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Similarly:

\begin{proposition}
  \label{min_deg_min_path}
  Let $G$ be a \gls{connected} \gls{graph} on $n$
  \glspl{vertex} ($n \ge 3$). Then $\mindeg(G) \ge
  \frac{k}{2}$ (for some $n$) $\implies$ $G$ has a
  \gls{path} of length $k$.
\end{proposition}

\begin{remark*}
  Do need $G$ \gls{connected} -- for example:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/69f87eaf90c94ce7.png}
  \end{center}
\end{remark*}

\begin{proof}
  Let $x_1, \ldots, x_l$ be a longest \gls{path}
  in $G$ ($l \ge 3$ since $G$ \gls{connected}, $|G| >
  2$). Suppose $l \le k$. Then $G$ has no
  $l$-\gls{cycle} (as before), and $\Gamma(x_1),
  \Gamma^+(x_l) \subset \{x_2, \ldots, x_l\}$ and
  are disjoint (as before). But $|\Gamma(x_1)|,
  |\Gamma^+(x_l)| \ge \frac{k}{2}$ and $|\{x_2,
  \ldots, x_l\}| < k$, \contradiction.
\end{proof}

\begin{theorem}
  \label{number_of_edges_gives_path}
  Let $G$ a \gls{graph} on $n$ \glspl{vertex}.
  Then $e(G) > \frac{n(k - 1)}{2} \implies G
  \supset P_k$.
\end{theorem}

\begin{note*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Equivalently: if $G \not\supset P_k$
      then $e(G) \le \frac{n(k - 1)}{2}$.
    \item Bound is best possible, e.g. if $k \mid
      n$:
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/9a5b4e5f4d874a97.png}
      \end{center}
  \end{enumerate}
\end{note*}

\begin{proof}
  Induction on $n$, $n \ge 2$.

  Given $G$ on $n \ge 3$ \glspl{vertex}, with $G
  \not\supset P_k$: want $e(G) \le \frac{n(k -
  1)}{2}$. Without loss of generality, $G$
  \gls{connected}: if $G$
  \glsref[connected]{disconnected}, with
  \glspl{conn_comp} $G_1, \ldots, G_r$ on $n_1,
  \ldots, n_r$ \glspl{vertex}, then $e(G_i) \le
  \frac{n_i(k - 1)}{2}$ (induction), so
  \[ e(G) \le \sum \frac{n_i(k - 2)}{2} =
  \frac{n(k - 1)}{2} .\]
  Hence, by \cref{min_deg_min_path}, $G$ has a
  \gls{vertex} $x$ of
  degee $\le \frac{k - 1}{2}$ ($e(G) \le \frac{n(k
  - 1)}{2}$ trivial if $k > n$, so without loss of
  generality $k < n$).

  Let $G' = G - x$, then $|G'| = n - 1$, $G'
  \not\supset P_k$, so $e(G') \le \frac{(n -
  1)(k - 1)}{2}$. So $e(G) = e(G') + \vdeg(x) \le
  \frac{(n - 1)(k - 1)}{2} + \frac{k - 1}{2} =
  \frac{n(k - 1)}{2}$.
\end{proof}

Often we ask: How ``large'' can a \gls{graph} be
with a certain property? These are called
\emph{extremal problems}. Often, this property is
non-containment of a particular \gls{subgraph}.

For example, \cref{min_deg_implies_ham} is about
$\cycle_n$, and
\cref{number_of_edges_gives_path}. What about
\glspl{complete_graph}?

\subsection{Tur\'an's theorem}

How many \glspl{edge} can a \gls{graph} (on $k$
\glspl{vertex}) have without containing a
$\complete_k$?

\begin{example*}
  We'd try $G$ \gls{bipartite}, indeed
  \gls{complete_bipartite}, so $G = \bipartite_{a, b}$
  where $a + b = n$.
  We'd want $a + b = \frac{n}{2}$ ($n$ even), $a =
  \frac{n + 1}{2}$, $b = \frac{n - 1}{2}$ if $n$
  odd.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/5caf10c7f172483a.png}
  \end{center}
  $k = 4$, might try:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e14bf1479d9845a9.png}
  \end{center}
\end{example*}

\begin{flashcard}[r-partite-dn]
\begin{definition*}[$r$-partite]
  \glsnoundefn{r_partite}{$r$-partite}{$r$-partite}
  \cloze{Say \gls{graph} $G$ is \emph{$r$-partite} if there
  exists a partition $V = V_1 \cup \cdots \cup
  V_r$ of $V$ such that $E(V_i) = \emptyset
  ~\forall n$.

  It is \emph{complete $r$-partite} if $E = \{xy :
  x \in V_i, y \in V_j, i \neq j\}$. So $G$ is $k
  - 1$-partite $\implies$ $G$ has no $\complete_k$ (else
  $2$ points of $\complete_k$ in same class,
  \contradiction).
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[turan-graph-defn]
\begin{definition*}[Tur\'an graph]
  \glssymboldefn{turan}{Turan}{graph}
  \cloze{The \emph{Tur\'an graph} $T_r(n)$ is the
  complete \gls{r_partite} \gls{graph} on
  $n$ \glspl{vertex} with classes as equal as
  possible. ($a_1, \ldots, a_r$ are as equal as
  possible if $|a_i - a_j| \le 1 ~\forall i, j$).}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/4bf0941b6b794837.png}
  \end{center}
  Certainly $\turan_{k - 1}(n) \not\supset
  \complete_k$. Also, $\turan_{k - 1}(n)$ is
  \emph{maximal} $\complete_k$-free: $\turan_{k -
  1}(n) + e$ always has a $\complete_k$.
\end{example*}
\vspace{-1em}

If $r \mid n$: all points of $\turan_r(n)$ have
\gls{v_deg} $n - \frac{n}{r} = \left( 1 -
\frac{1}{r} \right)n$.

If not, then all points have \gls{v_deg} $n -
\left\lfloor \frac{n}{r} \right\rfloor$ or $n -
\left\lceil \frac{n}{r} \right\rceil$.

\begin{note*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item To obtain $\turan_r(n - 1)$ from
      $\turan_r(n)$, remove a point from a largest
      class (i.e. a point of minimum \gls{v_deg}).
    \item To obtain $\turan_r(n + 1)$ from
      $\turan_r(n)$, add a point to a smallest
      class.
  \end{enumerate}
\end{note*}

\begin{flashcard}[turan-thm]
\begin{theorem}[Tur\'an's theorem]
  \label{turans_thm}
  \cloze{Let $G$ be a \gls{graph} on $n$ \glspl{vertex}
  with $\edges(G) > e(\turan_{k - 1}(n))$. Then $G
  \supset K_k$.}
\end{theorem}

\begin{remark*}
  \cloze{\phantom{}
  \begin{enumerate}[(1)]
    \item Equivalently, if $G \not\supset
      \complete_k$, then $\edges(G) \le
      \edges(\turan_{k - 1}(n))$.
    \item If we know $G$ is \glsref[r_partite]{$(k
      - 1)$-partite}, we'd be done by some form of
      AM-GM. But no reason why $G$ should be
      \glsref[r_partite]{$(k - 1)$-partite}. For
      example $C_5$ does not contain
      $\complete_3$, but is not \gls{bipartite}.
    \item Looks like proof has to be fiddly and
      nasty, because $\edges(\turan{k - 1}(n))$ is
      an unpleasant formula if $n$ is not a
      multiple of $k - 1$.
    \item We'll actually prove a stronger result:
      $\edges(G) = \edges(\turan_{k - 1}(n))$, $G
      \not\supset \complete_k$ $\implies$ $G \cong
      \turan_{k - 1}(n)$. (``$\turan_{k - 1}(n)$
      is the unique winner''). This does imply
      \nameref{turans_thm}, as cannot add an
      \gls{edge} to $\turan_{k - 1}(n)$.
  \end{enumerate}
}
\end{remark*}

\begin{proof}[Proof of \nameref{turans_thm}]
  \cloze{Induction on $n$, $n \le k - 1$. Given $G$, $|G|
  = n$, $\edges(G) = \edges(\turan(n))$, $G
  \not\supset \complete_k$,
  want $G \gcong \turan(n)$.

  \textbf{Claim:} $\mindeg(G) \le \mindeg(\turan(n))$.
  This is because we have $\sum_{x \in G}
  \vdeg_G(x) = \sum_{y \in \turan(n)} \vdeg_{\turan(n)}(y)$.
  But the $\vdeg_{\turan(n)}(y)$ are as equal as
  possible. So $\mindeg(G) \le
  \mindeg(\turan(n))$.

  Pick $x \in G$ of minimal \gls{v_deg}, and put
  $G' = G - x$. $\size{G'} = n - 1$, $G'
  \not\supset \complete_k$, and $\edges(G') =
  \edges(G) - \mindeg(G) \ge \edges(\turan(n)) -
  \mindeg(\turan(n)) = \edges(\turan(n - 1))$. So
  $G' \gcong \turan(n - 1)$ (induction) with
  $\mindeg(G) = \mindeg(\turan(n))$: say classes
  $V_1, \ldots, V_{k - 1}$.

  Must have, in $G$, that $\nbd(x)$ misses some
  $V_i$ (if $\nbd(x)$ meets each $V_i$ then $G
  \supset \complete_k$, \contradiction). But
  $|\nbd(x)| = n - 1 - \min\size{V_i}$, as this is
  $\mindeg(\turan(n))$.

  Hence $\nbd(x) = \bigcup_{j \neq i} V_j$ for
  some $i$ such that $\size{V_i}$ is of minimum
  size.

  So $G$ is complete \rpartite{(k - 1)},
  \gls{vertex}
  classes $V_j$ (each $j \neq i$) and $V_i \cup
  \{x\}$, so $G \gcong \turan(n)$.}
\end{proof}
\end{flashcard}