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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.
+ % 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
\section*{Clicker Questions}
\begin{itemize}

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