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  We can choose $X_1 Y_1$ in three equivalent
  ways:
  \begin{enumerate}[(1)]
    \item Pick an edge uniformly from all edges.
    \item Pick a vertex $x$ with probability
      proportional to its degree $d(x)$, and then
      pick a random neighbour $y$ of $x$.
    \item Same with $x$ and $y$ exchanged.
  \end{enumerate}
  It follows that $Y_1 = y$ with probability
  $\frac{d(y)}{|E(G)|}$, so $X_2 Y_1$ is
  uniform in $E(G)$, so $X_2 = x'$ with
  probability $\frac{d(x')}{|E(G)|}$, so $X_2 Y_2$
  is uniform in $E(G)$.

  Therefore,
  \begin{align*}
    \ent{X_1, Y_1, X_2, Y_2}
    &= \ent{X_1} + \cent{Y_1}{X_1} +
    \cent{X_2}{X_1, Y_1} + \cent{Y_2}{X_1, Y_1,
    X_2} \\
    &= \ent{X_1} + \cent{Y_1}{X_1} +
    \cent{X_2}{Y_1} + \cent{Y_2}{X_2} \\
    &= \ent{X_1} + \ent{X_1, Y_1} - \ent{X_1} +
    \ent{X_2, Y_1} - \ent{Y_1} + \ent{Y_2, X_2} -
    \ent{X_2} \\
    &= 3\ent{U_{E(G)}} - \ent{Y_1} - \ent{X_2} \\
    &\ge 3\ent{U_{E(G)}} - \ent{U_Y} - \ent{U_X} \\
    &= 3\log(\alpha mn) - \log m - \log n \\
    &= \log(\alpha^3 m^2 n^2)
  \end{align*}
  So we are done my \gls{maximality}.

  Alternative finish (to avoid using $\log$!):

  Let $X', Y'$ be uniform in $X, Y$ and
  independent of each other and $X_1, Y_1, X_2,
  Y_2$. Then:
  \begin{align*}
    \ent{X_1, Y_2, X_2, Y_2, X', Y'}
    &= \ent{X_1, Y_1, X_2, Y_2} + \ent{U_X} +
    \ent{U_Y} \\
    &\ge 3\ent{U_{E(G)}}
  \end{align*}
  So by \gls{maximality},
  \[
    \#P_3s \times |X| \times |Y| \ge |E(G)|^3
    .
    \qedhere
  \]
\end{proof}

\newpage
\section{Br\'egman's Theorem}

\begin{fcdefnstar}[Permanent of a matrix]
  \glssymboldefn{per}%
  Let $A$ be an $n \times n$ matrix over $\Rbb$.
  The \emph{permanent} of $A$, denoted
  $\perinternal(A)$, is
  \[
    \sum_{\sigma \in S_n} \prod_{i = 1}^n
    A_{i\sigma(i)}
  ,\]
  i.e. ``the determinant without the signs''.
\end{fcdefnstar}

Let $G$ be a bipartite graph with vertex sets $X,
Y$ of size $n$. Given $(x, y) \in X_Y$, let
\[
  A_{xy} = \begin{cases}
    1 & xy \in E(G) \\
    0 & xy \notin E(G)
  \end{cases}
\]
ie $A$ is the bipartite adjacency matrix of $G$.

Then $\per(A)$ is the number of perfect matchings
in $G$.

Br\'egman's theorem concerns how large $\per(A)$
can be if $A$ is a $01$-matrix and the sum of
entres in the $i$-th row is $d_i$.

Let $G$ be a disjoint union of $K_{a_i a_i}$s for $i
= 1, \ldots, k$, with $a_1 + \cdots + a_k = n$.

Then the number of perfect matchings in $G$ is
\[
  \prod_{i = 1}^k a_i!
.\]

\begin{fcthm}[Bregman]
  \label{thm:3.1}
  % [Br\'egman]
  Assuming:
  - $G$ a bipartite graph with vertex sets $X, Y$
  of size $n$
  Then:
  the number of perfect matchings in $G$ is at
  most
  \[
    \prod_{x \in X} (d(x)!)^{\frac{1}{d(x)}}
  .\]
\end{fcthm}

\begin{proof}[Proof (Radhakrishnan)]
  Each matching corresponds to a bijection $\sigma
  : X \to Y$ such that $x\sigma(x) \in E(G)$ for
  every $x$. Let $\sigma$ be chosen uniformly from
  all such bijections.
  \[
    \ent{\sigma} = \ent{\sigma(x_1)} +
    \cent{\sigma(x_2)}{\sigma(x_1)}
    + \cdots
    + \cent{\sigma(x_n)}{\sigma(x_1), \ldots,
    \sigma(x_{n - 1})}
  ,\]
  where $x_1, \ldots, x_n$ is some enumeration of
  $X$.

  Then
  \begin{align*}
    \ent{\sigma(x_1)}
    &\le \log d(x_1) \\
    \cent{\sigma(x_2)}{\sigma(x_1)}
    &\le \Ebb_\sigma \log d_{x_1}^\sigma(x_2)
  \end{align*}
  where
  \[
    d_{x_i}^\sigma(x_2) = |N(x_2) \setminus
    \{\sigma(x_1)\}|
  .\]
  In general,
  \[
    \ent{\sigma(x_i)}{\sigma(x_1), \ldots,
    \sigma(x_{i - 1})}
    \le \Ebb_\sigma \log d_{x_1, \ldots, x_{i -
    1}}^\sigma(x_i)
  ,\]
  where
  \[
    d_{x_1, \ldots, x_{i - 1}}^\sigma(x_i)
    = |N(x_i) \setminus \{\sigma(x_1), \ldots,
    \sigma(x_{i - 1})\}|
  .\]