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  Let $X_v$, for $v \in V(G)$, be defined by
  \[
    X_v
    = d \cdot \indicator{v \in I}
    + |N(v) \cap I|
  .\]
  Note, for fixed $I$,
  \[
    \sum_{v \in V(G)} X_v
    = \sum_{v \in V(G)} d \cdot \indicator{v \in I} + |N(v) \cap I|
    \le d|I| + d|I|
    = 2d|I|
  .\]
  So
  \[
    \sum_v \Ebb X_v
    \le 2d \cdot \Ebb |I|
  ,\]
  and so enough to show
  \[
    \Ebb X_v
    \ge c \log d
  ,\]
  for $\forall v \in V(G)$ and some $c > 0$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/6375d7146ab04f12.png}
  \end{center}
  Now fix $v \in V(G)$ and define
  \[
    L = N(v) \cup \{v\},
    \qquad
    F = V(G) - v - N(v)
  .\]
  We now bound
  \[
    \Ebb X_v
    \ge \min_{J \subset F} \Ebb(X_v \mid I \cap F = J)
  .\]
  So now fix some $J \subset F$. We consider $I \cap L'$ where $L' = L \setminus
  N(J)$ and $N(J)$ means $\{y : \exists x \in J, x \sim y\}$.
  Let $l = |L'|$.

  Now observe that $I \cap L'$ is uniform over all independent sets in $G[L']$.
  So since $I \cap L'$ is either $\{v\}$ or any subset of $L' \setminus \{v\}$,
  we have
  \[
    \Ebb(X_v \mid I \cap F = J)
    = \frac{d}{2^{l - 1} + 1}
    + \left( \frac{l - 1}{2} \right) \left( \frac{2^{l - 1}}{2^{l - 1} + 1} \right)
  .\]
  We are happy with this inequality, because if $l$ is large, then the second
  term is large, and if $l$ is small then the first term is large.

  Continuing the above calculation, we have
  \[
    \Ebb(X_v \mid I \cap F = J)
    \ge \max \left\{ \frac{d}{2^{l - 1} + 1}, \frac{l - 1}{4} \right\}
  .\]
  Now solve 
  \[
    \frac{d}{2^{l - 1} + 1}
    = \frac{l - 1}{4}
  ,\]
  which is $4d = (l - 1)(2^{l - 1} + 1)$, so $l = c\log d$.
  So
  \[
    \Ebb(X_v \mid I \cap F = J)
    \ge c \log d
  ,\]
  and thus $\Ebb X_v \ge c \log d$ as desired.
\end{proof}

\subsection{Recap of $R(3, k)$ bounds proved in this course}

We have shown
\[
  c \frac{k^2}{(\log k)^2}
  \le R(3, k)
  \le (1 + o(1)) \frac{k^2}{ \log k}
.\]

Kim showed $R(3, k) = \Theta \left( \frac{k^2}{\log k} \right)$.

\textbf{Triangle-free process:}
Define the random graph process $(G_i)_i$.
Given $G_i$ define
\[
  O_i
  = \{e \in [n]^{(2)} : G_i + e \not\supset K_3\}
.\]
Now sample $e_{i + 1} \sim O_i$ uniformly at random, and define $G_{i + 1} = G_i
+ e_{i + 1}$.

This method is not what Kim used, but it was used by others later to improve the
lower bound.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/17e0ea7ee8e84b27.png}
\end{center}

Up until very recently, every proof of lower bound on $R(3, k)$ has used a
variant of the triangle-free process.

\begin{remark*}
  There is some similarity between the triangle-free process and the first proof
  of \cref{thm:shearermaxdegtrianglefree} that we saw: both are about a
  process where each step is uniformly random once we condition on a desired
  property.
  So we might expect a related theorem to hold for the triangle-free process,
  and indeed this is the case.
\end{remark*}

The Ramsey numbers $R(3, k)$ are still of interest, as there is now interest in
determining the right constant.

We now finish this section by using \cref{thm:shearermaxdegtrianglefree} to
deduce the upper bound on $R(3, k)$:

\begin{fcthm}[Shearer upper bound on $R(3, k)$]
  $R(3, k) \le (1 + o(1)) \frac{k^2}{\log k}$.
\end{fcthm}

\begin{proof}
  Let $n = (1 + \delta) \frac{k^2}{\log k}$.
  If there is a vertex of degree $\ge k$ in the blue graph then we have an
  independent set of size $k$ because the blue graph must be triangle-free,
  otherwise done.
  So the max degree of blue graph is $\le k$.

  So apply \nameref{thm:shearermaxdegtrianglefree} to get (for large $k$) that
  \begin{align*}
    \indnum(G)
    &\ge (1 + o(1)) \frac{n}{k} \log k \\
    &= (1 + o(1))(1 + \delta) \frac{k^2}{(\log k) k} \log k \\
    &\ge k
  \end{align*}
  (where $G$ is the blue graph).
\end{proof}

\newpage
\section{Dependent Random Choice}

\begin{fcdefn}[$(s, k)$-rich]
  \glsadjdefn{skrich}{$(s, k)$-rich}{set}%
  Let $G$ be a graph.
  Say that $R \subset V(G)$ is $(s, k)$-rich if for every $x_1, \ldots, x_s \in
  R$ we have
  \[
    |N(x_1) \cap \cdots \cap N(x_s)|
    \ge k
  .\]
\end{fcdefn}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/250be422cde94c5c.png}
\end{center}


\begin{fcthm}[Dependent Random Choice]
  \label{lemma:drc}
  Assuming:
  - $G$ a graph on $n$ vertices with $e(G) = m$
  - $s, t, r, k \ge 0$
  - $\frac{(2m)^t}{n^{2t - 1}} - {n \choose s} \left( \frac{k}{n} \right)^t \ge
  r$
  Then:
  there exists $R \subset V(G)$, $|R| \ge r$ which is \gls{skrich}.
\end{fcthm}

Note that $t$ doesn't appear in the conclusion.
It is a parameter that we can tune appropriately in applications.

\begin{proof}
  We sample $x_1, \ldots, x_t \in V(G)$ uniformly at random (with replacement).
  Then
  \begin{align*}
    \Ebb_{x_1, \ldots, x_t} |N(x_1) \cap \cdots \cap N(x_t)|
    &= \Ebb_{x_1, \ldots, x_t} \sum_{y \in V(G)} \indicator{x_1 \sim y \wedge
    x_2 \sim y \wedge \cdots \wedge x_t \sim y} \\
    &= \sum_{y \in V(G)} [\Pbb_{x_1}(x_1 \sim y)]^t \\
    &= \sum_{y \in V(G)} \left( \frac{d(y)}{n} \right)^t \\
    &\ge n \left( \frac{2m}{n^2} \right)^t \\
    &= \frac{(2m)^t}{n^{2t - 1}}
  \end{align*}
  Let $Y$ be the random variable that counts the number of $\{y_1, \ldots, y_s\}
  \in (N(x_1) \cap \cdots \cap N(x_t))^{(s)}$ with
  \[
    |N(y_1) \cap \cdots \cap N(y_s)|
    < k
  .\]
  We want $Y$ to be not too big.
  This will be the case (on average) because $\Pbb(y_1, \ldots, y_s \in N(x_1)
  \cap \cdots \cap N(x_t))$ is an increasing function of $|N(y_1) \cap \cdots
  \cap N(y_s)|$ (so our probability distribution is biased away from fully
  selecting tuples that will increase $Y$).
  We make this precise as follows:

  % We show
  % \[
  %   \Ebb Y
  %   \le {n \choose s} \left( \frac{k}{n} \right)^t
  % .\]
  % Note that if we sample $x_1, \ldots, x_t$ as above and delete a vertex from
  % each $y_1, \ldots, y_s$ that violates the rich condition, what remains has
  % size at least
  % \[
  %   |N(x_1) \cap \cdots \cap N(x_t)|
  %   - Y(x_1, \ldots, x_t)
  % \]
  % and is \gls{skrich}.
% \end{proof}