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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
    \rule{\textwidth}{0.5pt} \\
    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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    \textbf{Lecture #1:} #2 \\
    \textbf{Date:} #3 \hfill     \textbf{Scribe:} #4
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%%% -+-+-+-+-+-+- BEGIN HERE -+-+-+-+-+-+- %%%
\newcommand{\lecturenumber}{20}
\newcommand{\lecturetitle}{More on Planer Graphs.}
\newcommand{\scribename}{Noura Allugmani}
\newcommand{\lecturedate}{April 8, 2026}

\begin{document}

\thispagestyle{firstpage}
\scribebox{\lecturenumber}{\lecturetitle}{\lecturedate}{\scribename}

\noindent 
We will start with a duality theorem w.r.t cycles in a plane graph and bonds (i.e., minimal edge cuts) in its dual.

\begin{theorem}
   Edges in a plane graph $G$ form a cycle if and only if the corresponding edges in $G^*$ form a bond.
\end{theorem}

\begin{proof}
 Let $D \subseteq E(G)$ and consider its corresponding set $D^* \subseteq E\left(G^*\right)$.\\
$\Rightarrow$ Suppose $D$ forms a cycle in $G$. Let $S$ be the faces of $G$ inside of $D$ $(S \neq \phi$, because of the Jordan Curve Theorem). Then, the edge cut $\delta(S)$ in $G^*$ Consists of all the edges of $G^*$ that cross edges of $D$, i.e., $D^*$.  
  \begin{figure}[H]
      \centering
      \includegraphics[width=0.5\linewidth]{./figures/lecture_20_IMG_0132.PNG}
    \caption*{\textcolor{red!70!black}{$D$ forms a cycle.}}
    \caption*{\textcolor{blue!70!black}{$D^{*}$ is an edge cut in the dual, i.e., $D^{*} = \delta(S)$.}}
  \end{figure}

% \begin{center}
 
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% \end{center}

\noindent Conversely, suppose $D$ contains no cycle in $G$. \\
$\Rightarrow D$ encloses no region.\\
The unbounded face of $G$ is reachable from any where without crossing $D$.
\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\linewidth]{./figures/lecture_20_IMG_0131.PNG}
\caption*{$D$ does not contain a cycle (imagine the diagonal edge on the bottom right is not there).}
\end{figure}
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% \node[v] (R4) at (4,1) {};

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% \draw (R4)--(R1);

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% % Inner hexagon
% % \draw[red,decorate, decoration={snake, amplitude=0.8mm, segment length=4mm}]


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% % \fill[red] (2.1,0.02) circle (1.6pt);
% % \fill[red] (3.3,0.4) circle (1.6pt);
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\noindent $\Rightarrow G^*-D^*$ is connected. \\
$\Rightarrow D^*$ is not an edge cut.\\
Hence, when $D$ is a cycle, $D^*$ is a minimal edge cut, i.e., $D^*$ is a bond. 
\end{proof}

% Graph! (additional Explination)!

\noindent There is also duality w.r.t deletion and contraction.
\begin{figure}[H]
\centering
    \includegraphics[width=0.5\linewidth]{./figures/lecture_20_IMG_0127.PNG}
     \caption*{The original graph with primal edge $e$ and dual edge $e^*$}
    % \label{fig:placeholder}
\end{figure}
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% ]

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% \node[v] (B) at (0,-1.5) {};

% % Edges
% \draw (T)--(L)--(B)--(R)--(T);
% \draw (T)--(B);

% \end{tikzpicture}
% \end{center}

%%
\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\linewidth]{./figures/lecture_20_IMG_0126.PNG}
    \caption*{We deleted $e$ and contracted $e^*$.}
    % \label{fig:placeholder}
\end{figure}

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%     v/.style={circle, draw, minimum size=7mm, inner sep=0pt}
% ]

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% \node[v] (B) at (0,-1.5) {};

% % Edges
% \draw (T)--(L)--(B)--(R)--(T);
% %\draw (T)--(B);

% \end{tikzpicture}
% \end{center}


\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\linewidth]{./figures/lecture_20_IMG_0129.PNG}
    \caption*{We contracted $e$ and deleted $e^*$}
    \label{fig:placeholder}
\end{figure}

% \begin{center}
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%     v/.style={circle, draw, minimum size=7mm, inner sep=0pt}
% ]

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% % \draw (B) to[out=70, in=110, looseness=1.3] (C);
% \end{tikzpicture}
% \end{center}

\noindent Euler's Formula the \# of nodes, edges, and faces of a plane graph.


\begin{theorem}[Euler's Formula]
 Let $G=(V, E)$ be a connected plane graph with $|V|=n,|E|=m$, and $r$ faces. Then $n-m+r=2$.  
\end{theorem}
\begin{proof}
By induction on $n$. \\
\underline{Base Case} $(n=1)$: \\
$
\begin{aligned}
\hookrightarrow \text{If} \quad m=0 &\Rightarrow r=1 \\
&\Rightarrow n-m+r=1-0+1=2 .
\end{aligned}
$


\noindent $\hookrightarrow$ Otherwise, since $n=1, m>0, G$ looks like a bouquet of loops. 
\begin{center}
\begin{tikzpicture}[scale=1.5]
    \fill (0,0) circle (2.2pt);

    % left loop
    \draw (0,0) .. controls (-0.6,0.6) and (-0.6,-0.6) .. (0,0);

    % three right loops
    \draw (0,0) .. controls (0.6,0.6) and (0.6,-0.6) .. (0,0);
    \draw (0,0) .. controls (1.2,1) and (1.2,-1) .. (0,0);
    \draw (0,0) .. controls (2,1.5) and (2,-1.5) .. (0,0);
\end{tikzpicture}
\end{center}
\noindent Each new loop takes a face and splits it into two faces. $\Rightarrow m$ loops, $r=m+1$
$$
\Rightarrow n-m+r=n-m+m+1=1+1=2.
$$

\noindent \underline{Induction step} $(n>1)$ : since $G$ is connected and $n>1$, there exists some edge $e$ that is not a loop contract $e$, to obtain a new plane graph $G^{\prime}$ with $n^{\prime}=n-1,~ m^{\prime}=m-1$, and $r^{\prime}=r$ since contracting an edge shortens the boundary of a face but does not  remove it. By the inductive hypothesis:

$$
\begin{aligned}
2=n^{\prime}-m^{\prime}+r^{\prime} & =n^{\prime}+1-\left(m^{\prime}+1\right)+r^{\prime} \\
& =n-m+r
\end{aligned}
$$

\end{proof}
\begin{corollary}
 All plane embeddings of planar graph have the same number of faces.
 
\end{corollary}
    
\begin{proof}
 $n, m$ and 2 are fixed $\Rightarrow r$ is fixed.    
\end{proof}

\noindent We can use Euler's Formula as a sufficient condition for non planarity.

\begin{theorem}
 Let $G$ be a simple connected planar graph, with $n \geqslant 3$ nodes, and $m$ edges. Then:
\begin{enumerate}
     \item $m \leqslant 3 n-6$.
     \item If $G$ is triangle-free (i.e., no 3-cycles), $m \leqslant 2 n-4$
 \end{enumerate}
\end{theorem}

\begin{proof}
   \begin{enumerate}
     
\item If $n \geq 3, \ell \left(F_i\right) \geqslant 3$ for all faces of its plane embedding. By the dual degree sum formula,
$2 m=\displaystyle \sum_{i=1} \ell\left(F_i\right) \geqslant 3 r $. \\
$
\begin{aligned}
 \Rightarrow \quad & r \leq \frac{2 m}{3}  \\
& n-m+r
\end{aligned} ~ \Longrightarrow m \leq 3 n-6
$

\item If $G$ is triangle-free, $\ell\left(F_i\right) \geqslant 4, \forall F_i$

$
\begin{aligned}
& 2 m \geq 4 r  \\
& n-m+r = 2
\end{aligned} ~ \Longrightarrow m \leq 2n-4
$

 \end{enumerate}
\end{proof}

\noindent As an application, we can show that $K_5$ is not planar. As another application, we can show that the complete bipartite graph $K_{3,3}$ is not planar.

\begin{figure}[H] 
    \centering
\begin{tikzpicture}[scale=1.2, every node/.style={circle, draw, minimum size=7mm}]
    % Left vertices
    \node (a) at (0,2) {};
    \node (b) at (0,1) {};
    \node (c) at (0,0) {};

    % Right vertices
    \node (d) at (3,2) {};
    \node (e) at (3,1) {};
    \node (f) at (3,0) {};

    % Edges
    \draw (a) -- (d);
    \draw (a) -- (e);
    \draw (a) -- (f);

    \draw (b) -- (d);
    \draw (b) -- (e);
    \draw (b) -- (f);

    \draw (c) -- (d);
    \draw (c) -- (e);
    \draw (c) -- (f);
\end{tikzpicture}
\caption*{$K_{3, 3}$}
\end{figure}
%%





\begin{enumerate}
     

\item $n=6,$
\item $m=3.3=9,$
\item $K_{3,3}$ is triangle free, because it is bipartite (all cycles have even length).

$$
m=9 \nleqslant 2 n-4=2.6-4=8
$$
\end{enumerate}

\noindent So far, we know that $K_5$ and $K_{3,3}$ are not planar. In fact, these are the main ingredients that lead to non planarity.

\noindent A subdivision of a graph $G$ is a graph obtained from $G$ by subdividing its edges with new nodes from paths.

%%

\noindent \textbf{Example}



\begin{center}
\begin{tikzpicture}[scale=1.0,
    every node/.style={circle, draw, minimum size=6mm, inner sep=0pt}
]

% --- Left graph ---
\node (a1) at (0,0) {$a$};
\node (b1) at (2,0) {$b$};
\draw (a1) -- (b1);

% Arrow 
\node[draw=none] at (3.5,0) {$\Rightarrow$};

% --- Right graph ---
\node (a2) at (5,0) {$a$};
\node (x1) at (7,0) {};
\node (x2) at (9,0) {};
\node (x3) at (11,0) {};
\node (b2) at (13,0) {$b$};

\draw (a2) -- (x1) -- (x2) -- (x3) -- (b2);

\end{tikzpicture}
\end{center}


\begin{proposition}
If a graph $G$ Contains a subdivision of $K_5$ or $K_{3,3}$, then $G$ is not planar.
\end{proposition}

\begin{proof}
A subgraph of a planar graph is planar. Therefore, it suffices to show that a subdivision of $K_5$ or $K_{3,3}$ is not planar.
But $K_5$ and $K_{3,3}$ are not planar, and subdividing their edges has no effect on their planarity / non planarity.   
\end{proof}

\noindent In fact, this is also sufficient.
\begin{theorem}[Kuratowski] A graph is planar if and only if it dose not contain a subdivision of $K_5$ or $K_{3,3}$
\end{theorem}
%

 \noindent \textbf{Note:} This theorem does not directly yield an efficient planarity testing algorithm: there are
exponentially - many subgraph to check.
However, there exists polynomial-time algorithms to test planarity.    

\end{document}