# christhomson/lecture-notes

MATH 239: added March 20, 2013 lecture.

 @@ -3784,7 +3784,241 @@ 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}