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