\bar{\epsilon }}=\{1-{\epsilon }_1\}\times \cdots \times \{1-{\epsilon }_i\}\times I^{s-1-i},$$

$$\overline{B_{\epsilon }}=\overline{X_{|\epsilon |}\cap (I^{d-2s+1}\times A_{\epsilon }})=\overline{X_{|\epsilon |}}\cap (I^{d-2s+1}\times \overline{A_{\epsilon }})=X_{|\bar{\epsilon }|}\cap (I^{d-2s+1}\times A_{\bar{\epsilon }})=B_{\bar{\epsilon }}.$$

$$\mathrm{diam}(B_{\epsilon },B_{\gamma })⩽\min \{d-s-⟨\epsilon ,\gamma ⟩,d-s-⟨\bar{\epsilon },\bar{\gamma }⟩\}.$$

$$\mathrm{dist}(x,y)$$

$$\mathrm{dist}(x,y)⩽d-s-⟨\epsilon ,\gamma ⟩.$$

$$\mathrm{dist}(x,y)=d-s-(s-|\epsilon |-|\gamma |+⟨\epsilon ,\gamma ⟩)+(⟨\epsilon ,\gamma ⟩-2⟨x,y⟩)$$

$$\mathrm{dist}(x,y)⩽d-s-⟨\bar{\epsilon },\bar{\gamma }⟩,$$

$$\mathrm{diam}(B_{\epsilon },B_{\gamma })$$

$$\mathrm{diam}(C_{\epsilon },C_{\gamma })=\mathrm{diam}(B_{\epsilon },B_{\gamma })+\mathrm{diam}({\ast }^{\epsilon },{\ast }^{\gamma })$$

Full text

In our main result we give a new construction of families of $k$-neighborly boxes which improves the lower bound for $n(k,d)$ when $k$ is close to $d$. Together with some recent upper bounds on $n(k,d)$, it gives the asymptotic equality $n(d-s,d)\thicksim\frac{2^{s}+1}{2^{s+1}}\cdot 2^{d}$, for every fixed $s\leqslant d/2$. In our constructions we use a familiar interpretation of the problem in the language of Hamming cubes represented by binary strings with a special blank symbol, called joker.

1. Introduction

A box in the space $\mathbb{R}^{d}$ is any axis-parallel $d$-dimensional cuboid. The intersection of two boxes is a again a box of possibly smaller dimension. Two boxes are said to be $k$-neighborly, for a fixed positive integer $k\leqslant d$, if their intersection is a box of dimension strictly smaller than $d$ but not less than $d-k$. Let $n(k,d)$ denote the maximum size of a family of pairwise $k$-neighborly boxes in $\mathbb{R}^{d}$.

The exact values of $n(k,d)$ are known only in the following three cases. In the case $k=d$ we trivially have $n(d,d)=2^{d}$. For $k=1$, Zaks [16] proved that $n(1,d)=d+1$ by relating the problem to the celebrated theorem of Graham and Pollak [7] on bipartite partitions of cliques. Finally, for $k=d-1$, the equality $n(d-1,d)=3\cdot 2^{d-2}$ was derived only recently in [4]. The first general bounds were determined by Alon [2] who proved that for all $1\leqslant k\leqslant d$:

[TABLE]

These bounds were recently improved in [4] and [5]. In particular, in [4], the following lower bound for $n(k,d)$ was obtained:

[TABLE]

Moreover, the following conjecture was posed in [4].

Conjecture 1**.**

For every fixed integer $k\geqslant 1$, there exists a real number $\gamma_{k}$ such that

[TABLE]

By the result of Zaks, $\gamma_{1}=1$, but for every $k\geqslant 2$ the conjecture is widely open. It is however tempting to guess that perhaps $\gamma_{k}=\frac{1}{k!}$ for every $k\geqslant 1$. Indeed, in [4] we made another supposition, which (if true) would imply this guess.

Conjecture 2**.**

For every $1\leqslant k\leqslant d$,

[TABLE]

accepting where necessary that $n(0,d)=1$ for all $d\geqslant 1$, and $n(k,d)=2^{d}$ for $k>d$.

For instance, for $k=2$ this would give $n(2,d)\leqslant d^{2}/2+O(d)$, implying that $\gamma_{2}=1/2$. In general, by induction and the well known formula for the sum $1^{k}+2^{k}+\cdots+d^{k}$, one easily gets $n(k,d)\leqslant d^{k}/k!+O(d^{k-1})$, which shows that Conjecture 2 implies Conjecture 1 with $\gamma_{k}=1/k!$. Our main result in this paper reads as follows.

Theorem 1**.**

Let $s$ and $d$ be positive integers, with $s\leqslant d/2$ and $d$ even. Then

[TABLE]

The proof is by a construction conducted in the related setting of strings (see [4] or [8]). It is given in the next section. In the last section we derive, as consequence of this result and the upper bound found in [5], the following asymptotic estimates.

Theorem 2**.**

Let $s$ and $d$ be positive integers, with $s\leqslant d/2$. Then

[TABLE]

In consequence, for every fixed $s\leqslant d/2$, we have

[TABLE]

1.1. Binary strings with jokers

The problem of neighborly boxes in $\mathbb{R}^{d}$ can be encoded in a purely combinatorial setting using strings over alphabet with just three symbols. Let $J=[0,1]$ be the unit segment and let $J_{0}=[0,1/2]$ and $J_{1}=[1/2,1]$ be its left and right half, respectively. A normalized box is a $d$-dimensional cuboid of the form $A=A_{1}\times\cdots\times A_{d}$, where $A_{i}\in\{J_{0},J_{1},J\}$ for all $i=1,2,\ldots,d$.

Suppose that we are given two normalized boxes, $A=A_{1}\times\cdots\times A_{d}$ and $B=B_{1}\times\cdots\times B_{d}$. If for some fixed coordinate $i$, we have $\{A_{i},B_{i}\}=\{J_{0},J_{1}\}$, then we say that $A$ and $B$ pass each other in dimension $i$. Otherwise, we say that $A$ and $B$ overlap in dimension $i$. Clearly, the intersection $A\cap B$ is a cuboid whose dimension equals exactly the number of dimensions in which $A$ and $B$ overlap. For instance, if $A$ and $B$ overlap in exactly $d-1$ dimensions, or the same, if they pass each other in exactly one dimension, then $A$ and $B$ are neighborly. In general, two normalized boxes are $k$-neighborly if and only if they pass in at least one and at most $k$ dimensions.

It is not hard to imagine that any family of boxes in $\mathbb{R}^{d}$ can be transformed to a family of normalized boxes preserving dimensions of all intersecting pairs. Therefore in investigating the function $n(k,d)$ one may only restrict to normalized boxes.

To further simplify the setting, let $S=\{0,1,\ast\}$ be an alphabet consisting of two binary digits and one special symbol called joker. Let $S^{d}$ be the set of all strings of length $d$ over $S$. Clearly, a normalized box $A=A_{1}\times\cdots\times A_{d}$ can be identified with a string $u=u_{1}u_{2}\cdots u_{d}$ so that $u_{i}=0$ if $A_{i}=J_{0}$, $u_{i}=1$ if $A_{i}=J_{1}$, and $u_{i}=\ast$ if $A_{i}=J$.

The distance between two strings $u,v\in S^{d}$ is defined as the number of positions where they differ, but none of them is occupied by a joker. It is denoted by $\operatorname{dist}(u,v)$. More formally, if $u=u_{1}\cdots u_{d}$ and $v=v_{1}\cdots v_{d}$, then

[TABLE]

Notice that the distance may be zero even if the two strings are not the same. For instance, if $u=0\ast 1\ast$ and $v=\ast 1\ast 0$, then $\operatorname{dist}(u,v)=0$. However, if $x,y\in\{0,1\}^{n}$, then $\operatorname{dist}(x,y)$ is equal to the Hamming distance between $x$ and $y$.

Clearly, $\operatorname{dist}(u,v)$ is exactly the number of dimensions in which the two corresponding boxes pass. Thus, by the above discussion, $n(k,d)$ is just the maximum number of strings in $S^{d}$ such that every two of them satisfy $1\leqslant\operatorname{dist}(u,v)\leqslant k$.

2. Proof of Theorem 1

Let us introduce some notation that will be helpful in our construction. For any two strings $x=x_{1}x_{2}\cdots x_{k}$ and $y=y_{1}y_{2}\cdots y_{l}$, the ordered pair $(x,y)$ is identified with their concatenation $xy=x_{1}x_{2}\cdots x_{k}y_{1}y_{2}\cdots y_{l}$. Let us set $I=\{0,1\}$, $*^{0}=\{*\}$, $*^{1}=I$. If $s$ is a positive integer, then for any member $\varepsilon$ of the $s$-fold Cartesian product $I^{s}$ of $I$, we set $*^{\varepsilon}=*^{\varepsilon_{1}}\times*^{\varepsilon_{2}}\times\cdots\times*^{\varepsilon_{s}}$. If $x\in I^{n}$, then $\bar{x}\in I^{n}$ denotes the antipode of $x$; that is $\bar{x}_{i}=1-x_{i}$, for $i=1,2,\ldots,n$. If $A\subseteq I^{n}$, then $\bar{A}=\{\bar{x}\colon x\in A\}$ is the antipode of the set $A$. If $x,y\in I^{n}$, then $|x|=\sum_{i=1}^{n}x_{i}$ and $\langle x,y\rangle=\sum_{i}^{n}x_{i}y_{i}$.

The diameter $\operatorname{diam}(A)$ of $A\subseteq\{0,1,*\}^{n}$ is defined in the standard way, as the maximum distance $\operatorname{dist}(x,y)$ between any pair of elements $x,y\in A$. If $\operatorname{diam}A\leqslant k$, then we will call $A$ a $k$-neighborly code. Let in addition $B\subseteq\{0,1,*\}^{n}$. It is convenient to consider the diameter of a pair $A$, $B$ of sets:

[TABLE]

Clearly, $\operatorname{diam}(A)=\operatorname{diam}(A,A)$.

Let $\mathscr{Z}=\{Z_{u}\colon u\in U\}$ be an indexed family of subsets of a set $Z$. The family $\mathscr{Z}$ defines a partition of $Z$ if $\bigcup\mathscr{Z}=Z$ and for every pair $u,v\in U$, if $u\neq v$, then $Z_{u}$ and $Z_{v}$ are disjoint. We do not rule out that the empty set is a member of $\mathscr{Z}$.

For $\varepsilon\in I^{s}$, if $\varepsilon\neq\alpha=(0,0,\ldots,0)$ and $\varepsilon\neq\omega=(1,1,\ldots,1)$, then there is a unique $i$ such that $\varepsilon_{i}\neq\varepsilon_{i+1}=\ldots=\varepsilon_{s}$. Let us set

[TABLE]

Let us remark that the uniqueness of $i$ determines a function $t$ on the set of admissible $\varepsilon$’s: $\varepsilon\mapsto t(\varepsilon)=i$.

Lemma 1**.**

Let $s>1$, and $k\in\{1,2,\ldots,s-1\}$. The indexed family $\{A_{\varepsilon}\colon|\varepsilon|=k\}$ defines a partition of $I^{s-1}$.

Proof.

Let $\varepsilon$ and $\gamma$ be different members of $I^{s}$ and let $|\varepsilon|=|\gamma|=k$. Let us fix the minimum index $i$ for which $\varepsilon_{i}\neq\gamma_{i}$. Suppose that $t(\varepsilon)<i$. Then $\varepsilon_{i}=\varepsilon_{i+1}=\ldots=\varepsilon_{s}$. Suppose first that $\varepsilon_{i}=0$. Then, by the definition of $i$, one has $k=|\varepsilon|=\sum_{n=1}^{i-1}\varepsilon_{n}$ and $\gamma_{j}=\varepsilon_{j},$ whenever $j\leq i-1$. Since $\gamma_{i}$ has to be $1$, one gets $|\gamma|>k$, which is impossible. Suppose now that $\varepsilon_{i}=1$. Then

[TABLE]

which again is impossible Consequently, $t(\varepsilon)\geqslant i$ and, by the same reason, $t(\gamma)\geqslant i$. By the definition of $\varepsilon\mapsto A_{\varepsilon}$, it follows that the $i$-th factor of $A_{\varepsilon}$ is $\{\varepsilon_{i}\}$, while the corresponding factor of $A_{\gamma}$ is $\{\gamma_{i}\}$. Therefore, $A_{\varepsilon}$ and $A_{\gamma}$ are disjoint.

It remains to show that $\{A_{\varepsilon}\colon|\varepsilon|=k\}$ is a covering of $I^{s-1}$. Let $x\in I^{s-1}$. If $|x|<k$, then let us replace subsequent zeros appearing in $x$ by ones, starting from the right end, until we reach the element $u$ with the norm $|u|=k-1$. Set $\varepsilon=(u,1)$. If $|x|\geq k$, then let us replace subsequent ones appearing in $x$ by zeros, starting from the right end, until we reach the element $v$ with the norm $|v|=k$. Now, set $\varepsilon=(v,0)$. It is easily seen that in each case under consideration $x$ is in the resulting $A_{\varepsilon}$. $\square$

Let $d$ and $s$ be as in Theorem 1. For $k=0,1,\ldots,s$ we define sets $X_{k}\subseteq I^{d-s}$ as follows:

[TABLE]

Clearly, these sets define a partition of the discrete box $I^{d-s}$.

For $\varepsilon\in I^{s}\setminus\{\alpha,\omega\}$, let us set

[TABLE]

Clearly,

[TABLE]

As an immediate consequence of Lemma 1, we obtain

Corollary 1**.**

Let $k\in\{1,2,\ldots,s-1\}$. The indexed family $\{B_{\varepsilon}\colon|\varepsilon|=k\}$ defines a partition of $X_{k}$.

Lemma 2**.**

Let $\varepsilon$ and $\gamma$ belong to $I^{s}\setminus\{\alpha,\omega\}$. Let $i=t(\varepsilon)$ and $j=t(\gamma)$. Let $\varepsilon_{s}=0$. If one of the two possibilities

1. (1)

$\varepsilon_{s}=\gamma_{s}=0$, 2. (2)

$\gamma_{s}=1$* and $j\geqslant i$*

holds true, then for every $x\in B_{\varepsilon}$ and $y\in B_{\gamma}$

[TABLE]

Proof.

For a string $x=x_{1}\cdots x_{n}$ and an index $k\leqslant n$, let $x|k=x_{1}\cdots x_{k}$. Then, by the assumptions on $\varepsilon$ and $\gamma$, we have

[TABLE]

We also have

[TABLE]

where $x=(x^{\prime},x^{\prime\prime})$, $y=(y^{\prime},y^{\prime\prime})$ are the decompositions of $x$ and $y$ as described in the definition of $\varepsilon\mapsto B_{\varepsilon}$. Since, in addition, $\varepsilon|i\leqslant x^{\prime\prime}|i$ and $\gamma|i\leqslant y^{\prime\prime}|i$ with respect to the coordinatewise order, our inequality can be easily deduced. $\square$

For the elements $\alpha$ and $\omega$ of $I^{s}$, let us set $B_{\alpha}=X_{|\alpha|}=X_{0}$ and $B_{\omega}=X_{|\omega|}=X_{s}$. It is clear that each of the sets $X_{0}$ and $X_{1}$ is antipodal to the other. Therefore,

[TABLE]

This observation is a part of the following lemma:

Lemma 3**.**

For every positive integer $s$ and every $\varepsilon\in I^{s}$

[TABLE]

Proof.

Due to the comment preceding our lemma, we may assume that $|\varepsilon|\in\{1,\ldots,s-1\}$. It is easily seen that $\overline{X_{|\varepsilon|}}=X_{s-|\varepsilon|}=X_{|\bar{\varepsilon}|}$. Observe now that if $i=t(\varepsilon)$, then $i=t(\bar{\varepsilon})$. Therefore,

[TABLE]

which readily implies $\overline{A_{\varepsilon}}=A_{\bar{\varepsilon}}$. Consequently,

[TABLE]

$\square$

Lemma 4**.**

For every $\varepsilon,\gamma\in I^{s}\setminus\{\alpha,\omega\}$

[TABLE]

Proof.

Let $x\in B_{\varepsilon}$ and $y\in B_{\gamma}$. Since $x\in X_{|\varepsilon|}$ and $y\in X_{|\gamma|}$, we have

[TABLE]

Since the expression in parentheses has a value not exceeding $\langle\varepsilon,\gamma\rangle$, we obtain

[TABLE]

Suppose from now on that $\varepsilon$ and $\gamma$ are as assumed in Lemma 2. Then the expression $\langle\varepsilon,\gamma\rangle-\langle x,y\rangle)$ has non-positive value and

[TABLE]

Similarly,

[TABLE]

The first term in parentheses is equal to $\langle\bar{\varepsilon},\bar{\gamma}\rangle$, while the second one is, according to Lemma 2, non-positive, consequently

[TABLE]

Combining (2.2) with the last inequality, we arrive to the desired result. However, to complete the proof, we need to remove the constraints on $\varepsilon$ and $\gamma$.

Note first that the constraints imposed in Lemma 2 on $\varepsilon$ and $\gamma$ are not symmetric. If the pair $\varepsilon,\gamma$ does not satisfy them, then one of the pairs $\gamma$, $\varepsilon$ or $\bar{\varepsilon},\bar{\gamma}$ or $\bar{\gamma},\bar{\varepsilon}$ has to. It is obvious that it is enough to consider the case when the pair $\bar{\varepsilon}$, $\bar{\gamma}$ satisfies the assumptions of our lemma. By Lemma 4, the preceding part and the fact that $\varepsilon\mapsto\bar{\varepsilon}$ is an involution, we obtain

[TABLE]

$\square$

For $\varepsilon\in I^{s}$, let us set $C_{\varepsilon}=B_{\varepsilon}\times*^{\varepsilon}$. Since the members of the indexed family $\{B_{\varepsilon}\colon\varepsilon\in I^{s}\}$ are pairwise disjoint subsets of the discrete box $I^{d-s}$, the members of the family $\mathscr{C}=\{C_{\varepsilon}\colon\varepsilon\in I^{s}\}$ are also pairwise disjoint. Moreover, $\bigcup\mathscr{C}$ is a $1$-neighborly code. In fact we have the following lemma.

Lemma 5**.**

Let $d$ be a positive even integer and Let $s$ be a positive integer so that $2s\leqslant d$ The union of $\mathscr{C}=\{C_{\varepsilon}\colon\varepsilon\in I^{s}\}$ is a $(d-s)$-neighborly code.

Proof.

Since $\mathscr{C}$ is a $1$-neighborly code, it suffices to show that for every $\varepsilon,\gamma\in I^{s}$, we have $\operatorname{diam}(C_{\varepsilon},C_{\gamma})\leqslant d-s$. By the definition of $\mathscr{C}$,

[TABLE]

As $\operatorname{diam}(*^{\varepsilon},*^{\gamma}))=\langle\varepsilon,\gamma\rangle$, it follows from Lemma 4 that indeed $\operatorname{diam}(C_{\varepsilon},C_{\gamma})$ does not exceed $d-s$ as long as $\varepsilon$ and $\gamma$ are different from $\alpha$ and $\omega$.

Suppose first that at least one of the two $\varepsilon$ and $\gamma$ is $\alpha$. Then $\operatorname{diam}(*^{\varepsilon},*^{\gamma}))=0$. Moreover, $\operatorname{diam}(B_{\varepsilon},B_{\gamma})\leqslant d-s$, as $B_{\varepsilon}$ and $B_{\gamma}$ are subsets of $I^{d-s}$. Therefore, also in this case $\operatorname{diam}(C_{\varepsilon},C_{\gamma})\leqslant d-s$.

Finally, suppose that one of the two $\varepsilon$ and $\gamma$ is $\omega$, while the other is in $I^{d-s}\setminus\{\alpha,\omega\}$. We may assume that $\varepsilon=\omega$. In this case,

[TABLE]

In order to estimate $D=\operatorname{diam}(B_{\varepsilon},B_{\gamma})$, we employ Lemma 3:

[TABLE]

Let $x\in B_{\alpha}$ and $y\in B_{\bar{\gamma}}$.Since $B_{\alpha}\subseteq X_{0}$ and $B_{\bar{\gamma}}\subseteq X_{|\bar{\gamma}|}$, we have

[TABLE]

Consequently, $D\leqslant d-s-|\gamma|$. This, combined with (2.3), implies the expected inequality. $\square$

Proof of Theorem 1. Clearly, $n(d-s,d)\geqslant|\mathscr{C}|$. We have

[TABLE]

Since $|*^{\varepsilon}|=2^{|\varepsilon|}$, we obtain

[TABLE]

Since

• •

$B_{\alpha}=X_{0}$ and $|X_{0}|=\sum_{i=0}^{\frac{d}{2}-s-1}{d-s\choose i}$,

• •

$(B_{\varepsilon}\colon|\varepsilon|=k)$ is a partition of $X_{k}$, whenever $k=1,2,\ldots,s-1$ and $|X_{k}|={d-s\choose\frac{d}{2}-s+k}$,

• •

$B_{\omega}=X_{s}$ and $|X_{s}|=\sum_{i=\frac{d}{2}}^{d-s}{d-s\choose i}$,

we easily come to the conclusion that the right side of the inequality under consideration expresses the cardinality of $\mathscr{C}$. $\square$

3. Proof of Theorem 2

We shall assume for simplicity that both $d$ and $s$ are even. The remaining cases can be proven in a similar manner. To get the left inequality, we can use our Theorem 1. In fact, we need less than that. It is not so easy to prove Theorem 1 but it is almost trivial to show that $C_{\alpha}\cup C_{\omega}$ is a $(d-s)$-neighborly. Thus, the cardinality of this set does not exceed $n(d-s,d)$. Consequently,

[TABLE]

By the symmetry of the binomial coefficients,

[TABLE]

Therefore,

[TABLE]

Since the central binomial coefficient is the largest and satisfies the inequality

[TABLE]

we obtain

[TABLE]

Finally,

[TABLE]

The estimate on the right follows easily from the result by Cheng, Wang, Xu, and Yip [arxiv:2301.06485v1, Corollary 1.6], which we formulate below.

Theorem 3**.**

Let $1\leqslant k\leq d-2$ be integers. If $k=2t$, then

[TABLE]

and if $k=2t+1$, then

[TABLE]

Observe first that $\sum_{i=0}^{t}{d\choose i}\leqslant\frac{1}{2}2^{d}$ in the first case, while $\sum_{i=0}^{t}{d-1\choose i}\leqslant\frac{1}{2}2^{d-1}$. Therefore, in both cases we have

[TABLE]

and it remains to replace $k$ by $d-s$.

Let us remark that the published version [5] of the paper by Cheng, Wang, Xu, and Yip (as well as its later arxiv versions) appears to be slightly different than the one we used. In particular, Corollary 1.6 from arxiv:2301.06485v1 is not reproduced there and it is not quite clear how it can be deduced without additional work.