263 lines
5.9 KiB
TeX
263 lines
5.9 KiB
TeX
% vim: set ts=2 sw=2 tw=80 et:
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\documentclass[12pt]{article}
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\usepackage[margin=3cm]{geometry}
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\usepackage{xcolor}
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\usepackage{lmodern}
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\usepackage{listings}
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\title{Graded Assignment 2 -- DSA}
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\author{Claudio Maggioni}
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% listings configuration
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\lstset{
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basicstyle=\small\ttfamily,
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frame=shadowbox,
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rulesepcolor=\color{black},
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columns=fullflexible,
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commentstyle=\color{gray},
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keywordstyle=\bfseries,
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keywords={,while,if,elif,else,FUNCTION,return,for,from,to,TRUE,FALSE},
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mathescape=true,
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aboveskip=2em,
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captionpos=b,
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abovecaptionskip=1em,
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belowcaptionskip=1em
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}
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\begin{document}
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\maketitle
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\tableofcontents
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\lstlistoflistings
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\newpage
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\section{Exercise 1}
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\subsection{Mergesort}
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\begin{verbatim}
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[5, 6, 12, 8, 4, 10, 3, 12, 11, 1]
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[5, 6, 12, 8, 4] [10, 3, 12, 11, 1]
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[5, 6, 12, 8, 4]
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[5, 6], [12, 8, 4]
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[5, 6]
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[5], [6]
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[5]
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[6]
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[5, 6]
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[12, 8, 4]
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[12] [8, 4]
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[12]
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[8, 4]
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[8], [4]
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[8]
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[4]
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[4, 8]
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[4, 8, 12]
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[4, 5, 6, 8, 12]
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[10, 3, 12, 11, 1]
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[10, 3], [12, 11, 1]
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[10], [3]
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[10]
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[3]
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[3, 10]
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[12, 11, 1]
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[12], [11, 1]
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[12]
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[11, 1]
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[11], [1]
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[11]
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[1]
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[1, 11]
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[1, 11, 12]
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[1, 3, 10, 11, 12]
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[1, 3, 4, 5, 6, 8, 10, 11, 12, 12]
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\end{verbatim}
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\subsection{Selection sort}
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\begin{verbatim}
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[4, 6, 12, 8, 5, 10, 3, 12, 11, 1]
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[3, 6, 12, 8, 5, 10, 4, 12, 11, 1]
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[1, 6, 12, 8, 5, 10, 4, 12, 11, 3]
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[1, 5, 12, 8, 6, 10, 4, 12, 11, 3]
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[1, 4, 12, 8, 6, 10, 5, 12, 11, 3]
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[1, 3, 12, 8, 6, 10, 5, 12, 11, 4]
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[1, 3, 8, 12, 6, 10, 5, 12, 11, 4]
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[1, 3, 6, 12, 8, 10, 5, 12, 11, 4]
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[1, 3, 5, 12, 8, 10, 6, 12, 11, 4]
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[1, 3, 4, 12, 8, 10, 6, 12, 11, 5]
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[1, 3, 4, 8, 12, 10, 6, 12, 11, 5]
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[1, 3, 4, 6, 12, 10, 8, 12, 11, 5]
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[1, 3, 4, 5, 12, 10, 8, 12, 11, 6]
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[1, 3, 4, 5, 10, 12, 8, 12, 11, 6]
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[1, 3, 4, 5, 8, 12, 10, 12, 11, 6]
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[1, 3, 4, 5, 6, 12, 10, 12, 11, 8]
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[1, 3, 4, 5, 6, 10, 12, 12, 11, 8]
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[1, 3, 4, 5, 6, 8, 12, 12, 11, 10]
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[1, 3, 4, 5, 6, 8, 11, 12, 12, 10]
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[1, 3, 4, 5, 6, 8, 10, 12, 12, 11]
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[1, 3, 4, 5, 6, 8, 10, 11, 12, 12]
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[1, 3, 4, 5, 6, 8, 10, 11, 12, 12]
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\end{verbatim}
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\subsection{Quicksort}
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\begin{verbatim}
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[5, 6, 12, 8, 4, 10, 3, 12, 11, 1]
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[5, 6, 3, 1, 4] 8 [12, 12, 11, 10]
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[] 1 [6, 3, 4, 5]
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[] 3 [5, 4, 6]
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[5, 4] 6 []
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[4] 5 []
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[4] 5 []
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[4, 5] 6 []
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[] 3 [4, 5, 6]
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[] 1 [3, 4, 5, 6]
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[10] 11 [12, 12]
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[12] 12 []
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[12] 12 []
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[10] 11 [12, 12]
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[1, 3, 4, 5, 6] 8 [10, 11, 12, 12]
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[1, 3, 4, 5, 6, 8, 10, 11, 12, 12]
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\end{verbatim}
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\subsection{Insertion sort}
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\begin{verbatim}
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[5, 6, 12, 8, 4, 10, 3, 12, 11, 1]
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[5, 6, 8, 12, 4, 10, 3, 12, 11, 1]
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[5, 6, 8, 4, 12, 10, 3, 12, 11, 1]
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[5, 6, 4, 8, 12, 10, 3, 12, 11, 1]
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[5, 4, 6, 8, 12, 10, 3, 12, 11, 1]
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[4, 5, 6, 8, 12, 10, 3, 12, 11, 1]
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[4, 5, 6, 8, 10, 12, 3, 12, 11, 1]
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[4, 5, 6, 8, 10, 3, 12, 12, 11, 1]
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[4, 5, 6, 8, 3, 10, 12, 12, 11, 1]
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[4, 5, 6, 3, 8, 10, 12, 12, 11, 1]
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[4, 5, 3, 6, 8, 10, 12, 12, 11, 1]
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[4, 3, 5, 6, 8, 10, 12, 12, 11, 1]
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[3, 4, 5, 6, 8, 10, 12, 12, 11, 1]
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[3, 4, 5, 6, 8, 10, 12, 11, 12, 1]
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[3, 4, 5, 6, 8, 10, 11, 12, 12, 1]
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[3, 4, 5, 6, 8, 10, 11, 12, 1, 12]
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[3, 4, 5, 6, 8, 10, 11, 1, 12, 12]
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[3, 4, 5, 6, 8, 10, 1, 11, 12, 12]
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[3, 4, 5, 6, 8, 1, 10, 11, 12, 12]
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[3, 4, 5, 6, 1, 8, 10, 11, 12, 12]
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[3, 4, 5, 1, 6, 8, 10, 11, 12, 12]
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[3, 4, 1, 5, 6, 8, 10, 11, 12, 12]
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[3, 1, 4, 5, 6, 8, 10, 11, 12, 12]
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[1, 3, 4, 5, 6, 8, 10, 11, 12, 12]
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\end{verbatim}
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\section{Exercise 2}
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\subsection{Exercise a}
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The pseudocode for \textit{Sum of two} can be found in listing \ref{lst:sum2}.
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The total cost of this algorithm in the worst case is the sum of the worst case
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of mergesort ($O(n log(n))$) and the cost of the worst case in the partition
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done afterwards (which is equivalent to not finding a sum close to the median,
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i.e. $2 n = O(n)$). Therefore, the total cost is $\theta(n log(n))$.
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\begin{lstlisting}[caption=Sum of two in pseudocode, label={lst:sum2}]
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FUNCTION SUM-OF-TWO(A, s):
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A $\gets$ mergesort(A)
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i $\gets$ 1
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j $\gets$ A.length
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while i < j:
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sum $\gets$ $A_i$ + $A_j$
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if sum = s:
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return TRUE
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elif sum > s:
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j $\gets$ j - 1
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else:
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i $\gets$ i + 1
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return FALSE
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\end{lstlisting}
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\subsection{Exercise b}
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The pseudocode for \textit{Sum of three} can be found in listing \ref{lst:sum3}.
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\textsc{search-two} has a time cost of $O(n)$ in the worst case (if no elements
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are found), and the loop of \textsc{search} has an added cost of $O(n)$. The
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total cost in the worst case then, including mergesort, is $n^2 + n log(n)
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= \theta(n^2)$.
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\begin{lstlisting}[caption=Sum of three in pseudocode, label={lst:sum3}]
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FUNCTION SEARCH-TWO(A, sum2, i_skip):
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i $\gets$ 1
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j $\gets$ A.length
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while i < j:
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if i = i_skip:
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i $\gets$ i + 1
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elif j = i_skip:
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j $\gets$ j - 1
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else:
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sum $\gets$ $A_i$ + $A_j$
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if sum = sum2:
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return TRUE
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elif sum > sum2:
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j $\gets$ j - 1
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else:
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i $\gets$ i + 1
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return FALSE
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FUNCTION SUM-OF-THREE(A, s):
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A $\gets$ mergesort(A)
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l $\gets$ A.length
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for i from 1 to l:
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if SEARCH-TWO(A, s - $A_i$, i):
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return TRUE
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return FALSE
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\end{lstlisting}
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\subsection{Exercise c}
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The \textit{Python} code used to implement \textit{Sum of three} can be found in
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the listing \ref{lst:sum3py}.
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\begin{lstlisting}[caption=Sum of three in Python, label={lst:sum3py},%
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language=python]
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#!/usr/bin/env python3
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import sys
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def search_two(A, sum2, i_skip):
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i = 0
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j = len(A) - 1
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while i < j:
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if i == i_skip:
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i = i + 1
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elif j == i_skip:
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j = j - 1
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else:
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cs = A[i] + A[j]
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if cs == sum2:
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return True
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elif cs > sum2:
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j = j - 1
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else:
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i = i + 1
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return False
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def sum_of_three(A, sum3):
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A.sort() # assume using mergesort for worst case of O(n*log(n))
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l = len(A)
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for i in range(l):
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if search_two(A, sum3 - A[i], i):
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return True
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return False
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if __name__ == "__main__":
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args = [int(x) for x in sys.argv[1:]]
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print(sum_of_three(args[1:], args[0]))
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\end{lstlisting}
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\end{document}
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