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\begin{flashcard}
\begin{theorem}[5-colour Theorem]
  \label{5_col_thm}
  \cloze{$G$ \gls{planar} $\implies$ $G$
  \glsref[r_col]{$5$-colourable}.}
\end{theorem}

\begin{proof}
  \cloze{
  Induction on $\size{G}$. $\size{G} \le 5$ trivial.

  Given \gls{planar} $G$, $\size{G} \ge 6$: have
  $x \in G$ with $\vdeg(x) \le 5$ (as before). So
  remove $x$. By induction, we can \rcol{5} $G -
  x$. Can now colour $x$, unless $\vdeg(x) = 5$
  and all colours in appear in $\nbd(x)$. Say
  $\nbd(x) = \{x_1, x_2, \ldots, x_3\}$
  (clockwise) with $x_i$ of colour $i$ for all
  $i$.

  \textbf{Question:} Is there a $1-3$ path from
  $x_3$ to $x_1$ (a $1-3$ \gls{path} is a
  \gls{path} that
  alternates between colours $1$ and $3$)?

  If no, let $H$ be the $1-3$ component of
  $x_3$ (all the vertices that are part of a $1-3$
  \gls{path} starting at $x_3$). Then $x_1 \not\in
  H$. So flip colours $1$ and $3$ on $H$. Still a
  legal \gls{colring} of $G - x$, but now can use
  colour $3$ for $x$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/2280341957834706.png}
  \end{center}

  If yes, then there is no $2-4$ path from $x_2$
  to $x_4$ (else it would meet the $1-3$ path from
  $x_1$ to $x_3$ as our drawing is planar).
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/cac0094c402f492e.png}
  \end{center}
  (even if our paths go the other way round, we
  can see that in any of the 4 cases, they must
  intersect).
  So done as before: swap $2$ and $4$ on the $2-4$
  component of $x_2$, leaving $2$ as a colour for
  $x$.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item These $i-j$ \glspl{path} are called
      `Kempe chains'.
    \item What if we wanted to colour the
      \glspl{face} of $G$ (so that \glspl{face} sharing
      an \gls{edge} get different colours)? Called
      `colouring a plane map'.

      \begin{definition*}[Dual graph]
        \vspace{0.5em}
        For a \gls{plane} \gls{graph} $G$, the
        \emph{dual graph} $G'$ has \glspl{vertex}
        being the \glspl{face} of $G$, with two
        joined if they share an \gls{edge}.
        Clearly $G'$ \gls{planar}.
      \end{definition*}
      \vspace{-1em}

      Then a colouring of the \glspl{face} of $G$
      is precisely a \gls{colring} of $G'$. So
      \cref{5_col_thm} tells us that every
      \gls{plane} map is \rcolable{5}.
    \item Ofcourse, can have \gls{planar} $G$ with
      $\chromnum(G) = 4$. For example,
      $\complete_4$.
  \end{enumerate}
\end{remark*}
\vspace{-1em}

The following theorem is non-examinable.

\begin{theorem*}[4-colour Theorem]
  \label{4_col_thm}
  $G$ \gls{planar} $\implies$ $G$ \rcolable{4}.
\end{theorem*}

\begin{proof}[``Proof'']
  Induction on $\size{G}$: $\size{G} \le 4$
  trivial.

  Given \gls{planar} $G$, $\size{G} \ge 5$: have
  $x \in G$ with $\vdeg(x) \le 5$ (as before). So
  remove $x$: can \rcol{4} $G - x$ (induction). So
  done unless all $4$ colours appear in $\nbd(x)$.

  If $\vdeg(x) = 4$: Have say $\nbd(x) = \{x_1,
  \ldots, x_4\}$ (clockwise), with $c(x_i) = i$
  for all $i$. If no $1-3$ path from $x_1$ to
  $x_3$, swap $1$ and $3$ on the $1-3$ component
  of $x_3$ (leaving colour $3$ for $x$). If there
  exists $1-3$ path from $x_1$ to $x_3$, then no
  $2-4$ path from $x_4$ to $x_2$ (as $G$
  \gls{planar}), so done as before.

  If $\vdeg(x) = 5$, could have:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/c36b30c8edb543cd.png}
  \end{center}
  First case: if no $2-4$ \gls{path} from $x_2$ to
  $x_4$, then done as usual. If there exists a
  $2-4$ \gls{path} from $x_2$ to $x_4$, then no $1-3$
  \gls{path} from $x_3$ to $x_1$ or $x_1'$. So
  swap $1$ and $3$ on $1-3$ component of $x_3$,
  leaving colour $3$ available for $x$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/96761f1df022462a.png}
  \end{center}
  Second case: without loss of generality $\exists$ $2-3$ path from
  $x_2$ to $x_3$ (else done), so no $1-4$ path
  from $x_1'$ to $x_4$. Also without loss of
  generality $\exists$ $2-4$ path from $x_2$ to
  $x_4$ (else done), so there is no $1-3$ path
  from $x_1$ to $x_3$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/1b2d3213cfa740c2.png}
  \end{center}
  So done: swap $1$ and $4$ on the $1-4$ component
  of $x_1'$ and $1$ and $3$ on the $1-3$ component
  of $x_1$. Then we can use colour $1$ for $x$.
\end{proof}

This ``Proof'' was given by Kempe in 1879. In
1890, he found a mistake in the proof. Where is
the mistake?

The \nameref{4_col_thm} was proved in 1976 by
Appel and Haken.

In the proof of \cref{5_col_thm}, we showed that
``a vertex of degree $5$ or $4$ or \ldots or
$0$''. Formed an unavoidable set of reducible
configurations.

Appel and Haken found such a set of about 1900
configurations (computers used).

The following content is now examinable again.

\bigskip
\bigskip

We know any graph $G$ has $\chromnum(G) \le
\maxdeg(G) + 1$. Can have equality, for example
$\complete_n$, or $\cycle_\text{odd}$.

\textbf{Aim:} $\chromnum(G) \le \maxdeg(G)$ unless
$G = \complete_n$ or $\cycle_\text{odd}$ (for $G$
\gls{connected}).

\begin{remark*}
  \refsteplabel[remark above]{lec12_final_remark}
  If (\gls{connected}) $G$ not \gls{regular}, then can
  always \gls{col} in $\maxdeg(G)$ colours.
  Indeed, choose $x_n \in G$ with $\vdeg(x) \le
  \maxdeg(G) - 1$. Then choose $x_{n - 1}$ with
  $x_n \in \nbd(x_{n - 1})$ ($G$ \gls{connected}),
  choose $x_{n - 2}$ with $\nbd(x_{n - 2})$
  meeting $\{x_{n - 1}, x_n\}$ ($G$
  \gls{connected})\ldots Keep going. We obtain
  $x_n, \ldots, x_1$ such that every $x_i$ ($i \le
  n$) has a `forward edge', i.e. there exists $j >
  i$ with $x_i x_j \in E$. Now just run greedy on
  $x_1, \ldots, x_n$. At $x_i$ we have $\le
  \maxdeg(G) - 1$ \glspl{neighbour}
  \glsref[col]{coloured} (all $i$), so greedy uses
  $\le \maxdeg(G)$ \glsref[col]{colours}.
\end{remark*}