%! TEX root = EMC.tex
% vim: tw=50
% 06/02/2025 09AM

  \vspace{-1em}
  \textbf{Key idea:} we now regard $x_1, \ldots,
  x_n$ as a \emph{random} enumeration of $X$ and
  take the average.

  For each $x \in X$, define the
  \emph{contribution} of $x$ to be
  \[
    \log(d_{x_1, \ldots, x_{i - 1}}^\sigma(x_i))
  \]
  where $x_i = x$ (note that this ``contribution''
  is a random variable rather than a constant).

  We shall now fix $\sigma$. Let the neighbours of
  $x$ be $y_1, \ldots, y_k$.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/0b09257615d24c01.png}
  \end{center}
  Then one of the $y_j$ will be $\sigma(x)$, say
  $y_h$. Note that $d_{x_1, \ldots, x_{i -
  1}}^\sigma(x_i)$ (given that $x_i = x$) is
  \[
    d(x) - |\{j : \text{$\sigma^{-1}(y_j)$ comes
    earlier than $x = \sigma^{-1}(y_h)$}\}|
  .\]
  All positions of $\sigma^{-1}(y_h)$ are equally
  likely, so the average contribution of $x$ is
  \[
    \frac{1}{d(x)} \left( \log d(x) + \log (d(x) -
    1) + \cdots + \log 1 \right)
    = \frac{1}{d(x)} \log (d(x)!)
  .\]
  By linearity of expectation,
  \[
    \ent \sigma \le \sum_{x \in X} \frac{1}{d(x)}
    \log(d(x)!)
  ,\]
  so the number of matchings is at most
  \[
    \prod_{x \in X} (d(x)!)^{\frac{1}{d(x)}}
    .
    \qedhere
  \]
\end{proof}

\begin{fcdefnstar}[$1$-factor]
  \glsnoundefn{onefac}{$1$-factor}{$1$-factors}%
  Let $G$ be a graph with $2n$ vertices. A
  \emph{$1$-factor} in $G$ is a collection of $n$
  disjoint edges.
\end{fcdefnstar}

\begin{fcthm}[Kahn-Lovasz]
  \label{thm:3.2}
  % [Kahn-Lov\'asz]
  Assuming:
  - $G$ a graph with $2n$ vertices
  Then:
  the number of \glspl{onefac} in $G$ is at most
  \[
    \prod_{x \in V(G)} (d(x)!)^{\frac{1}{2d(x)}}
  .\]
\end{fcthm}

\begin{proof}[Proof (Alon, Friedman)]
  Let $\mathcal{M}$ be the set of $1$-factors of
  $G$, and let $(M_1, M_2)$ be a uniform random
  element of $\mathcal{M}^2$. For each $M_1,
  M_2$, the union $M_1 \cup M_2$ is a collection
  of disjoint edges and even cycles that covers
  all the vertices of $G$.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/a92d21cb6dfe425a.png}
  \end{center}
  Call such a union a \emph{cover of $G$ by edges
  and even cycles}.

  If we are given such a cover, then the number of
  pairs $(M_1, M_2)$ that could give rise to it is
  $2^k$, where $k$ is the number of even cycles.

  Now let's build a bipartite graph $G_2$ out of
  $G$. $G_2$ has two vertex sets (call them $V_1,
  V_2$), both copies of $V(G)$. Join $x \in V_1$
  to $y \in V_2$ if and only if $xy \in E(G)$.

  For example:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/446a168023094c19.png}
  \end{center}
  By \nameref{thm:3.1}, the number of perfect
  matchings in $G_2$ is $\le \prod_{x \in V(G)}
  (d(x)!)^{\frac{1}{d(x)}}$. Each matching gives a
  permutation $\sigma$ of $V(G)$, such that
  $x\sigma(x) \in E(G)$ for every $x \in V(G)$.

  Each such $\sigma$ has a cycle decomposition,
  and each cycle gives a cycle in $G$. So $\sigma$
  gives a cover of $V(G)$ by isolated vertices,
  edges and cycles.

  Given such a cover with $k$ cycles, each edge
  can be directed in two ways, so the number of
  $\sigma$ that give rise to is is $2^k$, where
  $k$ is the number of cycles.

  So there is an injection from $\mathcal{M}^2$ to
  the set of matchings of $G_2$, since every cover
  by edges and even cycles is a cover by vertices,
  edges and cycles.

  So
  \[
    |\mathcal{M}|^2 \le \prod_{x \in V(G)}
    (d(x)!)^{\frac{1}{d(x)}}
    .
    \qedhere
  \]
\end{proof}

\newpage
\section{Shearer's lemma and applications}

\begin{notation*}
  Given a random variable $X = (X_1, \ldots, X_n)$
  and $A = \{a_1, \ldots, a_k\} \subset [n]$ with
  $a_1 < a_2 < \cdots < a_k$, write $X_A$ for the
  random variable $(X_{a_1}, X_{a_2}, \ldots,
  X_{a_k})$.
\end{notation*}

\begin{fclemma}[Shearer]
  \label{lemma:4.1}
  Assuming:
  - $X = (X_1, \ldots, X_n)$ a random variable
  - $\mathcal{A}$ a family of subsets of $[n]$
  such that every $i \in [n]$ belongs to at least
  $r$ of the sets $A \in \mathcal{A}$
  Then:
  \[
    \ent{X_1, \ldots, X_n}
    \le \frac{1}{r} \sum_{A \in \mathcal{A}}
    \ent{X_A}
  .\]
\end{fclemma}

\begin{proof}
  For each $a \in [n]$, write $X_{< a}$ for $(X_1,
  \ldots, X_{a - 1})$.

  For each $A \in \mathcal{A}$, $A = \{a_1,
  \ldots, a_k\}$ with $a_1 < \cdots < a_k$, we
  have
  \begin{align*}
    \ent{X_A}
    &= \ent{X_{a_1}} + \cent{X_{a_2}}{X_{a_1}}
    + \cdots
    + \cent{X_{a_k}}{X_{a_1}, \ldots, X_{a_k}} \\
    &\ge \cent{X_{a_1}}{X_{<a_1}}
    + \cent{X_{a_2}}{X_{<a_2}}
    + \cdots
    + \cent{X_{a_k}}{X_{<a_k}}
    &&\text{(\cref{lemma:1.15})} \\
    &= \sum_{a \in A} \cent{X_a}{X_{<a}}
  \end{align*}
  Therefore,
  \begin{align*}
    \sum_{A \in \mathcal{A}} \ent{X_A}
    &\ge \sum_{A \in \mathcal{A}} \sum_{a \in A}
    \cent{X_a}{X_{<a}} \\
    &\ge r \sum_{a = 1}^n \cent{X_a}{X_{<a}} \\
    &= r \ent X
    \qedhere
  \end{align*}
\end{proof}