Proof of an entropy conjecture of Leighton and Moitra

TL;DR

This paper proves a conjecture by Leighton and Moitra, establishing an entropy bound for probability distributions on permutations that favor certain tournament arcs, with improvements for transitive tournaments.

Contribution

The paper confirms the conjecture, providing a general proof and a sharper entropy bound for transitive tournaments, advancing understanding of permutation distributions in combinatorics.

Findings

Entropy of distributions exceeds a threshold if they favor tournament arcs.

For transitive tournaments, a shorter proof with a better entropy bound is provided.

The entropy bound depends on a fixed positive parameter related to arc bias.

Abstract

We prove the following conjecture of Leighton and Moitra. Let $T$ be a tournament on $[n]$ and $S_n$ the set of permutations of $[n]$. For an arc $uv$ of $T$, let $A_{uv}=\{\sigma \in S_n \, : \, \sigma(u)<\sigma(v) \}$. $\textbf{Theorem.}$ For a fixed $\varepsilon>0$, if $\mathbb{P}$ is a probability distribution on $S_n$ such that $\mathbb{P}(A_{uv})>1/2+\varepsilon$ for every arc $uv$ of $T$, then the binary entropy of $\mathbb{P}$ is at most $(1-\vartheta_{\varepsilon})\log_2 n!$ for some (fixed) positive $\vartheta_\varepsilon$. When $T$ is transitive the theorem is due to Leighton and Moitra; for this case we give a short proof with a better $\vartheta_\varepsilon$.

Equations

$$uv$$

$$H(\sigma )\le (1-\vartheta )\log n!,$$

$$H(\sigma )\le (1-{\epsilon }^2/8)n\log n.$$

$$D\cap (X\times Y)$$

$$\text{fit}(\sigma ,D)=|D\cap T_{\sigma }|-|D^r\cap T_{\sigma }|$$

$$|d_H(X^{\prime },Y^{\prime })-d_H(X,Y)|<\delta$$

$$T\cap (X\times Y)$$

$$\mu ({\cap }_{i\in I}{A}_i)<\exp [-\sum _{i\in I}\Omega (\Vert S_i\Vert )]$$

$$([(2s-2)2^{-j}n+1,(2s-1)2^{-j}n],[(2s-1)2^{-j}n+1,2s{2}^{-j}n]),$$

$$\Lambda =\sum _{i=1}^m|V_i|$$

$$D(p\Vert q)=\sum p_i\log (q_i/p_i)\le 0$$

$$\sum (u_i/s)d_i\log b$$

$$\mathbb{E}[\text{fit}(\sigma ,S_i)]\le \mathbb{P}(A_i)|S_i|+(1-\mathbb{P}(A_i))\epsilon |S_i|\le (\mathbb{P}(A_i)+\epsilon )|S_i|$$

$$H(\sigma )$$

$$=1+\log n!+\mathbb{P}(Q)\log \mu (Q)\le 1+\log n!+(\epsilon /2)\log \mu (Q)$$

$$\mathcal{J}=\{I\subseteq [m]:{\sum }_{i\in I}|V_i|\ge \epsilon \Lambda /2\}$$

$$b={\epsilon }^2\delta {\beta }^3/33$$

$$\mu (A_I)\le e^{-b\epsilon \Lambda /2},$$

$$\log \mu (Q)=\log \mu ({\cup }_{I\in \mathcal{J}}{A}_I)\le \log |\mathcal{J}|-(b\epsilon \Lambda \log e)/2\le n-(b\epsilon \Lambda \log e)/2,$$

$$(1-\epsilon c/2)\log n!,$$

$$\text{fit}(\tau ,D)<\epsilon |L||R|/4$$

$$|X\cap I_j|=(\gamma \lambda \pm \zeta )l\forall j\in [r],$$

$$|\text{fit}(\tau ,D)|\le \sum _{1\le i<j\le r}||D\cap (L_i\times R_j)|-|D\cap (L_j\times R_i)||+\gamma (1-\gamma )\lambda l^2.$$

$$\sum {\gamma }_j(1-{\gamma }_j)\le \sum {\gamma }_j-(\sum {\gamma }_j{)}^2/r=(\gamma -{\gamma }^2)/\lambda$$

$$\sum |L_j||R_j|=\sum {\gamma }_j(1-{\gamma }_j){\lambda }^2{l}^2\le \gamma (1-\gamma )\lambda l^2.$$

$$|D\cap (L_i\times R_j)|=(d\pm \delta )|L_i||R_j|,$$

$$[(d+\delta )(\gamma \lambda +\zeta )((1-\gamma )\lambda +\zeta )-(d-\delta )(\gamma \lambda -\zeta )((1-\gamma )\lambda -\zeta )]l^2$$

$$=2[\lambda \zeta d+\delta (\gamma (1-\gamma ){\lambda }^2+{\zeta }^2)]l^2$$

$$\left\{2\left(\genfrac{}{}{0ex}{}{r}{2}\right)[\lambda \zeta d+\delta (\gamma (1-\gamma ){\lambda }^2+{\zeta }^2)]+\gamma (1-\gamma )\lambda \right\}{l}^2<\epsilon \gamma (1-\gamma )l^2/4.$$

$$\Pr(\mbox{$\sigma$ is unsafe for $D$})<2r\exp[-2\zeta^{2}l/\lambda].$$

$$B_{i}=\{\sigma\in\mathfrak{S}_{n}\,:\,\mbox{$\sigma$ is unsafe for $S_{i}$}\}$$

Full text

Theorem. For a fixed $\varepsilon>0$, if $\mathbb{P}$ is a probability distribution on $\mathfrak{S}_{n}$ such that $\mathbb{P}(A_{uv})>1/2+\varepsilon$ for every arc $uv$ of $T$, then the binary entropy of $\mathbb{P}$ is at most $(1-\vartheta_{\varepsilon})\log_{2}n!$ for some (fixed) positive $\vartheta_{\varepsilon}$.

When $T$ is transitive the theorem is due to Leighton and Moitra; for this case we give a short proof with a better $\vartheta_{\varepsilon}$.

1 Introduction

In what follows we use $\log$ for $\log_{2}$ and $H(\cdot)$ for binary entropy. The purpose of this note is to prove the following natural statement, which was conjectured by Tom Leighton and Ankur Moitra [6] (and told to the third author by Moitra in 2008).

Theorem 1**.**

Let $T$ be a tournament on $[n]$ and $\sigma$ a random (not necessarily uniform) permutation of $[n]$ satisfying:

[TABLE]

Then

[TABLE]

where $\vartheta>0$ depends only on $\varepsilon$.

(We will usually think of permutations as bijections $\sigma:[n]\rightarrow[n]$). The original motivation for Leighton and Moitra came mostly from questions about sorting partially ordered sets; see [6] for more on this.

For the special case of transitive $T$, Theorem 1 was proved in [6] with $\vartheta_{\varepsilon}=C\varepsilon^{4}$. Note that for a typical (a.k.a. random) $T$, the conjecture’s hypothesis is unachievable, since, as shown long ago by Erdős and Moon [2], no $\sigma$ agrees with $T$ on more than a $(1/2+o(1))$-fraction of its arcs. In fact, it seems natural to expect that transitive tournaments are the worst instances, being the ones for which the hypothesized agreement is easiest to achieve. From this standpoint, what we do here may be considered somewhat unsatisfactory, as our $\vartheta$’s are quite a bit worse than those in [6]. For transitive $T$ it’s easy to see [6, Claim 4.14] that one can’t take $\vartheta$ greater than $2\varepsilon$, which seems likely to be close to the truth. We make some progress on this, giving a surprisingly simple proof of the following improvement of [6].

Theorem 2**.**

For $T$, $\mathbb{P}$, $\sigma$ as Theorem 1 with $T$ transitive,

[TABLE]

The proof of Theorem 1 is given in Section 3 following brief preliminaries in Section 2. The underlying idea is similar to that of [6], which in turn was based on the beautiful tournament ranking bound of W. Fernandez de la Vega [1]; see Section 3 (end of “Sketch”) for an indication of the relation to [6]. Theorem 2 is proved in Section 4.

2 Preliminaries

Usage

In what follows we assume $n$ is large enough to support our arguments and pretend all large numbers are integers.

As usual $G[X]$ is the subgraph of $G$ induced by $X$; we use $G[X,Y]$ for the bipartite subgraph induced (in the obvious sense) by disjoint $X$ and $Y$. For a digraph $D$, $D[X]$ and $D[X,Y]$ are used analogously. For both graphs and digraphs, we use $|\cdot|$ for number of edges (or arcs).

Also as usual, the density of a pair $(X,Y)$ of disjoint subsets of $V(G)$ is $d(X,Y)=d_{G}(X,Y)=|G[X,Y]|/(|X||Y|)$, and we extend this to bipartite digraphs $D$ in which

[TABLE]

For a digraph $D$, $D^{r}$ is the digraph gotten from $D$ by reversing its arcs.

Write $\mathfrak{S}_{n}$ for the set of permutations of $[n]$. For $\sigma\in\mathfrak{S}_{n}$, we use $T_{\sigma}$ for the corresponding (transitive) tournament on $[n]$ (that is, $uv\in T_{\sigma}$ iff $\sigma(u)<\sigma(v)$) and for a digraph $D$ (on $[n]$) define

[TABLE]

(e.g. when $D$ is a tournament, this is a measure of the quality of $\sigma$ as a ranking of $D$).

Regularity

Here we need just Szemerédi’s basic notion [7] of a regular pair and a very weak version (Lemma 3) of his Regularity Lemma. As usual a bipartite graph $H$ on disjoint $X\cup Y$ is $\delta$-regular if

[TABLE]

whenever $X^{\prime}\subseteq X$, $Y^{\prime}\subseteq Y$, $|X^{\prime}|>\delta|X|$ and $|Y^{\prime}|>\delta|Y|$, and we extend this in the obvious way to the situation in (3). It is easy to see that if a bigraph $H$ is $\delta$-regular then its bipartite complement is as well; this implies that for a tournament $T$ on $[n]$ and $X$, $Y$ disjoint subsets of $[n]$,

[TABLE]

The following statement should perhaps be considered folklore, though similar results were proved by János Komlós, circa 1991 (see [5, Sec. 7.3]).

Lemma 3**.**

For each $\delta>0$ there is a $\beta>2^{-\delta^{-O(1)}}$ such that for any bigraph $H$ on $X\cup Y$ with $|X|,|Y|\geq n$, there is a $\delta$-regular pair $(X^{\prime},Y^{\prime})$ with $X^{\prime}\subseteq X,Y^{\prime}\subseteq Y$ and each of $|X^{\prime}|,|Y^{\prime}|$ at least $\beta n$.

Corollary 4**.**

For each $\delta>0$, $\beta$ as in Lemma 3 and digraph $G=(V,E)$, there is a partition $L\cup R\cup W$ of $V$ such that $E\cap(L\times R)$ is $\delta$-regular and $\min\{|L|,|R|\}\geq\beta|V|/2.$

Proof.

Let $X\cup Y$ be an (arbitrary) equipartition of $V$ and apply Lemma 3 to the undirected graph $H$ underlying the digraph $G\cap(X\times Y)$.∎

3 Proof of Theorem 1

We now assume that $\sigma$ drawn from the probability distribution $\mathbb{P}$ on $\mathfrak{S}_{n}$ satisfies (1) and try to show (2) (with $\vartheta$ TBA). We use ${\mathbb{E}}$ for expectation w.r.t. $\mathbb{P}$ and $\mu$ for uniform distribution on $\mathfrak{S}_{n}$.

Sketch and connection with [6]

We will produce $S_{1},\ldots,S_{m}\subseteq T$ with $S_{i}\subseteq L_{i}\times R_{i}$ for some disjoint $L_{i},R_{i}\subseteq[n]$, satisfying:

• (i)

with $\|S_{i}\|:=\min\{|L_{i}|,|R_{i}|\}$, $\sum\|S_{i}\|=\Omega(n\log n)$ (where the implied constant depends on $\varepsilon$);

• (ii)

each $S_{i}$ is $\delta$-regular (with $\delta=\delta_{\varepsilon}$ TBA);

• (iii)

for all $i<j$, either $(L_{i}\cup R_{i})\cap(L_{j}\cup R_{j})=\varnothing$ or $L_{j}\cup R_{j}$ is contained in one of $L_{i},R_{i}$ (note this implies the $S_{i}$’s are disjoint).

Let $A_{i}=\{\text{fit}(\sigma,S_{i})>\varepsilon|S_{i}|\}$ and $Q=\{\sum\{\|S_{i}\|:A_{i}~{}\text{occurs}\}=\Omega(n\log n)\}$. The main points are then:

• (a)

$\mathbb{P}(Q)$ is bounded below by a positive function of $\varepsilon$. (This is just (i) together with a couple applications of Markov’s Inequality.)

• (b)

Regularity of $S_{i}$ implies $\mu(A_{i})\leq\exp[-\Omega(\|S_{i}\|)]$.

• (c)

Under (iii), for any $I\subseteq[m]$,

[TABLE]

(a weak version of independence of the $A_{i}$’s under $\mu$).

And these points easily combine to give (2) (see (3) and (8)).

For the transitive case in [6] most of this argument is unnecessary; in particular, regularity disappears and there is a natural decomposition of $T$ into $S_{i}$’s: Supposing $T=\{ab:a<b\}$ and (for simplicity) $n=2^{k}$, we may take the $S_{i}$’s to be the sets $L_{i}\times R_{i}$ with $(L_{i},R_{i})$ running over pairs

[TABLE]

with $j\in[k]$ and $s\in[2^{j-1}]$. (As mentioned earlier, this decomposition of the (identity) permutation $(1,\ldots,n)$ also provides the framework for [1].) After some translation, our argument (really, a fairly small subset thereof) then specializes to essentially what’s done in [6].∎

Set $\delta=.03\varepsilon$ and let $\beta$ be half the $\beta$ of Lemma 3 and Corollary 4. We use the corollary to find a rooted tree $\cal T$ each of whose internal nodes has degree (number of children) 2 or 3, together with disjoint subsets $S_{1},S_{2},\ldots,S_{m}$ of (the arc set of) $T$, corresponding to the internal nodes of ${\cal T}$. The nodes of $\cal T$ will be subsets of $[n]$ (so the size, $|U|$, of a node $U$ is its size as a set).

To construct ${\cal T}$, start with root $V_{1}=[n]$ and repeat the following for $k=1,\ldots$ until each unprocessed node has size less than (say) $t:=\sqrt{n}$. Let $V_{k}$ be an unprocessed node of size at least $t$ and apply Corollary 4 to $T[V_{k}]$ to produce a partition $V_{k}=L_{k}\cup R_{k}\cup W_{k}$, with $|L_{k}|,|R_{k}|>\beta|V_{k}|$ and $S_{k}:=T\cap(L_{k}\times R_{k})$ $\delta$-regular of density at least 1/2. (Note (4) says we can reverse the roles of $L_{k}$ and $R_{k}$ if the density of $T\cap(L_{k}\times R_{k})$ is less than 1/2.) Add $L_{k},R_{k},W_{k}$ to ${\cal T}$ as the children of $V_{k}$ and mark $V_{k}$ “processed.” (Note the $V_{k}$’s are the internal nodes of ${\cal T}$; nodes of size less then $t$ are not processed and are automatically leaves. Note also that there is no restriction on $|W_{k}|$ and that, for $k>1$, $V_{k}$ is equal to one of $L_{i}$, $R_{i}$, $W_{i}$ for some $i<k$.)

Let $m$ be the number of internal nodes of ${\cal T}$ (the final tree). Note that the leaves of ${\cal T}$ have size at most $t$ and that the $S_{i}$’s satisfy (ii) and (iii) of the proof sketch; that they also satisfy (i) is shown by the next lemma.

Set

[TABLE]

this quantity will play a central role in what follows.

Lemma 5**.**

$\Lambda\geq\frac{1}{2}n\log_{3}n$* *

Proof.

This will follow easily from the next general (presumably known) observation, for which we assume ${\cal T}$ is a tree satisfying:

• •

the nodes of ${\cal T}$ are subsets of $S$, an $s$-set which is also the root of ${\cal T}$;

• •

the children of each internal node $U$ of ${\cal T}$ form a partition of $U$ with at most $b$ blocks;

• •

the leaves of ${\cal T}$ are $U_{1},\ldots,U_{r}$, with $|U_{i}|=u_{i}\leq t$ (any $t$) and depth $d_{i}$.

Lemma 6**.**

With the setup above, $\sum u_{i}d_{i}\geq s\log_{b}(s/t)$.

(Of course this is exact if ${\cal T}$ is the complete $b$-ary tree of depth $d$ and all leaves have size $2^{-b}s$).

Proof.

Recall that the relative entropy between probability distributions $p$ and $q$ on $[r]$ is

[TABLE]

(the inequality given by the concavity of the logarithm). We apply this with $p_{i}=u_{i}/s$ and $q_{i}$ the probability that the ordinary random walk down the tree ends at $u_{i}$. In particular $q_{i}\geq b^{-d_{i}}$, which, with nonpositivity of $D(p\|q)$ and the assumption $u_{i}\leq t$, gives

[TABLE]

The lemma follows.∎

This gives Lemma 5 since $\sum|V_{i}|=\sum_{U}|U|d(U)$, with $U$ ranging over leaves of ${\cal T}$ (and $d(\cdot)$ again denoting depth).∎

Lemma 7**.**

The number m of internal nodes of ${\cal T}$ is less than $n$.

Proof.

A straightforward induction shows that the number of leaves of a rooted tree is $1+\sum(b(w)-1)$, where $w$ ranges over internal nodes and $b$ denotes number of children. The lemma follows since here the number of leaves is at most $n$ (actually at most $3\sqrt{n}$) and each $d(w)$ is at least 2. ∎

Recalling that $A_{i}=\{\sigma\in\mathfrak{S}_{n}\,:\text{fit}(\sigma,S_{i})\geq\varepsilon|S_{i}|\}$ and that ${\mathbb{E}}$ refers to $\mathbb{P}$, we have $\mathbb{E}[\text{fit}(\sigma,S_{i})]\geq 2\varepsilon|S_{i}|,$ which with

[TABLE]

gives $\mathbb{P}(A_{i})\geq\varepsilon$ (essentially Markov’s Inequality applied to $|S_{i}|-\text{fit}(\sigma,S_{i})$).

Set $\xi_{i}=|V_{i}|\boldsymbol{1}_{A_{i}}$ and $\xi=\sum_{i}\xi_{i}$, and let $Q$ be the event $\{\xi\geq\varepsilon\Lambda/2\}$. Then $\mathbb{E}[\xi_{i}]=|V_{i}|\mathbb{P}(A_{i})\geq\varepsilon|V_{i}|,$ implying $\mathbb{E}[\xi]=\sum\mathbb{E}[\xi_{i}]\geq\varepsilon\Lambda,$ and (since $\xi_{i}\leq|V_{i}|$) $\xi\leq\Lambda$; so using Markov’s Inequality as above gives $\mathbb{P}(Q)\geq\varepsilon/2$.

Thus, with $\sigma$ chosen from $\mathfrak{S}_{n}$ according to $\mathbb{P}$, we have

[TABLE]

(recall $\mu$ is the uniform measure on $\mathfrak{S}_{n}$).

Let

[TABLE]

and, for $I\in\cal J$, let $A_{I}=\cap_{i\in I}A_{i}$. Set

[TABLE]

(see (12) for the reason for the choice of $b$). We will show, for each $I\in{\cal J}$,

[TABLE]

which implies

[TABLE]

the second inequality following from $|{\cal J}|\leq 2^{m}$ together with Lemma 7. With c=\varepsilon^{\color[rgb]{0,0,0}{3}}\delta\beta^{3}/150<(b\varepsilon\log_{3}e)/4, this bounds (for large $n$) the r.h.s. of (3) by

[TABLE]

which proves Theorem 1 with $\vartheta=\varepsilon^{4}\delta\beta^{3}/300=\exp[-\varepsilon^{-O(1)}]$. ∎

The rest of our discussion is devoted to the proof of (8). For a digraph $D\subseteq L\times R$ with $L,R$ disjoint subsets of $V$, say a pair $(X,Y)$ of disjoint subsets of $[n]$ with $|X|=|L|$, $|Y|=|R|$ is safe for $D$ if

[TABLE]

for every bijection $\tau:L\cup R\rightarrow X\cup Y$ with $\tau(L)=X$ (where $\text{fit}(\tau,D)$ has the obvious meaning). We also say $\sigma\in\mathfrak{S}_{n}$ is safe for $D$ if $(\sigma(L),\sigma(R))$ is. Note that since $S_{i}$ has density at least 1/2 in $L_{i}\times R_{i}$, the $\sigma$’s in $A_{i}$ are unsafe for $S_{i}$.

Lemma 8**.**

Assume the above setup with $|L|+|R|=l$ and $|L|=\gamma l$, and set $\lambda=2\delta$ and $\zeta=\varepsilon\delta\gamma(1-\gamma)/4$. Let $I_{1}\cup\cdots\cup I_{r}$ be the natural partition of $X\cup Y$ into intervals of size $\lambda l$. If D is $\delta$-regular and

[TABLE]

then $(X,Y)$ is safe for $D$.

(Of course an interval of $Z=\{i_{1}<\cdots<i_{u}\}$ is one of the sets $\{i_{s},\ldots,i_{s+t}\}$.)

Proof.

For $\tau$ as in the line after (9), let $L_{j}=L\cap\tau^{-1}(I_{j})$ and $R_{j}=R\cap\tau^{-1}(I_{j})$ ($j\in[r]$). Then

[TABLE]

Here the last term is an upper bound on the contribution of pairs contained in the $I_{j}$’s: if $|L_{j}|=\gamma_{j}|I_{j}|=\gamma_{j}\lambda l$ (so $|R_{j}|=(1-\gamma_{j})\lambda l$ and $\sum\gamma_{j}=\gamma/\lambda$), then

[TABLE]

gives

[TABLE]

On the other hand, regularity and (10) (which implies $|L_{i}|>\delta|L|$ ($=\delta\gamma l$) since $\gamma\lambda-\zeta>\gamma\delta$, and similarly $|R_{i}|>\delta|R|$) give, for all $i\neq j$,

[TABLE]

where $d$ is the density of $D$. Combining this with (10) bounds each of the summands in (11) by

[TABLE]

[TABLE]

and the r.h.s. of (11) by

[TABLE]

(The main term on the l.h.s. is the one with $\lambda\zeta d$, which, since $r^{-1}=\lambda=2\delta$, is less than half the r.h.s. The second and third terms are much smaller (the second since $\delta$ is much smaller than $\varepsilon$).) ∎

Corollary 9**.**

For D and parameters as in Lemma 8, and $\sigma$ uniform from $\mathfrak{S}_{n}$,

[TABLE]

Proof.

Let $(X,Y)=(\sigma(L),\sigma(R))$. Once we’ve chosen $X\cup Y$ (determining $I_{1},\ldots,I_{r}$), $2\exp[-2\zeta^{2}l/\lambda]$ is the usual Hoeffding bound [3, Eq. (2.3)] on the probability that $X$ violates (10) for a given $j$. (The bound may be more familiar when elements of $X\cup Y$ are in $X$ independently, but also applies to the hypergeometric r.v. $|X\cap I_{j}|$; see e.g. [4, Thm. 2.10 and (2.12)].) ∎

Proof of (8)..

Let

[TABLE]

and $B_{I}=\cap_{i\in I}B_{i}$. Then $A_{i}\subseteq B_{i}$ (as noted above) and (therefore) $A_{I}\subseteq B_{I}$. Moreover—perhaps the central point—the $B_{i}$’s are independent, since $B_{i}$ depends only on the relative positions of $\sigma(L_{i})$ and $\sigma(R_{i})$ within $\sigma(V_{i})$.

On the other hand, Corollary 9, applied with $D=S_{i}$ (so $L=L_{i}$, $R=R_{i}$, $l=|L_{i}|+|R_{i}|$ and $\gamma=|L_{i}|/l\in(\beta,1-\beta)$) gives

[TABLE]

(Recall $b$ was defined in (7); since we assume $|V_{i}|$ is large ($|V_{i}|>t=\sqrt{n}$), the choice leaves a little room to absorb the $2r$.) And of course (12) and the independence of the $B_{i}$’s give (8). ∎

4 Back to the transitive case

Theorem 2 is an easy consequence of the next observation.

Lemma 10**.**

Let ${\bf Y}$ a random $m$-subset of $[2m]$ satisfying

[TABLE]

Then $H({\bf Y})<(1-\varepsilon^{2}/8)2m$.

To get Theorem 2 from this, let $T=\{ab:a<b\}$ and, for simplicity, $n=2^{k}$, and decompose $T=\bigcup(L_{i}\times R_{i})$ as in (5). For each $i$, say with $|L_{i}|$ ($=|R_{i}|$) $=m_{i}$, let ${\bf Y}_{i}\subseteq[2m_{i}]$ consist of the indices of positions within $\sigma(L_{i}\cup R_{i})$ occupied by $\sigma(R_{i})$; that is, if $\sigma(L_{i}\cup R_{i})=\{j_{1}<\cdots<j_{2m_{i}}\}$, then ${\bf Y}_{i}=\{l:j_{l}\in\sigma(R_{i})\}$. Then Lemma 10 (its hypothesis provided by (1)) gives

[TABLE]

so, since $\sigma$ is determined by the ${\bf Y}_{i}$’s, we have

[TABLE]

Remark. Note that the $\Omega(\varepsilon^{2})$ of Theorem 2 is the best one can do without more fully exploiting (1) (that is, beyond (13) for the $(L_{i},R_{i},Y_{i})$’s, which is all we are using).

Proof of Lemma 10..

For $a\in[2m]$, set $\mathbb{P}(a\in{\bf Y})=1/2+\delta_{a}$. Then

[TABLE]

(where the 2 could actually be $2\log e$); so it is enough to show

[TABLE]

For a given $m$-subset $Y$ of $[2m]$, we have

[TABLE]

(the first sum counts pairs $(a,b)$ with $a<b$ and $b\in Y$, and ${{m}\choose{{2}}}$ is the number of such pairs with $a$ also in $Y$); so we have

[TABLE]

implying $\sum\delta_{b}b>\varepsilon m^{2}$. Combining this with $2m\sum_{\delta_{b}>0}\delta_{b}\geq\sum\delta_{b}b$, we have $\sum_{\delta_{b}>0}\delta_{b}>\varepsilon m/2$ and then, using Cauchy-Schwarz,

[TABLE]