adding all but the q2 solutions... those suck

q1/a1q1.aux

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q1/a1q1.blg

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+                                                   1 \div b \lt \infty$ , we...
+The control sequence at the end of the top line
+of your error message was never \def'ed. If you have
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+and I'll forget about whatever was undefined.
+
+! Undefined control sequence.
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+                                                   \infty$ , we have a \thet...
+The control sequence at the end of the top line
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+
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+                $
+l.46 ...lt 1 \div b \lt \infty$ , we have a \theta
+                                                   bound.
+I've inserted a begin-math/end-math symbol since I think
+you left one out. Proceed, with fingers crossed.
+
+! Missing $ inserted.
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+                $
+l.47 \end{solution}
+                   
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+With luck, this will get me unwedged. But if you
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q1/a1q1.tex

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+\documentclass[11pt]{article}
+\usepackage{amsthm,amsmath,amsfonts,amssymb}
+\usepackage[ruled,noline,noend]{algorithm2e}
+\usepackage[margin=1in]{geometry}
+\usepackage{enumerate}
+
+%%% Update the following commands with your name, ID, and acknowledgments.
+\newcommand{\YourName}{Tareef Dedhar}
+\newcommand{\YourStudentID}{20621325}
+\newcommand{\YourAck}{This question was completed only with the teachings from CS courses here at uWaterloo.}
+
+\newenvironment{solution}{\paragraph{Solution.}}{}
+
+
+\title{
+\vspace{-0.8in} \normalsize
+\begin{flushright} \YourName \\ \YourStudentID \end{flushright}
+\rule{\textwidth}{.5pt}\\[0.4cm] \huge Assignment 1: Question 1 \\
+\rule{\textwidth}{2pt}
+\vspace{-1in}
+}
+\date{}
+
+\begin{document}
+\maketitle
+\paragraph{Acknowledgments.} \YourAck
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section*{Asymptotics}
+
+Prove or disprove each of the following statements.
+
+\begin{enumerate}[(a)]
+\item For any constant $b > 0$, the function $f : n \mapsto 1 + b + b^2 + b^3 + \cdots + b^n$ satisfies
+\[
+f(n) = \begin{cases}
+\Theta(b^n) & \mbox{if } b > 1 \\
+\Theta(1) & \mbox{if } b \le 1.
+\end{cases}
+\]
+
+\begin{solution}
+Since f(n) is a geometric series, we can express it as $\sum_{i=1}^{n} {b^i}$ = $\frac{1 - {b^N}}{1 - b}$ = $\frac{{b^n} - 1}{b - 1}$
+Using this:\\
+When b=1: $\lim{n\to\infty}{\dfrac{\tfrac{{b^n} - 1}{b - 1}}{b^n}}$=$\frac{{b^n} - 1}{{b^{n+1}} - b}$, which, by :'Hopital's rule = $1 \div b$. Since $0 \lt 1 \div b \lt \infty$ , we have a \theta bound.
+\end{solution}
+
+\item For every pair of functions $f, g : \mathbb{Z}^+ \to \mathbb{R}^{\ge 1}$ that satisfy $f = \Theta(g)$, the functions $F : n \mapsto 2^{f(n)}$ and $G : n \mapsto 2^{g(n)}$ also satisfy $F = \Theta(G)$.
+
+\begin{solution}
+(ENTER YOUR SOLUTION HERE.)
+\end{solution}
+
+
+\item For every pair of functions $f, g : \mathbb{Z}^+ \to \mathbb{R}^{\ge 1}$ that satisfy $f = o(g)$, the functions $F : n \mapsto 2^{f(n)}$ and $G : n \mapsto 2^{g(n)}$ also satisfy $F = o(G)$.
+
+\begin{solution}
+(ENTER YOUR SOLUTION HERE.)
+\end{solution}
+
+\end{enumerate}
+
+
+\end{document}

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+\@writefile{toc}{\contentsline {paragraph}{Solution.}{1}}

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+********************************************************
+Package `algorithm2e' Release 5.1 -- october 19 2015 --
+- algorithm2e-announce@lirmm.fr mailing list for announcement about releases
+- algorithm2e-discussion@lirmm.fr mailing list for discussion about package
+subscribe by emailing sympa@lirmm.fr with 'subscribe <list> <firstname name>'
+- Author: Christophe Fiorio (christophe.fiorio@umontpellier.fr)
+********************************************************
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q2/q2.tex

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+\documentclass[11pt]{article}
+\usepackage{amsthm,amsmath,amsfonts,amssymb}
+\usepackage[ruled,noline,noend]{algorithm2e}
+\usepackage[margin=1in]{geometry}
+\usepackage{enumerate}
+
+%%% Update the following commands with your name, ID, and acknowledgments.
+\newcommand{\YourName}{REPLACE WITH YOUR NAME}
+\newcommand{\YourStudentID}{REPLACE WITH ID NUMBER}
+\newcommand{\YourAck}{REPLACE this text with a full acknowledgement of all sources (people you discussed the question with and/or online/text sources you consulted) used while completing this question. If you completed the question without consulting any sources, say so here explicitly.}
+
+
+\newenvironment{solution}{\paragraph{Solution.}}{}
+
+
+\title{
+\vspace{-0.8in} \normalsize
+\begin{flushright} \YourName \\ \YourStudentID \end{flushright}
+\rule{\textwidth}{.5pt}\\[0.4cm] \huge Assignment 1: Question 2 \\
+\rule{\textwidth}{2pt}
+\vspace{-1in}
+}
+\date{}
+
+\begin{document}
+\maketitle
+\paragraph{Acknowledgments.} \YourAck
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section*{Solving recurrences}
+
+Solve the following recurrence relations to obtain a closed-form big-$\Theta$ expression for $T(n)$.
+In each question, assume $T(c)$ is bounded by a constant for any small constant $c$.
+
+\begin{enumerate}[(a)]
+\item $T(n) = 9\,T(\frac{n}{3}) + n^2$
+
+\begin{solution}
+(ENTER YOUR SOLUTION HERE.)
+\end{solution}
+
+\item $T(n) = 4\,T(\frac{n}{4}) + n \log n$
+
+\begin{solution}
+(ENTER YOUR SOLUTION HERE.)
+\end{solution}
+
+\item $T(n) = T(\frac{n}{4}) + T(\frac{3n}{4}) + n$
+
+\begin{solution}
+(ENTER YOUR SOLUTION HERE.)
+\end{solution}
+
+\item $T(n) = \sqrt{n} \cdot T(\sqrt{n}) + n$ \\
+\emph{Hint.} The correct expression is somewhere between $\Omega(n)$ and $O(n \log n)$.
+
+\begin{solution}
+(ENTER YOUR SOLUTION HERE.)
+\end{solution}
+
+\end{enumerate}
+
+\end{document}

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q3/q3.tex

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+\documentclass[11pt]{article}
+\usepackage{amsthm,amsmath,amsfonts,amssymb}
+\usepackage[ruled,noline,noend]{algorithm2e}
+\usepackage[margin=1in]{geometry}
+\usepackage{enumerate}
+
+%%% Update the following commands with your name, ID, and acknowledgments.
+\newcommand{\YourName}{Tareef Dedhar}
+\newcommand{\YourStudentID}{20621325}
+\newcommand{\YourAck}{I have completed this question with no outside sources.}
+
+
+\newenvironment{solution}{\paragraph{Solution.}}{}
+
+
+\title{
+\vspace{-0.8in} \normalsize
+\begin{flushright} \YourName \\ \YourStudentID \end{flushright}
+\rule{\textwidth}{.5pt}\\[0.4cm] \huge Assignment 1: Question 3 \\
+\rule{\textwidth}{2pt}
+\vspace{-1in}
+}
+\date{}
+
+\begin{document}
+\maketitle
+\paragraph{Acknowledgments.} \YourAck
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section*{Testing primality [10 marks]}
+
+Analyze the time complexity of the following pseudocode in terms of $n$ 
+using big-$O$ notation. 
+For this analysis, each operation on integers (including multiplication and squaring) takes constant time.
+
+\begin{algorithm}
+\caption{{\sc IsPrime($n$)}}
+$j \gets 2$\;
+\While{$j^2 \le n$}{
+  $k \gets 2$\;
+  \While{$j * k \le n$}{
+    \If{$j * k = n$}{
+      \Return False\;
+    }
+    $k \gets k+1$\;
+  }
+  $j \gets j+1$\;
+}
+\Return True\;
+\end{algorithm}
+
+\begin{solution}
+We shall analyze the algorithm in order, by scope blocks. First, we assign j the initial value of 2. This takes constant time. Then, we have a loop. This loop iterates on j, and if we reduce ${j^2} \leq n$ to $j \leq \sqrt{n}$, and notice that j increments one at a time, we can see this loop will iterate $\sqrt{n}$ times in the worst case.
+
+As for what we are doing $\sqrt{n}$ times, we have another loop, from k=2 to j*k$\leq$n (by 1 each time) in the worst case. Since j is at least 2, in the worst case, this loop will iterate n/2 times. This loop performs an if check and multiplication, which are constant time operations, so the inner loop is O(n).
+
+So, since we are looping $\sqrt{n}$ times, and in each of those loops entering another loop of O(n), our total worst-case runtime is O($n\sqrt{n}$).
+\end{solution}
+
+
+\end{document}

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q4/q4.tex

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+\documentclass[11pt]{article}
+\usepackage{amsthm,amsmath,amsfonts,amssymb}
+\usepackage[ruled,noline,noend]{algorithm2e}
+\usepackage[margin=1in]{geometry}
+\usepackage{enumerate}
+
+%%% Update the following commands with your name, ID, and acknowledgments.
+\newcommand{\YourName}{Tareef Dedhar}
+\newcommand{\YourStudentID}{20621325}
+\newcommand{\YourAck}{I have completed this question without outside sources.}
+
+
+\newenvironment{solution}{\paragraph{Solution.}}{}
+
+
+\title{
+\vspace{-0.8in} \normalsize
+\begin{flushright} \YourName \\ \YourStudentID \end{flushright}
+\rule{\textwidth}{.5pt}\\[0.4cm] \huge Assignment 1: Question 4 \\
+\rule{\textwidth}{2pt}
+\vspace{-1in}
+}
+\date{}
+
+\begin{document}
+\maketitle
+\paragraph{Acknowledgments.} \YourAck
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section*{Common sum [10 marks]}
+
+In the \textsc{Common\,Sum} problem, we are given two arrays $A$ and $B$ of length $n$ containing
+non-negative (not necessarily distinct) integers, and we must determine whether there are indices
+$i_1, i_2, j_1, j_2 \in \{1,2,\ldots,n\}$ for which
+\[
+A[i_1] + A[i_2] = B[j_1] + B[j_2].
+\]
+Design an algorithm that solves the \textsc{Common\,Sum} problem and has time complexity $O(n^2 \log n)$ in the setting where operations on individual integers take constant time.
+
+\bigskip
+
+Your solution must include a description of the algorithm in words, the pseudocode for the algorithm,
+a proof of its correctness, and an analysis of its time complexity in big-$\Theta$ notation.
+
+\begin{solution}
+\begin{section}{Description}
+First, we take our two arrays of length n (a and b), and new empty array of length ((n)(n+1)/2) named c. We will fill c with all the possible sums in b (which number ((n)(n+1)/2). Then, we will sort c using an efficient sort such a mergesort. Then, for each sum in a, we will check if that sum is located in c, using a binary search. If we get a match, return true, otherwise return false.
+\end{section}
+
+\begin{section}{Pseudocode}
+\begin{verbatim}
+int[n] a,b;
+int[n(n+1)/2)] c;
+
+for (i from 0 to n)
+    for (j from i to n)
+    c.insert([b[i]+b[j])
+
+sort(c)
+
+for (i from 0 to n)
+    for (j from i to n)
+    if (binary_search((a[i]+a[j]), c)
+    return true
+
+return false
+\end{verbatim}
+\end{section}
+
+\begin{section}{Proof}
+First, we will show the number of possible sums in an array numbers (n)(n+1)/2. Intuitively, if you want all the sums possible using b[0], you will add b[0] with b[i], where i = 0 to n. However, if you do this with every index, you will get repeats. For example, if we look at b[1], we don't need to add b[1] with b[0] since we already summed these in the previous step. The same is true for b[2] with b[0] and b[1], so on and so forth. We can express this as a sum $\sum_{i=0}^{n} \sum_{j=i}^{n} b[i]+b[j]$, since each time we start finding sums using b[i], we already added b[i] with b[0] through [i]. So, we can also justify how the array c is populated using this logic.
+
+We sort c to allow for binary search to function, as this allows us to be more efficient.
+
+As for the last loop, we use the same looping logic as above to generate all the sums in array a as we did with array b. However, instead of generating an array containing all these sums, for each of them, we will search our existing array of sums in b for the value of a[i] + a[j]. If it exists, we have a match, and there is a common sum. Otherwise, we return false if all possibilities have been exhausted.
+\end{section}
+
+\begin{section}{Time Complexity}
+This is $\Theta$(${n^2}\log{n}$), because:
+\begin{enumerate}
+
+\begin{item}
+Looping from i=0 to n is obviously done n times, and this is executing another loop that is up to n recurrences, where each internal operation is a constant time insertion. This is executing something of constant time up to n times, n times over. So our first step is $\Theta$(${n^2}$).
+\end{item}
+
+\begin{item}
+Sorting something of length ((n)(n+1)/2), which is of the order ${n^2}$, using an efficient sort such as mergesort is known to be of time complexity $\Theta$(${n^2}\log{n^2}$), which reduces to $\Theta$(${n^2}\log{n}$).
+\end{item}
+
+\begin{item}
+This is the same loop as in step 1, except for the innermost step; instead of a constant time insertion, we are performing a binary search on an array of length ${n^2}$. So, each internal step is $\Theta$($\log{n^2}$), which reduces to $\Theta$($\log{n}$). So, our total loop time efficiency is $\Theta$(${n^2}\log{n}$).
+\end{item}
+
+\begin{item}
+Since our longest step is $\Theta$(${n^2}\log{n}$), and the three steps are independent of each other, we can conclude our total efficiency is $\Theta$(${n^2}\log{n}$).
+\end{item}
+\end{enumerate}
+\end{section}
+\end{solution}
+
+\end{document}