Add the rest of the exercises
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README.md
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README.md
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Using styling files from [oysteikt/texmf](https://gitlab.stud.idi.ntnu.no/oysteikt/texmf)
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| Num | Exercise PDF | My Solution PDF | Answer Sheet PDF |
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| --- | ------------------------ | --------------- | ---------------- |
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| 1 | [wiki.math.ntnu.no][ex1] | [ex1.pdf][so1] | [wiki.math.ntnu.no][as1] |
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| 2 | [wiki.math.ntnu.no][ex2] | [ex2.pdf][so2] | [wiki.math.ntnu.no][as2] |
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| 3 | [wiki.math.ntnu.no][ex3] | [ex3.pdf][so3] | [wiki.math.ntnu.no][as3] |
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| 4 | [wiki.math.ntnu.no][ex4] | [ex4.pdf][so4] | [wiki.math.ntnu.no][as4] |
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| 5 | [wiki.math.ntnu.no][ex5] | [ex5.pdf][so5] | <!--[wiki.math.ntnu.no][as5]--> |
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| 6 | [wiki.math.ntnu.no][ex6] | [ex6.pdf][so6] | <!--[wiki.math.ntnu.no][as6]--> |
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| 7 | [wiki.math.ntnu.no][ex7] | [ex7.pdf][so7] | <!--[wiki.math.ntnu.no][as7]--> |
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| 8 | [wiki.math.ntnu.no][ex8] | [ex8.pdf][so8] | <!--[wiki.math.ntnu.no][as8]--> |
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| Num | Exercise PDF | Answer PDF | Solutions PDF |
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| --- | ------------------------- | ---------------- | ------------------------- |
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| 1 | [wiki.math.ntnu.no][ex1] | [ex1.pdf][as1] | [wiki.math.ntnu.no][so1] |
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| 2 | [wiki.math.ntnu.no][ex2] | [ex2.pdf][as2] | [wiki.math.ntnu.no][so2] |
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| 3 | [wiki.math.ntnu.no][ex3] | [ex3.pdf][as3] | [wiki.math.ntnu.no][so3] |
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| 4 | [wiki.math.ntnu.no][ex4] | [ex4.pdf][as4] | [wiki.math.ntnu.no][so4] |
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| 5 | [wiki.math.ntnu.no][ex5] | [ex5.pdf][as5] | [wiki.math.ntnu.no][as5] |
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| 6 | [wiki.math.ntnu.no][ex6] | [ex6.pdf][as6] | [wiki.math.ntnu.no][as6] |
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| 7 | [wiki.math.ntnu.no][ex7] | [ex7.pdf][as7] | [wiki.math.ntnu.no][as7] |
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| 8 | [wiki.math.ntnu.no][ex8] | [ex8.pdf][as8] | [wiki.math.ntnu.no][as8] |
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| 9 | [wiki.math.ntnu.no][ex9] | [ex9.pdf][as9] | [wiki.math.ntnu.no][as9] |
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| 10 | [wiki.math.ntnu.no][ex10] | [ex10.pdf][as10] | [wiki.math.ntnu.no][as10] |
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| 11 | [wiki.math.ntnu.no][ex11] | [ex11.pdf][as11] | N/A <!--[wiki.math.ntnu.no][as11]--> |
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| 12 | [wiki.math.ntnu.no][ex12] | [ex12.pdf][as12] | N/A <!--[wiki.math.ntnu.no][as12]--> |
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[ex1]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-1-2021-new.pdf "Exercise 1 Questions"
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[ex2]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-2-2021-new.pdf "Exercise 2 Questions"
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@ -23,17 +27,31 @@ Using styling files from [oysteikt/texmf](https://gitlab.stud.idi.ntnu.no/oystei
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[ex6]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-6-2021.pdf "Exercise 6 Questions"
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[ex7]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-7-2021.pdf "Exercise 7 Questions"
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[ex8]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-8-2021.pdf "Exercise 8 Questions"
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[ex9]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-9-2021.pdf "Exercise 9 Questions"
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[ex10]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-10-2021.pdf "Exercise 10 Questions"
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[ex11]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-11-2021.pdf "Exercise 11 Questions"
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[ex12]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-12-2021.pdf "Exercise 12 Questions"
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[so1]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise1.pdf "Exercise 1 Solutions"
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[so2]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise2.pdf "Exercise 2 Solutions"
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[so3]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise3.pdf "Exercise 3 Solutions"
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[so4]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise4.pdf "Exercise 4 Solutions"
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[so5]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise5.pdf "Exercise 5 Solutions"
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[so6]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise6.pdf "Exercise 6 Solutions"
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[so7]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise7.pdf "Exercise 7 Solutions"
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[so8]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise8.pdf "Exercise 8 Solutions"
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[as1]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise1.pdf "Exercise 1 Answers"
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[as2]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise2.pdf "Exercise 2 Answers"
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[as3]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise3.pdf "Exercise 3 Answers"
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[as4]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise4.pdf "Exercise 4 Answers"
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[as5]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise5.pdf "Exercise 5 Answers"
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[as6]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise6.pdf "Exercise 6 Answers"
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[as7]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise7.pdf "Exercise 7 Answers"
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[as8]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise8.pdf "Exercise 8 Answers"
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[as9]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise9.pdf "Exercise 9 Answers"
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[as10]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise10.pdf "Exercise 10 Answers"
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[as11]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise11.pdf "Exercise 11 Answers"
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[as12]: http://oysteikt.pages.stud.idi.ntnu.no/v21-ma0301/exercise12.pdf "Exercise 12 Answers"
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[as1]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-1-2021-solutions.pdf "Exercise 1 Answer sheet"
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[as2]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-2-2021-solutions.pdf "Exercise 2 Answer sheet"
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[as3]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-3-2021-solutions.pdf "Exercise 3 Answer sheet"
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[as4]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-4-2021-solutions.pdf "Exercise 4 Answer sheet"
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[so1]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-1-2021-solutions.pdf "Exercise 1 Solutions"
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[so2]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-2-2021-solutions.pdf "Exercise 2 Solutions"
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[so3]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-3-2021-solutions.pdf "Exercise 3 Solutions"
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[so4]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-4-2021-solutions.pdf "Exercise 4 Solutions"
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[so5]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-5-2021-solutions.pdf "Exercise 5 Solutions"
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[so6]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-6-2021-solutions.pdf "Exercise 6 Solutions"
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[so7]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-7-2021-solutions.pdf "Exercise 7 Solutions"
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[so8]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-8-2021-solutions.pdf "Exercise 8 Solutions"
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[so9]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-9-2021-solutions.pdf "Exercise 9 Solutions"
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[so10]: https://wiki.math.ntnu.no/_media/ma0301/2021v/set-10-2021-solutions.pdf "Exercise 10 Solutions"
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@ -0,0 +1,116 @@
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\documentclass[12pt]{article}
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\usepackage{ntnu}
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\usepackage{ntnu-math}
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\author{Øystein Tveit}
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\title{MA0301 Exercise 10}
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\usepackage{amsthm}
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\usepackage{mathabx}
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\usetikzlibrary{arrows.meta}
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\begin{document}
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\ntnuTitle{}
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\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}
|
||||
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\exc{}
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||||
\begin{subexcs}
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\subexc{}
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\[ bcbcd \]
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\subexc{}
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\[ bacbed \]
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\subexc{}
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\subexc{}
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\[ bacb \]
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\end{subexcs}
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\exc{}
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\begin{figure}[H]
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\center
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\scalebox{2}{
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\input{diagrams/ex2.tex}
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}
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\end{figure}
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\exc{}
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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}
|
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|
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The red vertices represent the guards
|
||||
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\exc{}
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\begin{subexcs}
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\subexc{}
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The graphs are not isomorphic because the shortest cycle between the vertices with a degree of 3 has a different length.
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\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}
|
||||
|
|
@ -0,0 +1,24 @@
|
|||
\newcommand{\point}[3]{
|
||||
\node [label=#3:$#1$] (#1) at #2 {};
|
||||
}
|
||||
|
||||
\begin{tikzpicture}[]
|
||||
\begin{scope}[every node/.style={fill=black, shape=circle, inner sep=1pt}]
|
||||
\point{a}{(0,1)}{left}
|
||||
\point{b}{(1,2)}{above}
|
||||
\point{c}{(2,1)}{above right}
|
||||
\point{d}{(1,0)}{below}
|
||||
|
||||
\point{e}{(4,1)}{above left}
|
||||
\point{f}{(5,2)}{above}
|
||||
\point{g}{(6,1)}{right}
|
||||
\point{h}{(5,0)}{below}
|
||||
\end{scope}
|
||||
|
||||
\draw (a) -- (b) -- (c) -- (d) -- (a);
|
||||
\draw (e) -- (f) -- (g) -- (h) -- (e);
|
||||
\draw (b) -- (f);
|
||||
\draw (c) -- (e);
|
||||
\draw (d) -- (h);
|
||||
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,27 @@
|
|||
\newcommand{\point}[3]{
|
||||
\node [label=#3:$#1$] (#1) at #2 {};
|
||||
}
|
||||
|
||||
\begin{tikzpicture}[]
|
||||
\begin{scope}[every node/.style={fill=black, shape=circle, inner sep=1pt}]
|
||||
\point{a}{(0,1)}{left}
|
||||
\point{b}{(1,2)}{above}
|
||||
\point{c}{(2,1)}{above right}
|
||||
\point{d}{(1,0)}{below}
|
||||
|
||||
\point{e}{(4,1)}{above left}
|
||||
\point{f}{(5,2)}{above}
|
||||
\point{g}{(6,1)}{right}
|
||||
\point{h}{(5,0)}{below}
|
||||
\end{scope}
|
||||
|
||||
|
||||
\draw (a) -- (b);
|
||||
\draw (e) -- (f);
|
||||
\draw (c) -- (d);
|
||||
\draw (h) -- (e);
|
||||
\draw (b) -- (c);
|
||||
\draw (f) -- (g);
|
||||
\draw (c) -- (e);
|
||||
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,211 @@
|
|||
\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}
|
||||
|
|
@ -0,0 +1,40 @@
|
|||
def kruskal(vs, es):
|
||||
f = []
|
||||
sets = [set(v) for v in vs]
|
||||
|
||||
find_set = lambda v: [x for x in sets if v in x][0]
|
||||
|
||||
def merge_sets_that_contains(u, v):
|
||||
setv = find_set(v)
|
||||
newsets = [x for x in sets if v not in x]
|
||||
newsets = [setv.union(x) if u in x else x for x in newsets]
|
||||
return newsets
|
||||
|
||||
sorted_es = [e for e,w in sorted(es, key=lambda e: e[1])]
|
||||
|
||||
for (u, v) in sorted_es:
|
||||
if find_set(u) != find_set(v):
|
||||
f += [(u, v)]
|
||||
sets = merge_sets_that_contains(u,v)
|
||||
|
||||
return f
|
||||
|
||||
if __name__ == '__main__':
|
||||
vs = [chr(i) for i in range(ord('a'), ord('h') + 1)]
|
||||
es = [
|
||||
(('a', 'b'), 1),
|
||||
(('b', 'c'), 5),
|
||||
(('c', 'd'), 3),
|
||||
(('d', 'a'), 7),
|
||||
|
||||
(('e', 'f'), 2),
|
||||
(('f', 'g'), 6),
|
||||
(('g', 'h'), 8),
|
||||
(('h', 'e'), 4),
|
||||
|
||||
(('b', 'f'), 10),
|
||||
(('c', 'e'), 9),
|
||||
(('d', 'h'), 11)
|
||||
]
|
||||
|
||||
print(kruskal(vs,es))
|
||||
|
|
@ -0,0 +1,21 @@
|
|||
% https://www3.nd.edu/~kogge/courses/cse30151-fa17/Public/other/tikz_tutorial.pdf
|
||||
|
||||
\begin{tikzpicture}
|
||||
\tikzset{
|
||||
->, % makes the edges directed
|
||||
>=Stealth, % makes the arrow heads bold
|
||||
node distance=3cm, % specifies the minimum distance between two nodes. Change if necessary.
|
||||
every state/.style={thick, fill=white}, % sets the properties for each ’state’ node
|
||||
initial text=$ $, % sets the text that appears on the start arrow
|
||||
}
|
||||
|
||||
\node[state, initial] (s0) {$s_0$};
|
||||
\node[state, below of=s0] (s1) {$s_1$};
|
||||
\node[state, accepting, right of=s0] (s2) {$s_2$};
|
||||
|
||||
\draw (s0) edge[right] node{b} (s1)
|
||||
(s0) edge[above] node{a} (s2)
|
||||
(s1) edge[loop below] node{a, b} (s1)
|
||||
(s2) edge[bend left, right] node{a} (s1)
|
||||
(s2) edge[loop above] node{b} (s2);
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,26 @@
|
|||
|
||||
\begin{tikzpicture}
|
||||
\tikzset{
|
||||
->, % makes the edges directed
|
||||
>=Stealth, % makes the arrow heads bold
|
||||
node distance=5cm, % specifies the minimum distance between two nodes. Change if necessary.
|
||||
every state/.style={thick, fill=white}, % sets the properties for each ’state’ node
|
||||
initial text=$ $, % sets the text that appears on the start arrow
|
||||
}
|
||||
|
||||
\node[state] (s0) {$s_0$};
|
||||
\node[state, right of=s0] (s2) {$s_2$};
|
||||
\node[state, below of=s0] (s3) {$s_3$};
|
||||
\node[state, right of=s3] (s1) {$s_1$};
|
||||
|
||||
\draw (s0) edge[loop above, above] node{a,0} (s0)
|
||||
(s0) edge[right] node{b,1} (s3)
|
||||
(s0) edge[above] node{c,1} (s2)
|
||||
(s1) edge[below, loop below] node{(a,0), (b,0)} (s1)
|
||||
(s1) edge[below] node{c,1} (s3)
|
||||
(s2) edge[right] node{(a,1), (b,1)} (s1)
|
||||
(s2) edge[below right, bend left] node{c,0} (s3)
|
||||
(s3) edge[above left, bend left] node{a,1} (s2)
|
||||
(s3) edge[below, loop below] node{b,0} (s3)
|
||||
(s3) edge[left, bend left] node{c,1} (s0);
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,110 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 12}
|
||||
|
||||
\usetikzlibrary{automata, positioning, arrows.meta}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
|
||||
\exc{}
|
||||
|
||||
\[ r = \{a,b\}^* a \{a,b\}^* a \{a,b\}^* a \{a,b\}^* \]
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\[ \{ab\} \{ab\}^* \]
|
||||
|
||||
\subexc{}
|
||||
\[ a (a | \lambda) b (b | \lambda)\]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
|
||||
\[ M = (Q, \Sigma, \delta, s, F) \]
|
||||
|
||||
\begin{align*}
|
||||
Q &= \{ s_0, s_1, s_2 \} \\
|
||||
\Sigma &= \{a, b\} \\
|
||||
\delta &= \begin{Bmatrix}
|
||||
s_0 \xrightarrow{b} s_1, \\
|
||||
s_0 \xrightarrow{a} s_2, \\
|
||||
s_1 \xrightarrow{a,b} s_1, \\
|
||||
s_2 \xrightarrow{a} s_1, \\
|
||||
s_2 \xrightarrow{b} s_2
|
||||
\end{Bmatrix} \\
|
||||
s &= s_0 \\
|
||||
F &= \{ s_2 \}
|
||||
\end{align*}
|
||||
|
||||
\includeDiagram[scale=1.6, width=10cm]{diagrams/ex3.tex}
|
||||
|
||||
\exc{}
|
||||
|
||||
The words $L$ can be described by the regular expression $r$ where
|
||||
|
||||
\[ r = a^* b b^* a \{a,b\}^* \]
|
||||
|
||||
\exc{}
|
||||
|
||||
The words in $L$ can be described by the regular expression $r$ where
|
||||
|
||||
\[ r = (a^* b)^3 \{ (a^* b)^4 \} \]
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{align*}
|
||||
s_0 &\xrightarrow{a, 0} s_0 \\
|
||||
s_0 &\xrightarrow{a, 0} s_0 \\
|
||||
s_0 &\xrightarrow{b, 1} s_3 \\
|
||||
s_3 &\xrightarrow{b, 0} s_3 \\
|
||||
s_3 &\xrightarrow{c, 1} s_0 \\
|
||||
s_0 &\xrightarrow{c, 1} s_2
|
||||
\end{align*}
|
||||
|
||||
The output would be $001011$
|
||||
|
||||
\subexc{}
|
||||
\includeDiagram[scale=1.2, width=13cm]{diagrams/ex6_b.tex}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
Suppose we have $a \in A, b \in B$
|
||||
|
||||
\begin{align*}
|
||||
AB^* &= \{a, ab, ab^2, ab^3, \ldots \} \\
|
||||
&= \{a\} \cup \{ab, ab^2, ab^3, \ldots \} \\
|
||||
&= A \cup \{ab, ab^2, ab^3, \ldots \} \\
|
||||
&\Rightarrow A \subseteq AB^*
|
||||
\end{align*}
|
||||
\qed
|
||||
|
||||
\subexc{}
|
||||
Since $A \subseteq B$, we can rewrite $B$ as $A \cup \overline{A}$ where $\overline{A} = \{b \mid b \in B, b \notin A \}$
|
||||
|
||||
\begin{align*}
|
||||
B^* &= (A \cup \overline{A})^* \\
|
||||
&= A^* \cap \overline{A}^* \cap B_1, \qquad B_1 = \{(B^*\ a\ B^*\ a_1\ B^*) \vee (B^*\ a_1\ B^*\ a\ B^*) \mid a \in A, a_1 \in \overline{A}\} \\
|
||||
&\Rightarrow A^* \subseteq B^*
|
||||
\end{align*}
|
||||
\qed
|
||||
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
|
||||
|
||||
\end{excs}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,29 @@
|
|||
|
||||
\def\cone{(90:2cm) circle (2.5cm)}
|
||||
\def\ctwo{(180:2cm) circle (2.5cm)}
|
||||
\def\cthree{(270:2cm) circle (2.5cm)}
|
||||
\def\cfour{(360:2cm) circle (2.5cm)}
|
||||
\def\universe{(-5, -5) rectangle (5,5)}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
|
||||
\fill[cyan] \universe;
|
||||
\begin{scope}
|
||||
\clip \ctwo;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cthree;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cfour;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
|
||||
\draw \cone node[text=black,above] {$c_1$};
|
||||
\draw \ctwo node [text=black,left] {$c_2$};
|
||||
\draw \cthree node [text=black,below] {$c_3$};
|
||||
\draw \cfour node [text=black,right] {$c_4$};
|
||||
\draw \universe;
|
||||
\draw (0, 5) node [text=black,above] {$N$};
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,30 @@
|
|||
|
||||
\def\cone{(90:2cm) circle (2.5cm)}
|
||||
\def\ctwo{(180:2cm) circle (2.5cm)}
|
||||
\def\cthree{(270:2cm) circle (2.5cm)}
|
||||
\def\cfour{(360:2cm) circle (2.5cm)}
|
||||
\def\universe{(-5, -5) rectangle (5,5)}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
|
||||
\fill[white] \universe;
|
||||
\fill[red] \cone;
|
||||
\begin{scope}
|
||||
\clip \ctwo;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cthree;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cfour;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
|
||||
\draw \cone node[text=black,above] {$c_1$};
|
||||
\draw \ctwo node [text=black,left] {$c_2$};
|
||||
\draw \cthree node [text=black,below] {$c_3$};
|
||||
\draw \cfour node [text=black,right] {$c_4$};
|
||||
\draw \universe;
|
||||
\draw (0, 5) node [text=black,above] {$N$};
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,33 @@
|
|||
|
||||
\def\cone{(90:2cm) circle (2.5cm)}
|
||||
\def\ctwo{(180:2cm) circle (2.5cm)}
|
||||
\def\cthree{(270:2cm) circle (2.5cm)}
|
||||
\def\cfour{(360:2cm) circle (2.5cm)}
|
||||
\def\universe{(-5, -5) rectangle (5,5)}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
|
||||
\fill[ForestGreen] \universe;
|
||||
\begin{scope}
|
||||
\clip \cone;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \ctwo;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cthree;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cfour;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
|
||||
\draw \cone node[text=black,above] {$c_1$};
|
||||
\draw \ctwo node [text=black,left] {$c_2$};
|
||||
\draw \cthree node [text=black,below] {$c_3$};
|
||||
\draw \cfour node [text=black,right] {$c_4$};
|
||||
\draw \universe;
|
||||
\draw (0, 5) node [text=black,above] {$N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$};
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,33 @@
|
|||
|
||||
\def\cone{(90:2cm) circle (2.5cm)}
|
||||
\def\ctwo{(180:2cm) circle (2.5cm)}
|
||||
\def\cthree{(270:2cm) circle (2.5cm)}
|
||||
\def\cfour{(360:2cm) circle (2.5cm)}
|
||||
\def\universe{(-5, -5) rectangle (5,5)}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
|
||||
\fill[ForestGreen] \universe;
|
||||
\begin{scope}
|
||||
\clip \cone;
|
||||
\fill[red] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \ctwo;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cthree;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
\begin{scope}
|
||||
\clip \cfour;
|
||||
\fill[white] \universe;
|
||||
\end{scope}
|
||||
|
||||
\draw \cone node[text=black,above] {$c_1$};
|
||||
\draw \ctwo node [text=black,left] {$c_2$};
|
||||
\draw \cthree node [text=black,below] {$c_3$};
|
||||
\draw \cfour node [text=black,right] {$c_4$};
|
||||
\draw \universe;
|
||||
\draw (0, 5) node [text=black,above] {$N$};
|
||||
\end{tikzpicture}
|
||||
|
|
@ -0,0 +1,227 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 9}
|
||||
|
||||
\usepackage{amsthm}
|
||||
\usepackage{mathabx}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
p \rightarrow (q \vee r) &\equiv \neg p \vee (q \vee r) \\
|
||||
&\equiv \neg p \vee q \vee r \\
|
||||
&\equiv \neg (p \vee \neg q) \vee r \\
|
||||
&\equiv (p \vee \neg q) \rightarrow r \\
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
|
||||
R does not define a partial orderering, because it is not transitive.
|
||||
|
||||
$aRb$ and $bRc$, however $\neg aRc$
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
xyz + xy\overline{z}+\overline{x}y \\
|
||||
xy + \overline{x}y \\
|
||||
y
|
||||
\end{gather*}
|
||||
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
y + \overline{x}z + x\overline{y} \\
|
||||
y + \overline{x}z + x \\
|
||||
y + z + x \\
|
||||
x + y + z
|
||||
\end{gather*}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
|
||||
Step 1:
|
||||
|
||||
\begin{align*}
|
||||
\sum^1_{n=1}\frac{1}{(2n-1)(2n+1)} &= \frac{1}{2\cdot1 + 1} \\
|
||||
\frac{1}{(2\cdot1-1)(2\cdot1+1)} &= \frac{1}{3} \\
|
||||
\frac{1}{(1)(3)} &= \frac{1}{3} \\
|
||||
\frac{1}{3} &= \frac{1}{3} \\
|
||||
\end{align*}
|
||||
|
||||
Step 2:
|
||||
|
||||
Assume
|
||||
|
||||
\[ \sum^k_{n=1} \frac{1}{(2n-1)(2n+1)} = \frac{k}{2k+1} \]
|
||||
|
||||
then
|
||||
|
||||
\begin{align*}
|
||||
\sum^{k+1}_{n=1} \frac{1}{(2n-1)(2n+1)} &= \frac{k}{2k+1} \\[2ex]
|
||||
&= \frac{1}{(2\cdot1 - 1)(2\cdot1+1)} + \frac{1}{(2\cdot2 - 1)(2\cdot2+1)} + \ldots \\[2ex]
|
||||
&\qquad + \frac{1}{(2\cdot k - 1)(2\cdot k+1)} + \frac{1}{(2\cdot(k+1) - 1)(2\cdot(k+1)+1)} \\[2ex]
|
||||
&= \frac{k}{2k+1} + \frac{1}{(2\cdot(k+1) - 1)(2\cdot(k+1)+1)} \\[2ex]
|
||||
&= \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} \\[2ex]
|
||||
&= \frac{k(2k1)(2k+3) + (2k+1)}{(2k+1)^2(2k+3)} \\[2ex]
|
||||
&= \frac{k(2k+3) + 1}{(2k+1)(2k+3)} \\[2ex]
|
||||
&= \frac{2k^2+3k + 1}{(2k+1)(2k+3)} \\[2ex]
|
||||
&= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} \\[2ex]
|
||||
&= \frac{(k+1)}{(2k+3)} \\[2ex]
|
||||
&= \frac{k+1}{2(k+1)+1}
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
|
||||
\textbf{Injective:}
|
||||
|
||||
Suppose $a,b \in \R$
|
||||
|
||||
\begin{align*}
|
||||
f(a) &= f(b) \\
|
||||
2a-3 &= 2b-3 \\
|
||||
2a &= 2b \\
|
||||
a &= b \\
|
||||
\end{align*}
|
||||
|
||||
thus
|
||||
|
||||
\[ f(a) = f(b) \Leftrightarrow a = b \]
|
||||
|
||||
which means that $f$ is injective \\
|
||||
|
||||
\textbf{Surjective:}
|
||||
|
||||
Suppose $a \in \R$
|
||||
|
||||
\begin{align*}
|
||||
a &= 2x-3 \\
|
||||
x &= \frac{a+3}{2} \\
|
||||
a &\in \R
|
||||
\end{align*}
|
||||
|
||||
therefore $f$ is surjective \\
|
||||
|
||||
\textbf{Inverse:}
|
||||
|
||||
\begin{align*}
|
||||
y &= 2x-3 \\[2ex]
|
||||
x &= \frac{y+3}{2} \\[2ex]
|
||||
f^{-1}(y) &= \frac{y+3}{2}
|
||||
\end{align*}
|
||||
|
||||
|
||||
\exc{}
|
||||
|
||||
\begin{gather*}
|
||||
(\overline{X \cap Y \cap Z}) \\
|
||||
\{ x \mid x \in \overline{X \cap Y \cap Z} \} \\
|
||||
\{ x \mid x \notin X \cap Y \cap Z \} \\
|
||||
\{ x \mid x \notin X \wedge x \notin Y \wedge x \notin Z \} \\
|
||||
\{ x \mid x \in \overline{X} \vee x \in \overline{Y} \vee x \in \overline{Z} \} \\
|
||||
\{ x \mid x \in \overline{X} \cup \overline{Y} \cup \overline{Z} \} \\
|
||||
\overline{X} \cup \overline{Y} \cup \overline{Z}
|
||||
\end{gather*}
|
||||
|
||||
\exc{}
|
||||
|
||||
Assuming 'dozen' is to be interpreted as 12
|
||||
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
|
||||
\[ \nPr{31}{12} = 67\ 596\ 957\ 267\ 840\ 000 \]
|
||||
|
||||
\subexc{}
|
||||
|
||||
\[ 31^{12} = 787\ 662\ 783\ 788\ 549\ 761 \]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
If we imagine a row of numbers going from 1 to 40, we can rephrase the question as how many ways we can split
|
||||
the numbers into $5$ chunks. Imagine a chunk as inserting $4$ delimiters like this:
|
||||
|
||||
\[ 1\ 2\ 3\ |\ 4\ 5\ 6\ 7\ |\ 8\ 9\ 10\ |\ 11\ \ldots\ 39\ |\ 40 \]
|
||||
|
||||
In this case, we split the amount of numbers so that $x_1 = 3, x_2 = 4, x_3 = 3, x_4 = 29, x_5 = 1$
|
||||
|
||||
By doing $\nCr{n}{r}$ where n is the number of numbers and delimiters, and r is the number of delimiters,
|
||||
we will get all combinations of $x_1 + x_2 + x_3 + x_4 + x_5 = 40$A
|
||||
|
||||
In order to make it $x_1 + x_2 + x_3 + x_4 + x_5 \leq 40$, we will add a fifth delimiter, indicating the last block of unused numbers.
|
||||
|
||||
$x_1 + x_2 + x_3 + x_4 + x_5 < 40 \Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 \leq 39$
|
||||
|
||||
This leaves us with $\nCr{39 + 5}{5} = 1086008$ different combinations.
|
||||
|
||||
\subexc{}
|
||||
|
||||
In this case, we modify the problem by adjusting the inequality of $x_i$ like the following
|
||||
|
||||
\begin{align*}
|
||||
x_1 + x_2 + x_3 + x_4 + x_5 &< 40 \qquad &&x_i \geq -3 \\
|
||||
y_1 - 3 + y_2 - 3 + y_3 - 3 + y_4 - 3 + y_5 - 3 &< 40 &&(y_i - 3 = x) \\
|
||||
y_1 + y_2 + y_3 + y_4 + y_5 &< 40 + 5 \cdot 3 \\
|
||||
y_1 + y_2 + y_3 + y_4 + y_5 &< 55
|
||||
\end{align*}
|
||||
|
||||
\[ (y_i - 3 = x) \Rightarrow (y_i - 3 \geq - 3) \Leftrightarrow (y_i \geq 0) \]
|
||||
|
||||
With this information, we use the same way of solving as in \textbf{a)}
|
||||
|
||||
\[\nCr{54 + 5}{5} = 5006386 \]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\break{}
|
||||
|
||||
\exc{}
|
||||
|
||||
This is a Venn diagram of the set containing the elements that satisfy $\overline{c}_1$, $\overline{c}_2$ and $\overline{c}_3$.
|
||||
|
||||
\includeDiagram[width=11cm, caption={$N(\overline{c}_2\overline{c}_3\overline{c}_4$)}]{diagrams/ex9_1.tex}
|
||||
|
||||
The following diagrams show the terms of the RHS.
|
||||
|
||||
\includeDiagram[width=11cm, caption={$N(c_1\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_2.tex}
|
||||
|
||||
\includeDiagram[width=11cm, caption={$N(\overline{c_1}\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_3.tex}
|
||||
|
||||
When we lay these two diagrams on top of each other, we can see that $N(c_1\overline{c}_2\overline{c}_3\overline{c}_4) + N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$ is equal to $N(\overline{c}_2\overline{c}_3\overline{c}_4)$
|
||||
|
||||
\includeDiagram[width=11cm, caption={$N(c_1\overline{c}_2\overline{c}_3\overline{c}_4) + N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_4.tex}
|
||||
|
||||
\exc{}
|
||||
|
||||
let
|
||||
|
||||
$c_1 = 2 \mid n $ \\
|
||||
$c_2 = 3 \mid n $ \\
|
||||
$c_3 = 5 \mid n $ \\
|
||||
$c_4 = 7 \mid n $
|
||||
|
||||
\begin{align*}
|
||||
N(\overline{c}_1\overline{c}_2\overline{c}_3c_4) &= N(c_4) - \left( N(c_1c_7) + N(c_2c_7) + N(c_3c_7) \right) \\
|
||||
&\quad + \left( N(c_1c_2c_4) + N(c_1c_3c_4) + N(c_2c_3c_4) \right) \\
|
||||
&\quad - N(c_1c_2c_3c_4) \\[2ex]
|
||||
&= \left\lfloor \frac{2000}{7} \right\rfloor - \left( \left\lfloor \frac{2000}{2 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{3 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{5 \cdot 7} \right\rfloor \right) \\[2ex]
|
||||
&\quad + \left( \left\lfloor \frac{2000}{2 \cdot 3 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{2 \cdot 5 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{3 \cdot 5 \cdot 7} \right\rfloor \right) \\[2ex]
|
||||
&\quad - \left\lfloor \frac{2000}{2 \cdot 3 \cdot 5 \cdot 7} \right\rfloor \\[2ex]
|
||||
&= 76
|
||||
\end{align*}
|
||||
|
||||
|
||||
\end{excs}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,8 @@
|
|||
main :: IO ()
|
||||
main = print
|
||||
$ length
|
||||
$ [a | a <- [0..2000],
|
||||
a `mod` 2 /= 0,
|
||||
a `mod` 3 /= 0,
|
||||
a `mod` 5 /= 0,
|
||||
a `mod` 7 == 0]
|
||||
Loading…
Reference in New Issue