%! TEX root = GT.tex
% vim: tw=50
% 28/10/2023 09AM

\glssymboldefn{blown_up_complete}{$K_r(t)$}{$K_r(t)$}
Write $K_r(t) = K_r(rt)$ ($\complete_r$ but with each
point `blown up' to a set of size $t$).
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/4ee84fd561ac4def.png}
\end{center}
\nameref{turans_thm} says
\[ \frac{e(G)}{{n \choose 2}} > 1 - \frac{1}{r -
1} \implies G \supset \complete_r \]
What if $\frac{e(G)}{{n \choose 2}} > 1 -
\frac{1}{r - 1} + 0.01$? Remarkably, this implies
that $G \supset \bucomplete_r(1000)$ ($n$ large).

\begin{theorem*}[Erd\H{o}s-Stone Theorem]
  \label{erdos_stone_thm}
  For all integers $r$, $t$, and $\eps > 0$, there
  exists $n_0(r, \eps, t)$ such that for all $n
  \ge n_0$,
  \[ |G| = n, \frac{e(G)}{{n \choose 2}} > 1 -
  \frac{1}{r - 1} + \eps \implies G \supset
  \bucomplete_r(t) .\]
\end{theorem*}

\begin{proof}[Sketch proof]
  Have $G$, $|G| = n$, average \gls{v_deg} $> \left(1 -
  \frac{1}{r - 1} + \eps\right)n$.
  \begin{enumerate}[(1)]
    \item Pass to a \gls{subgraph} $H$ on $n'$
      \glspl{vertex} ($n'$ still large), with
      $\mindeg(H) > \left( 1 - \frac{1}{r - 1} +
      \eps \right) n'$ (similar to \es[7]{1}).
    \item $H$ contains a $\bucomplete_{r - 1}(t')$
      ($t'$ large), = $K$ say (induction on $r$).
    \item Each $x \in H - K$ has $\ge t$
      neighbours in each class of $K$ (by
      $\mindeg(H)$).
    \item Get $t$ points $x_1, \ldots, x_t \in H -
      K$, each joined to same $t$-set in each
      class of $K$ (by pigeonhole). So have
      $\bucomplete_r(t)$. \qedhere
  \end{enumerate}
\end{proof}

Given $H$, let $r$ be the least such that $H$ is
\rpartite{r}. Then $H \not\supset \turan_{r -
1}(n)$ (as $\turan_{r - 1}(n)$ is \rpartite{(r -
1)}), so
\[ \frac{\ex(n, H)}{{n \choose 2}} \ge 1 -
\frac{1}{r - 1} .\]
However, $H \subset \bucomplete_r(t)$, some $t$
(as $H$ is \rpartite{r}), so
\nameref{erdos_stone_thm} tells us that
\[ \frac{e(G)}{{n \choose 2}} > 1 - \frac{1}{r -
1} + \eps \implies G > H .\]
\textbf{Conclusion:}
\[ \frac{\ex(n, H)}{{n \choose 2}} \to 1 -
\frac{1}{r - 1} \]
as $n \to \infty$.

\begin{remark*}
  If $H$ bipartite, this says that $\frac{\ex(n,
  H)}{{n \choose 2}} \to 0$. How fast does $\ex(n,
  H)$ grow? This is unknown even for many very
  simple $H$, for example $C_{2k}$, $k \ge 6$.
\end{remark*}
\vspace{-1em}

This marks the end of this subsection of
non-examinable content.

\newpage
\mychapter{Colourings}

\newpage

\begin{flashcard}[r-colouring-defn]
\begin{definition*}[$r$-colouring]
  \glssymboldefn{colset}{$[r]$}{$[r]$}
  \glsnoundefn{r_col}{$r$-colouring}{$r$-colouring}
  \glspropdefn{chrom_num}{chromatic
  number}{\gls{graph}}
  \glssymboldefn{chrom_num}{$\chi(G)$}{$\chi(G)$}
  \glsnoundefn{colring}{colouring}{colourings}
  \glsnoundefn{col}{colour}{colours}
  \cloze{
  An \emph{$r$-clouring} of a \gls{graph} $G$ is a
  function $c : V(G) \to [r] \defeq = \{1, \ldots,
  r\}$ such that $xy \in E(G) \implies c(x) \neq
  c(y)$.

  The \emph{chromatic number} $\chromnum(G)$ of
  $G$ is the least $r$ for which $G$ has an
  \gls{r_col}.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\chromnum(\Ppath_n) = 2$.
    \item $\chromnum(C_n) = \begin{cases}
        2 & \text{$n$ even} \\
        3 & \text{$n$ odd}
      \end{cases}$.
    \item $\chromnum(K_n) = n$ (each point gets a
      different colour).
    \item $T$ a \gls{tree} $\implies \chromnum(T) = 2$
      (for example remove a \gls{leaf} + induction).
    \item $\chromnum(\bipartite_{m, n}) = 2$.
  \end{enumerate}
\end{example*}

\begin{note*}
  $G$ \gls{bipartite} $\implies$ $G$ is
  \glsref[r_col]{$2$-colouring}. Conversely, $G$
  \glsref[r_col]{$2$-colourable} $\implies$ $G$
  \gls{bipartite} (as can take $X = \{x : c(x) =
  1\}$, $Y = \{x : c(x) = 2\}$). So, $G$
  \gls{bipartite} $\iff$ $\chromnum(G) \le 2$.

  In general, $G$ \rpartite{r} $\iff$
  $\chromnum(G) \le r$. So $\chromnum(G)$ is the
  least $r$ such that $G$ is \rpartite{r}. Thus
  \hyperlink{erdos_stone_thm}{Erd\H{o}s-Stone
  Corollary} says
  \[ \frac{\ex(n, H)}{{n \choose 2}} \to 1 -
  \frac{1}{\chromnum(H) - 1} \]
\end{note*}
\vspace{-1em}

For any $G$ on $n$ \glspl{vertex}, have $\chromnum(G)
\le n$. But can often improve this:

\begin{flashcard}[colouring-max-degree-prop]
\begin{proposition}
  \label{colouring_max_deg}
  For any \gls{graph} $G$, have $\chromnum(G) \le
  \cloze{\maxdeg(G) + 1}$.
\end{proposition}

\begin{proof}
  \cloze{
  Order $\vset(G)$ as $x_1, \ldots, x_n$, and
  colour each $x_i$ in turn. When we come to
  colour $x_i$, it has $\le \maxdeg(G)$
  \glspl{neighbour}, so we have $\le \maxdeg(G)$
  colours to avoid.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Can have equality, for example
      \gls{complete_graph} or odd \gls{cycle}.
    \item Often $\chromnum(G)$ is much less than
      $\maxdeg(G)$, for example $\bipartite_{1,n -
      1}$.
    \item Could view \cref{colouring_max_deg} as
      an application of the \emph{greedy
      algorithm} on ordering $x_1, \ldots, x_n$:
      when we came to colour \gls{vertex} $x_i$,
      we use least colour available.
    \item Greedy might use $> \chromnum(G)$
      colours, for example
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/c7195def88484716.png}
      \end{center}
    \item No `formula' for $\chromnum(G)$.
  \end{enumerate}
\end{remark*}

\newpage
\section{Colouring Planar Graphs}

\begin{flashcard}[6-colour-thm]
\begin{proposition}[$6$-colour Theorem]
  Let $G$ be \gls{planar}. Then $\chromnum(G) \le
  6$.
\end{proposition}

\begin{proof}
  \cloze{
  Clearly this works if $|G| \le 6$. Hence consider
  $|G| = n \ge 7$.

  We claim $\mindeg(G) \le 5$. Indeed, we have
  $\edges(G) \le 3n - 6$, so $\sum_{x \in G}
  \vdeg(x) \le 6n - 12$, so there is an $x$ with
  $\vdeg(x) \le 5$.

  Choose $x \in G$ with $\vdeg(x) \le 5$, and let
  $G' = G - x$. We can \glsref[r_col]{$6$-colour}
  $G'$ by induction on $n$. Now, $|\nbd(x)| \le
  5$, so $\nbd(x)$ has at most $5$ colours. Hence
  we can colour $x$ with a $6$-th colour.
  }
\end{proof}
\end{flashcard}

How about $5$ colours? If $\text{G \gls{planar}}
\implies \mindeg(G) \le 4$, then the same proof
would work. However this is not true, for example
the icosahedron has $\mindeg(G) = 5$ (or a
football).