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\begin{flashcard}[general-nmf-ineq]
\prompt{Inequality for plane $G$ on general
surface?

}
\cloze{For $G$ drawn on the \gls{surf} of \gls{euler_char}
$\eulerchar$, have $n - m + f \ge \eulerchar$.}
\end{flashcard}

\begin{flashcard}[general-planar-simple-ineq]
\prompt{Prove planar graph $m, n$ inequality for
general surface.

}
\cloze{We know $3f \le 2m$ (as usual by counting
edges via faces / vertices), for $m \ge 3$. So $n
- m + \frac{2m}{3} \ge \eulerchar$, i.e. $n - \frac{m}{3}
\ge \eulerchar$, so $m \le (n - \eulerchar)$.}
\end{flashcard}

\begin{example*}
  We can draw $\complete_5$ on a torus:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/860fa665bdea4e69.png}
  \end{center}
  Exercise: draw $\complete_6$ and $\complete_7$
  on a torus.
\end{example*}

What can we say about $\chromnum(G)$?

\begin{flashcard}[heawood-thm]
\begin{theorem}[Heawood's Theorem]
  % Theorem 8
  \glssymboldefn{heawood}{$H$}{$H$}
  \label{heawood_thm}
  \cloze{
  Let $G$ be a \gls{graph} drawn on a \gls{surf}
  of \gls{euler_char} $\eulerchar \le 0$. Then
  \[ \chromnum(G) \le \heawood(\eulerchar) \le \left\lfloor
  \frac{7 + \sqrt{49 - 24\eulerchar}}{2}
  \right\rfloor \]
  }
\end{theorem}
\fcscrap{
\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\heawood(0) = 7$. So our $\complete_7$
      cannot be improved, and nor can the bound
      $\chromnum(G) \le 7$.
    \item Amusingly, $\heawood(2) = 4$.
  \end{enumerate}
\end{remark*}
}
\begin{proof}
  \cloze{Let $G$ have $\chromnum(G) = k$. Need to show $k
  \le \heawood(\eulerchar)$. Pick $G$ minimal (least
  $n$) with $\chromnum(G) = k$. So certainly $n
  \ge k$. Also, $\mindeg(G) \ge k - 1$ (by
  minimality of $G$). Now, from $m \le 3(n -
  \eulerchar)$ we have $2m \le 6(n - \eulerchar)$,
  so average \gls{v_deg} $\le 6 -
  \frac{\eulerchar}{2}$, and in particular
  $\mindeg(G) \le 6 - \frac{6\eulerchar}{2}$.
  Hence $k - 1 \le 6 - \frac{\eulerchar}{2} \le 6
  - \frac{6\eulerchar}{k}$ (as $n \ge k$ and
  $\eulerchar \le 0$). So $k^2 - k \le 6k -
  6\eulerchar$, i.e. $k^2 - 7k + 6\eulerchar \le
  0$ -- whence bound (solve quadratic).}
\end{proof}
\end{flashcard}

\begin{remark*}
  \nameref{heawood_thm} is best possible -- can
  draw $\complete_{\heawood(\eulerchar)}$ on
  surface (this is the ``Map Color Theorem'',
  1960s).
\end{remark*}

\newpage
\mychapter{Ramsey Theory}

The philosophical question guiding this chapter is:
\begin{center}
  ``Can we find some order in enough disorder?''
\end{center}

\begin{example*}
  Suppose we have $c : \eulerchar(\complete_6) \to
  \{1, 2\}$ (2-coloured $\complete_6$). Can we
  find a triangle, all \glspl{edge} red or all
  \glspl{edge} blue (a \emph{monochromatic} triangle)?
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/eb67a64a72934cbb.png}
  \end{center}
  Answer: yes. Pick \gls{vertex} $x$. Have
  $\vdeg(x) = 5$, so there exists $\ge 3$
  \glspl{edge} out of $x$ of the same colour: say
  $xy_1, xy_2, xy_3$ red. If any $y_i y_j$ red
  then we have found a red triangle $(x, y_i,
  y_j)$. If all $y_i y_j$ blue then we have a blue
  triangle $(y_1, y_2, y_3)$.
\end{example*}
\vspace{-1em}

How about a monochromatic $\complete_4$?

\begin{flashcard}[ramsey-number-defn]
\begin{definition*}
  \glsnoundefn{ramsey_num}{Ramsey number}{Ramsey
  numbers}
  \glssymboldefn{ramsey_num}{$R(s)$}{$R(s)$}
  \cloze{Write} $\ramsey(s)$ \cloze{for the least $n$
  \fcscrap{(if it exists) }such that whenever
  $\complete_n$ is 2-coloured, there exists a
  monochromatic $\complete_s$.}\fcscrap{

  (Equivalently, least $n$ such that every
  \gls{graph} $G$ on $n$ vertices has
  $\complete_s \subset G$ or $\complete_s \subset
  \complement{G}$).
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

So the above proof shows $\ramsey(3) \le 6$. In
fact, $\ramsey(3) = 6$, since there is a colouring
of $\complete_5$ which has no monochromatic
triangle:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/32993601ed8146d3.png}
\end{center}
What about $\ramsey(4)$? We'll go for a `halfway
house' of finding a red $\complete_3$ or a blue
$\complete_4$.

\begin{flashcard}[ramseyd-number-defn]
\begin{definition*}
  \glssymboldefn{ramseyd_num}{$R$}{$R$}
  For \cloze{$s, t \ge 2$} write $\ramseyd(s, t)$
  for \cloze{for the least $n$ (if it exists) such
  that whenever $\complete_n$ is 2-coloured, there
  exists red $\complete_s$ or blue $\complete_t$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\ramseyd(s, s) = \ramsey(d)$.
    \item $\ramseyd(s, t) = \ramsey(t, s)$.
    \item $\ramseyd(s, 2) = s$.
  \end{enumerate}
\end{example*}

\begin{flashcard}[ramseys-thm]
\begin{theorem}[Ramsey's Theorem]
  % Theorem 1
  \label{ramseys_thm}
  \cloze{$\ramseyd(s, t)$ exists $\forall s, t$.
  Moreover, $\ramseyd(s, t) \le \ramseyd(s - 1, t)
  + \ramseyd(s, t - 1) ~\forall s, t \ge 3$. }
\end{theorem}

\begin{proof}
  \cloze{Enough to show that, if $\ramseyd(s - 1, t)$ and
  $\ramseyd(s, t - 1)$ are finite, then
  \[ \ramseyd(s, t) \le \ramseyd(s - 1, t) +
  \ramseyd(s, t - 1) \]
  As then $\ramsey(s, t)$ finite for all $s, t$
  (induction on $s + t$)!

  Let $a = \ramseyd(s - 1, t)$, $b = \ramseyd(s, t
  - 1)$. Given a 2-colouring of $\complete_{a +
  b}$: pick a \gls{vertex} $x$. Have $\vdeg(x) = a + b -
  1$. So there exists $a$ red \glspl{edge} or there
  exists $b$ blue \glspl{edge} from $x$.

  If $a$ red \glspl{edge}: say $xy_1, \ldots, x y_a$ are
  all red. In the $\complete_a$ given by
  \glspl{vertex} $y_1, \ldots, y_a$. Have red
  $\complete_{s - 1}$ or blue $\complete_t$. If we
  have a blue $\complete_t$, then done. If we have
  a red $\complete_{s - 1}$, then when combined
  with $x$, we have a $\complete_s$.

  If $b$ blue \glspl{edge}, similar.}
\end{proof}
\end{flashcard}

\begin{remark*}
  Very few of these `\glspl{ramsey_num}' $\ramseyd(s,
  t)$ are known exactly (see later). Question: how
  fast does $\ramsey(s)$ grow?
\end{remark*}