Template of the Answer Sheet of the Coursework for the module "Languages and Computation" COMP2012 at the University of Nottingham, United Kingdom.
\begin
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%-----------------------Homework------------------------------------
%-------------------Arman Shokrollahi---------------------------------
\documentclass[a4 paper]{article}
% Set target color model to RGB
\usepackage{tikz}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{comment}
\usepackage{algorithmic}
\DeclareMathOperator{\st}{\backepsilon^{'}}
\usetikzlibrary{decorations.text,calc,arrows.meta}
\usetikzlibrary{automata,positioning}% Set target color model to RGB
\input{preamble3.tex}
\begin{document}
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%ADD ACADEMIC YEAR, YOUR DATA
\homework{Coursework}{20XX/20XX}{WRITE YOUR SURNAME}{WRITE YOUR FIRST NAME}{WRITE YOUR ID NUMBER}%{XX$^{\mathbf{th}}$ Month 20XX at 5PM (GMT/BST)}{XX}
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%\begin{center}\rule{\textwidth}{0.4pt}\end{center}
%\begin{center}\textbf{\textit{Answer all FIVE questions}
%}\end{center}
\vspace{5mm}
\problem{1}
This is how an Automaton is drawn
\begin{center}
\begin{tikzpicture}[shorten >=2pt,node distance=3cm,on grid,auto]
\node[state, initial] (A) {$A$};
\node[state, accepting] (B) [right =of A] {$B$};
\node[state, accepting] (C) [right =of B] {$C$};
\node[state, accepting] (D) [below =of A] {$D$};
\node[state, accepting] (E) [right =of D] {$E$};
\node[state] (F) [right =of E] {$F$};
\node[state, accepting] (G) [above right=of C] {$G$};
\node[state, accepting] (H) [below =of E] {$H$};
%\node[state] (q_1) [above right=of q_0] {$q_1$};
% \node[state] (B) [above=of A] {$DE$};
% \node[state] (q2) [below =of s1] {$q_2$};
% \node[state] (q3) [right =of s2] {$q_3$};
% \node[state] (q4) [above =of q3] {$q_4$};
\path[->]
(G) edge [loop above] node {$1$} ()
(H) edge [loop below] node {$\epsilon$} ()
(F) edge [loop right] node {$0,1$} ()
(A) edge [above] node {$0$} (B)
(A) edge [left] node {$1$} (D)
(B) edge [above] node {$0$} (C)
(D) edge [above] node {$\epsilon$} (E)
(E) edge [above] node {$0,1$} (F)
(C) edge [left] node {$0,1$} (F)
(D) edge [bend right, left] node {$1$} (H)
(H) edge [bend right, right] node {$0$} (F)
(B) edge [bend left, above] node {$1$} (G)
(G) edge [bend left] node {$0$} (F)
;
% (q1) edge [bend left, right] node {$b$} (q2)
% (q2) edge [bend left, left] node {$b$} (q1)
% (q2) edge [below] node {$a$} (q3)
% (q3) edge [bend left, left] node {$a$} (q4)
% (q4) edge [bend left, right] node {$a$} (q3)
% (q4) edge [above] node {$b$} (q2);
\end{tikzpicture}
\end{center}
\noindent and a function of transitions is a table of the type
\begin{center}
$\delta$: \begin{tabular}{c|c|c}
& $0$ & $1$\\
\hline
$A$ & $A,B$ & $B,C$ \\
$B$ & $\emptyset$& $B,C$ \\
$C$ & $\emptyset$ & $\emptyset$
\end{tabular}
\end{center}
\begin{flushright}
\boxed{\textbf{20 marks}}
\end{flushright}
\newpage
\vspace{5mm}
\problem{2}
The table to check the equivalence of two Automata is of the type
\begin{center}
\begin{tabular}{c|c|c}
&$a$&$b$\\
\hline
$s_1,q_1$ & $s_1,q_1$ & \colorbox{yellow}{$s_2,q_2$}\\
& \textcolor{orange}{\textbf{FS, FS}}&\textcolor{blue}{\textbf{IS, IS}} \\
\hline
$s_2,q_2$ &\colorbox{yellow}{$s_3,q_3$} &$s_1,q_1$\\
& \textcolor{orange}{\textbf{FS, FS}} & \textcolor{blue}{\textbf{IS, IS}}\\
\hline
$s_3,q_3$ &\colorbox{yellow}{$s_2,q_4$} &$s_3,q_3$\\
& \textcolor{blue}{\textbf{IS, IS}} & \textcolor{blue}{\textbf{IS, IS}}\\
\hline
$s_2,q_4$ &$s_3,q_3$ &$s_1,q_1$\\
& \textcolor{blue}{\textbf{IS, IS}} &\textcolor{orange}{\textbf{FS, FS}}\\
\end{tabular}
\end{center}
A Turing Machine is also drawn as an Automaton
\begin{center}
\begin{tikzpicture}[shorten >=2pt,node distance=3cm,on grid,auto]
\node[state,initial] (Q1) {$Q_1$};
\node[state] (Q2) [right=of Q1] {$Q_2$};
\node[state] (Q3) [right=of Q2] {$Q_3$};
\node[state] (Q4) [right =of Q3] {$Q_4$};
\node[state] (Q5) [below =of Q1] {$Q_5$};
\node[state, accepting] (Q6) [right =of Q5] {$Accept$};
\path[->]
(Q2) edge [loop above] node {$\begin{array}{l}
a \to a, R \\
y \to y, R
\end{array}$} ()
(Q3) edge [loop above] node {$\begin{array}{l}
b \to b, R \\
z \to z, R
\end{array}$} ()
(Q4) edge [loop above] node {$\begin{array}{l}
b \to b, L \\
z \to z, L \\
a \to a, L \\
y \to y, L
\end{array}$} ()
(Q5) edge [loop below] node {$\begin{array}{l}
y \to y, R \\
z \to z, R
\end{array}$} ()
(Q1) edge node {$a \to x, R$} (Q2)
(Q2) edge node {$b \to y, R$} (Q3)
(Q3) edge node {$c \to z, L$} (Q4)
(Q4) edge [bend left] node {$x \to x, R$} (Q1)
(Q1) edge [left] node {$y \to y, R$} (Q5)
(Q5) edge [above] node {$\sqcup \to \sqcup, R$} (Q6);
\end{tikzpicture}
\end{center}
\begin{flushright}
\boxed{\textbf{20 marks}}
\end{flushright}
\newpage
\vspace{5mm}
\problem{3}
This is the notation to indicate a Grammar
\begin{equation*}
G=\left(\lbrace S,A,B\rbrace, \lbrace a,b \rbrace, S, P \right)
\end{equation*}
\noindent with production rules $P$ equal to
\begin{equation*}
\begin{array}{l}
S \to ASA|aB\\
A\to B|S\\
B\to b| \epsilon
\end{array}
\end{equation*}
\noindent and when transformations are made, we may indicate them as
\noindent \textbf{step 1: } Let us remove the null productions.
\begin{center}
\begin{algorithmic}
\STATE \textbf{step 0 :}$W_0=\lbrace B\rbrace$
\STATE \textbf{step 1 :}$W_1=\lbrace S, A, B\rbrace$
\STATE \textbf{step 2 :}$W_2=\lbrace S, A, B\rbrace$
%\STATE \textbf{step 3 :}$W_3=\lbrace S, A, B\rbrace$
\end{algorithmic}
\end{center}
\noindent The set of nullable variables is $W=\lbrace S, A, B\rbrace$.
We may also present the algorithm in the following way
\begin{equation*}
\begin{array}{l}
\textbf{step 0 :}W_0=\lbrace B\rbrace \\
\textbf{step 1 :}W_1=\lbrace S, A, B\rbrace\\
\textbf{step 2 :}W_2=\lbrace S, A, B\rbrace
\end{array}
\end{equation*}
\noindent \textbf{step 2: } Let us remove the unit productions. $\ldots$
\begin{flushright}
\boxed{\textbf{20 marks}}
\end{flushright}
\newpage
\vspace{5mm}
\problem{4}
A single state Pushdown Automaton is drawn like a Finite Automaton
\begin{center}
\begin{tikzpicture}[shorten >=2pt,node distance=3cm,on grid,auto]
\node[state, initial] (Q) {$Q$};
\path[->]
(Q) edge [loop above] node {
%\begin{equation*}
$\begin{array}{l}
\epsilon, C \to 0S|1S|0 \\
\epsilon, S \to 0CC
\end{array}$
%\end{equation*}
} ()
(Q) edge [loop below] node {
% \begin{equation*}
$\begin{array}{l}
0, 0 \to \epsilon\\
1, 1 \to \epsilon
\end{array}$
%\end{equation*}
} ();
\end{tikzpicture}
\end{center}
This is an example of acceptance procedure by $\vdash$ notation
\begin{equation*}
\begin{array}{l}
\mathbf{\epsilon,S\to 0CC}: \left(Q,010000,\textcolor{blue}{S}\right)\vdash \left(Q,010000,\textcolor{red}{0CC}\right)\\
\mathbf{0,0\to \epsilon}:\left(Q,\textcolor{blue}{0}10000,\textcolor{blue}{0}CC\right) \vdash \left(Q,10000,\textcolor{red}{CC}\right)\\
\mathbf{\epsilon,C\to 1S}:\left(Q,10000,\textcolor{blue}{C}C\right)\vdash \left(Q,10000,\textcolor{red}{1S}C\right) \\
\mathbf{1,1\to \epsilon}:\left(Q,\textcolor{blue}{1}0000,\textcolor{blue}{1}SC\right) \vdash \left(Q,0000,\textcolor{red}{SC}\right)\\
\mathbf{\epsilon,S\to 0CC}: \left(Q,0000,\textcolor{blue}{S}C\right)\vdash \left(Q,0000,\textcolor{red}{0CC}C\right)\\
\mathbf{0,0\to \epsilon}:\left(Q,\textcolor{blue}{0}000,\textcolor{blue}{0}CCC\right) \vdash \left(Q,000,\textcolor{red}{CCC}\right)\\
\mathbf{\epsilon,C\to 0}:\left(Q,000,\textcolor{blue}{C}CC\right)\vdash \left(Q,000,\textcolor{red}{0}CC\right) \\
\mathbf{0,0\to \epsilon}:\left(Q,\textcolor{blue}{0}00,\textcolor{blue}{0}CC\right) \vdash \left(Q,00,\textcolor{red}{CC}\right)\\
\mathbf{\epsilon,C\to 0}:\left(Q,00,\textcolor{blue}{C}C\right)\vdash \left(Q,00,\textcolor{red}{0}C\right) \\
\mathbf{0,0\to \epsilon}:\left(Q,\textcolor{blue}{0}0,\textcolor{blue}{0}C\right) \vdash \left(Q,0,\textcolor{red}{C}\right)\\
\mathbf{\epsilon,C\to 0}:\left(Q,0,\textcolor{blue}{C}\right)\vdash \left(Q,0,\textcolor{red}{0}\right) \\
\mathbf{0,0\to \epsilon}:\left(Q,\textcolor{blue}{0},\textcolor{blue}{0}\right) \vdash \left(Q,\epsilon,\textcolor{red}{\epsilon}\right)
\end{array}
\end{equation*}
\begin{flushright}
\boxed{\textbf{20 marks}}
\end{flushright}
\newpage
\vspace{5mm}
\problem{5}
Theoretical questions may need inline equations that is $\sum_{i=1}^n x_i$ or a regular expression like $\left(a+b\right)^*$. A separate line equation is of the type
\begin{equation*}
\left(a+b\right)^*
\end{equation*}
If a proof has to be provided then
\begin{proof}
Let us consider the following language $L\left(P\right)= \lbrace a \rbrace$, and equation
\begin{equation*}
L\left(P^*\right)=\lbrace \epsilon, a, aa, aaa, \ldots \rbrace,
\end{equation*}
\noindent which can make a point. Mathematical expressions can be of the type $w \in L\left(QP^*\right)$ or something of the type
\begin{equation*}
\exists k \in \mathbb{N} \st w \in L\left(QP^k\right)
\end{equation*}
\noindent or something or the type
\begin{equation*}
\forall x \in A \exists ! y \in B \st \left(x, y \right) \in \mathcal{R}
\end{equation*}
\noindent with $f:A \to B$ and $\mathcal{R} \subseteq A \times B$.
\end{proof}
\begin{flushright}
\boxed{\textbf{20 marks}}
\end{flushright}
\end{document}