MA0301/exercise11/main.tex

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\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}
\usepackage{ntnu-code}
\author{Øystein Tveit}
\title{MA0301 Exercise 11}
\begin{document}
\ntnuTitle{}
\break{}
\begin{excs}
\exc{}
\[n^{n-2} = 4^{4-2} = 4^{2} = 16\]
\exc{}
To solve this exercise, I chose to implement the algorithm in python
In order to keep track of the nodes, I have given them the following labels
\includeDiagram[width=13cm, scale=1.6]{diagrams/ex2_1.tex}
\break
\codeFile{scripts/Kruskal.py}{python}
Output:
\begin{verbatim}
[('a', 'b'), ('e', 'f'), ('c', 'd'), ('h', 'e'), ('b', 'c'), ('f', 'g'),
('c', 'e')]
\end{verbatim}
When we connect the nodes, we get the minimal spanning tree:
\includeDiagram[width=13cm, scale=1.6]{diagrams/ex2_2.tex}
\exc{}
\begin{subexcs}
\subexc{}
By counting the vertices, edges and regions, we can see that
\begin{align*}
|V| &= 17 \\
|E| &= 34 \\
|R| &= 19
\end{align*}
By applying Eulers theorem, we can confirm that this is a possible graph
\begin{align*}
V + R - E &= 2 \\
17 + 19 - 34 &= 2 \\
36 - 34 &= 2 \\
2 &= 2
\end{align*}
\subexc{}
By counting the vertices, edges and regions, we can see that
\begin{align*}
|V| &= 10 \\
|E| &= 24 \\
|R| &= 16
\end{align*}
By applying Eulers theorem, we can confirm that this is a possible graph
\begin{align*}
V + R - E &= 2 \\
10 + 16 - 24 &= 2 \\
26 - 24 &= 2 \\
2 &= 2
\end{align*}
\end{subexcs}
\exc{}
Every edge touches 2 regions. And every is connected to at least 5 edges. Therefore the amount of edges will be
\[ E \geq \frac{53 \cdot 5}{2} = 132.5 \]
Since the amount of edges has to be an integer, we can round it up to $E \geq 133$
Now we can use Eulers theorem for planar graphs to determine the amount of vertices
\begin{align*}
V + R - E &= 2 \\
V &= 2 - R + E \\
V &\geq 2 - 53 + 133 \\
V &\geq 82
\end{align*}
Therefore $|V| \geq 82$
\exc{}
\begin{subexcs}
\subexc{}
By flipping the matrix once vertically and once horizontally, the matrix will equal the other matrix.
Because flipping a matrix is a bijective function, composing two of them will also make a bijective function.
After checking that the last matrix is a valid undirected graph, it is safe to conclude that the graphs are isomorphic
\[
\begin{bmatrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
1 & 1 & 0
\end{bmatrix}
\cong
\begin{bmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 1
\end{bmatrix}
\cong
\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix}
\]
\subexc{}
By the same reasoning as \textbf{a)}, we have the following
\[
\begin{bmatrix}
0 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
1 & 1 & 1 & 0
\end{bmatrix}
\cong
\begin{bmatrix}
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 1
\end{bmatrix}
\cong
\begin{bmatrix}
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 0
\end{bmatrix}
\]
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
\[ uv = ababbab \]
\[ |uv| = 7 \]
\subexc{}
\[ vu = bababab \]
\[ |vu| = 7 \]
\subexc{}
\[ v^2 = babbab \]
\[ |v^2| = 6 \]
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
\[ KL = \{ ab^2, abb^2, a^2b^2, aaba, ababa, a^2aba \} \]
\subexc{}
\[ LL = \{ b^2b^2, b^2aba, abab^2, abaaba \} \]
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
\[ L^* = \{b^2\}^* \]
\subexc{}
\[ L^* = \{a,b\}^* \]
\subexc{}
\[ L^* = \{a,b,c^3\}^* \]
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
$w$ does not belong to $r$ because $w$ is does neither fit $a^*$ nor $(b \vee c)^*$
\subexc{}
$w$ does belong to $r$ because $w$ is exactly $(a \cdot 1) (b \vee c \cdot 2)$
\end{subexcs}
\end{excs}
\end{document}