Add CT331 Programming Paradigms

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+%! TeX program = lualatex
+\documentclass[a4paper]{article} 
+
+% packages
+\usepackage{microtype}      % Slightly tweak font spacing for aesthetics
+\usepackage[english]{babel} % Language hyphenation and typographical rules
+\usepackage[final, colorlinks = false, urlcolor = cyan]{hyperref} 
+\usepackage{changepage}     % adjust margins on the fly
+\usepackage{fontspec}
+
+\usepackage{minted}
+% \usemintedstyle{algol_nu}
+\usepackage{xcolor}
+
+\usepackage{pgfplots}
+\pgfplotsset{width=\textwidth,compat=1.9}
+
+\usepackage{caption}
+\newenvironment{code}{\captionsetup{type=listing}}{}
+\captionsetup[listing]{skip=0pt}
+\setlength{\abovecaptionskip}{5pt}
+\setlength{\belowcaptionskip}{5pt}
+
+\usepackage[yyyymmdd]{datetime}
+\renewcommand{\dateseparator}{--}
+\setmainfont{EB Garamond}
+\setmonofont[Scale=MatchLowercase]{Deja Vu Sans Mono}
+
+\usepackage{titlesec}
+% \titleformat{\section}{\LARGE\bfseries}{}{}{}[\titlerule]
+% \titleformat{\subsection}{\Large\bfseries}{}{0em}{}
+% \titlespacing{\subsection}{0em}{-0.7em}{0em}
+%
+% \titleformat{\subsubsection}{\large\bfseries}{}{0em}{$\bullet$ }
+% \titlespacing{\subsubsection}{1em}{-0.7em}{0em}
+
+% margins
+\addtolength{\hoffset}{-2.25cm}
+\addtolength{\textwidth}{4.5cm}
+\addtolength{\voffset}{-3.25cm}
+\addtolength{\textheight}{5cm}
+\setlength{\parskip}{0pt}
+\setlength{\parindent}{0in}
+% \setcounter{secnumdepth}{0}
+
+\begin{document}
+\hrule \medskip
+\begin{minipage}{0.295\textwidth} 
+    \raggedright
+    \footnotesize 
+    Name: Andrew Hayes \\
+    E-mail: \href{mailto://a.hayes18@universityofgalway.ie}{\texttt{a.hayes18@universityofgalway.ie}}  \hfill\\   
+    ID: 21321503 \hfill
+\end{minipage}
+\begin{minipage}{0.4\textwidth} 
+    \centering 
+    \vspace{0.4em}
+    \Large 
+    \textbf{CT331} \\ 
+\end{minipage}
+\begin{minipage}{0.295\textwidth} 
+    \raggedleft
+    \today
+\end{minipage}
+\medskip\hrule 
+\begin{center}
+    \normalsize
+    Assignment 3: Declarative Programming with Prolog
+\end{center}
+\hrule
+
+\section{Question 1}
+\subsection{Rule that returns true if a given instructor teaches a given student}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+teaches(Instructor, Student) :- instructs(Instructor, Course), takes(Student, Course).
+\end{minted}
+
+\subsection{Query that uses the \mintinline{prolog}{teaches} rule to show all students instructed by \mintinline{prolog}{bob}}
+For this, I wasn't sure if the desired answer was a query that returned a student instructed by \mintinline{prolog}{bob}, followed by 
+a couple semi-colons to get every student instructed by \mintinline{prolog}{bob}, or if the desired answer was a single query that 
+returned a list of students taught by \mintinline{prolog}{bob}, so I did both.
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- teaches(bob, Student).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q1.2.png}
+    \caption{Using the \mintinline{prolog}{teaches} rule to show all students instructed by \mintinline{prolog}{bob}}
+\end{figure}
+
+Alternatively, this could be done using the \mintinline{prolog}{findall()} predicate:
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- findall(Student, teaches(bob, Student), Students).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q1_2_findall.png}
+    \caption{Using the \mintinline{prolog}{teaches} rule \& the \mintinline{prolog}{findall} predicate to show all students instructed by \mintinline{prolog}{bob}}
+\end{figure}
+
+\subsection{Query that uses the \mintinline{prolog}{teaches} rule to show all instructors that instruct \mintinline{prolog}{mary}}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- teaches(Instructor, mary).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q1_3.png}
+    \caption{Using the \mintinline{prolog}{teaches} rule to show all instructors that instruct \mintinline{prolog}{mary}}
+\end{figure}
+
+Alternatively, this could be done using the \mintinline{prolog}{findall()} predicate:
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- findall(Instructor, teaches(Instructor, mary), Instructors).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q1_3_findall.png}
+    \caption{Using the \mintinline{prolog}{teaches()} rule \& the \mintinline{prolog}{findall()} predicate to show all instructors that instruct \mintinline{prolog}{mary}}
+\end{figure}
+
+\subsection{Result of query \mintinline{prolog}{teaches(ann,joe).}}
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q1_4.png}
+    \caption{Result of query \mintinline{prolog}{teaches(ann,joe).}}
+\end{figure}
+
+The result of the query \mintinline{prolog}{teaches(ann,joe).} is \mintinline{prolog}{false.} because \mintinline{prolog}{ann}
+only instructs \mintinline{prolog}{ct345} and \mintinline{prolog}{joe} only takes \mintinline{prolog}{ct331}, and therefore 
+\mintinline{prolog}{ann} does not teach \mintinline{prolog}{joe} because \mintinline{prolog}{ann} does not teach a course
+that \mintinline{prolog}{joe} takes.
+
+\subsection{Rule that returns \mintinline{prolog}{true} if two students take the same course}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+takesSameCourse(Student1, Student2) :- takes(Student1, Course), takes(Student2, Course).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q1_5.png}
+    \caption{Queries to test \mintinline{prolog}{takesSameCourse()}}
+\end{figure}
+
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- takesSameCourse(tom,mary).
+?- takesSameCourse(joe,mary).
+?- takesSameCourse(joe,tom).
+?- takesSameCourse(bob, mary).
+\end{minted}
+
+\section{Question 2}
+\subsection{Query that displays the head \& tail of a list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- [Head | Tail] = [1,2,3].
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q2_1.png}
+    \caption{Query to display the head \& tail of the list \mintinline{prolog}{[1,2,3]}}
+\end{figure}
+
+\subsection{Display the head of a list, the head of the tail of the list, \& the tail of the tail of the list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- [Head | [HeadOfTail | TailOfTail]] = [1,2,3,4,5].
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q2_2.png}
+    \caption{Query to display the head of the list, the head of the tail of the list, \& the tail of the tail of the list \mintinline{prolog}{[1,2,3,4,5]}}
+\end{figure}
+
+\subsection{Rule that returns \mintinline{prolog}{true} if a given element is the first element of a given list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+contains1(Element, [Element | Tail]).
+
+?- contains1(1, [1,2,3,4]).
+?- contains1(3, [1,2,3,4]).
+?- contains1(1, [2,3,4]).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q2_3.png}
+    \caption{\mintinline{prolog}{contains1()} testing}
+\end{figure}
+
+\subsection{Rule that returns \mintinline{prolog}{true} if a given list is the same as the tail of another given list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+contains2(Sublist, [Head | Sublist]).
+
+?- contains2([2,3,4], [1,2,3,4]).
+?- contains2([2,3,4], [1,2,3,4,5]).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q2_4.png}
+    \caption{\mintinline{prolog}{contains2()} testing}
+\end{figure}
+
+\subsection{Query to display the first element of a given list using \mintinline{prolog}{contains1()}}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+?- contains1(FirstElement, [1,2,3,4,5]).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q2_5.png}
+    \caption{Query to display the first element of a given list using \mintinline{prolog}{contains1()}}
+\end{figure}
+
+\section{Determine if a given element is not in a given list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+% base case: any element is not in an empty list
+isNotElementInList(_, []).
+
+% return true if Element is not the Head of the list and it's not found recursively searching the rest of the list
+isNotElementInList(Element, [Head | Tail]) :- Element \= Head, isNotElementInList(Element, Tail).
+
+% testing
+isNotElementInList(1, []).  
+isNotElementInList(1, [1]). 
+isNotElementInList(1, [2]).  
+isNotElementInList(2, [1, 2, 3]).  
+isNotElementInList(7, [1, 2, 9, 4, 5]). 
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q3.png}
+    \caption{Testing \mintinline{prolog}{isNotElementInList()}}
+\end{figure}
+
+\section{Facts \& rules to merge three lists}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+% predicate to merge two lists
+% base case: if the first list is empty, just return the second
+mergeTwoLists([], List, List).
+
+% recursive predicate to merge two lists
+% split the first list into head and tail, and recurse with its tail and the second list until the first list is empty (base case)
+% then merge the original head of the first list with the resulting tail
+mergeTwoLists([Head | Tail], List2, [Head | ResultTail]) :- mergeTwoLists(Tail, List2, ResultTail).
+
+% predicate to merge 3 lists 
+% base case: merging an empty list and two others is the same as merging two lists
+mergeLists([], List2, List3, Merged) :- mergeTwoLists(List2, List3, Merged).
+
+% split the first list into head and tail, and recurse with its tail and the other two lists until the first list is empty (base case)
+mergeLists([Head1 | Tail1], List2, List3, [Head1 | MergedTail]) :- mergeLists(Tail1, List2, List3, MergedTail).
+
+?- mergeLists([7],[1,2,3],[6,7,8], X).
+?- mergeLists([2], [1], [0], X).
+?- mergeLists([1], [], [], X).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q4.png}
+    \caption{Testing \mintinline{prolog}{mergeLists()}}
+\end{figure}
+
+\section{Facts \& rules to reverse a given list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+% call the helper predicate with the list to be reversed and an empty Accumulator to build up
+reverseList(List, Reversed) :- reverseListHelper(List, [], Reversed).
+
+% base case fact: when the list to reverse is empty, the accumulator is the reversed list
+reverseListHelper([], Accumulator, Accumulator).
+
+% recurse with the tail after prepending the head to the accumulator
+reverseListHelper([Head | Tail], Accumulator, Reversed) :- reverseListHelper(Tail, [Head | Accumulator], Reversed).
+
+?- reverseList([1,2,3], X).  
+?- reverseList([1], X).  
+?- reverseList([], X).  
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q5.png}
+    \caption{Testing \mintinline{prolog}{reverseList()}}
+\end{figure}
+
+\section{Facts \& rules to insert an element into its correct position in a given list}
+\begin{minted}[linenos, breaklines, frame=single]{prolog}
+% base fact: if the list is empty, the list to be returned is just the element
+insertInOrder(Element, [], [Element]).
+
+% if the element to be inserted is <= the head of the list, insert it at the head of the list
+insertInOrder(Element, [Head | Tail], [Element, Head | Tail]) :- Element =< Head.
+
+% if the element to be inserted is greater than the head of the list, recurse with the tail of the list until 
+insertInOrder(Element, [Head | Tail], [Head | NewTail]) :- Element > Head, insertInOrder(Element, Tail, NewTail).
+\end{minted}
+
+\begin{figure}[H]
+    \includegraphics[width=\textwidth]{./images/q6.png}
+    \caption{Testing \mintinline{prolog}{insertInOrder()}}
+\end{figure}
+
+
+\end{document}