%! TEX root = GT.tex
% vim: tw=50
% 18/11/2023 09AM

\subsection{Graphs with high chromatic number}

To ensure $\chromnum(G) \ge k$, could just have $G
\supset \complete_k$. Not necessary to have
$G \supset \complete_k$ -- for example $\cycle_5$
has no $\complete_3$, but $\chromnum(\cycle_5) =
3$.

\glssymboldefn{CL}{CL$(G)$}{CL$(G)$}
\glspropdefn{cliquen}{clique number}{\gls{graph}}
Can have $\chromnum(G)$ much greater than
$\cliquen(G)$, the \emph{clique number} of $G$,
namely $\max\{k : G \supset \complete_k\}$.

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The $G$ of \cref{ramsey_lower_bound}:
      $G$ on $2^{s/2}$ \glspl{vertex}, no
      $\complete_s \subset G$, so $\cliquen(G) \le
      s - 1$. Also, each \emph{independent set}
      (set with no \glspl{edge}) has size $\le s -
      1$. But in any vertex-colouring of a graph,
      each colour class is an independent set.
      Hence $\chromnum(G) \ge \frac{2^{s/2}}{s -
      1}$ -- much more than $\cliquen(G)$. Better:
    \item Can find $G$ which is triangle-free
      ($\cliquen(G) = 2$), but $\chromnum(G)$
      arbitrarily large -- quite hard (on \es{3}).
  \end{enumerate}
\end{example*}
\vspace{-1em}

Could we even ask for \gls{girth} $\ge 5$? Or more --
like $G$ with \gls{girth} $\ge 10$, $\chromnum(G)
\ge 100$? Sounds unlikely, but\ldots

\begin{flashcard}[random-girth-chromnum-thm]
\begin{theorem}
  $\forall k, g$ there exists \gls{graph} $G$ with
  \gls{girth} $\ge g$ and $\chromnum(G) \ge k$.
\end{theorem}
\vspace{-1em}
\cloze{
\textbf{Idea:} Find $G$ on $n$ \glspl{vertex}
such that the number of short \glspl{cycle} is $\le
\frac{n}{2}$ and each independent set has size
$\le \frac{n}{2k}$ -- then done, by removing a
\gls{vertex} from each short \gls{cycle} to obtain
a \gls{graph} $H$ with \gls{girth} $\ge g$ and
$\chromnum(G) \ge \frac{n/2}{n/2k} = k$.

\begin{proof}
  For large $n$, choose $G \in \psp G(n, p)$ where
  $p = n^{-1 + \frac{1}{g}}$. For $i = 3, 4,
  \ldots, g - 1$, let $X_i = \#\text{$i$
    \glspl{cycle} in
  $G$}$. Let $X = X_3 + \cdots + X_{g - 1}$ be the
  number of \glspl{cycle} in $G$ of length $< g$.
  Then
  \begin{align*}
    \EE(X_i)
    &\le (\#\text{possible $i$-\glspl{cycle}})
    \PP(\text{given $i$-\gls{cycle} $\subset G$})
    \\
    &\le n^i p^i \\
    &= n^{i/g} \\
    &\le n^{\frac{g - 1}{g}}
  \end{align*}
  Hence $\EE(X) \le g \cdot \frac{n}{n^{1/g}} <
  \frac{n}{4}$ for $n$ large (as $n^{1/g} \to 0$
  as $n \to \infty$). So $\PP \left( X >
  \frac{n}{2} \right) < \half$ (else $\EE(X) \ge
  \frac{n}{2} \cdot \half = \frac{n}{4}$
  \contradiction). Write $t = \frac{n}{2k}$ ($n$ a
  multiple of $2k$), and let $Y =
  \#\text{independent $t$-sets in $G$}$. Then
  \begin{align*}
    \EE(Y)
    &\le n^t (1 - p)^{{t \choose 2}} \\
    &\le n^t e^{-p{t \choose 2}}
    &&\text{(using $1 - x \le e^{-x}$)} \\
    &\le \exp\left(\frac{n}{2k} \log n - n^{-1 +
    \frac{1}{g}} \cdot \frac{n^2}{16k^2} \right)
    \\
    &\to 0
  \end{align*}
  as $n \to \infty$ ($n^{1 + \frac{1}{g}}$ grows
  faster than $n\log n$). So $\PP(X = 0) > \half$
  if $n$ large enough. Hence there exists $G$ on
  $n$ vertices with $\le \frac{n}{2}$ short cycles
  and no independent set of $\frac{n}{2k}$.
\end{proof}
}
\end{flashcard}

\subsection{The structure of a random graph}

What does $G \in \psp G(n, p)$ look like? How do
the properties of $G$ vary as $p$ invreases? For
example, how to $\PP(\text{no isolated
\gls{vertex}})$ behave?

\glsadjdefn{isolated}{isolated}{\gls{vertex}}
(A \gls{vertex} is \emph{isolated} if it has no
\glspl{neighbour}).

We might guess:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/ce7842842fe043a1.png}
\end{center}
But in fact:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/919db8529d3948d0.png}
\end{center}
Why does this happen? Where is the threshold?

\subsubsection*{Probability Digression / Reminder}

Let $X$ be a random variable taking values in $0,
1, 2, \ldots$.

\textbf{To show $\PP(X = 0)$ is big:} enough to show $\mu =
\EE(X)$ is small. Indeed, for any $t > 0$ have
$\PP(X \ge t) t \le \mu$. So
$\PP(X \ge t) \le \frac{\mu}{t}$
(Markov). In particular, $\PP(X \ge 1) \le \mu$,
so $\PP(X = 0) \ge 1 - \mu$.

\textbf{To show $\PP(X = 0)$ is small:} not enough
to have $\mu$ large, for example
\[ X = \begin{cases}
  0 & \text{probability $\frac{99}{100}$} \\
  10^{10} & \text{probability $\frac{1}{100}$}
\end{cases} \]
So instead we look at the variance $V = \Var(X) =
\EE((X - \mu)^2) = \EE(X^2) - \EE(X)^2$. Then
\[ \PP(|X - \mu| \ge t) = \PP(|X - \mu|^2 \ge t^2)
\le \frac{V}{t^2} \]
by Markov (this is known as Chebyshev inequality).
So $\PP(|X - \mu| \ge \mu) \le \frac{V}{\mu^2}$,
whence $\PP(X = 0) \le \frac{V}{\mu^2}$.
Conclusion: to show $\PP(X = 0)$ is small, check
$\frac{V}{\mu^2}$ is small.
\begin{hiddenflashcard}[show-Px-0-big]
\prompt{How to show $\PP(X = 0)$ is big?}

\cloze{
Show $\EE(X)$ is small (sufficient by Markov's
inequality).
}
\end{hiddenflashcard}
\begin{hiddenflashcard}[show-Px-0-small]
\prompt{How to show $\PP(X = 0)$ is small?}

\cloze{
Show $\frac{\Var(X)}{\EE(X)^2}$ is small (using
Chebyshev's inequality).
}
\end{hiddenflashcard}

Suppose $X$ counts the number of some events $A$
that occur. Then $\mu = \EE(X) = \sum_A \PP(A)$.
Variance? Have
\begin{align*}
  \EE(X)^2
  &= \sum_A \sum_B \PP(A) \PP(B) \\
  \EE(X^2)
  &= \EE \left( \left( \sum_A \indicator{A}
  \right)^2 \right) \\
  &= \EE \left( \sum_A \sum_B \indicator{A}
  \indicator{B} \right) \\
  &= \sum_A \sum_B \EE(\indicator{A}
  \indicator{B}) \\
  &= \sum_A \sum_B \PP(A \cap B) \\
  &= \sum_A \sum_B \PP(A) PP(B \mid A)
\end{align*}
\begin{flashcard}[var-nice-formula]
\fcscrap{So }$\Var(X) = \cloze{\sum_A \PP(A) \sum_B (\PP(B \mid A)
- \PP(B))}$\cloze{. Key: $\PP(B \mid A) - \P(B)$ is $0$ if
$A$ and $B$ are independent.}
\end{flashcard}
