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%\usepackage{opensans}          % Can't make this work so far. Shame. Would be lovely.
\usepackage[normalem]{ulem}     % For underlining links
% Most of the following are not required for the majority
% of cheat sheets but are needed for some symbol support.
\usepackage{amsmath}            % Symbols
\usepackage{MnSymbol}           % Symbols
\usepackage{wasysym}            % Symbols
%\usepackage[english,german,french,spanish,italian]{babel}              % Languages

% Document Info
\author{mrbhalerao}
\pdfinfo{
  /Title (b.pdf)
  /Creator (Cheatography)
  /Author (mrbhalerao)
  /Subject (b Cheat Sheet)
}

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% Header and Footer
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\fancyhead{} % Set header to blank
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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
    \SetRowColor{DarkBackground}
    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
    }
\end{tabulary}
\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{b Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{mrbhalerao} via \textcolor{DarkBackground}{\uline{cheatography.com/141437/cs/31834/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}mrbhalerao \\
  \uline{cheatography.com/mrbhalerao} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 24th April, 2022.\\
   Updated 24th April, 2022.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
  \SetRowColor{white}
  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{2.33919 cm} x{2.63781 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{DFA (Deterministic Finite Automaton)}}  \tn
% Row 0
\SetRowColor{LightBackground}
Automaton Repres- entation & M=(Q,Σ,δ,q0,F) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Q: Set of states & \{q0,q1,q2\} \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
Σ: input alphabet & \{a,b\} \& ε ∉ Σ \tn 
% Row Count 4 (+ 1)
% Row 3
\SetRowColor{white}
δ: transition function & δ(q,x)=q' \tn 
% Row Count 6 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{q0: initial state} \tn 
% Row Count 7 (+ 1)
% Row 5
\SetRowColor{white}
F: set of accepting states & \{q2\} \tn 
% Row Count 9 (+ 2)
% Row 6
\SetRowColor{LightBackground}
Language of Automaton & L(M)=\{ w∈Σ{\emph{ : δ}} (q0,w)∈F\} \tn 
% Row Count 11 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Languages are regular if a DFA exists for \newline that language. \newline DFA: Each state has one transition for \newline evrey alphabet}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.23965 cm} x{2.73735 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{NFA (Non-deterministic Finite Automaton)}}  \tn
% Row 0
\SetRowColor{LightBackground}
Formal Definition & M=(Q,Σ,Δ,S,F) \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
Q: Set of states & \{q0,q1,q2\} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
Σ: input alphabet & \{a,b\} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
Δ: transition function & Δ(q,x)=\{q1,q2,...\} (include ε) \tn 
% Row Count 5 (+ 2)
% Row 4
\SetRowColor{LightBackground}
S: initial states & \{q0\} \tn 
% Row Count 6 (+ 1)
% Row 5
\SetRowColor{white}
F: set of accepting states & \{q2\} \tn 
% Row Count 8 (+ 2)
% Row 6
\SetRowColor{LightBackground}
Language of Automaton & L(M) = \{w1,w2,...,wn\} \tn 
% Row Count 10 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{NFA: Each state can have different \newline transition with the same language output}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{NFA to DFA Conversion}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{1. Set initial state of NFA to initial state of DFA} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{2. Take the states in the DFA, and compute in the NFA what the union Δ of those states for each letter in the alphabet and add them as new states in the DFA.} \tn 
% Row Count 6 (+ 4)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{For example if (q0,a) takes you to \{q1,q2\} add a state \{q1,q2\}. If there isn't one, the add state null} \tn 
% Row Count 9 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{NFA to DFA Conversion (cont)}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{3. Set every DFA state as accepting if it contains an accepting state from the NFA} \tn 
% Row Count 2 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The language for the NFA (M) and DFA (M') are equivalent L(M)=L(M')}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.44333 cm} x{3.53367 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Properties of Regex}}  \tn
% Row 0
\SetRowColor{LightBackground}
L1 ∪ L2 & Initial state has two ε transi- tions, one to L1 and one to L2 \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
L1L2L1L2 & L1 accept state tranitions (ε) to L2 initial state \tn 
% Row Count 5 (+ 2)
% Row 2
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L1* & New initial state transitions to L1 initial state. New accept state transitions (ε) from L1 accept state. Initial to accept state transition transitions (ε) and vice versa. \tn 
% Row Count 12 (+ 7)
% Row 3
\SetRowColor{white}
L1R (reverse) & Reverse all transitions. Make initial state accepting state and vice versa. \tn 
% Row Count 15 (+ 3)
% Row 4
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!(L1) Complement & Accepting states become non-accepting and vice versa \tn 
% Row Count 17 (+ 2)
% Row 5
\SetRowColor{white}
L1 ∩ L2 & !( !(L1) ∪ !(L2) ) \tn 
% Row Count 18 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{P.S. To turn multiple states to one accept \newline state in an NFA, just add a new accept \newline state, and add transition to the old accept \newline states with language ε.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Intersection DFA1 ∩ DFA2}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Transitions}} \{\{nl\}\}M1 :\{\{nl\}\} q1→ →q2 \{\{nl\}\}M2 : \{\{nl\}\}p1→ →p2 & DFA M: (q1,p1)→ a →- \{\{nl\}\}(q2,p2) \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
{\bf{Initial State}}\{\{nl\}\} M1: q0 \{\{nl\}\}M2 : p0 & DFA M: (q0,p0) \tn 
% Row Count 7 (+ 3)
% Row 2
\SetRowColor{LightBackground}
{\bf{Accept State}}\{\{nl\}\} M1 : qi \{\{nl\}\}M2 : pj,pk & DFA M: (qi,pj) , (qi,pk) \tn 
% Row Count 10 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.84149 cm} x{3.13551 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Regular Expressions}}  \tn
% Row 0
\SetRowColor{LightBackground}
L(r1+r2) & = L(r1) ∪ L(r2) \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
L(r1•r2) & = L(r1)L(r2) \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
L(r1*) & = (L(r1))* \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
L(a) & = \{a\} \tn 
% Row Count 4 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Precedence: * → → • → +}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{NFA to Regular Language}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{1. Transform each transition into regex (e.g. (a,b) is a+b} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{2. Remove each state one by one, until you are left with the initial and accepting state} \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{3. Resulting regular expression: r = r1{\emph{ r2(r4+r3r1}}r2)* where:\{\{nl\}\} r1: initial → initial \{\{nl\}\} r2: initial → accepting \{\{nl\}\}r3: accepting → initial \{\{nl\}\}r4: accepting → accepting} \tn 
% Row Count 8 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Proving Regularity with Pumping Lemma}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Prove than an infinite language L is not regular:} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{1. Assume L is regular} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{2. The pumping lemma should hold for L} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{3. Use the pumping lemma to obtain a contradiction:} \tn 
% Row Count 5 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{~a. Let m be the critical length for L\{\{nl\}\} ~b. Choose a particular string w∈L which} \tn 
% Row Count 7 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{satisfies the length condition |w|≥m} \tn 
% Row Count 8 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{~c. Write w=xyz \{\{nl\}\}~d. Show that w'=xy z∉L for some i≠1} \tn 
% Row Count 10 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}