The sandglass conjecture beyond cancellative pairs

TL;DR

This paper advances the understanding of the sandglass conjecture by establishing a new upper bound on the product of sizes of recovering families, surpassing previous bounds and addressing a key theoretical barrier.

Contribution

It introduces an exponential separation between bounds on recovering pairs and cancellative pairs, leading to a tighter upper bound on the sandglass conjecture.

Findings

New upper bound: |A| |B| ≤ 2.2543^n

Exponential separation between recovering and cancellative pairs

Improved understanding of the sandglass conjecture's limits

Abstract

The sandglass conjecture, posed by Simonyi, states that if a pair $(A, B)$ of families of subsets of $[n]$ is recovering then $|A| |B| \leq 2^n$. We improve the best known upper bound to $|A| |B| \leq 2.2543^n$. To do this we overcome a significant barrier by exponentially separating the upper bounds on recovering pairs from cancellative pairs, a related notion.

Equations

$$A\setminus B$$

$$B\setminus A$$

$$|\mathcal{A}||\mathcal{B}|⩽2.2543^n.$$

$$A\setminus B$$

$$B\setminus A$$

$$\displaystyle\mu_{\operatorname{rec}}\mathrel{\coloneqq}\lim_{n\to\infty}\max\{\big{\lparen}\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert\big{\rparen}^{1/n}\colon(\mathcal{A},\mathcal{B})\text{ is a recovering pair over }[n]\},$$

$$\displaystyle\mu_{\operatorname{can}}\mathrel{\coloneqq}\lim_{n\to\infty}\max\{\big{\lparen}\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert\big{\rparen}^{1/n}\colon(\mathcal{A},\mathcal{B})\text{ is a cancellative pair over }[n]\}.$$

$$|\mathcal{A}||\mathcal{B}|⩽\max \{2.2499,{\theta }^{\alpha }\cdot {\mu }_{\mathrm{can}}^{1-\alpha }{\}}^n.$$

$${\mathcal{A}}^{\prime \prime }:=\{A\cup A^{\prime }:A\in \mathcal{A},A^{\prime }\in {\mathcal{A}}^{\prime }\},$$

$$H(Z):=-\sum _{i\in [m]}p(z_i){\log }_{2}p(z_i).$$

$${\log }_{2}p(z_1)<-H(Z)=\sum _{i=1}^{m}p(z_i){\log }_{2}p(z_i)⩽{\log }_{2}p(z_1)\sum _{i=1}^{m}p(z_i)={\log }_{2}p(z_1),$$

$$H(Z_1,Z_2)⩽H(Z_1)+H(Z_2),$$

$$h(p):=-p{\log }_{2}p-(1-p){\log }_2(1-p).$$

$$\displaystyle\mathcal{F}_{i}\mathrel{\coloneqq}\big{\{}F\in\mathcal{F}\colon i\in F\big{\}},\qquad\mathcal{F}^{\prime}_{i}\mathrel{\coloneqq}\big{\{}F\in\mathcal{F}\colon i\notin F\big{\}}.$$

$$a_i:=\frac{|{\mathcal{A}}_i|}{|\mathcal{A}|},a_i^{\prime }:=\frac{|{\mathcal{A}}_i^{\prime }|}{|\mathcal{A}|},b_i:=\frac{|{\mathcal{B}}_i|}{|\mathcal{B}|},b_i^{\prime }:=\frac{|{\mathcal{B}}_i^{\prime }|}{|\mathcal{B}|}.$$

$${\mathcal{A}}_{C,S,P}$$

$${\mathcal{B}}_{C,S}$$

$$|{\mathcal{A}}_{C,S,P}|$$

$$|{\mathcal{B}}_{C,S}|$$

$$A_1$$

$$A_{1}\setminus B=\big{\lparen}(A_{1}\setminus C)\setminus B\big{\rparen}\sqcup P=\big{\lparen}(A_{2}\setminus C)\setminus B\big{\rparen}\sqcup P=A_{2}\setminus B.$$

$$A_1\setminus B_1$$

$$\displaystyle=(A^{\prime}_{1}\setminus B^{\prime}_{1})\sqcup\big{\lparen}(A_{1}\cap C)\setminus(C\setminus S)\big{\rparen}$$

$$=(A_1^{\prime }\setminus B_1^{\prime })\bigsqcup (A_1\cap S)$$

$$=(A_1^{\prime }\setminus B_1^{\prime })\bigsqcup P.$$

$$A_1\setminus B_1=(A_1^{\prime }\setminus B_1^{\prime })\bigsqcup P=(A_2^{\prime }\setminus B_2^{\prime })\bigsqcup P=A_2\setminus B_2.$$

$$\displaystyle B_{1}\setminus A=\big{\lparen}(B_{1}\setminus C)\setminus(A\setminus C)\big{\rparen}\sqcup\big{\lparen}(B_{1}\cap C)\setminus(A\cap C)\big{\rparen}=(B^{\prime}_{1}\setminus A^{\prime})\sqcup\big{\lparen}(C\setminus S)\setminus A\big{\rparen},$$

$$\displaystyle B_{1}\setminus A=(B^{\prime}_{1}\setminus A^{\prime})\sqcup\big{\lparen}(C\setminus S)\setminus A\big{\rparen}=(B^{\prime}_{2}\setminus A^{\prime})\sqcup\big{\lparen}(C\setminus S)\setminus A\big{\rparen}=B_{2}\setminus A.$$

$$\log_{2}\Bigg{\lparen}\frac{\lvert\mathcal{A}_{C,S,P}\rvert}{\lvert\mathcal{A}\rvert}\Bigg{\rparen}\geqslant-\sum_{i\in S}h(a_{i}).$$

$$H(X^S)⩽\sum _{i\in S}H(X_i)=\sum _{i\in S}h(a_i).$$

$$\frac{|{\mathcal{A}}_{C,S,P}|}{|\mathcal{A}|}=P(A\cap S=P)⩾2^{-H(X^S)}⩾2^{-{\sum }_{i\in S}h(a_i)}.\text{qed}$$

$$\frac{|{\mathcal{A}}_{A,A\setminus B,P}||{\mathcal{B}}_{A,A\setminus B}|}{|\mathcal{A}||\mathcal{B}|}⩽{\theta }^{-k},$$

$${\log }_2|\mathcal{A}|⩽\sum _{i\in X}f(a_i,b_i,\theta ),$$

$$P(Z_{\mathcal{A}}\setminus Z_{\mathcal{B}}=A\setminus B)=\frac{|\{(A^{\prime },B^{\prime })\in \mathcal{A}\times \mathcal{B}:A^{\prime }\setminus B^{\prime }=A\setminus B\}|}{|\mathcal{A}||\mathcal{B}|}.$$

Full text

1. Introduction

In this paper we study the notion of recovering pairs, introduced by Ahlswede and Simonyi [Ahlswede1994-st]. Denote by $2^{X}$ the family of all subsets of a set $X$. Given two families $\mathcal{A},\mathcal{B}\subseteq 2^{X}$, we say that $(\mathcal{A},\mathcal{B})$ is a recovering pair over $X$ if for all $A,A^{\prime}\in\mathcal{A}$ and all $B,B^{\prime}\in\mathcal{B}$,

[TABLE]

In 1989, Simonyi made the following seminal conjecture about the maximal product size of recovering pairs, called the “sandglass conjecture”. This conjecture111More precisely, they use “sandglass conjecture” to refer to a generalisation where the lattice $2^{[n]}$ is replaced by the product of finite chains. Still, Simonyi’s original conjecture is generally referred to as the “sandglass conjecture”. was presented in print by Ahlswede and Simonyi [Ahlswede1994-st].

Conjecture 1.1**.**

Any recovering pair $(\mathcal{A},\mathcal{B})$ over $[n]$ satisfies $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert\leqslant 2^{n}$.

If ˜1.1 holds, then it is sharp. To see this, fix $S\subseteq[n]$, take $\mathcal{A}$ to be the family of all subsets of $S$, and take $\mathcal{B}$ to be the family of all subsets of $[n]\setminus S$. Then $(\mathcal{A},\mathcal{B})$ is a recovering pair satisfying $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert=2^{\lvert S\rvert}2^{n-\lvert S\rvert}=2^{n}$.

In this paper, we improve the best known upper bound for ˜1.1.

Theorem 1.2**.**

If $(\mathcal{A},\mathcal{B})$ is a recovering pair over $[n]$, then

[TABLE]

This result is a corollary of our main result, Theorem˜1.3, where we overcome a significant obstacle in the study of the sandglass conjecture by showing an exponential separation between recovering pairs and cancellative pairs – a related notion that was introduced by Holzman and Körner [Holzman1995-xb] in their work on Simomyi’s conjecture.

Given two families $\mathcal{A},\mathcal{B}\subseteq 2^{X}$, we say that $(\mathcal{A},\mathcal{B})$ is a cancellative pair over $X$ if for all $A,A^{\prime}\in\mathcal{A}$ and all $B,B^{\prime}\in\mathcal{B}$,

[TABLE]

Note that any recovering pair is also a cancellative pair, and therefore any upper bound on cancellative pairs implies an upper bound for Simonyi’s conjecture. Holzman and Körner [Holzman1995-xb] proved that for any cancellative pair $(\mathcal{A},\mathcal{B})$ over a ground set of size $n$, satisfies $|\mathcal{A}||\mathcal{B}|\leqslant 2.3264^{n}$, implying the same upper bound for recovering pairs.

Following [Holzman1995-xb], Soltész [Soltesz2018-sf] improved the upper bound to be $2.284^{n}$ for recovering pairs only. More recently B. Janzer [Janzer2018-pb] improved the upper bound for cancellative pairs to $2.2682^{n}$ and hence also for recovering pairs, which was later improved by Nair and Yazdanpanah [nair2020and] to $2.2663^{n}$ for recovering pairs only, using techniques from coding theory.

In [Holzman1995-xb] the authors also observed that there exist222This is done by taking a product of the pair $(\mathcal{A},\mathcal{B})$ with itself, over the ground set $[3]$, where $\mathcal{A}=\mathcal{B}=\{\{1\},\{2\},\{3\}\}$. cancellative pairs $(\mathcal{A},\mathcal{B})$ satisfying $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert\approx 2.08^{n}$ for infinitely many values of $n$, thus showing that the study of cancellative pairs is not enough to prove ˜1.1. Later, Tolhuizen [Tolhuizen2000-ww] improved this lower bound by constructing cancellative pairs $(\mathcal{A},\mathcal{B})$ over $[n]$ with $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert=2.25^{n-o(n)}$, for arbitrarily large $n$. It is an enticing open problem to determine if this construction is optimal for cancellative pairs.

Define the optimal rate333Using supermultiplicativity, one can show that both limits exist (see Section 2). for recovering pairs as

[TABLE]

and, analogously for cancellative pairs, define

[TABLE]

It is clear that $2\leqslant\mu_{\operatorname{rec}}\leqslant\mu_{\operatorname{can}}$. The current state of art, due the works of Tolhuizen [Tolhuizen2000-ww], B. Janzer [Janzer2018-pb], and Nair and Yazdanpanah [nair2020and], is that $2.25\leqslant\mu_{\operatorname{can}}\leqslant 2.2682$, and $2\leqslant\mu_{\operatorname{rec}}\leqslant 2.2663$.

The main result of this paper shows that $\mu_{\operatorname{rec}}$ and $\mu_{\operatorname{can}}$ are not equal, exponentially separating the maximal product size of recovering pairs from the one of cancellative pairs.

Theorem 1.3**.**

Let $(\mathcal{A},\mathcal{B})$ be a recovering pair over a ground set $X$ of size $n$, and let $\alpha=0.27$ and $\theta=2.222$. Then we have

[TABLE]

In particular, $\mu_{\operatorname{rec}}<\mu_{\operatorname{can}}$.

Theorem˜1.2 follows immediately as a corollary from Theorem˜1.3 giving the new upper bound for the maximal product size of recovering pairs.

Given Theorem˜1.3, any improvement of the upper bound on cancellative pairs to about $2.26^{n}$ would imply that $\mu_{\operatorname{rec}}<2.25$. Moreover, if we had $\mu_{\operatorname{can}}=2.25$, then Theorem˜1.3 would imply $\mu_{\operatorname{rec}}\leqslant 2.243$.

We remark that in the original paper [Ahlswede1994-st], recovering pairs are actually defined in a slightly different way. Instead of (1.1) and (1.2), the required conditions are that $A\cup B=A^{\prime}\cup B^{\prime}\implies A=A^{\prime}$ and that $A\cap B=A^{\prime}\cap B^{\prime}\implies B=B^{\prime}$, for all $A,A^{\prime}\in\mathcal{A}$ and $B,B^{\prime}\in\mathcal{B}$. It is easily shown that a pair $(\mathcal{A},\mathcal{B})$ is recovering in this sense if and only if the pair $(\{A^{c}\colon A\in\mathcal{A}\},\mathcal{B})$ is recovering as defined by (1.1) and (1.2). Therefore, the two definitions are interchangeable for the objective of maximizing $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert$. In this formulation, the sharp examples are given by taking $\mathcal{A}$ to be all supersets of a fixed set $S\subseteq[n]$, and $\mathcal{B}$ to be all its subsets, giving rise to the name “the sandglass conjecture”.

In [Ahlswede1994-st] the authors also discuss a one-sided variant of the sandglass problem, in which one aims to maximize $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert$ over all pairs satisfying (1.3) but not necessarily (1.4). We call such pairs left-cancellative. They observe an upper bound of $3^{n}$ and a lower bound of $6^{n/2}$ for this quantity. Using Tolhuizen’s construction for large cancellative pairs, we determine the maximum size of a one-sided cancellative pair up to a sub-exponential error.

Theorem 1.4**.**

The maximum of $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert$ over all left-cancellative pairs $(\mathcal{A},\mathcal{B})$ is $3^{n-o(n)}$.

The organisation of the remainder of the paper is as follows. In Section˜2, we present preliminaries results we will use in the proof of Theorem˜1.3. In Section˜3, we define an operation we perform to a recovering pair. In Section˜4, we prove the main lemma we need for the proof of Theorem˜1.3, and in Section˜5, we prove Theorem˜1.3. Lastly, in Section˜6, we prove Theorem˜1.4.

2. Preliminaries

Given two recovering (cancellative) pairs $(\mathcal{A},\mathcal{B})$ and $(\mathcal{A}^{\prime},\mathcal{B}^{\prime})$ over disjoint ground sets $X$ and $X^{\prime}$, respectively, we define their product to be the pair $(\mathcal{A}^{\prime\prime},\mathcal{B}^{\prime\prime})$ given by

[TABLE]

Note that the pair $(\mathcal{A}^{\prime\prime},\mathcal{B}^{\prime\prime})$ is also recovering (cancellative) over the ground set $X\cup X^{\prime}$. Moreover, we have $|\mathcal{A}^{\prime\prime}||\mathcal{B}^{\prime\prime}|=|\mathcal{A}||\mathcal{B}||\mathcal{A}^{\prime}||\mathcal{B}^{\prime}|$, implying that in (1.5) and in (1.6) the limits exist.

We say that a pair of sets $(\mathcal{A},\mathcal{B})$ over a ground set of $n$ elements is $k$-uniform if for every $A\in\mathcal{A}$ and every $B\in\mathcal{B}$, we have $\lvert A\rvert=\lvert B\rvert=k$. We rely on the the following lemma (originally from [Soltesz2018-sf] and then reformulated in [Janzer2018-pb]), which shows, using a product argument, that in order to prove an upper bound for recovering (or cancellative) pairs in general, it is enough to consider $k$-uniform pairs only.

Lemma 2.1** (Lemma 2 in [Janzer2018-pb]).**

Let $\mu>0$, and suppose that for all $n,k\in\mathbb{N}$ with $k\leqslant n$, every $k$-uniform recovering pair over $n$-element ground set satisfies $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert\leqslant\mu^{n}$ for some constant $\mu$. Then the same inequality holds for any (not necessarily uniform) recovering pair.

2.1. Entropy of discrete random variables

We will make use of some basic facts about entropy, which we recall below. See [Cover2005-xs, Csiszar2011-lr] for standard references in information theory. For a random variable $Z$ taking finitely many values $z_{1},\dotsc,z_{m}$, for some $m\in\mathbb{N}$, with probabilities $p(z_{1}),\dotsc,p(z_{m})$, respectively, its entropy $H(Z)$ is defined as $H(Z)\mathrel{\coloneqq}\operatorname{\mathbb{E}}\lparen-\log_{2}p(Z)\rparen$, or equivalently,

[TABLE]

Observation 2.2**.**

If $Z$ is a discrete random variable taking finitely many values, for some $m\in\mathbb{N}$, with probabilities $p(z_{1})\leqslant\ldots\leqslant p(z_{m})$. Then we have $p(z_{1})\geqslant 2^{-H(Z)}$.

Proof.

If $p(z_{1})<2^{-H(Z)}$, this yields

[TABLE]

which is a contradiction. ∎

We also recall that the entropy function is subadditive in the following sense. If $Z_{1}$ and $Z_{2}$ are two discrete random variables, then their joint entropy (that is, the entropy of the joint random variable $(Z_{1},Z_{2})$) satisfy

[TABLE]

with equality only when $Z_{1}$ and $Z_{2}$ are independent.

If $Z$ attains only two values, one with probability $p\in(0,1)$ and the other with probability $1-p$, then its entropy is given by the binary entropy function

[TABLE]

We extend $h(p)$ to $p\in[0,1]$ continuously, setting $0\log_{2}0\mathrel{\coloneqq}0$.

3. Filtered pairs

In this section, we discuss some operations one can perform on a recovering or cancellative pair $(\mathcal{A},\mathcal{B})$ on $X$ to obtain new pairs $(\mathcal{A}^{\prime},\mathcal{B}^{\prime})$ on a subset of $X$. This will allow us to perform inductive arguments with these families.

The first such operation is a simple restriction, already considered in the work of Holzman and Körner [Holzman1995-xb]. If $\mathcal{F}$ is a family of subsets of $X$ and $i\in X$, define

[TABLE]

It is then clear that the following holds.

Observation 3.1**.**

If $(\mathcal{A},\mathcal{B})$ is a recovering pair over $X$, then for every $i\in X$:

1. (i)

$(\mathcal{A}_{i},\mathcal{B}_{i})$ is a recovering pair over $X$; 2. (ii)

$(\mathcal{A}^{\prime}_{i},\mathcal{B}^{\prime}_{i})$ is a recovering pair over $X$ and over $X\setminus\{i\}$; 3. (iii)

\big{\lparen}\{A\setminus\{i\}\colon A\in\mathcal{A}_{i}\},\{B\setminus\{i\}\colon A\in\mathcal{B}_{i}\}\big{\rparen} is a recovering pair over $X\setminus\{i\}$.

Moreover, the same is true if we replace ‘recovering’ by ‘cancellative’.

We remark that ˜3.1 holds also for left-cancellative pairs and for left-recovering pairs (which are defined analogously to left-cancellative pairs).

For the remainder of the paper, we denote the relative size of each restriction for $i\in X$ by

[TABLE]

It follows then that for any $i\in X$, we have $a_{i}+a^{\prime}_{i}=1$ and $b_{i}+b^{\prime}_{i}=1$. Furthermore, note that if the pair $(\mathcal{A},\mathcal{B})$ is $k$-uniform, for some integer $k$, then $\sum_{i\in X}a_{1}=\sum_{i\in X}b_{i}=k$. The proofs in [Holzman1995-xb, Janzer2018-pb] rely on an inductive argument based on this type of restriction and its properties, as given in ˜3.1. In order to obtain a separation between the bounds $\mu_{\operatorname{can}}$ and $\mu_{\operatorname{rec}}$ however, we need to a different type of restriction, which exploits the recovering property.

We introduce a new type of restriction in order to differentiate recovering and cancellative pairs. Let $(\mathcal{A},\mathcal{B})$ be a pair of families over $X$. Then given $P\subseteq S\subseteq C\subset X$, the filtered pair $(\mathcal{A}_{C,S,P},\mathcal{B}_{C,S})$ of the pair $(\mathcal{A},\mathcal{B})$ is defined such that

[TABLE]

We use three important statements regarding filtered pairs as described above. Later we will only use filtered pairs where $C=A$ and $S=A\setminus B$, for some $A\in\mathcal{A}$ and $B\in\mathcal{B}$, but the properties described in the remainder of this section hold in larger generality.

Observation 3.2**.**

Let $(\mathcal{A},\mathcal{B})$ be a cancellative pair over a set $X$. Then for any sets $P\subseteq S\subseteq C\subset X$ such that there is $B\in\mathcal{B}$ with $S=C\setminus B$, the filtered pair $(\mathcal{A}_{C,S,P},\mathcal{B}_{C,S})$ satisfies

[TABLE]

Proof.

For the first equality, we need to check that if two distinct $A_{1},A_{2}\in\mathcal{A}$ satisfy $A_{1}\cap S=P=A_{2}\cap S$, then they give rise to distinct sets in $\mathcal{A}_{C,S,P}$, meaning that $A_{1}\setminus C\neq A_{2}\setminus C$. Suppose otherwise that $A_{1}\setminus C=A_{2}\setminus C$. Let $B\in\mathcal{B}$ have $C\setminus B=S$, or equivalently $C\setminus S=C\cap B$. We have

[TABLE]

and similarly, $A_{2}=(A_{2}\setminus C)\sqcup(A_{2}\cap C\cap B)\sqcup P$. Therefore,

[TABLE]

Since $(\mathcal{A},\mathcal{B})$ is a cancellative pair, we get $A_{1}=A_{2}$.

For the second inequality, if $B_{1},B_{2}\in\mathcal{B}$ are such that $B_{1}\cap C=C\setminus S=B_{2}\cap C$ and $B_{1}\setminus C=B_{2}\setminus C$, then $B_{1}=(B_{1}\setminus C)\sqcup(B_{1}\cap C)=(B_{2}\setminus C)\sqcup(B_{2}\cap C)=B_{2}$, completing the proof. ∎

˜3.2 tells us that for every $A^{\prime}\in\mathcal{A}_{C,S,P}$, there is unique set $A\in\mathcal{A}$ such that $A^{\prime}=A\setminus C$. We say that $A$ is the associated set of $A^{\prime}$. Similarly, the associated set for $B^{\prime}\in\mathcal{B}_{C,S}$ is the unique set $B\in\mathcal{B}$ satisfying $B^{\prime}=B\setminus C$.

Unfortunately, considering the filtered pair of a recovering pair may destroy the recovering property, but as we see bellow, it preserves the cancellation property.

Observation 3.3**.**

Let $(\mathcal{A},\mathcal{B})$ be a recovering pair over a set $X$. Then for any sets $P\subseteq S\subseteq C\subseteq X$, the filtered pair $(\mathcal{A}_{C,S,P},\mathcal{B}_{C,S})$ is cancellative over $X\setminus C$.

Proof.

In fact, we show that for sets $A^{\prime}_{1},A^{\prime}_{2}\in\mathcal{A}_{C,S,P}$ and $B^{\prime}_{1},B^{\prime}_{2}\in\mathcal{B}_{C,S}$ the recovering property is preserved on one side, namely, they satisfy (1.1). However, while they satisfy (1.4), they need not satisfy (1.2), and thus the filtered pair is cancellative but not necessarily recovering.

Let $A^{\prime}_{1},A^{\prime}_{2}\in\mathcal{A}_{C,S,P}$ and $B^{\prime}_{1},B^{\prime}_{2}\in\mathcal{B}_{C,S}$ and let $A_{1},A_{2}\in\mathcal{A}$ and $B_{1},B_{2}\in\mathcal{B}$ be their associated sets. We have

[TABLE]

Similarly, we get $A_{2}\setminus B_{2}=(A^{\prime}_{2}\setminus B^{\prime}_{2})\sqcup P$. Thus, if we assume that $A^{\prime}_{1}\setminus B^{\prime}_{1}=A^{\prime}_{2}\setminus B^{\prime}_{2}$, we obtain

[TABLE]

Since $(\mathcal{A},\mathcal{B})$ is recovering, we get that $A_{1}=A_{2}$, and consequently $A^{\prime}_{1}=A^{\prime}_{2}$.

Now let $B^{\prime}_{1},B^{\prime}_{2}\in\mathcal{B}_{C,S}$ and $A^{\prime}\in\mathcal{A}_{C,S,P}$ and let $B_{1},B_{2}\in\mathcal{B}$ and $A\in\mathcal{A}$ be their associated sets. We have

[TABLE]

and similarly B_{2}\setminus A=(B^{\prime}_{2}\setminus A^{\prime})\sqcup\big{\lparen}(C\setminus S)\setminus A\big{\rparen}. Thus, if we assume that $B^{\prime}_{1}\setminus A^{\prime}=B^{\prime}_{2}\setminus A^{\prime}$, we obtain

[TABLE]

Since $(\mathcal{A},\mathcal{B})$ is recovering and in particular cancellative, we get that $B_{1}=B_{2}$, and consequently $B^{\prime}_{1}=B^{\prime}_{2}$. ∎

Finally, we now show that given $S\subseteq C\subseteq X$, there is always a choice of $P\subseteq S$ for which the family $\mathcal{A}_{C,S,P}$ in the filtered pair is reasonably sized. Recall the binary entropy function as defined in (2.1).

Proposition 3.4**.**

Let $(\mathcal{A},\mathcal{B})$ be a recovering pair over $X$ and recall the definition of $a_{i}$ from (3.1). Let $S\subseteq C\subseteq X$ be such that $S=C\setminus B$ for some $B\in\mathcal{B}$. Then there exists a subset $P\subseteq S$ such that

[TABLE]

Proof.

Let $A\in\mathcal{A}$ be chosen uniformly at random, and note that we have $a_{i}=\operatorname{\mathbb{P}}\lparen i\in A\rparen$ for all $i\in X$. Let $X^{S}\mathrel{\coloneqq}(X_{i})_{i\in S}$ be the characteristic function of $A\cap S$. By the subadditivity of entropy, we have

[TABLE]

Let $P\subseteq S$ be such that $\operatorname{\mathbb{P}}\lparen A\cap S=P\rparen$ is maximal. By ˜2.2 we get that $\operatorname{\mathbb{P}}\lparen A\cap S=P\rparen\geqslant 2^{-H(X^{S})}$. By ˜3.2, as recovering implies cancellative, and the above we get

[TABLE]

4. Uniform pairs and main lemma

The following proposition is the main technical tool in our argument. Roughly, it says that if a recovering pair $(\mathcal{A},\mathcal{B})$ does not have a relatively dense filtered pair, then it cannot be too large. For the remainder of this section, we only consider filtered pairs such that $C=A$ and $S=A\setminus B$ for some sets $A\in\mathcal{A}$ and $B\in\mathcal{B}$.

Proposition 4.1**.**

Let $(\mathcal{A},\mathcal{B})$ be a $k$-uniform recovering pair over $X$ and suppose that for every $A\in\mathcal{A}$, $B\in\mathcal{B}$ and $P\subseteq A\setminus B$ we have

[TABLE]

for some constant $\theta$. Then we have

[TABLE]

where f(x,y,t)\mathrel{\coloneqq}x(1-y)h(x)+h\big{\lparen}x(1-y)\big{\rparen}-x\log_{2}t.

Proof.

Let $Z_{\mathcal{A}}$ and $Z_{\mathcal{B}}$ be two independent random variables choosing uniformly at random sets from $\mathcal{A}$ and from $\mathcal{B}$, respectively. Fix $A\in\mathcal{A}$ and $B\in\mathcal{B}$, then we have

[TABLE]

Since the pair $(\mathcal{A},\mathcal{B})$ is recovering, we have

[TABLE]

Additionally, by ˜3.2 we have $\lvert\{B^{\prime}\in\mathcal{B}\colon A\setminus B^{\prime}=A\setminus B\}\rvert=\lvert\mathcal{B}_{A,A\setminus B}\rvert$. Therefore, (4.3) is equivalent to

[TABLE]

By Proposition˜3.4, there exists a subset $P\subseteq A\setminus B$ for which

[TABLE]

By rearranging condition (4.1) and by the above we get

[TABLE]

Therefore, (4.5) and (4.6) together imply

[TABLE]

Recall that $k$-uniformity implies that $k=\sum_{i\in X}a_{i}=\sum_{i\in X}b_{i}$, so we obtain

[TABLE]

Define the function

[TABLE]

Choosing $A\in\mathcal{A}$ and $B\in\mathcal{B}$ independently and uniformly at random, we get

[TABLE]

where H(Z_{\mathcal{A}}\setminus Z_{\mathcal{B}})\leqslant\sum_{i\in X}h\big{\lparen}a_{i}(1-b_{i})\big{\rparen} follows from the subadditivity of entropy. Considering $A_{0}\in\mathcal{A}$ and $B_{0}\in\mathcal{B}$ for which $F(A_{0},B_{0})\leqslant\operatorname{\mathbb{E}}\lparen F(A,B)\rparen$ in (4.7) we get

[TABLE]

implying the desired bound (4.2). ∎

We remark that the crux of the proof of Proposition˜4.1 is captured in (4.4). Indeed, unlike other parts of the argument, (4.4) holds due to the recovering property and does not hold for cancellative pairs in general. This is what allows us to attain the desired separation in Theorem˜1.3.

The proof of Proposition˜4.1 used only that $A\setminus B$ determines $A$ but not that $B\setminus A$ determines $A$. Using the full power of the recovering property, we can symmetrise Proposition˜4.1.

Corollary 4.2**.**

Let $(\mathcal{A},\mathcal{B})$ be a $k$-uniform recovering pair over $X$ and suppose that there exists a constant $\theta$ such that the following holds. For every $A\in\mathcal{A}$, $B\in\mathcal{B}$ and for every $P_{1}\subseteq A\setminus B$ and $P_{2}\subseteq B\setminus A$ we have $\lvert\mathcal{A}_{A,A\setminus B,P_{1}}\rvert\lvert\mathcal{B}_{A,A\setminus B}\rvert\leqslant\theta^{-k}\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert$, and $\lvert\mathcal{B}_{B,B\setminus A,P_{2}}\rvert\lvert\mathcal{A}_{B,B\setminus A}\rvert\leqslant\theta^{-k}\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert$. Then

[TABLE]

where $g(x,y,t)\mathrel{\coloneqq}f(x,y,t)+f(y,x,t)$, and $f$ is as in Proposition˜4.1. ∎

5. Beyond cancellation

We are now ready to prove Theorem˜1.3.

Proof of Theorem˜1.3.

We will show that (1.7) holds for every $k$-uniform recovering pair $(\mathcal{A},\mathcal{B})$, for any $1\leqslant k\leqslant n$, and then the result will follow by Lemma˜2.1. Set $\alpha=0.27$, $\theta=2.222$, and $\mu=\max\{2.2499,\theta^{\alpha}\cdot\mu_{\operatorname{can}}^{1-\alpha}\}$. We prove by induction that $\lvert\mathcal{A}\rvert\lvert\mathcal{B}\rvert\leqslant\mu^{n}$.

The base case $n=1$ is trivially true, so fix $n\geqslant 2$. Let $(\mathcal{A},\mathcal{B})$ be a $k$-uniform recovering pair over a ground set $X$ of size $n$ for some $1\leqslant k\leqslant n$, and for every $i\in X$, consider the restrictions $\mathcal{A}_{i}$, $\mathcal{A}^{\prime}_{i}$, $\mathcal{B}_{i}$ and $\mathcal{B}^{\prime}_{i}$, along with the parameters $a_{i}$, $a^{\prime}_{i}$, $b_{i}$ and $b^{\prime}_{i}$ as defined in (3.1).

Note that that $\theta<\mu_{\operatorname{can}}$, $0\leqslant\alpha\leqslant 1/2$, $1/(1-2\alpha)\leqslant\mu$ and $\theta^{\alpha}\mu_{\operatorname{can}}^{1-\alpha}\leqslant\mu$. If $k\leqslant\alpha n$, we proceed as in [Janzer2018-pb]. Indeed, since $(\mathcal{A},\mathcal{B})$ is $k$-uniform, we have

[TABLE]

so there is $i\in X$ with $a_{i}+b_{i}\leqslant 2\alpha$, and hence with $a^{\prime}_{i}+b^{\prime}_{i}\geqslant 2-2\alpha$. This implies that $a^{\prime}_{i}b^{\prime}_{i}\geqslant 1-2\alpha\geqslant 0$, as $\alpha\leqslant 1/2$. By ˜3.1, we have that $(\mathcal{A}^{\prime}_{i},\mathcal{B}^{\prime}_{i})$ is a recovering pair over $X\setminus\{i\}$. The induction hypothesis then gives that $\lvert\mathcal{A}^{\prime}_{i}\rvert\lvert\mathcal{B}^{\prime}_{i}\rvert\leqslant\mu^{n-1}$, thus

[TABLE]

since $1/(1-2\alpha)\leqslant\mu$, and we are done.

Henceforth, we assume $k\geqslant\alpha n$, and moreover, that $a^{\prime}_{i}b^{\prime}_{i}\leqslant 1-2\alpha$ holds for all $i\in X$. Assume further, that for every $A\in\mathcal{A}$, $B\in\mathcal{B}$ and for every $P_{1}\subseteq A\setminus B$ and $P_{2}\subseteq B\setminus A$, we have

[TABLE]

Therefore, by Corollary˜4.2 we get that

[TABLE]

where

[TABLE]

and

[TABLE]

Recall that $(1-a_{i})(1-b_{i})=a^{\prime}_{i}b^{\prime}_{i}\leqslant 1-2\alpha$ for all $i\in X$. Given this restriction, we use the following claim to bound form above the value of the function $g(a_{i},b_{i},\theta)$. As the proof of ˜5.1 is a straightforward but technical optimization argument, we include it as an appendix.

Claim 5.1**.**

For $x,y\in(0,1)$ and $\theta=2.222$ we have

[TABLE]

By (5.1) together with ˜5.1 we get

[TABLE]

proving the statement.

Hence, we may assume, without loss of generality, that there are sets $A\in\mathcal{A}$, $B\in\mathcal{B}$ and $P\subseteq A\setminus B$ such that

[TABLE]

˜3.3 implies that the filtered pair $(\mathcal{A}_{A,A\setminus B,P},\mathcal{B}_{A,A\setminus B})$ is cancellative over $X\setminus A$. Since $(\mathcal{A},\mathcal{B})$ is $k$-uniform and $|X|=n$, we have $\lvert X\setminus A\rvert=n-k$, and thus, by the induction hypothesis,

[TABLE]

Therefore, we get

[TABLE]

Since $k\geqslant\alpha n$ and $\theta<\mu_{\operatorname{can}}$, we get

[TABLE]

finishing the proof. ∎

6. One-sided pairs

As stated in the introduction, in establishing the lower bound Theorem˜1.4, we mostly follow Tolhuizen’s linear coding construction of a large cancellative pair [Tolhuizen2000-ww]. The proof of the upper bound was given in an argument by Ahlswede and Simonyi for one-sided recovering pairs from [Ahlswede1994-st], which we repeat here for the convenience of the reader.

Proof of Theorem˜1.4.

Let $(\mathcal{A},\mathcal{B})$ be a left-cancellative pair on $[n]$. For fixed $B\in\mathcal{B}$, each set $A\in\mathcal{A}$ has to have a different intersection with $[n]\setminus B$, whence $\lvert\mathcal{A}\rvert\leqslant 2^{n-\lvert B\rvert}$. Thus, we obtain

[TABLE]

For the lower bound, we construct a left-cancellative pair $(\mathcal{A},\mathcal{B})$ with $|\mathcal{A}||\mathcal{B}|\geqslant 3^{n-o(n)}$. Given $n$, let $k\in[n]$ and consider a matrix $M\in\mathbb{F}_{2}^{n\times(n-k)}$. We call $S\in[n]^{(n-k)}$ an information set if the square submatrix of $M$ given by the rows indexed by $S$ is invertible. Denote the family of information sets of $M$ by $\mathcal{I}(M)$. Associating vectors in $\mathbb{F}_{2}^{n}$ with subsets of $[n]$ in the usual manner, we let $\mathcal{A}_{M}$ be the image of $M$. Then, letting $\mathcal{B}_{M}=\{[n]\setminus S\colon S\in\mathcal{I}(M)\}$, the pair $(\mathcal{A}_{M},\mathcal{B}_{M})$ is left-cancellative.

Indeed, let $A,A^{\prime}\in\mathcal{A}_{M}$, $B\in\mathcal{B}_{M}$ and let $v_{A},v_{A^{\prime}}\in\mathbb{F}_{2}^{n-k}$ be vectors that give rise to $A$ and $A^{\prime}$, respectively. If $A\setminus B=A^{\prime}\setminus B$, then the projections of $Mv_{A}$ and $Mv_{A^{\prime}}$ to $\mathbb{F}_{2}^{[n]\setminus B}$ are equal. However, these projections are the same as the images of $v_{A}$ and $v_{A^{\prime}}$ under multiplication with the submatrix of $M$ given by the rows indexed by $[n]\setminus B$. Since this submatrix is invertible, the above can only happen when $v_{A}=v_{A^{\prime}}$, meaning that $A=A^{\prime}$.

It thus remains only to find $k$ and $M$ so that $\lvert\mathcal{A}_{M}\rvert\cdot\lvert\mathcal{B}_{M}\rvert=2^{n-k}\lvert\mathcal{I}(M)\rvert$ is at least $3^{n-o(n)}$. By considering the expected number of information sets for a uniformly random matrix, Tolhuizen showed (see Lemma 1 in [Tolhuizen2000-ww]), that for all $k$ there exists $M$ such that

[TABLE]

Thus, for every $k\in[n]$, there exists a left-cancelling pair of size at least

[TABLE]

Taking $k=\lfloor n/3\rfloor$ and applying Stirling’s formula (or by recognizing that $2^{2n/3}\binom{n}{n/3}$ is the largest term of the binomial expansion in (6.1)), we see that the above is $\Omega(3^{n}/\sqrt{n})=3^{n-o(n)}$, as desired. ∎

Acknowledgements.

The first author was funded by UK Research and Innovation grant MR/W007320/2, and the second author was partially supported by ERC Starting Grant 101163189 and UKRI Future Leaders Fellowship MR/X023583/1.

Appendix A Proof of ˜5.1

Recall that $f(x,y)\mathrel{\coloneqq}x(1-y)h(x)+h(x(1-y))$ and $g(x,y,t)\mathrel{\coloneqq}f(x,y)+f(y,x)-(x+y)\log_{2}t$. We restate the claim proved in this appendix.

Claim 5.1.

For $x,y\in(0,1)$ and $\theta=2.222$ we have

[TABLE]

Proof of ˜5.1.

We show that $\theta$ satisfies conditions of a different optimisation problem. Let $h^{*}(z)$ for $0<z<1$ be defined as follows.

[TABLE]

and note that $h(z)\leqslant h^{*}(z)$ for any $z\in[0,1]$. Further define

[TABLE]

and note that, given $t$, we have $g(x,y,t)\leqslant g^{*}(x,y,t)$ for any $x,y\in(0,1)$. Hence, it is sufficient to show that for any $x,y\in(0,1)$ we have

[TABLE]

It is routine to show that $g^{*}(x,y,\theta)$ is Lipschitz with a constant $25$ in either variable. Hence, if we fix an integer $k$ and maximise $g^{*}(i/k,j/k,\theta)$ over $0\leqslant i,j\leqslant k$, the maximum of $g^{*}(x,y,\theta)$ for all $0<x,y<1$ can be at most $25/k$ bigger. Choosing $k=30000$ and performing a computer check, we can hence bound $g^{*}(x,y,\theta)$ from above for all $0<x,y<1$ by $1.1687+\frac{25}{30000}<1.1696$, while $\log_{2}(2.2499)>1.1698$. The statement follows. ∎