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\documentclass[12pt]{article}

\usepackage[margin=3cm]{geometry}
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\usepackage{lmodern}
\usepackage{listings}

\title{Graded Assignment 2 -- DSA}
\author{Claudio Maggioni}

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\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}