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\begin{flashcard}[brooks-thm]
\begin{theorem}[Brooks' Theorem]
  \cloze{
  Let $G$ be a \gls{connected} \gls{graph}, not
  a \gls{complete_graph} or an odd length
  \gls{cycle}. Then $\chromnum(G) \le \maxdeg(G)$.
  }
\end{theorem}

\begin{proof}
  \cloze{
  Without loss of generality, assume $G$ is
  \kreg{\maxdeg} (by
  \nameref{lec12_final_remark}), and $\maxdeg(G)
  \ge 3$ ($\maxdeg(G) = 1$ is trivial, and
  $\maxdeg(G) = 2$ implies $G$ is a \gls{cycle},
  so trivial).

  If not true, let $G$ be a smallest
  counterexample (e.g. $\size{G}$ minimal).
  Without loss of generality, $G$ has no
  \gls{cutvertex}. Indeed, suppose $x$ is a
  \gls{cutvertex}. Let \glspl{conn_comp} of $G -
  x$, together with $x$, be $G_1, \ldots, G_k$.
  Then each $G_i$ is \rcolable{\maxdeg(G)} (since each
  $G_i$ is not \kreg{\maxdeg(G)}, as
  $\vdeg_{G_i}(x) \le \maxdeg(G) - 1$ $\forall n$)
  Hence $G$ is \rcol{\maxdeg(G)}.

  \textbf{Case 1:} $G$ is \con{3}: Want an
  ordering of the \glspl{vertex} as $x_1, \ldots,
  x_n$ such that for all $i < n$ there exists $j >
  i$ with $x_i x_j \in E$ and two members of
  $\nbd(x_n)$ get the same colour -- then done (as
  before, by running greedy). Pick any
  \gls{vertex} $x_n$. Cannot have $\nbd(x_n)$ a
  \gls{complete_graph}, elese $\nbd(x_n) \cup
  \{x_n\}$ is a $\complete_{\maxdeg(G) + 1}$, a
  contradiction as $G$ is not a
  \gls{complete_graph} (and is \gls{connected}),
  so some \gls{vertex} in this
  $\complete_{\maxdeg(G) + 1}$ is joined to some
  point outside it \contradiction.

  Choose $x_1, x_2 \in \nbd(x_n)$ with $x_1 x_2
  \not\in E$. Then $G - \{x_1, x_2\}$
  \gls{connected} (as $G$ is \con{3}), so can
  order it as $x_3, \ldots, x_n$ such that each
  $x_i$ has a forward edge (as before). Now run
  greedy algorithm on $x_1, \ldots, x_n$. This
  completes the first case.

  \textbf{Case 2:} $G$ not \gls{connected}: Let
  $\{x, y\}$ be a \gls{separator} of size $2$ (i.e. $G -
  \{x, y\}$ \glsref[connected]{disconnected}). Let
  $G_1, \ldots, G_k$ be \glspl{conn_comp} of $G -
  \{x, y\}$ with $x, y$ added. Then each $G_i$ is
  \rcol{\maxdeg(G)} (since $G_i$ not
  \kreg{\maxdeg(G)}, as $\vdeg_{G_i}(x),
  \vdeg_{G_i}(y) \le \maxdeg(G) - 1$ for all $i$).
  If $xy \in E$, then each of these \glspl{colring}
  give $x, y$ distinct \glspl{col}, so can
  \glsref[col]{recolour} them and combine to
  obtain a \rcolring{\maxdeg(G)} of $G$. So may
  assume $xy \not\in E$. Now, if each $G_i$ has
  $\vdeg_{G_k}(x) \le \maxdeg(G) - 2$ or
  $\vdeg_{G_i}(y) \le \maxdeg(G) - 2$. Then by
  \glsref[colring]{recolouring} we may assume that
  the \rcolring{\maxdeg(G)} of $G_i$ gives $x, y$
  distinct \glspl{col} -- so done as above. Hence
  some $G_i$ has $\vdeg_{G_i}(x) = \vdeg_{G_i}(y)
  = \maxdeg(G) - 1$ -- say $i = 1$. So $k = 2$,
  and $\vdeg_{G_2}(x) = \vdeg_{G_2}(y) = 1$. Let
  $\nbd_{G_2}(x) = \{u\}$. Then $\{y, u\}$ is a
  \gls{separator}, not of this form.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/57bee270797d459d.png}
  \end{center}
  }
\end{proof}
\end{flashcard}

\newpage
\section{Chromatic Polynomials}

\begin{flashcard}[chrom-poly-defn]
\begin{definition*}[Chromatic polynomial]
  \glsnoundefn{chrom_poly}{chromatic
  polynomial}{chromatic polynomials}
  \glssymboldefn{chrom_poly}{$P$}{$P$}
  \cloze{
  For a \gls{graph} $G$, and for $t = 1, 2, 3,
  \ldots$, write $P_G(t)$ for the number of
  \rcolrings{t} of $G$.\fcscrap{ (So $\chromnum(G)$
  is the least $t$ such that $P_G(t) \neq 0$).}
  $P_G$ is \emph{chromatic polynomial} of $G$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

Why is it a polynomial?

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\chrompoly_{\complete_n}(t) = t(t - 1)(t - 2)
      \cdots (t - n + 1)$.
    \item $\chrompoly_{\emptygraph_n}(t) = t^n$.
    \item $P_{\Ppath_n}(t) = t(t - 1)^n$.
      Similarly, for $T$ a \gls{tree} on $n$
      \glspl{vertex}, $\chrompoly_T(t) = t(t -
      1)^{n - 1}$ (remove a \gls{leaf} and
      induct).
    \item $\chrompoly_{\cycle_n}(t) = ?$. For the
      first \gls{vertex}, we have $t$ choices,
      then $t - 1$ for the next, then $t - 1$ for
      the next and so on, \ldots until we get to
      the final \gls{vertex}. Then the number of
      choices depends on whether the two
      \glspl{neighbour} are \glsref[col]{coloured}
      the same colour. It turns out that it is a
      bit complicated to work out the
      \gls{chrom_poly} in this case.
  \end{enumerate}
\end{example*}

\begin{flashcard}[contraction-defn]
\begin{definition*}[Contraction]
  For a \gls{graph} $G$ with \gls{edge} $e = xy$,
  the \emph{contraction} $G / e$ is \cloze{obtained by
  replacing the \glspl{vertex} $x$ and $y$ by a
  single \gls{vertex} $e$, joined to each
  \gls{neighbour} of $x$ or $y$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[deletion-contraction-lemma]
\begin{lemma}[deletion-contraction]
  \label{deletion_contraction}
  \cloze{
  Let $G$ be a \gls{graph}, $e$ an \gls{edge}.
  Then
  \[ \chrompoly_G
  = \chrompoly_{G - e} - \chrompoly_{G / e} .\]
  (This relation is sometimes
  called \emph{cut-fuse} relation).
  }
\end{lemma}

\begin{proof}
  \cloze{
  \glsref[col]{Colourings} of $G - e$ with
  endpoints of $e$ different \glspl{col} corresponds
  exactly to \glspl{colring} of $G$. Also,
  \glspl{colring} of $G - e$ with endpoints of $e$
  the same \gls{col} correspond to \glspl{colring}
  of $G / e$. Hence $\forall t$,
  \[ \chrompoly_{G - e}(t) = \chrompoly_G(t) +
  \chrompoly_{G / e}(t) \qedhere \]
}
\end{proof}
\end{flashcard}

\begin{note*}
  We cannot use \nameref{deletion_contraction}
  (together with $\chrompoly_{\emptygraph_n}(t) =
  t^n$) to define $\chrompoly_G$, as it might not
  be well-defined.
\end{note*}

\begin{proposition}
  Let $G$ be a \gls{graph} on $n$ \glspl{vertex},
  with $m$ \glspl{edge}. Then $\chrompoly_G$ is a
  polynomial in $t$, of \gls{v_deg} $n$, leading terms
  $t^n - mt^{n - 1} + \cdots$.
\end{proposition}

\begin{proof}
  Induction on $\edges(G)$. $\edges(G) = 0$
  trivial ($\chrompoly_{\emptygraph_n}(t) = t^n$).

  Given $G$, $\edges(G) > 0$, pick $e \in E$. Have
  for all $t$, (\nameref{deletion_contraction})
  \[ \chrompoly_G(t) = \chrompoly_{G - e}(t) -
  \chrompoly_{G / e}(t) .\]
  By induction,
  \begin{align*}
    \chrompoly_{G - e}(t)
    &= t^n - (m - 1)t^{n - 1} + \cdots \\
    \chrompoly_{G / e}
    &= t^{n - 1} - \cdots
  \end{align*}
  so
  \[ \chrompoly_G(t) = t^n - m t^{n - 1} + \cdots
  \qedhere \]
\end{proof}
