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Say I have a graph $H$, with max degree $d$ where $d$ fixed and $|V(H)| = k \to
\infty$.
Then it is known that
\[
  R(H)
  \le C_d \cdot k
  \tag{$*$}
  \label{lec10eq1}
.\]

\begin{fcprop}[]
  Assuming:
  - $d \ge 1$, $\eps > 0$
  - $k \ge k_0$, for some $k_0$ depending on $d$ and $\eps$
  - $H$ bipartite and max degree $\le d$
  Then:
  \[
    R(H)
    \le k^{1 + \eps}
  .\]
\end{fcprop}

\begin{proof}
  Apply \nameref{lemma:drc} and embedding \cref{lemma:maxdegembed} to the
  majority colour.
\end{proof}

So \nameref{lemma:drc} and \cref{lemma:maxdegembed} are enough to prove an upper
bound that is almost linear in $k$, but not quite powerful enough to prove the
linear bound as in \eqref{lec10eq1}.

% In order to get beyond this, we want to do a variant of this version of
% \nameref{lemma:drc}.
%
\begin{remark*}
  To improve this to get the bound $R(H) \le C_d n$, one idea might be to find a
  \skrich{d}{k} set where $k = \Theta(n)$ ($n = C_d k$), where the
  \glsref[skrich]{rich} set has size $\Theta(n)$ on $n$ vertices.
  But this is impossible.
\end{remark*}

\begin{fcprop}[]
  For all $c > 0$, there exists a graph $G$ with $\left( \half + o(1) \right) {n
  \choose 2}$ edges so that every set of size $\ge cn$ contains some $x, y$ with
  $o(n)$ common neighbours.
\end{fcprop}

\begin{center}
  \includegraphics[width=0.3\linewidth]{images/6e31ab5fde7f4d30.png}
\end{center}

\begin{proof}
  We won't prove.
\end{proof}

So we have to go back into the guts of the embedding idea.
In some sense, the embedding algorithm was a little bit wasteful: for example,
we dumped $A$ into the \glsref[skrich]{rich} set without any clever choice.

The key idea to improve the previous bound will be to perform a more clever
embedding algorithm that requires a weaker notion of rich set.
We want to use a weaker notion of rich set so that we can find large linear size
rich sets.

\begin{fcdefn}[Approximately $(s, k)$-rich]
  \glsadjdefn{approxskrich}{approximately $(s, k)$-rich}{set}%
  Let $G$ be a graph.
  Say $U \subset V(G)$ is approximately $(s, k)$-rich if
  \[
    |\{(x_1, \ldots, x_s) \in U^s : d(x_1, \ldots, x_s) < k\}|
    \le \left( \frac{1}{2s} \right)^s |U|^s
  .\]
  Here we use the notation
  \[
    d(x_1, \ldots, x_s)
    = |N(x_1) \cap \cdots \cap N(x_s)|
  .\]
\end{fcdefn}

% \begin{center}
%   \includegraphics[width=0.3\linewidth]{images/db0f9f1444554dfb.png}
% \end{center}

\begin{fclemma}[]
  \label{lemma:approxembed}
  Assuming:
  - $G$ a graph
  - $U \subset V(G)$ is \approxskrich{k}{k}
  - $|U| > 2k$
  - $H$ a bipartite graph on $k$ vertices with max degree $\le s$
  Then:
  $G \supset H$.
\end{fclemma}

\begin{fclemma}[]
  \label{lemma:secondrandomchoice}
  Assuming:
  - $\eps > 0$
  - $G$ a graph with $\frac{\eps n^2}{2}$ edges on $n$ vertices
  - $n \ge 4 s \eps^{-s} k$
  Then:
  there exists $U \subset V(G)$ that is \approxskrich{s}{k} and $|U| > 2k$.
\end{fclemma}

\begin{fcthm}[Fox, Sudakov, Conlon]
  \label{thm:foxsudakovconlon}
  Assuming:
  - $H$ a bipartite graph on $k$ vertices
  - max degree $\le d$
  Then:
  \[
    R(H)
    \le C d 2^d \cdot k
  .\]
\end{fcthm}

\begin{remark*}
  This also improves our bound on $Q_d$:
  \[
    R(Q_d)
    \le C d 2^{2d}
  .\]
\end{remark*}

The key difference in proving \cref{lemma:approxembed} compared to the previous
embedding algorithm is that we will need to embed $A$ in a more clever way.
We will need to ensure that for all $b \in B$, the neighbourhood of $b$ is not
sent to one of the bad subsets of $U$ (the ``bad'' subsets are those which have
$< k$ common neighbours).
\begin{center}
  \includegraphics[width=0.3\linewidth]{images/20835250015c4ba7.png}
\end{center}

\begin{proof}[Proof of \cref{lemma:approxembed}]
  For $x_1, \ldots, x_t$, $t < s$, let
  \[
    \beta(x_1, \ldots, x_t)
    = |\{(x_1, \ldots, x_t, x_{t + 1}, \ldots, x_s) \in U^s : d(x_1, \ldots,
    x_s) < k\}|
  .\]
  Say that $(x_1, \ldots, x_t)$ is healthy if
  \[
    \beta(x_1, \ldots, x_t)
    \le \left( \frac{1}{2s} \right)^{s - t} |U|^{s - t}
  .\]
  We will embed vertices of $A$ one after the other.
  We will use this notion of healthiness to make sure that every time we place a
  vertex, we keep plenty of good extensions.

  \textbf{Observation:} If $x_1, \ldots, x_t$ is healthy, then the number of $y$
  such that $(x_1, \ldots, x_t, y)$ is not healthy is $\le \frac{|U|}{2s}$.

  Proof of observation:
  Note
  \[
    \beta(x_1, \ldots, x_t)
    = \sum_{y \in U} \beta(x_1, \ldots, x_t, y)
  .\]
  Let $B_{x_1, \ldots, x_t} = \{y : \text{$(x_1, \ldots, x_t, y)$ is not
  healthy}\}$.
  Then
  \begin{align*}
    \beta(x_1, \ldots, x_t)
    &\ge \sum_{y \in B_{x_1, \ldots, x_t}} \beta(x_1, \ldots, x_t, y) \\
    &\ge |B_{x_1, \ldots, x_t}| \left( \frac{1}{2s} \right)^{s - t - 1} |U|^{s -
    t - 1}
  \end{align*}
  But if $|B_{x_1, \ldots, x_t}| > \frac{|U|}{2s}$, I contradict that $x_1,
  \ldots, x_t$ is healthy.
  So we have proved the observation.

  We now inject $f : A \to U$ inductively.
  Assume we have embedded $a_1, \ldots, a_i$ so far.
  We assume that we have
  \[
    f(N(b) \cap \{a_1, \ldots, a_i\})
  \]
  is healthy for all $b \in B$.
  We now embed $a_{i + 1}$.
  Let $b_1, \ldots, b_s$ be the neighbour of $a_{i + 1}$.
  We know $f(N(b_i) \cap \{a_1, .., a_i\})$ are healthy, so there are at most
  (by observation) $\le \frac{|U|}{2s}$ ways of choosing $f(a_{i + 1}) \in U$ so
  that one of the neighbours becomes unhealthy.
  But there are only $\le s$ neighbourhoods.
  So there are $> \frac{|U|}{2}$ choices that maintain all healthy neighbours.
  Since $|U| > 2k$, I can map $a_{i + 1}$ to a new vertex.

  We now inject $f : B \to V(G)$ exactly as we saw in the proof of
  \cref{lemma:maxdegembed}.
\end{proof}

\begin{proof}[Proof of \cref{lemma:secondrandomchoice}]
  Let $x_1, \ldots, x_s \in V(G)$ be chosen uniformly at random.
  Let $U = N(x_1) \cap \cdots \cap N(x_s)$.
  Let $X = |U|$.
  We have $\Ebb X \ge \eps^s \cdot n$.

  Let $Y = |\{(y_1, \ldots, y_s) \in U^s : d(y_1, \ldots, y_s) < k\}|$.
  $\Ebb Y \le n^s \left( \frac{k}{n} \right)^s$.
  Now note that by Jensen,
  \[
    \Ebb \left[ X^s - \frac{(\Ebb X)^s Y}{2\Ebb Y} - \frac{(\Ebb X)^s}{2}
    \right]
    = \Ebb X^s - (\Ebb X)^s
    \ge 0
  .\]
  So there is a $U$ such that
  \[
    |U|^s
    \ge \frac{(\Ebb X)^s}{2}
    \ge \frac{(\eps^s n)^s}{2}
  .\]
  So $|U| \ge \frac{\eps^s n}{2} > 2k$.
  \[
    Y \le \frac{2(\Ebb Y)}{(\Ebb X)^s} |U|^s
    \le \left( \frac{1}{2s} \right)^s |U|^s
    .
    \qedhere
  \]
\end{proof}
