\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}

\author{Øystein Tveit}
\title{MA0301 Exercise 10}

\usepackage{amsthm}
\usepackage{mathabx}

\usetikzlibrary{arrows.meta}

\begin{document}
  \ntnuTitle{}
  \break{}

  Because there are no exercise where there are multiple edges between two vertices, I will use strings of vertex names to represent a walk.
  
  \begin{excs}

  \exc{}
  \begin{subexcs}

    \subexc{}
    \[ bcbcd \]

    \subexc{}
    \[ bacbed \]

    \subexc{}
    \[ bcd \]

    \subexc{}
    \[ bcb \]

    \subexc{}
    \[ befgedcb \]
    
    \subexc{}
    \[ bacb \]
    
  \end{subexcs}

  \exc{}

    \begin{figure}[H]
      \center
      \scalebox{2}{
        \input{diagrams/ex2.tex}
      }
    \end{figure}

  \exc{}
  
  By using trial and error, starting with the nodes that had a higher degree, I managed to bring it down to three nodes.

  \includeDiagram[scale=2, width=12cm]{diagrams/ex3.tex}

  The red vertices represent the guards
  
  \exc{}
  \begin{subexcs}
    \subexc{}
    The graphs are not isomorphic because the shortest cycle between the vertices with a degree of 3 has a different length.

    \subexc{}
    The graphs are isomorphic

  \end{subexcs}

  \exc{}

  \begin{subexcs}
    \subexc{}

  \includeDiagram[scale=2, width=12cm]{diagrams/ex5_a.tex}

    \[ adhijkgcbgjfbefiedba \]

    \subexc{}
    Because $deg(e) = deg(f) = 3$ is now odd, they have to be the starting vertex and ending vertex.

  \includeDiagram[scale=2, width=12cm]{diagrams/ex5_b.tex}

    \[ dabdhijkgcbgjfbefie \]

  \end{subexcs}

  \exc{}

  \begin{subexcs}
    \subexc{}
    $G_1$ is not an induced subgraph if it's missing an edge $e_1$ between $v_1, v_2 \in G_1$ where $e_1 \in G$

    \subexc{}
    \includeDiagram[scale=0.8, width=6cm, pdf=true]{diagrams/ex6_b.pdf}

  $G_1$ contains the vertices $c$ and $d$ while it is missing the edge $cd$ even though $cd$ was present in $G$. Therefore, it is not an induced subgraph

  \end{subexcs}

  \exc{}

  \begin{align*}
    \sum_{deg(v) \in V} = 2 |E| \\
    3|V| \leq 2 |E| \\
    |V| \leq \frac{2 |E|}{3} \\
    |V| \leq \frac{2 \cdot 17}{3} \\
    |V| \leq \frac{34}{3} \\
    |V| \leq 11.33 \\
  \end{align*}

  The max amount of vertices in $G$ has to be $11$

  \end{excs}
\end{document}