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\begin{proof}
  Let $P$ be $v_1, v_2, \ldots, v_l$.
  Let $x \in X(P)$ and let $y \in V(P)$ and $xy \in G$.
  We show $y \in \ol{X}(P)$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/6716f2688a924815.png}
  \end{center}
  Let $P_x$ be a path $P_x \sim P$ with endpoint $x$.
  Let $y = v_i$.

  \textbf{Case 1:} $v_{i - 1} v_i$ and $v_i v_{i + 1} \in P_x$.
  We rotate through the edge $xy$ from $P_x$ to form a path $Q$.
  Of course $\ol{P} \sim Q$.
  Note in this path $Q$, either $v_{i - 1}$ or $v_{i + 1} \in X(P)$.
  This means $v_i \in \ol{X}(P)$.

  \textbf{Case 2:} If either $v_{i - 1} v_i$ or $v_i v_{i + 1} \notin P_x$ then
  there was a first path $P'$, when we were rotating from $P$ to $P_x$, where we
  removed one of $v_{i - 1} v_i$ or $v_i v_{i + 1}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/30ce6cb92b5e4a8b.png}
  \end{center}
  Note carefully that this means that in $P'$, one of $v_{i - 1}, v_i, v_{i +
  1}$ is an end of $P$.
  So $v_i \in \ol{X}(P)$.
\end{proof}

\begin{fcdefn}[Expander]
  \glsadjdefn{exp}{expander}{graph}%
  Say an $n$-vertex graph $G$ is an \emph{expander} if for all $A, B \subset
  V(G)$ disjoint with $1 \le |A| \le \frac{n}{6}$, $|B| = n - 3|A|$, we have
  $e(A, B) \neq 0$.
\end{fcdefn}

\begin{fclemma}[]
  Let $G$ be an \gls{exp} on $n$ vertices.
  Let $P \subset G$ be a longest path.
  Then $|X(P)| > \frac{n}{6}$ (note $|P| \ge |X(P)|$).
\end{fclemma}

\begin{proof}
  If $P$ is a longest path in $G$, we have that
  \[
    e(X(P), V(G) \setminus P)
    = 0
  ,\]
  otherwise we could rotate and extend $P$, contradicting maximality of $P$.
  By \cref{lemma:XPpathedges},
  \[
    e(X(P), V(P) \setminus \ol{X}(P))
    = 0
  .\]
  So
  \[
    e(X(P), V(G) \setminus \ol{X(P)})
    = 0
  .\]
  Note
  \[
    |\ol{X}(P)|
    \le 3|X(P)|
  \]
  so use expander condition to see
  \[
    |X(P)|
    > \frac{n}{6}
    .
    \qedhere
  \]
\end{proof}

\begin{fclemma}[]
  Let $p \ge \frac{12\log n}{n}$, $G \sim G(n, p)$.
  Then $G$ is an expander with probability $\ge 1 - n^{-2}$ for large enough
  $n$.
\end{fclemma}

\begin{proof}
  \begin{align*}
    \Pbb(\text{$G$ not an expander})
    &\le \sum_{\substack{A \subset [n] \\ |A| \le \frac{n}{6}}}
    \sum_{\substack{B \subset [n] \\ |B| = n - 3|A|}}
    \Pbb(e(A, B) = 0) \\
    &\le \sum_{k = 1}^{\frac{n}{6}} {n \choose k} {n - k \choose n - 3k} (1 -
    p)^{k(n - 3k)} \\
    &\le \sum_{k = 1}^{\frac{n}{6}} {n \choose k} {n \choose 2k} e^{-pk(n - 3k)}
    \\
    &\le \sum_{k = 1}^{\frac{n}{6}} \left( \frac{en}{k} \right)^{3k} e^{-pk(n -
    3k)} \\
    &\le \sum_{k = 1}^{\frac{n}{6}} \left( \frac{en^3}{k} e^{-p(n - 3k)}
    \right)^k \\
    &\le \sum_{k = 1}^{\frac{n}{6}} \left( \frac{en^3}{k} n^{-6} \right)^k \\
    &\le n^{-3}
    \qedhere
  \end{align*}
\end{proof}

\begin{fclemma}[]
  If $p \ge \frac{12 \log n}{n}$, $G \sim G(n, p)$.
  Then $G \supset P_n$ with high probability.
\end{fclemma}

\begin{proof}
  Let $E_i$ be the event that $G - v_i$ is an expander.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/3a4bfc439e17471c.png}
  \end{center}
  
  Let
  \[
    M_i
    = \text{``the longest path in $G$ has the same length as the longest path in
    $G - v_i$''}
  .\]
  Then
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/1533c10dd2bf4c43.png}
  \end{center}
  \begin{align*}
    \Pbb(G \not\supset P_n)
    &\le \Pbb \left( \bigcup_{i = 1}^n M_i \right) \\
    &\le n \Pbb(M_1) \\
    &\le n \Pbb(M_1 \cap E_1) + O(n^{-1}) \\
    &\le n \cdot \max_{G - v \in M_1 \cap E_1} \Pbb_{N(v_1)} (M_1) + O(n^{-1})
    \\
    &\le n (1 - p)^{\frac{n}{6}} + O(n^{-1})
  \end{align*}
  Since a longest path $P$ in $G - v_1$ has $|X(P)| \ge \frac{n}{6}$.
  So if $M_1$ is to hold, we cannot include any edges between $v_1$ and $X(P)$.
  To finish,
  \[
    ne^{-\frac{pn}{6}} + O(n^{-1})
    = O(n^{-1})
    .
    \qedhere
  \]
\end{proof}

\subsection{Sprinkling}

Let $G_1 \sim G(n, p_1)$, $G_2 \sim G(n, p_2)$ on the same vertex set.
Consider $G = G_1 \cup G_2$.
Then $G \sim G(n, p)$, where $p = 1 - (1 - p_1)(1 - p_2) \approx p_1 + p_2$ for
$p_1, p_2$ small.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/5c345fdffa944e22.png}
\end{center}