\include{preamble}
\setcounter{try}{48}




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\begin{document}
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\centerline{\bf Math 365 -- Monday 4/15/19 -- 10.3 \& 10.4 Subgraphs, connectivity, and walks}



\begin{try}
\begin{enumerate}[(a)]
\item Consider the graph
$$G= 
\begin{matrix}
\begin{tikzpicture}[scale=1]
\node[bV, label=above:{\tiny$a$}] (a) at (0,1) {};
\node[bV, label=below:{\tiny$b$}] (b) at (0,0) {};
\node[bV, label=below:{\tiny$c$}] (c) at (1,0) {};
\node[bV, label=above:{\tiny$d$}] (d) at (1,1) {};
\draw (a) to (b) to (c)  (d) to (a) to (c);
\end{tikzpicture}
\end{matrix}$$
\begin{enumerate}[(i)]
\item Give an example of a subgraph of $G$ that is not induced. \vfill
\item How many induced subgraphs does $G$ have?  List them.\vfill
\item How many subgraphs does $G$ have? \vfill
\item Let $e$ be the edge connecting $a$ and $d$. Draw $G - e$, $G/e$, and $(G/e)_{\mathrm{simple}}$. \vfill
\item Let $e$ be the edge connecting $a$ and $c$. Draw $G - e$, $G/e$, and $(G/e)_{\mathrm{simple}}$. \vfill
\item Draw $\bar{G}$. \vfill
\end{enumerate}

\vfill

\item Show that $$G= 
\begin{matrix}
\begin{tikzpicture}[scale=1]
\node[bV, label=above:{\tiny$a$}] (a) at (0,1) {};
\node[bV, label=below:{\tiny$b$}] (b) at (0,0) {};
\node[bV, label=below:{\tiny$c$}] (c) at (1,0) {};
\node[bV, label=above:{\tiny$d$}] (d) at (1,1) {};
\draw (a) to (b) to (c)  (d) to (a) ;
\end{tikzpicture}
\end{matrix}$$
is isomorphic to its complement. 
\vfill

\item Find a simple graph with 5 vertices that is isomorphic to its own complement. (Start with: how many edges must it have?)
\end{enumerate}

\end{try}


\begin{try} Let $G$ be the graph 
$$\begin{matrix}\begin{tikzpicture}
	\node[bV, label=above:{\tiny$b$}] (b) at (1,1) {};
	\node[bV, label=above:{\tiny$c$}] (c) at (2,1) {};
	\node[bV, label=above:{\tiny$d$}] (d) at (3,1) {};
	\node[bV, label=below:{\tiny$a$}] (a) at (0,0) {};
	\node[bV, label=above right:{\tiny$g$}] (g) at (1,0) {};
	\node[bV, label=below:{\tiny$f$}] (f) at (2,0) {};
	\node[bV, label=below:{\tiny$e$}] (e) at (3,0) {};
	\draw (a) to (b) to (c) to (d) to (e) to (c) to (f) to (d) (e) to (f) to (g) to (a) to [bend right] (f) (g) to (b);
\end{tikzpicture}\end{matrix}$$
Note that $G$ is simple, so that walks in $G$ are determined by sequences of vertices. 
\begin{enumerate}[(a)]
\item Decide which of the following sequences of vertices determine walks. For those that are walks, decide whether they can be classified more finely (is it a circuit, a path, a cycle, or a a trail?).
\begin{multicols}{2}
\begin{enumerate}[(i)]
\item $a,b,g,f,c,b$
\item $b,g,f,c,b,g,a$
\item $c,e,f,c$
\item $c,e,f,c,e$
\item $a,b,f,a$
\item $f,d,e,c,b$
\end{enumerate}
\end{multicols}
\item Verify that  the following sequences of vertices determine paths. Decide whether they are maximal. If not, extend the sequence to a maximal path. 

\centerline{
(i) $a,b,g,f,d$ \qquad\qquad (ii) $d,e,f,c,b,a,g$, \qquad\qquad (iii) $f,a,g,b,c$
}

\medskip

\item Give an example of a maximal path whose length is not maximal (a path that cannot be extended, but which is shorter than some other path in $G$).
\end{enumerate}
\end{try}

\pagebreak

\begin{try}~
\begin{enumerate}[(a)]
\item For each pair of graphs, either use paths or cycles to show that they are not isomorphic, or give an isomorphism.


\centerline{\includegraphics[width=5.5in]{path-graphs}}
\centerline{\includegraphics[width=5.5in]{path-graphs2}}


\bigskip

\item Explain why if $v$ is a vertex of odd degree, then there is a walk from $v$ to another vertex of odd degree.\vfill
\end{enumerate}
\end{try}



\end{document}

