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  \textbf{Analysis of Algorithm:}
  We define the index of a partition $V_1, \ldots, V_k$ to be
  \[
    \frac{1}{k^2} \sum_{i, j} (\density(V_i, V_j))^2
  .\]
  % Note that this quantity is $\le 1$.
  % We show at each loop of the algorithm, the index increases by at least a
  % $\eps^5$.
  %
  % \begin{center}
  %   \includegraphics[width=0.6\linewidth]{images/2ca165641d5e42f0.png}
  % \end{center}
  %
  % \begin{align*}
  %   \frac{1}{r^2} \sum_{s, t = 1}^r \density(V_{is}, V_{jt})^2 - \density(V_i,
  %   V_j)^2
  %   &= \frac{1}{r^2} \sum_{s, t} (\density(V_{is}, V_{jt}) - \density(V_i,
  %   V_j))^2 \\
  %   &\ge \eps^2 \cdot \eps^2
  % \end{align*}
  We claim that as we go from $V_1, \ldots, V_k$ to $\{V_{ij}\}$, we increase
  the index by at least $+\frac{\eps^5}{2}$.
  Fix $V_i$, $V_j$ and consider
  \begin{align*}
    (*)
    &= \frac{1}{r^2} \sum_{s, t = 1}^r \density(V_{is}, V_{jt})^2
    - \density(V_i, V_j)^2 \\
    &= \frac{1}{r^2} \sum_{s, t} (\density(V_{is}, V_{jt}) - \density(V_i, V_j))^2
  \end{align*}
  Here we are using
  \[
    \frac{1}{r^2} \sum_{s, t} \density(V_{is}, V_{jt})
    = \density(V_i, V_j)
  .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/17a770d9ec9841d7.png}
  \end{center}
  Note $(*) \ge 0$.
  Now assume that $V_i$, $V_j$ are not \gls{epsunif}.
  Define
  \begin{align*}
    S
    &= \{s : V_{is} \subset W_{ij}\} \\
    T
    &= \{t : V_{jt} \subset W_{ji}\}
  \end{align*}
  So
  \begin{align*}
    (*)
    &\ge \frac{1}{r^2} \sum_{\substack{s \in S \\ t \it T}} (\density(V_{is},
    V_{jt}) - \density(V_i, V_j))^2 \\
    &\ge \frac{|S||T|}{r^2} \frac{1}{|S||T|} \sum_{\substack{s \in S \\ t \in
    T}} (\density(V_{is}, V_{jt}) - \density(V_i, V_j))^2 \\
    &\ge \eps^2 \left( \frac{1}{|S||T|} \sum_{\substack{s \in S \\ t \in T}}
    \density(V_{is}, V_{jt}) - \density(V_i, V_j) \right)^2 \\
    &= \eps^2 ( \density(W_{ij}, W_{ji}) - \density(V_i, V_j) )^2 \\
    &\ge \eps^4
  \end{align*}
  So since the number of non \gls{epsunif} pairs is $\ge {k \choose 2}$, we get
  an $+\eps^4$ boost from $\eps$ proportion of the parts.
  Also, the other pairs don't hurt us.
  So index increases by $\ge \frac{\eps^5}{2}$.

  \textbf{How to fix the tiny lie at $*$:}
  \begin{center}
    \includegraphics[width=0.2\linewidth]{images/834a2ed6382c46a7.png}
  \end{center}
  We can assume that we start the algorithm with $\gg \frac{1}{\eps^2}$ parts.
  Now each time we encounter a ``rounding issue'', we just throw out the
  remainder of the cell to some bin I track.
  At a given step, we throw out at most
  \[
    \left( \frac{n}{4^k \cdot k} \right) \cdot 2^k \cdot k
    < \frac{n}{2^k}
    \ll\!\!\ll \eps n
  .\]
  Summing over all steps in algorithm, we still throw out at most $\ll \eps n$.
  We just redistribute these vertices equally at the end.
  \begin{itemize}
    \item This does not ruin the \glsref[epsunif]{$\eps$-uniformity}: your new
      partition is at most \epsuniform{2\eps}.
    \item Also need to check that throwing out these bits does not affect the
      calculations by more than a small amount.
      \qedhere
  \end{itemize}
\end{proof}

\newpage
\section{The binomial Random Graph}

\[
  G(n, p)
\]
We are interested in sliding $p$ from $0$ to $1$.
At what point do certain structures emerge?

Questions:
\begin{itemize}
  \item When do we expect to see a triangle?
  \item When do we expect $G(n, p)$ to be connected?
  \item When does $G(n, p)$ have a component that spans $\Omega(n)$ vertices?
  \item When does $G(n, p)$ have a Hamiltonian cycle?
\end{itemize}

Fix some $H$.
When do we expect $G \supset H$, $G \sim G(n, p)$?

\begin{fcdefn}[m]
  \glssymboldefn{m}%
  For $H$ a graph, define
  \[
    m(H)
    = \max \left\{ \frac{e(F)}{|F|} : F \subset H, |F| \ge 1 \right\}
  .\]
\end{fcdefn}

\begin{example*}
  $H = K_3$, $\mdense(K_3) = 1$.

  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/923a891bf775407d.png}
  \end{center}
\end{example*}

\begin{fcthm}[Bollobás, 80s]
  Assuming:
  - $H$ a graph
  - $G \sim G(n, p)$
  Then:
  \[
    \lim_{n \to \infty} \Pbb(G \supset H)
    = \begin{cases}
      1 & \text{when $pn^{\frac{1}{\mdense(H)}} \to \infty$} \\
      0 & \text{when $pn^{\frac{1}{\mdense(H)}} \to 0$}
    \end{cases}
  \]
\end{fcthm}

\begin{proof}
  We use a second moment argument.
  Let
  \[
    X
    \ \text{\# of copies of $H$ in $G$}
  ,\]
  where $G \sim G(n, p)$.
  Then
  \[
    \Ebb X
    = \Theta(n^{|H|} \cdot p^{e(H)})
  .\]
  Since $\Var X = \Ebb X^2 - (\Ebb X)^2$, we will compute $\Ebb X^2$.
  \begin{align*}
    \Ebb X^2
    &= \left( \sum_H \indicator{H \subset G} \right)^2 \\
    &= \sum_{H_1, H_2} \Pbb(H_1 \subset G, H_2 \subset G) \\
    &\le (\Ebb X)^2
    + (\Ebb X)
    + \sum_{\emptyset \neq F \subsetneq H} \sum_{\substack{H_1, H_2 \\ H_1 \cap
    H_2 = F}} p^{2e(H) - e(F)} \\
    &\le (\Ebb X)^2
    + (\Ebb X)
    + C \sum_{\emptyset F \subsetneq H} n^{2|H| - |F|} p^{2e(H) - e(F)} \\
    &\le (\Ebb X)^2
    + (\Ebb X)
    + Cn^{2|H|} p^{2e(H)} \sum_F \frac{1}{n^{|F|} p^{e(F)}}
  \end{align*}
  Now note
  \[
    \left( \frac{1}{np^{\frac{e(F)}{|F|}}} \right)^{|F|}
    \le \left( \frac{1}{np^{\mdense(H)}} \right)^{|F|}
    \to 0
  \]
  when $pn^{\frac{1}{\mdense(H)}} \to \infty$.
  So
  \[
    \Ebb X^2
    = (1 + o(1)) (\Ebb X)^2 + \Ebb X
    .
    \qedhere
  \]
\end{proof}