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