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+\documentclass[letterpaper,11pt]{article}
+
+\usepackage{geometry}
+\usepackage{pslatex}
+\usepackage{fancyhdr}
+\usepackage{graphicx}
+\usepackage{color}
+\usepackage{enumitem} % for ordered list labels
+\usepackage{amssymb} % for symbols
+\usepackage{scrextend} % for indentation
+\usepackage{tabto} % for tabs
+\usepackage{amsmath} % for text in equation
+\usepackage{listings} % to highlight code
+
+% Define a custom color
+\definecolor{backcolour}{rgb}{0.95,0.95,0.92}
+\definecolor{codegreen}{rgb}{0,0.6,0}
+
+% Define a custom style
+\lstdefinestyle{myStyle}{
+    backgroundcolor=\color{backcolour},   
+    commentstyle=\color{codegreen},
+    basicstyle=\ttfamily\footnotesize,
+    breakatwhitespace=false,         
+    breaklines=true,                 
+    keepspaces=true,                 
+    numbers=left,       
+    numbersep=5pt,                  
+    showspaces=false,                
+    showstringspaces=false,
+    showtabs=false,                  
+    tabsize=2,
+}
+
+% Use \lstset to make myStyle the global default
+\lstset{style=myStyle}
+
+\graphicspath{ {./} }
+\geometry{ margin = 1.0in }
+
+%%% TODO modify these variables %%%
+\def\homeworknum{4}
+\def\myname{Harshit Jain}
+\def\myaccessid{hmj5262}
+\def\myrecitation{8}
+%%%%
+
+\pagestyle{fancy}
+\lhead{{\bf CMPSC 465 Fall 2022}}
+\chead{{\bf Assignment~\homeworknum}}
+\rhead{{\bf \today}}
+
+\newcounter{problemid}
+%\stepcounter{problemid}
+\def\newproblem{\clearpage\newpage{\bf Problem~\arabic{problemid}\stepcounter{problemid}}\hfill\fbox{\parbox{0.16\textwidth}{\bf Points:}}\par}
+
+\setlength\parindent{0em}
+\setlength\parskip{8pt}
+\setlength{\fboxsep}{6pt}
+
+
+\begin{document}
+
+\framebox[\textwidth]{
+	\parbox{0.96\textwidth}{
+		\parbox{0.12\textwidth}{\bf Name:}\parbox{0.6\textwidth}{\myname}\\
+		\parbox{0.12\textwidth}{\bf Access ID:}\parbox{0.6\textwidth}{\myaccessid}\\
+		\parbox{0.12\textwidth}{\bf Recitation:}\parbox{0.6\textwidth}{\myrecitation}
+	}
+}
+
+
+%% your solutions %%%
+
+\newproblem
+\textbf{Acknowledgements}
+\begin{enumerate}[label=(\alph*)]
+    \item I did not work in a group.
+    \item I did not consult without anyone my group members.
+    \item I did not consult any non-class materials.
+\end{enumerate}
+
+
+% PROBLEM 1
+\newproblem
+\textbf{Divide-and-Conquer}
+\begin{enumerate}[label=(\alph*)]
+    
+    \item We will first have the function which returns the frequency of the element in the given list. This function will take $O(n)$ time.
+    
+    \begin{lstlisting}[language=Python]
+    def frequencyCalculator(Array, element):
+        count = 0
+        for ele in Array:
+            if (element == ele):
+                count += 1
+        return count
+    \end{lstlisting}
+
+    Now, to start, we will split the array $A$ into $2$ subarrays $A_1$ and $A_2$ of half the size. Then we will calculate the majority elements of $A_1$ and $A_2$. 
+    The algorithm is as follows$:$
+
+    \begin{lstlisting}[language=Python]
+    def majorityElement(Array, low, high):
+    
+        subArray = Array[low:high+1]
+        
+        # base case
+        if (len(subArray)==1):
+            return subArray[0]
+        
+        mid = (low+high)//2
+        leftMajorityElement = majorityElement(Array, low, mid)
+        rightMajorityElement = majorityElement(Array, mid+1, high)
+        
+        if (leftMajorityElement == rightMajorityElement):
+            return leftMajorityElement
+        
+        leftFrequency = frequencyCalculator(subArray, leftMajorityElement)
+        rightFrequency = frequencyCalculator(subArray, rightMajorityElement)
+        
+        if (leftFrequency > len(subArray)//2):
+            return leftMajorityElement
+        elif (rightFrequency > len(subArray)//2):
+            return rightMajorityElement
+        else:
+            return -1 # no majority element found
+    \end{lstlisting}
+    
+    \item 
+
+    \item Here, we are choosing the majority element of sub-arrays after dividing them in half.
+    Moreover, frequency calculation is using $O(n)$ time as showen above.
+    Therefore, the recurrence relation for this algorithm is given by$:$ \[T(n)=2T(n/2)+O(n)\]
+    According to Master's Theorem, the time complexity$: \underline{O(n logn)}$
+
+\end{enumerate}
+
+
+
+% PROBLEM 2
+\newproblem
+\textbf{Reverse graph}
+\begin{enumerate}[label=(\alph*)]
+    
+    \item 
+
+    \item 
+
+\end{enumerate}
+
+
+% PROBLEM 3
+\newproblem
+\textbf{Graph Basics}
+\begin{enumerate}[label=(\alph*)]
+    
+    \item 
+    \begin{minipage}{.22 \linewidth}
+		Adjacency-list
+
+		\begin{tabular}{l | l}
+
+		$A$ & $\rightarrow B \rightarrow E$\\
+		$B$ & $\rightarrow G \rightarrow D$\\
+		$C$ & $\rightarrow I \rightarrow H$\\
+		$D$ & $\rightarrow E$\\
+		$E$ & $\rightarrow D$\\
+		$F$ & $\rightarrow$\\
+		$G$ & $\rightarrow D \rightarrow F$\\
+		$H$ & $\rightarrow I$\\
+		$I$ & $\rightarrow$ \\
+		\end{tabular}\\ 
+	\end{minipage}
+
+    \item The most number of edges that an undirected graph can have are $: \frac{|V|(|V|-1)}{2}$
+
+    \item Since the degree of a vertex is the number of edges incident with that vertex, the sum of degree counts the total number of times an edge is incident with a vertex.
+    Since every edge is incident with exactly two vertices, each edge gets counted twice, once at each end. Thus the sum of the degrees is equal twice the number of edges.
+    Let $n$ be the number of edges in a simple graph $G(E,V)$. We proceed our proof 
+
+    \item 
+
+    \item 
+\end{enumerate}
+\end{document}