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MATH 239: added March 18, 2013 lecture.

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- There are also two more graphs ($k = 3, l = 5$ and $k = 5, l = 3$), which we won't show here for simplicity.
+ There are also two more graphs ($k = 3, l = 5$ and $k = 5, l = 3$), which we won't show here for simplicity.
+
+ % Eric Katz gave this lecture.
+ \subsubsection{Review of Planar Graphs} \lecture{March 18, 2013}
+ \begin{itemize}
+ \item $\deg(f)$ is the length of the boundary walk about the face $f$.
+ \item Handshake Lemma: $$ \sum_{v \in V} \deg(v) = 2q $$
+ \item Dual Handshake Lemma: $$ \sum_{f \in \text{faces}} \deg(f) = 2q $$
+ \item Euler's Formula: $$p - q + s = 1 + c$$ (where $p$ is the number of vertices, $q$ is the number of edges, $s$ is the number of faces (earlier, we denoted this as $f$), and $c$ is the number of components.)
+ \end{itemize}
+
+ \subsubsection{Planar Solids}
+ There are solids that correspond to the five platonic graphs. What do they have in common? Every vertex has the same degree \emph{and} every face has the same degree.
+ \\ \\
+ We'll now finally define planar graphs formally.
+ \begin{defn}
+ A \textbf{planar graph} is a graph such that there are integers $k, l \ge 3$ such that every vertex has degree $k$ and every face has degree $l$.
+ \end{defn}
+
+ Does the data of $(k, l)$ determine the graph? Yes! When drawing these graphs, they are unique because there are no ambiguous decisions to be made.
+ \\ \\
+ Which values of $(k, l)$ are possible? If we try drawing the graphs for $(4,4)$ and $(5,4)$, we'll see that as we're constructing them, they'd each need to be infinite in size to satisfy the $k$ and $l$ constraints. Since these graphs cannot be finite, we say $(4,4)$ and $(5,4)$ are two particular instances of $(k, l)$ for which the constraints are not satisfied.
+ \\ \\
+ \underline{Idea}: use the facts about planar graphs to constrain $(k, l)$.
+
+ \begin{itemize}
+ \item By the handshake lemma, $\sum \deg(v) = kp = 2q$.
+ \item By the dual handshake lemma, $\sum \deg(f) = ls = 2q$. This implies that $p = \frac{2q}{k} = s = \frac{2q}{l}$. This is known as the duality of graphs.
+ \item By Euler's formula, $p - q + s = 2$.
+ \end{itemize}
+
+ This gives us:
+ \begin{align*}
+ \frac{2q}{k} - q + \frac{2q}{l} = 2 &> 0 \\
+ \frac{2}{k} - 1 + \frac{2}{l} = \frac{2}{q} &> 0 \\
+ \frac{2}{k} + \frac{2}{l} &> 1 \\
+ 2l + 2k &> kl \\
+ kl - 2k - 2l &< 0 \\
+ kl - 2k - 2l + 4 - 4 &< 0 \\
+ (k-2)(l-2) &< 4
+ \end{align*}
+
+ So, we know that $k \ge 3$ and $l \ge 3$. This gives us:
+ \begin{align*}
+ &(k-2)(l-2) < 4 \\
+ \implies& k - 2, l - 2 \ge 1 \text{ and } k-2, l-2 < 4 \\
+ \implies& k, l \le 5
+ \end{align*}
+
+ All of this gives us:
+ \begin{itemize}
+ \item If $k = 3$, then $l$ can be 3, 4, or 5.
+ \item If $k = 4$, then $l$ can only be 3.
+ \item If $k = 5$, then $l$ can only be 3.
+ \end{itemize}
+
+ \begin{theorem}
+ If $G$ is a platonic graph, then $(k, l)$ must be one of (3, 3), (3, 4), (3, 5), (4, 3), or (5, 3).
+ \end{theorem}
+
+ Does each of these possibilities of $(k, l)$ occur for a platonic graph? That is, we know these are all \emph{possibilities}, but are they all \emph{actually} possible? For any choice of $(k, l)$, how many platonic graphs are there?
+ \\ \\
+ \underline{Claim}: each of the five possibilities for $(k, l)$ can be achieved by a platonic graph.
+ \begin{proof}
+ Just use the five platonic solids that we know. % This is literally the entire proof as written on the board.
+ \begin{itemize}
+ \item (3, 3): tetrahedron.
+ \item (3, 4): cube.
+ \item (3, 5): dodecahedron.
+ \item (4, 3): octahedron.
+ \item (5, 3): icosahedron.
+ \end{itemize}
+ \end{proof}
+
+ \underline{Claim}: the only platonic graphs are the five coming from platonic solids.
+ \begin{proof}
+ Just try to draw six platonic graphs. As you keep drawing, the graph will be uniquely determined.
+ \end{proof}
+
+ How many faces, edges, and vertices does each have?
+ \begin{align*}
+ q \left( \frac{2}{k} + \frac{2}{l} + 1 \right) &= 2 \\
+ p &= \frac{2q}{k} \\
+ s &= \frac{2q}{l} \\ \\
+ \implies q &= \frac{2}{\frac{2}{k} + \frac{2}{l} - 1} \\
+ &= \frac{2kl}{2l + 2k - kl}
+ \\ \\
+ \implies s &= \frac{2q}{l} \\
+ &= \frac{4k}{2l + 2k - kl}
+ \end{align*}
+
+ We can verify this. If $(k, l) = (3, 5)$, then $s = \frac{12}{10 + 6 - 15} = 12$, as expected.
+
\newpage
\section*{Clicker Questions}

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