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\newpage
\mychapter{Random Graphs}

We know $\ramsey(s) \le 4^s$. How fast does
$\ramsey(s)$ grow? Easy to see that $\ramsey(s) >
(s - 1)^2$:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/862f707a78d3497a.png}
\end{center}
It was generally believed (in the 19030s and
1940s), that $\ramsey(s) \sim cs^2$. But
remarkably:

\begin{flashcard}[erdos-ramsey-bound]
\begin{theorem}[Erd\H{o}s, 1947]
  \label{ramsey_lower_bound}
  $\ramsey(s) \ge (\sqrt{2})^s$ for all $s \ge 3$.
\end{theorem}

\begin{proof}
  \cloze{Choose a colouring of $\complete_n$ at random --
  each \gls{edge} coloured red or blue, with probability
  $\half$, independently.

  Then $\PP(\text{a fixed $s$-set is
  monochromatic}) = 2 \cdot \left( \half
  \right)^{{s \choose 2}}$ (the probability that
  it is coloured entirely red is $\left( \half
  \right)^{{s \choose 2}}$, and then we double
  since it could be coloured either all red or all
  blue). Also, the number of $s$-sets is ${n
  \choose s}$, so certainly
  \[ \PP(\text{exists a monochromatic $s$-set})
  \le {n \choose s} 2^{1 - {s \choose 2}} .\]
  Hence $\ramsey(s) > n$ (i.e. there exists a
  \rcolring{2} of $\complete_n$ with no
  monochromatic $s$-set) if ${n \choose s} 2^{1 -
  {s \choose 2}} < 1$, i.e. if ${n \choose 2} <
  2^{{s \choose 2} - 1}$. Now, ${n \choose s} \le
  \frac{n^s}{s!}$ and $s! \ge 2^{\frac{s}{2} +
  1}$ for all $s \ge 3$ (induction on $s$). So
  \[ {n \choose s} \le \frac{n^s}{2^{\frac{s}{2} +
  1}} .\]
  So $\ramsey(s) > n$ if $n^s <
  2^{\frac{s^2}{2}}$, i.e. if $n <
  2^{\frac{s}{2}}$.}
\end{proof}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The above is a `random graphs' argument.
    \item Could rephrase as: $\#\text{colourings}
      = 2^{{n \choose 2}}$ and
      \[ \#\text{colourings
      making a fixed $s$-set monochromatic} = 2
      \cdot 2^{{n \choose 2} - {s \choose 2}} ,\]
      so done if ${n \choose 2} 2^{{n \choose 2} - {s
      \choose 2} + 1} < 2^{{n \choose 2}}$. But
      this is not a helpful viewpoint -- for
      example, later we will pick \glspl{edge} with
      probability $\neq \half$.
    \item Proof gives no hint of how to
      \emph{construct} a bad colouring.
    \item In fact, no \emph{construction} is known
      giving $\ramsey(s)$ exponential in $s$.
  \end{enumerate}
\end{remark*}
\vspace{-1em}

We have $(\sqrt{2})^2 \le \ramsey(s) \le 4^s$. The
right growth speed is unknown. No lower bound of
the form $(\sqrt{2} + \eps)^s$ is known, but it
was proved in 2023 that $\ramsey(s) \le (4 -
\eps)^s$ (for some small $\eps$).

\begin{flashcard}[prob-space-defn]
\begin{definition*}[Probability space on a graph]
  \glssymboldefn{Gnp}{$G(n, p)$}{$G(n, p)$}
  \cloze{For $0 < p < 1$, the probability space $G(n, p)$
  is defined as follows: we form a \gls{graph} on
  $n$ \glspl{vertex} by selecting each \gls{edge},
  independently, with probability $p$\fcscrap{ (so in the
  proof of \cref{ramsey_lower_bound}, we worked in
  $G \left( n, \half \right)$)}.}
\end{definition*}
\end{flashcard}

\begin{example*}
  In $\psp G(5, p)$,
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/105fa5033f5c444e.png}
  \end{center}
\end{example*}
\vspace{-1em}

Can be useful to consider $p \neq \half$\ldots

Recall that for the `problem of Zarankiewicz' (see
\cref{sub3_2}), we had $\zaran(n, t) \le 2 n^{2 -
\frac{1}{t}}$. How about a lower bound --
preferably not of the form $c \cdot n^1$
(preferably non-trivial).

We could do the following:

Form a random \gls{bipartite} \gls{graph} (vertex
classes $X, Y$, each size $n$) by choosing
\glspl{edge}
independent with probability $p$. Then
\[ \PP(\text{a fixed $\bipartite_{t, t} \subset G$})
= p^{t^2} .\]
Also, $\#\bipartite_{t, t} = {n \choose t}^2$. So
\[ \EE(\text{$\#\bipartite_{t, t}$ in $G$}) = {n
\choose t}^2 p^{t^2} \le \quarter n^{2t} p^{t^2} .\]
So take $p = n^{-\frac{2}{t}}$ -- gives
$\EE(\#\bipartite_{t, t}) \le \quarter$. So
$\PP(\text{no $\bipartite_{t, t}$ in $G$}) \ge
\frac{3}{4}$. Also, $\EE(\text{\#\glspl{edge} of
$G$}) = pn^2$. So
\[ \PP\left(\text{\#\glspl{edge}} \ge
\half pn^2\right) \ge \half .\]
Hence there exists a \gls{graph} $G$ with no
$\bipartite_{t, t}$ and $\edges(G) \ge \half pn^2
= \half n^{2 - \frac{2}{t}}$. Thus $\zaran(n, t)
\ge \half n^{2 - \frac{2}{t}}$.

But we can do better.

\begin{flashcard}[zaran-lower-bound]
\begin{theorem}
  $\zaran(n, t) \ge \cloze{\half n^{2 - \frac{2}{t
  + 1}}}$.
\end{theorem}

\cloze{\textbf{Idea:} If $G$ has $n$ \glspl{edge} and $r$
$\bipartite_{t, t}$s, remove an \gls{edge} from each
$\bipartite_{t, t}$ to obtain a \gls{graph} with $m - r$
\glspl{edge} and no $\bipartite_{t, t}$.}

\begin{proof}
  \cloze{
  Form a random \gls{bipartite} \gls{graph}
  (vertex classes size $n$), by choosing each edge
  with probability $p$ independently. Let $M =
  \#\text{\glspl{edge}}$, $R = \#\bipartite_{t,
  t}$. Then
  \[ \EE(M) = pn^2 \qquad \text{and} \qquad \EE(R)
  = {n \choose t}^2 p^{t^2} \le \half n^{2t} p^{t^2} ,\]
  so
  \[ \EE(M - R) \ge pn^2 - \half n^{2t} p^{t^2} .\]
  Pick $p = n^{-\frac{2}{t + 1}}$: so
  $pn^2 = n^{2 - \frac{2}{t + 1}}$ and $n^{2t}
  p^{t^2} = n^{2t - \frac{2t^2}{t + 1}} = n^{2 -
  \frac{2}{t + 1}}$. So
  \[ \EE(M - R) \ge \half n^{2 - \frac{2}{t + 1}}
  .\]
  Hence there exists $G$ with $m$ \glspl{edge},
  $r$ $\bipartite_{t, t}$s and $m - r \ge \half
  n^{2 - \frac{2}{t + 1}}$. Whence $\zaran(n, t)
  \ge \half n^{2 - \frac{2}{t + 1}}$.
}
\end{proof}
\end{flashcard}

This method is called `modifying a random
\gls{graph}'.