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Recall that
\[
  \setexp(S)
  = \q_{\tilde{L}_G}(\indicator{S})
\]
and
\[
  \graphexp(G)
  = \min_{\substack{S \subseteq V \\ 1 \le |S| \le |V|/2}}
  \q_{\tilde{L}_G}(\indicator{S})
.\]

\subsubsection*{Fiedler's Algorithm}

\textbf{Input:} $G, \varphi : V \to \Rbb$.
\begin{itemize}
  \item Sort vertices $x_1, \ldots, x_n$ such that $\varphi(x_1) \le \cdots \le
    \varphi(x_n)$.
  \item Find cut $K$ that minimises $\Phi(\{x_2, \ldots, x_k\}, \{x_{k + 1},
    \ldots, x_n\})$.
\end{itemize}
\textbf{Output}: The cut.

Running time: $O(|V| \log |V| + |E|)$.

\begin{fclemma}[]
  Assuming:
  - $\psi : V \to \Rbb$
  - $\langle \psi, 1 \rangle = 0$
  - let $(S, V \setminus S) = \operatorname{Fiedler}(G, \psi)$
  Then:
  \[
    \graphexp(S, V \setminus S)
    \le \sqrt{2\q_{\tilde{L}_G}(\psi)}
  .\]
\end{fclemma}

If $\psi : V \to \Rbb$, call a cut $(\{x : \psi(x) \ge \tau)\}, \{x : \psi(x) <
\tau\})$ a threshold cut for $\psi$.

\begin{fclemma}[]
  \label{lemma:lec5l2}
  Assuming:
  - $\varphi : V \to \Rbb$
  - $\langle \varphi, 1 \rangle = 0$
  Then:
  there is $\psi : V \to \Rbb_{\ge 0}$ such that $\q_{\tilde{L}_G}(\psi) \le
  \q_{\tilde{L}_G}(\psi)$, $|\supp \psi| = |\{x : \psi(x) > 0\}| \le
  \frac{|V|}{Q}$ and any threshold cut for $\psi$ is a threshold cut for
  $\varphi$.
\end{fclemma}

\begin{fclemma}[]
  \label{lemma:lec5l3}
  Assuming:
  - $\psi : V \to \Rbb_{\ge 0}$
  Then:
  there is $0 < t \le \|\psi\|_\infty$ such that
  \[
    \setexp(\{x : \psi(x) \ge t\})
    \le \sqrt{2\q_{\tilde{L}_G}(\psi)}
  .\]
\end{fclemma}

\begin{proof}[Proof of \cref{lemma:lec5l2}]
  If $\langle \varphi, 1 \rangle = 0$ then
  \begin{align*}
    \q_{\tilde{L}_G}(\varphi + \alpha 1)
    &= \frac{\Q_{\tilde{L}_G}(\varphi + \alpha 1)}{\|\varphi + \alpha 1\|^2} \\
    &= \frac{\Q_{\tilde{L}_G}(\varphi)}{\|\varphi\|^2 + \alpha^2} \\
    &\le \frac{\Q_{\tilde{L}_G}(\varphi)}{\|\varphi\|^2} \\
    &= \q_{\tilde{L}_G}(\varphi)
  \end{align*}
  Let $m \in \Rbb$ be the median of $\varphi$.
  \begin{align*}
    |\{x \in V : \varphi(x) > m\}|
    &\le \frac{|V|}{2} \\
    |\{x \in V : \varphi(x) < m\}|
    &\le \frac{|V|}{2}
  \end{align*}
  $\ol{\varphi} = \varphi - m1$, $\q_{\tilde{L}_G}(\ol{\varphi}) \le
  \q_{\tilde{L}_G}(\varphi)$.
  Let $\ol{\varphi} = \ol{\varphi}^+ - \ol{\varphi}^-$, where $\ol{\varphi}^+,
  \ol{\varphi}^- : V \to \Rbb_{\ge 0}$.
  So
  \[
    \ol{\varphi}(x)
    = \begin{cases}
      \ol{\varphi}^+(x)
      & \varphi(x) > m \\
      -\ol{\varphi}^-(x)
      & \varphi(x) < m \\
      0
      & \varphi(x) = m
    \end{cases}
  .\]
  Note $\langle \ol{\varphi}^-, \ol{\varphi}^+ \rangle = 0$.

  Claim: either $\ol{\varphi}^+$ or $\ol{\varphi}^-$ suffices.
  \begin{align*}
    \q_{\tilde{L}_G}(\varphi)
    &\ge \q_{\tilde{L}_G}(\ol{\varphi}) \\
    &= \q_{\tilde{L}_G}(\ol{\varphi}^+ - \ol{\varphi}^-) \\
    &= \frac{\sum_{x \sim y} (\ol{\varphi}^+(x) - \ol{\varphi}^-(x) -
    \ol{\varphi}^+(y) + \ol{\varphi}^-(y))^2}{\|\ol{\varphi}^+ -
    \ol{\varphi}^-\|^2} \\
    &= \frac{\sum_{x \sim y} ((\ol{\varphi}^+(x) - \ol{\varphi}^+(y)) -
    (\ol{\varphi}^-(x) - \ol{\varphi}^-(x))^2}{\|\ol{\varphi}^+\|^2 +
    \|\ol{\varphi}^-\|^2} \\
    &\ge \frac{\sum_{x \sim y} (\ol{\varphi}^+(x) - \ol{\varphi}^+(y))^2 +
    (\ol{\varphi}^-(x) - \ol{\varphi}^-(y))^2}{\|\ol{\varphi}^+\|^2 +
    \|\ol{\varphi}^-\|^2} \\
    &= \frac{\q_{\tilde{L}_G}(\ol{\varphi}^+) \|\ol{\varphi}\|^2 +
    \q_{\tilde{L}_G}(\ol{\varphi}^-) \|\ol{\varphi}^-\|^2}{\|\ol{\varphi}\|^2 +
    \|\ol{\varphi}^-\|^2}
  \end{align*}
\end{proof}

\begin{proof}[Proof of \cref{lemma:lec5l3}]
  Assume $\|\psi\|_\infty = 1$.
  We find $0 < t \le 1$.
  Choose $t$ at random, such that $t^2 \sim \Unif([0, 1])$.
  Let
  \[
    S_t = \{x \in V : \psi(x) > t\}
  .\]
  Then
  \[
    \Ebb|S_t|
    = \sum_x \Pbb(x \in S_t)
    = \sum_x \Pbb(\psi(x)^2 > t^2)
    = \sum_x \psi(x)^2
  .\]
  Have
  \[
    \setexp(S_t) = \frac{e(S_t, V \setminus S_t)}{d|S_t|}
  .\]
  Also, can calculate:
  \begin{align*}
    \Ebb d|S_t|
    &= d\sum_x \psi(x)^2 \\
    \Ebb e(S_t, V \setminus S_t)
    &= \sum_{x \sim y} \Pbb(\text{$xy$ is cut by $(S_t, V \setminus S_t)$}) \\
    &= \sum_{x \sim y} |\psi(x)^2 - \psi(y)^2| \\
    &= \sum_{x \sim y} |\psi(x) - \psi(y)| (\psi(x) + \psi(y)) \\
    &\le \sqrt{\sum_{x \sim y} (\psi(x) - \psi(y))^2}
    \cdot \sqrt{\sum_{x \sim y} (\psi(x) + \psi(y))^2}
  \end{align*}
  Then
  \begin{align*}
    \frac{\Ebb e(S_t, V \setminus S_t)}{\Ebb d|S_t|}
    &\le \sqrt{\q_{\tilde{L}_G}(\psi)}
    \cdot \sqrt{\frac{\sum_{x \sim y} (\psi(x) + \psi(y))^2}{d\sum_x \psi(x)^2}}
    \\
    &\le \sqrt{2\q_{\tilde{L}_G}(\psi)}
  \end{align*}
  Now use the following fact to finish:

  \textbf{Fact}: If $X$ and $Y$ are random variables with $\Pbb(Y > 0) = 1$,
  then $\Pbb \left( \frac{X}{Y} \le \frac{\Ebb X}{\Ebb Y} \right)$.

  (Proof: let $R = \frac{\Ebb X}{\Ebb Y}$. Then $\Ebb(X - RY) = 0$, so $\Pbb(X -
  RY \le 0)) > 0$, hence $\Pbb(\frac{X}{Y} \le R) > 0$).
\end{proof}