cheat6.tex

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    {\TextCheatsheetEng: Automata Theory}%
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%% BNF grammar
\usepackage{syntax}


\begin{document}

\selectlanguage{english}

\setcounter{section}{5}% +1 = actual
\section{Automata Theory Cheatsheet}

\begin{terms}
    \item \textbf{Alphabet}\Href{https://en.wikipedia.org/wiki/Alphabet_(formal_languages)} is a finite set of symbols, commonly denoted $\Sigma$.

    \item \textbf{Word} $w \in \Sigma^*$ is a finite sequence of symbols from alphabet $\Sigma$
    For example, $w = abacaba \in \Set{a, b, c}^*$.

    \item \textbf{Length} of a word: $|w| = n$, where $n$ is the number of symbols in word $w$.
    For example, $|abacaba| = 7$.

    \item \textbf{Empty word} $\varepsilon$ is a word of length 0.

    \item \textbf{Concatenation} of words $w_1$ and $w_2$ is $w_1 \cdot w_2 = w_1 w_2$.

    \item \textbf{Power} of a word $w$ is $w^n = w \cdot w \cdot \ldots \cdot w$ ($n$ times).
    % Note that $\Sigma^0 = \Set{\varepsilon} \neq \emptyset$.

    \item \textbf{Reverse} of a word $w$ is $w^R$.

    % \item \textbf{Subword} of a word $w$ is $w[i:j] = w_i \dots w_j$.

    % \item \textbf{Prefix} of a word $w$ is $w[1:i] = w_1 \dots w_i$.

    % \item \textbf{Suffix} of a word $w$ is $w[i:n] = w_i \dots w_n$.

    \item \textbf{Language}\Href{https://en.wikipedia.org/wiki/Formal_language} $L$ over an alphabet $\Sigma$ is a set of words $L \subseteq \Sigma^*$.

    \item \textbf{Empty language} $\emptyset$ is a language that contains no words.

    \item \textbf{Singleton\Href{https://en.wikipedia.org/wiki/Singleton_(mathematics)} language} $\Set{w}$ is a language that contains only one word $w$.

    \item \textbf{Empty string language} $\Set{\varepsilon}$ is a language that contains only one empty word $\varepsilon$.

    \item \textbf{Operations} on languages:

    \begin{terms}
        \item \textbf{Union}: $L_1 \union L_2 = \Set{w \given w \in L_1 \lor w \in L_2}$

        \item \textbf{Intersection}: $L_1 \intersection L_2 = \Set{w \given w \in L_1 \land w \in L_2}$

        \item \textbf{Complement}: $\neg{L} = \Set{w \given w \notin L}$

        \item \textbf{Concatenation}\Href{https://en.wikipedia.org/wiki/Concatenation}: $L_1 \cdot L_2 = \Set{ab \given a \in L_1, b \in L_2}$

        \item \textbf{Kleene star (Kleene closure)}\Href{https://en.wikipedia.org/wiki/Kleene_star}: $L^* = \bigunion_{k = 0}^{\infty}\Sigma^k$
    \end{terms}

    % TODO: move
    \item \textbf{Equivalence}: $L_1 \equiv L_2 \iff (L_1 \intersection \overline{L_2}) \union (\overline{L_1} \intersection L_2) = \emptyset$

    % TODO: rewrite
    % \item $\mathrm{REG}$ \textbf{(family of regular languages)}\Href{https://en.wikipedia.org/wiki/Regular_language} is set over an alphabet $\Sigma$.

    \item \textbf{Regular language}\Href{https://en.wikipedia.org/wiki/Regular_language} is a language that can be defined by a regular expression.

    Regular languages are defined inductively (recursively):

    \begin{terms}
        \item The empty language $\emptyset$ is regular.

        \item For any $a \in \Sigma$, the singleton language $\Set{a}$ is regular.

        \item If $A$ is a regular language, then $A^*$ (Kleene star) is also regular.

        \item If $A$ and $B$ are regular languages, then $A \union B$ (union) is also regular.

        \item If $A$ and $B$ are regular languages, then $A \cdot B$ (concatenation) is also regular.

        \item No other languages over $\Sigma$ are regular.
    \end{terms}

    % TODO: remove/rewrite
    \item \textbf{REG (set of regular languages)} is set over an alphabet $\Sigma$\\ $\mathrm{REG} = \bigunion_{k = 0}^{\infty} \mathrm{Reg}_{k} = \mathrm{Reg}_{\infty}$.

    % TODO: remove/rewrite
    \begin{terms}
        \item $\mathrm{Reg}_{0} = \Set{\emptyset, \Set{\varepsilon}} \union \Set{\Set{c} \given c \in \Sigma}$.

        \item $\mathrm{Reg}_{i + 1} = \mathrm{Reg}_{i} \union \Set{A \cdot B, A \union B \given A,B \in \mathrm{Reg}_{i}} \union \Set{A^* \given A \in \mathrm{Reg}_{i}}$.
    \end{terms}

    \item $\mathrm{REG}$ is closed under union, concatenation, and Kleene star operations.

    \item \textbf{Regular expressions (regex)}\Href{https://en.wikipedia.org/wiki/Regular_expression} is a sequence of special characters that define a regular language or an operation over regular languages.
    The table below illustrates the correspondence between regular languages and regular expressions.
    Here, $c \in \Sigma$ denotes the symbol of a given alphabet, $A \subseteq \Sigma^*$ and $B \subseteq \Sigma^*$ are some regular languages, $\alpha$ and $\beta$ are regular expressions.
    In regular expressions, concatenation is denoted by~$\cdot$ (can be omitted in regex), union by~$|$, Kleene star by~$*$, and the grouping is made by parentheses.

    \begingroup
    \setlength{\tabcolsep}{0.5em}
    \renewcommand{\arraystretch}{0.8}
    \begin{tabular}{cc}
        \toprule
        \thead{Language} & \thead{Regex} \\
        \midrule
        $\emptyset$ & $\emptyset$ \\
        $\Set{\varepsilon}$ & $\varepsilon$ \\
        $\Set{c}$ & $c$ \\
        $A \union B$ & $\alpha | \beta$ \\
        $A \cdot B$ & $\alpha \beta$ \\
        $A^*$ & $\alpha^*$ \\
        $A\cdot A^*$ & $\alpha^+$ \\
        $A \union \Set{\varepsilon}$ & $\alpha?$ \\
        \bottomrule
    \end{tabular}
    \endgroup

    \item \textbf{Deterministic Finite Automaton (DFA)}\Href{https://en.wikipedia.org/wiki/Deterministic_finite_automaton} is 5-tuple $\mathcal{A} = (\Sigma, Q, q_0, F, \delta)$, where:

    \begin{terms}
        \item $\Sigma$ is an alphabet;
        \item $Q = \Set{q_1, \dots, q_n}$ is a finite set of states;
        \item $q_0 \in Q$ is an initial state;
        \item $F \subseteq Q$ is set of final (terminal, accepting) states;
        \item $\delta \colon Q \times \Sigma \to Q$ is transition function.
    \end{terms}

    \item Language \textbf{accepted} by an automaton $\mathcal{A}$ is the set $L(\mathcal{A}) = \Set{w \given \delta(q_0, w) \in F}$.

    \item \textbf{Nondeterministic Finite Automaton (NFA)}\Href{https://en.wikipedia.org/wiki/Nondeterministic_finite_automaton} is 5-tuple $\mathcal{A} = (\Sigma, Q, q_0, F, \delta)$, where:

    \begin{terms}
        \item $\Sigma$ is an alphabet;

        \item $Q = \Set{q_1, \dots, q_n}$ is a finite set of states;

        \item $q_0 \in Q$ is an initial state;

        \item $F \subseteq Q$ is set of final (terminal, accepting) states;

        \item $\delta: Q \times \Sigma \to 2^Q$ is a transition function.

        % TODO: ?
        % $\delta: (q, c) \mapsto \Set{q^_1, q_2, \dots, q_n}$, $c \in \Sigma$, $q \in Q$ (Nondeterminism).
    \end{terms}

    % TODO: explain \vdash first
    % \item Language accepted by an NFA $\mathcal{A}$ is the set $L(\mathcal{A}) = \Set{w \given \Pair{q_0, w} \vdash^*_{\text{NFA}} \Pair{f, \epsilon}, f \in F}$.

    \item \textbf{NFA to DFA} conversion algorithm:

    \begin{enumerate}
        \item Set initial state of NFA to initial state of DFA.

        \item Take the states in the DFA, and compute in the NFA what the union $\delta$ of those states for each letter in the alphabet and add them as new states in the DFA.

        \item Set every DFA state as accepting if it contains an accepting state from the NFA
    \end{enumerate}

    \item \textbf{Epsilon-NFA ($\varepsilon$-NFA)} is an NFA which allows $\varepsilon$-moves, that is, the automaton can change state without consuming input.

    % TODO: inline function definition into the definition above.
    \begin{terms}
        \item $\delta \colon Q \times (\Sigma \union \{\varepsilon\}) \to 2^Q$.
    \end{terms}

    \item \textbf{$\varepsilon$-NFA to NFA}:
    % TODO: prove that each step is correct (does not change the semantics of the automaton, i.e. the language it accepts is the same after each step, including the last one)
    \begin{enumerate}
        \item Find transitive-closure of $\varepsilon$.

        \item Back-propagate accepting states over $\varepsilon$-transitions.

        \item Perform symbol-transition back-closure over $\varepsilon$-transitions.

        \item Remove $\varepsilon$-transitions.
    \end{enumerate}

    % TODO: rewrite
    % TODO: theorem environment
    \item \textbf{Pumping lemma}\Href{https://en.wikipedia.org/wiki/Pumping_lemma_for_regular_languages} states that if $L$ if a regular language, then there exists an integer $n > 1$ depending only on $L$, such that $\forall w \in L$, $|w| > n$ can be written as $w = xyz$, such that:

    \begin{enumerate}
        \item $|y| > 0$, i.e. $y \neq \varepsilon$

        \item $|xy| \leq n$

        \item $\forall k \geq 0$, word $x y^{k} z$ is also in language $L$
    \end{enumerate}


    \item \textbf{Mealy\footnotemark{} machine}\Href{https://en.wikipedia.org/wiki/Mealy_machine} is a finite-state machine whose output is determined both by the current state and the current input.
    \footnotetext{\textsc{Mealy, George H.} (1955).
    \href{https://doi.org/10.1002/j.1538-7305.1955.tb03788.x}{A Method for Synthesizing Sequential Circuits}.
    \textit{The Bell System Technical Journal}, 34(5), 1045--79.}

    \begin{minipage}{\linewidth}
    \begin{wrapfigure}{r}{0pt}
        \tikz[myautomatonstyle]{
            \node[state, initial] (si) {$s_i$};
            \node[state, above right=1cm and 2cm of si] (s0) {$s_0$};
            \node[state, below right=1cm and 2cm of si] (s1) {$s_1$};
            \path[->]
                (si) edge [bend left, sloped, above] node {0/0} (s0)
                     edge [bend right, sloped, below] node {1/0} (s1)
                (s0) edge [loop right, right] node {0/0} (s0)
                     edge [bend left, right] node {1/1} (s1)
                (s1) edge [loop right, right] node {1/0} (s1)
                     edge [bend left, left] node {0/1} (s0)
            ;
            \node at (5.7,0) [align=left] {This Mealy machine's \\
            output is 1 whenever both \\
            previous and current symbols \\
            are equal, and 0 otherwise};
        }
        \vspace{-\intextsep}
    \end{wrapfigure}

    Formally, $\mathcal{M}_\text{Mealy} = \Set{\Sigma, \Omega, Q, q_0, \delta, \lambda_\text{Mealy}}$, where:

    \begin{terms}
        \item $\Sigma$ is an input alphabet;

        \item $\Omega$ is an output alphabet;

        \item $Q = \Set{q_1, \dots, q_n}$ is finite set of states;

        \item $q_0 \in Q$ is an initial state;

        \item $\delta \colon Q \times \Sigma \to Q$ is a transition function;

        \item $\lambda_\text{Mealy} \colon Q \times \Sigma \to \Omega$ is an output function.
    \end{terms}

    \end{minipage}

    \item \textbf{Moore\footnotemark{} machine}\Href{https://en.wikipedia.org/wiki/Moore_machine} is a finite-state machine whose output is determines only by the current state.
    \footnotetext{\textsc{Moore, Edward F.} (1956).
    \href{https://doi.org/10.1515/9781400882618-006}{Gedanken-Experiments on Sequential Machines}.
    \textit{Automata Studies, Annals of Mathematical Studies} (34), 129--153.}

    \begin{minipage}{\linewidth}
    \begin{wrapfigure}{r}{0pt}
        \tikz[myautomatonstyle]{
            \node[state, initial] (s0) {$0$};
            \node[state, right=2cm of s0] (s1) {$1$};
            \node[state, right=2cm of s1] (s2) {$2$};
            \path[->]
                (s0) edge [loop above, above] node {0} (s0)
                     edge [bend left, above] node {1} (s1)
                (s1) edge [bend left, below] node {1} (s0)
                     edge [bend right, below] node {0} (s2)
                (s2) edge [bend right, above] node {1} (s1)
                     edge [loop above, above] node {0} (s2)
            ;
            \node at (2,-1) [align=center] {This Moore machine's output \\ is modulo 3 of a binary number};
        }
        \vspace{-\intextsep}
    \end{wrapfigure}

    Formally, $\mathcal{M}_\text{Moore} = (\Sigma, \Omega, Q, q_0, \delta, \lambda_\text{Moore})$, where:

    \begin{terms}
        \item $\Sigma$ is an input alphabet;

        \item $\Omega$ is an output alphabet;

        \item $Q = \Set{q_1, \dots, q_n}$ is a finite set of states;

        \item $q_0 \in Q$ is an initial state;

        \item $\delta \colon Q \times \Sigma \to Q$ is a transition function;

        \item $\lambda_\text{Moore} \colon Q \to \Omega$ is an output function.
    \end{terms}

    \end{minipage}

    % TODO: what is M?
    % TODO: theorem
    \item \textbf{Emptiness}.
    Language $L(M)$ is not empty ($L \neq \emptyset$) if $M$ accepts a word $w$ such that $|w| \leq n$.

    % TODO: what is M?
    % TODO: theorem
    \item \textbf{Infiniteness}.
    Language $L(M)$ is infinite $(|L| = \infty)$ if $M$ accepts a word $w$ such that $n \leq |w| < 2n$.

    % TODO: theorem environment
    \item \textbf{Myhill-Nerode theorem}\Href{https://en.wikipedia.org/wiki/Myhill–Nerode_theorem} states that the following three statement are equivalent:

    \begin{enumerate}
        \item $L \subseteq \Sigma^*$ is accepted by some finite automaton ($L$ is regular).

        \item $L$ is the union of some equivalence classes of right invariant equivalence relation of finite index.

        \item Let $R_L$ be a relation over words: $x \rel[R_L] y$ iff $\forall z \in \Sigma : xz \in L \equiv yz \in L$.
        Then the quotient $\quotient[R_L]{\Sigma^*}$ is finite.
    \end{enumerate}

\newpage

    \item \textbf{Formal grammar}\Href{https://en.wikipedia.org/wiki/Formal_grammar} is 4-tuple $\mathcal{G} = (V, T, S, \mathcal{P})$, where:

    \begin{terms}
        \item $\mathcal{V}$ is vocabulary, set of variables or non-terminal symbols.

        \item $T$ is set of terminal symbols disjoint from $\mathcal{V}$.

        \item $S$ is start symbol, also called sentence symbol.

        \item $\mathcal{P}$ is set of production rules, each rule of the form:
        $\mathcal{V}^*S\mathcal{V}^* \xrightarrow{} \mathcal{V}^*$.
    \end{terms}

    \item Binary relation $\mathbf{\Rightarrow}$ over an grammar $\mathcal{G}$ is defined by:

    $x \Rightarrow y \Longleftrightarrow \exists u,v,p,q \in \mathcal{V}: (x = upv) \land (p \rightarrow{} q \in \mathcal{P}) \land (y = uqv)$.

    Pronounce as \enquote{$y$ is directly derivable from $x$}.

    \item Binary relation $\mathbf{\Rightarrow^*}$ over a grammar $\mathcal{G}$ is defined as reflexive transitive closure of $\Rightarrow$.

    Pronounced as \enquote{$y$ is derivable from $x$}.

    \item \textbf{Backus-Naur Form (BNF)}\Href{https://en.wikipedia.org/wiki/Backus–Naur_form} is notation to describe the syntax of formal language.
    A BNF specification is a set of derivation rules, written as follows:
    \setlength{\grammarparsep}{0pt plus 4pt}
    \setlength{\grammarindent}{6em}
    \begin{grammar}
        <symbol> ::= <expression>
    \end{grammar}
    where:
    \begin{terms}
        \item \synt{symbol} is a non-terminal symbol that is enclosed in angle brackets.

        \item \synt{expression} consists of one or more sequences of either terminal or non-terminal symbols where each sequence is separated by a vertical bar indicating a choice.

        \item $::=$ is a symbol that separates the production rule for a non-terminal symbol.
    \end{terms}

    %TODO? EBNF.


\end{terms}

\end{document}