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MATH 239: added March 20, 2013 lecture.
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We can verify this. If $(k, l) = (3, 5)$, then $s = \frac{12}{10 + 6 - 15} = 12$, as expected. We can verify this. If $(k, l) = (3, 5)$, then $s = \frac{12}{10 + 6 - 15} = 12$, as expected.


% Eric Katz gave this lecture.
\subsubsection{Non-Planar Graphs} \lecture{March 20, 2013}
\begin{center}
\begin{tikzpicture}[scale=1.5]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (2,0) {};
\node [dot={0}{}] at (0,2) {};
\node [dot={0}{}] at (2,2) {};
\node [dot={0}{}] at (1,3) {};

\draw (0,0) -- (2,0) -- (2,2) -- (1,3) -- (0,2) -- (0,0);
\draw (2,0) -- (0,2) -- (2,2);
\draw (0,0) -- (2,2);
\draw (0,0) to[in=180,out=120] (1,3);
\draw (2,0) to[in=0,out=60] (1,3);
\node[draw=none] at (1,-0.25) {$K_5$};
\end{tikzpicture}
\begin{tikzpicture}[scale=1.5]
\node [dot={0}{}] at (0,1) {};
\node [dot={0}{}] at (1,0) {};
\node [dot={0}{}] at (1,2) {};
\node [dot={0}{}] at (3,0) {};
\node [dot={0}{}] at (3,2) {};
\node [dot={0}{}] at (4,1) {};

\draw (1,2) -- (0,1) -- (1,0) -- (3,0) -- (4,1) -- (3,2) -- (1,2);
\draw (1,0) -- (3,2);
\draw (1,2) -- (3,0);
\draw (0,1) to[out=90,in=180] (2,2.5) to[out=0,in=90] (4,1);
\node[draw=none] at (2, -0.25) {$K_{3, 3}$};
\end{tikzpicture}
\end{center}
$K_5$ and $K_{3,3}$ are non-planar because they have edges that unavoidably must cross. Are there any other planar graphs? Yes! You can just add more edges to $K_5$ or $K_{3,3}$. We could also add some vertices in the middle or edges, if we'd like.
\\ \\
The two following graphs are also non-planar, and are built by adding more edges and vertices to $K_5$, respectively.
\begin{center}
\begin{tikzpicture}[scale=1.5]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (2,0) {};
\node [dot={0}{}] at (0,2) {};
\node [dot={0}{}] at (2,2) {};
\node [dot={0}{}] at (1,2) {};
\node [dot={0}{}] at (1,3) {};

\draw (0,0) -- (2,0) -- (2,2) -- (1,3) -- (0,2) -- (0,0);
\draw (2,0) -- (0,2) -- (2,2);
\draw (0,0) -- (2,2);
\draw (0,0) to[in=180,out=120] (1,3);
\draw (2,0) to[in=0,out=60] (1,3);
\draw (1,3) -- (1,2) -- (0,0);
\draw (1,2) -- (2,0);
\end{tikzpicture}
\begin{tikzpicture}[scale=1.5]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (2,0) {};
\node [dot={0}{}] at (0,2) {};
\node [dot={0}{}] at (2,2) {};
\node [dot={0}{}] at (1,3) {};
\node [dot={0}{}] at (1,2) {};
\node [dot={0}{}] at (0,0.25) {};
\node [dot={0}{}] at (0,0.5) {};
\node [dot={0}{}] at (0,0.75) {};
\node [dot={0}{}] at (0,1) {};
\node [dot={0}{}] at (0,1.25) {};
\node [dot={0}{}] at (0,1.5) {};
\node [dot={0}{}] at (0,1.75) {};
\node [dot={0}{}] at (2,0.25) {};
\node [dot={0}{}] at (2,0.5) {};
\node [dot={0}{}] at (2,0.75) {};
\node [dot={0}{}] at (2,1) {};
\node [dot={0}{}] at (2,1.25) {};
\node [dot={0}{}] at (2,1.5) {};
\node [dot={0}{}] at (2,1.75) {};

\draw (0,0) -- (2,0) -- (2,2) -- (1,3) -- (0,2) -- (0,0);
\draw (2,0) -- (0,2) -- (2,2);
\draw (0,0) -- (2,2);
\draw (0,0) to[in=180,out=120] (1,3);
\draw (2,0) to[in=0,out=60] (1,3);
\end{tikzpicture}
\end{center}

\begin{defn}
A graph $H$ is a \textbf{subdivision} of a graph $G$ if $H$ is obtained from $G$ by replacing each edge by a path.
\end{defn}

\begin{ex}
Here's a graph $G$ and a subdivision of $G$.
\begin{center}
\begin{tikzpicture}[scale=2]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (0,1) {};
\node [dot={0}{}] at (1,0) {};
\node [dot={0}{}] at (1,1) {};

\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\draw (1,0) -- (0,1);
\draw (0,0) to[out=150,in=190] (0,1.25) to[out=10,in=90] (1,1);
\end{tikzpicture}
\begin{tikzpicture}[scale=2]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (0,1) {};
\node [dot={0}{}] at (1,0) {};
\node [dot={0}{}] at (1,1) {};
\node [dot={0}{}] at (0,0.25) {};
\node [dot={0}{}] at (0,0.5) {};
\node [dot={0}{}] at (0,0.75) {};
\node [dot={0}{}] at (0.33,0) {};
\node [dot={0}{}] at (0.66,0) {};
\node [dot={0}{}] at (0.85,0) {};
\node [dot={0}{}] at (0.5,0.5) {};
\node [dot={0}{}] at (1,0.25) {};
\node [dot={0}{}] at (1,0.5) {};
\node [dot={0}{}] at (1,0.75) {};
\node [dot={0}{}] at (0.75,1) {};
\node [dot={0}{}] at (0.25,1) {};
\node [dot={0}{}] at (0.15,1) {};
\node [dot={0}{}] at (0,1.25) {};

\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\draw (1,0) -- (0,1);
\draw (0,0) to[out=150,in=190] (0,1.25) to[out=10,in=90] (1,1);
\end{tikzpicture}
\end{center}

In addition, the following is \emph{not} a subdivision of $G$ because a vertex was introduced where there was previously a crossing:
\begin{center}
\begin{tikzpicture}[scale=2]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (0,1) {};
\node [dot={0}{}] at (1,0) {};
\node [dot={0}{}] at (1,1) {};
\node [dot={0}{}] at (0.5,0.5) {};

\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\draw (1,0) -- (0,1);
\draw (0,0) -- (1,1);
\end{tikzpicture}
\end{center}
\end{ex}

\begin{theorem}[Lemma]
If $G$ has a subgraph that is a subdivision of $K_5$ or $K_{3,3}$, then $G$ is not planar.
\end{theorem}

\begin{ex}
Consider the following graph, $G$.
\begin{center}
\begin{tikzpicture}[scale=1.5]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (1,0) {};
\node [dot={0}{}] at (2,0) {};
\node [dot={0}{}] at (3,0) {};
\node [dot={0}{}] at (4,0) {};
\node [dot={0}{}] at (5,0) {};
\node [dot={0}{}] at (6,0) {};

\node [dot={0}{}] at (0,1) {};
\node [dot={0}{}] at (1,1) {};
\node [dot={0}{}] at (2,1) {};
\node [dot={0}{}] at (3,1) {};
\node [dot={0}{}] at (4,1) {};
\node [dot={0}{}] at (5,1) {};
\node [dot={0}{}] at (6,1) {};

\draw (0,0) -- (6,0) -- (6,1) -- (0,1) -- (0,0);
\draw (1,0) -- (1,1);
\draw (2,0) -- (2,1);
\draw (3,0) -- (3,1);
\draw (4,0) -- (4,1);
\draw (5,0) -- (5,1);

\draw (0,0) .. controls (7,-0.5) .. (6,1);
\draw (0,1) .. controls (7,1.5) .. (6,0);
\end{tikzpicture}
\end{center}
Katz claims he has a bridge in New York and some cheap land in Florida to sell you, and he claims he has a planar embedding for $G$. Why must he be lying?
\\ \\
Essentially, we know he's lying about having a planar embedding for $G$, since then he could erase subdivisions and get a planar embedding for $K_{3, 3}$. Let's state that more formally.

\begin{proof}
Suppose $G$ is planar. Then, take a planar embedding of $G$. Since $H$ is a subgraph of $G$, we get a planar embedding of $H$. Undo that subdivision. Then we get a planar embedding of $K_{3, 3}$ or $K_5$, and that's ``unpossible.''
\end{proof}
\end{ex}
\subsubsection{Kuratowski's Theorem}
\begin{theorem}[Kuratowski's Theorem]
A graph $G$ is non-planar if and only if it contains a subdivision of $K_{3,3}$ or $K_5$.
\end{theorem}

\begin{defn}
The \textbf{crossing number} of a graph $G$ is the minimum number of crossings in drawings of $G$.
\end{defn}

That is, there may be some ways to draw non-planar graphs that are smarter than other ways. Smarter ways involve drawing the graph in such a way that the crossing number is minimized.
\\ \\
\underline{Observation}: a graph is planar if and only if it has crossing number zero.
\\ \\
It's generally hard to compute crossing numbers. It's an open question to choose the crossing number of $K_{m, n}$, since we'd have to check all possible drawings.
\\ \\
\subsection{Graph Coloring}
\begin{defn}
Let $n$ be a positive integer. A \textbf{graph coloring} is an assignment of one of $\set{1, 2, \ldots, n}$ to each vertex such that no edge connects vertices with the same label.
\end{defn}

\begin{ex}
This is $K_4$:
\begin{center}
\begin{tikzpicture}[scale=2]
\node [dot={0}{}] at (0,0) {};
\node [dot={0}{}] at (1,1) {};
\node [dot={0}{}] at (2,0) {};
\node [dot={0}{}] at (1,2) {};


\draw (0,0) -- (1,2) {};
\draw (1,2) -- (2,0) {};
\draw (0,0) -- (2,0) {};
\draw (0,0) -- (1,1) {};
\draw (1,1) -- (1,2) {};
\draw (1,1) -- (2,0) {};
\end{tikzpicture}
\end{center}

Since every vertex of $K_4$ has an edge to every other vertex, every edge needs its own color.
\end{ex}

\underline{Note}: $K_n$ needs $n$ colors. You must color every vertex differently.
\\ \\
\underline{Note}: $K_{m, n}$ needs two colors, because it's a bipartite graph.

\begin{theorem}[Four Color Theorem]
Every planar graph is 4-colorable.
\end{theorem}

The four color theorem is known to be true, but it's not known to be true in a nice way. The proof is basically a bunch of counterexamples. The proof also involved a computer to generate various cases.

\newpage \newpage
\section*{Clicker Questions} \section*{Clicker Questions}
\begin{itemize} \begin{itemize}
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