update with dynamic solution for problem 1

[ar1516.git] / 1.tex

diff --git a/1.tex b/1.tex
index 098a2e3..585e0cd 100644 (file)
--- a/1.tex
+++ b/1.tex
@@ -79,13 +79,13 @@ going over all pallets.
 
 \subsection{SMT format solution}
 The formula is easily convertable to SMT format by literal conversion and is
-listed in Listing~\ref{listing:a1.smt}. The maximization is done by
+listed in Listing~\ref{listing:a1.bash}. The maximization is done by
 incrementing a special variable called \texttt{<REP>} which is an instantiation
 of $n(p)$.  When the iteration yields unsat we know that the current number
 minus one is the maximum amount of prittles the trucks can carry. The
 maximization is done with a bash script shown in Listing~\ref{listing:1.bash}.
 
-\lstinputlisting[language=bash,
+\lstinputlisting[language=bash,firstline=66,
        caption={Iteratively find the largest solution for problem 1},
        label={listing:1.bash}]{src/a1.bash}
 
@@ -95,21 +95,7 @@ this solution takes less then $0.12$ seconds and is shown in Table~\ref{tab:s1}.
 
 \begin{table}[H]
        \centering
-       \begin{tabular}{l|lllll|ll}
-               \toprule
-               Truck & Nuzzles & Prittles & Skipples & Crottles & Dupples &
-                       Weight & Pallets\\
-               \midrule
-               1 & 0 & 0 & 4 & 2 & 1 & 7100 & 7\\
-               2 & 0 & 3 & 4 & 0 & 1 & 6500 & 8\\
-               3 & 0 & 8 & 0 & 0 & 0 & 6400 & 7\\
-               4 & 0 & 7 & 0 & 0 & 1 & 5700 & 8\\
-               5 & 0 & 0 & 0 & 5 & 1 & 7600 & 6\\
-               6 & 4 & 0 & 0 & 3 & 1 & 7400 & 8\\
-               \midrule
-               total & 4 & 18 & 8 & 10 & 5 & &\\
-               \bottomrule
-       \end{tabular}
+\input{a1.tex}
        \caption{Solution visualization for problem 1}
 \label{tab:s1}
 \end{table}