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author | Koen van der Heijden <koen.vd.heijden1@gmail.com> | 2017-01-15 23:33:25 +0100 |
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committer | Koen van der Heijden <koen.vd.heijden1@gmail.com> | 2017-01-15 23:33:25 +0100 |
commit | 66dad01dd33494a885b5577d92dc4bc619c36cf9 (patch) | |
tree | b18663093b358be7aa5f92426327671d93254999 /part2.tex | |
parent | 9a9ddd99405329ccfa605f9125e33ac968ff0c99 (diff) | |
download | 2IMF25-AR-66dad01dd33494a885b5577d92dc4bc619c36cf9.tar.gz |
Part 4 problem definition and horrible implementation
Diffstat (limited to 'part2.tex')
-rw-r--r-- | part2.tex | 31 |
1 files changed, 29 insertions, 2 deletions
@@ -447,7 +447,7 @@ Running this input through \textit{prover9} generated the following proofs for \ \textit{prover9} was not able to prover that $x * y = y * x$ holds for all finite groups. Therefor, we ran \ref{itm:logic_b_4} through \textit{mace4} to find the smallest group $G(I, *, inv)$ for which $x * y \neq y * x$. The following input was used: -\begin{verbatim} +\begin{lstlisting} formulas(sos). (x*(y*z)=(x*y)*z). (x*I=x). @@ -457,7 +457,7 @@ end_of_list. formulas(goals). all x all y (x*y=y*x). end_of_list. -\end{verbatim} +\end{lstlisting} After running this input through \textit{mace4}, we retrieved the following finite group $G(I, *, inv)$: \begin{verbatim} I: 0 @@ -489,5 +489,32 @@ single solutions of simply specified puzzles like sudokus. In case of doubt please contact the lecturer. \subsection*{Solution:} +We decided to choose a highscool room scheduling problem. + +\begin{itemize} + \item The highschool has $n$ classes + \item The highschool only has $ceil(n * 0.7)$ rooms though + \item Every class has to attend at least 1040 hours a year, which is equivalent to 26 hours per week + \item Every day contains 8 timeslots of 1 hour, in which lectures can take place + \item Lectures take place 5 days a week, giving a total of 40 timeslots per week + \item There are 10 different courses:\\\\ + \begin{tabular}{l c r} + English & 3 hours per week & \\ + German & 2 hours per week & \\ + Spanish & 2 hours per week & \\ + Math & 3 hours per week & \\ + Physics & 3 hours per week & \\ + Chemistry & 3 hours per week & \\ + Biology & 3 hours per week & \\ + Art & 1 hour per week & \\ + Geography & 3 hours per week & \\ + History & 3 hours per week & \\ + \end{tabular} +\end{itemize} + +We want to schedule the rooms in such way that every class attends all required lessons per week, and the total amount of free hours inbetween 2 lectures is minimized. + +\subsubsection*{Solution:} + \end{document} |