| author | Christian Urban <urbanc@in.tum.de> | 
| Sun, 21 May 2017 00:46:21 +0100 | |
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changeset | 1 | \documentclass{article}
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changeset | 2 | \usepackage{../style}
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changeset | 3 | \usepackage{../langs}
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changeset | 4 | \usepackage{../graphics}
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| 480 | 6 | |
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changeset | 7 | \begin{document}
 | 
| 480 | 8 | \fnote{\copyright{} Christian Urban, King's College London, 2014, 2015, 2016, 2017}
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changeset | 9 | |
| 482 | 10 | \section*{Handout 3 (Finite Automata)}
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changeset | 11 | |
| 488 | 12 | |
| 480 | 13 | Every formal language and compiler course I know of bombards you first | 
| 14 | with automata and then to a much, much smaller extend with regular | |
| 15 | expressions. As you can see, this course is turned upside down: | |
| 16 | regular expressions come first. The reason is that regular expressions | |
| 17 | are easier to reason about and the notion of derivatives, although | |
| 18 | already quite old, only became more widely known rather | |
| 488 | 19 | recently. Still, let us in this lecture have a closer look at automata | 
| 480 | 20 | and their relation to regular expressions. This will help us with | 
| 21 | understanding why the regular expression matchers in Python, Ruby and | |
| 485 | 22 | Java are so slow with certain regular expressions. On the way we will | 
| 488 | 23 | also see what are the limitations of regular | 
| 24 | expressions. Unfortunately, they cannot be used for \emph{everything}.
 | |
| 482 | 25 | |
| 26 | ||
| 27 | \subsection*{Deterministic Finite Automata}
 | |
| 28 | ||
| 485 | 29 | Lets start\ldots the central definition is:\medskip | 
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changeset | 30 | |
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changeset | 31 | \noindent | 
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changeset | 32 | A \emph{deterministic finite automaton} (DFA), say $A$, is
 | 
| 484 | 33 | given by a five-tuple written ${\cal A}(\varSigma, Qs, Q_0, F, \delta)$ where
 | 
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changeset | 34 | |
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changeset | 35 | \begin{itemize}
 | 
| 484 | 36 | \item $\varSigma$ is an alphabet, | 
| 482 | 37 | \item $Qs$ is a finite set of states, | 
| 38 | \item $Q_0 \in Qs$ is the start state, | |
| 39 | \item $F \subseteq Qs$ are the accepting states, and | |
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changeset | 40 | \item $\delta$ is the transition function. | 
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changeset | 41 | \end{itemize}
 | 
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changeset | 42 | |
| 490 | 43 | \noindent I am sure you have seen this definition already | 
| 488 | 44 | before. The transition function determines how to ``transition'' from | 
| 45 | one state to the next state with respect to a character. We have the | |
| 46 | assumption that these transition functions do not need to be defined | |
| 47 | everywhere: so it can be the case that given a character there is no | |
| 48 | next state, in which case we need to raise a kind of ``failure | |
| 483 | 49 | exception''.  That means actually we have \emph{partial} functions as
 | 
| 484 | 50 | transitions---see the Scala implementation of DFAs later on. A | 
| 51 | typical example of a DFA is | |
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changeset | 52 | |
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changeset | 53 | \begin{center}
 | 
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changeset | 54 | \begin{tikzpicture}[>=stealth',very thick,auto,
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changeset | 55 |                     every state/.style={minimum size=0pt,
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changeset | 56 | inner sep=2pt,draw=blue!50,very thick, | 
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changeset | 57 | fill=blue!20},scale=2] | 
| 482 | 58 | \node[state,initial]  (Q_0)  {$Q_0$};
 | 
| 59 | \node[state] (Q_1) [right=of Q_0] {$Q_1$};
 | |
| 60 | \node[state] (Q_2) [below right=of Q_0] {$Q_2$};
 | |
| 61 | \node[state] (Q_3) [right=of Q_2] {$Q_3$};
 | |
| 62 | \node[state, accepting] (Q_4) [right=of Q_1] {$Q_4$};
 | |
| 63 | \path[->] (Q_0) edge node [above]  {$a$} (Q_1);
 | |
| 64 | \path[->] (Q_1) edge node [above]  {$a$} (Q_4);
 | |
| 65 | \path[->] (Q_4) edge [loop right] node  {$a, b$} ();
 | |
| 66 | \path[->] (Q_3) edge node [right]  {$a$} (Q_4);
 | |
| 67 | \path[->] (Q_2) edge node [above]  {$a$} (Q_3);
 | |
| 68 | \path[->] (Q_1) edge node [right]  {$b$} (Q_2);
 | |
| 69 | \path[->] (Q_0) edge node [above]  {$b$} (Q_2);
 | |
| 70 | \path[->] (Q_2) edge [loop left] node  {$b$} ();
 | |
| 71 | \path[->] (Q_3) edge [bend left=95, looseness=1.3] node [below]  {$b$} (Q_0);
 | |
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changeset | 72 | \end{tikzpicture}
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changeset | 73 | \end{center}
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changeset | 74 | |
| 482 | 75 | \noindent In this graphical notation, the accepting state $Q_4$ is | 
| 76 | indicated with double circles. Note that there can be more than one | |
| 77 | accepting state. It is also possible that a DFA has no accepting | |
| 485 | 78 | state at all, or that the starting state is also an accepting | 
| 482 | 79 | state. In the case above the transition function is defined everywhere | 
| 80 | and can also be given as a table as follows: | |
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changeset | 81 | |
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changeset | 82 | \[ | 
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changeset | 83 | \begin{array}{lcl}
 | 
| 482 | 84 | (Q_0, a) &\rightarrow& Q_1\\ | 
| 85 | (Q_0, b) &\rightarrow& Q_2\\ | |
| 86 | (Q_1, a) &\rightarrow& Q_4\\ | |
| 87 | (Q_1, b) &\rightarrow& Q_2\\ | |
| 88 | (Q_2, a) &\rightarrow& Q_3\\ | |
| 89 | (Q_2, b) &\rightarrow& Q_2\\ | |
| 90 | (Q_3, a) &\rightarrow& Q_4\\ | |
| 91 | (Q_3, b) &\rightarrow& Q_0\\ | |
| 92 | (Q_4, a) &\rightarrow& Q_4\\ | |
| 93 | (Q_4, b) &\rightarrow& Q_4\\ | |
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changeset | 94 | \end{array}
 | 
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changeset | 95 | \] | 
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changeset | 97 | We need to define the notion of what language is accepted by | 
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changeset | 98 | an automaton. For this we lift the transition function | 
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changeset | 99 | $\delta$ from characters to strings as follows: | 
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changeset | 100 | |
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changeset | 101 | \[ | 
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changeset | 102 | \begin{array}{lcl}
 | 
| 484 | 103 | \widehat{\delta}(q, [])        & \dn & q\\
 | 
| 104 | \widehat{\delta}(q, c\!::\!s) & \dn & \widehat{\delta}(\delta(q, c), s)\\
 | |
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changeset | 105 | \end{array}
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changeset | 106 | \] | 
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changeset | 108 | \noindent This lifted transition function is often called | 
| 485 | 109 | \emph{delta-hat}. Given a string, we start in the starting state and
 | 
| 110 | take the first character of the string, follow to the next state, then | |
| 111 | take the second character and so on. Once the string is exhausted and | |
| 112 | we end up in an accepting state, then this string is accepted by the | |
| 480 | 113 | automaton. Otherwise it is not accepted. This also means that if along | 
| 114 | the way we hit the case where the transition function $\delta$ is not | |
| 115 | defined, we need to raise an error. In our implementation we will deal | |
| 485 | 116 | with this case elegantly by using Scala's \texttt{Try}.  Summing up: a
 | 
| 117 | string $s$ is in the \emph{language accepted by the automaton} ${\cal
 | |
| 484 | 118 | A}(\varSigma, Q, Q_0, F, \delta)$ iff | 
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changeset | 120 | \[ | 
| 484 | 121 | \widehat{\delta}(Q_0, s) \in F 
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changeset | 122 | \] | 
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changeset | 123 | |
| 488 | 124 | \noindent I let you think about a definition that describes the set of | 
| 490 | 125 | all strings accepted by a deterministic finite automaton. | 
| 480 | 126 | |
| 127 | \begin{figure}[p]
 | |
| 128 | \small | |
| 490 | 129 | \lstinputlisting[numbers=left]{../progs/display/dfa.scala}
 | 
| 482 | 130 | \caption{A Scala implementation of DFAs using partial functions.
 | 
| 488 | 131 |   Note some subtleties: \texttt{deltas} implements the delta-hat
 | 
| 487 | 132 | construction by lifting the (partial) transition function to lists | 
| 485 | 133 |   of characters. Since \texttt{delta} is given as a partial function,
 | 
| 134 |   it can obviously go ``wrong'' in which case the \texttt{Try} in
 | |
| 135 |   \texttt{accepts} catches the error and returns \texttt{false}---that
 | |
| 136 |   means the string is not accepted.  The example \texttt{delta} in
 | |
| 137 | Line 28--38 implements the DFA example shown earlier in the | |
| 138 |   handout.\label{dfa}}
 | |
| 480 | 139 | \end{figure}
 | 
| 140 | ||
| 485 | 141 | My take on an implementation of DFAs in Scala is given in | 
| 483 | 142 | Figure~\ref{dfa}. As you can see, there are many features of the
 | 
| 482 | 143 | mathematical definition that are quite closely reflected in the | 
| 144 | code. In the DFA-class, there is a starting state, called | |
| 483 | 145 | \texttt{start}, with the polymorphic type \texttt{A}.  There is a
 | 
| 146 | partial function \texttt{delta} for specifying the transitions---these
 | |
| 147 | partial functions take a state (of polymorphic type \texttt{A}) and an
 | |
| 148 | input (of polymorphic type \texttt{C}) and produce a new state (of
 | |
| 149 | type \texttt{A}). For the moment it is OK to assume that \texttt{A} is
 | |
| 150 | some arbitrary type for states and the input is just characters. (The | |
| 485 | 151 | reason for not having concrete types, but polymorphic types for the | 
| 152 | states and the input of DFAs will become clearer later on.) | |
| 483 | 153 | |
| 485 | 154 | The DFA-class has also an argument for specifying final states. In the | 
| 155 | implementation it is not a set of states, as in the mathematical | |
| 156 | definition, but a function from states to booleans (this function is | |
| 157 | supposed to return true whenever a state is final; false | |
| 158 | otherwise). While this boolean function is different from the sets of | |
| 159 | states, Scala allows to use sets for such functions (see Line 40 where | |
| 160 | the DFA is initialised). Again it will become clear later on why I use | |
| 161 | functions for final states, rather than sets. | |
| 162 | ||
| 163 | The most important point in the implementation is that I use Scala's | |
| 484 | 164 | partial functions for representing the transitions; alternatives would | 
| 165 | have been \texttt{Maps} or even \texttt{Lists}. One of the main
 | |
| 166 | advantages of using partial functions is that transitions can be quite | |
| 167 | nicely defined by a series of \texttt{case} statements (see Lines 28
 | |
| 168 | -- 38 for an example). If you need to represent an automaton with a | |
| 169 | sink state (catch-all-state), you can use Scala's pattern matching and | |
| 170 | write something like | |
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| 490 | 172 | {\small\begin{lstlisting}[language=Scala]
 | 
| 480 | 173 | abstract class State | 
| 174 | ... | |
| 175 | case object Sink extends State | |
| 176 | ||
| 177 | val delta : (State, Char) :=> State = | |
| 178 |   { case (S0, 'a') => S1
 | |
| 179 | case (S1, 'a') => S2 | |
| 180 | case _ => Sink | |
| 181 | } | |
| 182 | \end{lstlisting}}  
 | |
| 183 | ||
| 485 | 184 | \noindent I let you think what the corresponding DFA looks like in the | 
| 185 | graph notation. | |
| 484 | 186 | |
| 485 | 187 | Finally, I let you ponder whether this is a good implementation of | 
| 188 | DFAs or not. In doing so I hope you notice that the $\varSigma$ and | |
| 189 | $Qs$ components (the alphabet and the set of finite states, | |
| 190 | respectively) are missing from the class definition. This means that | |
| 191 | the implementation allows you to do some fishy things you are not | |
| 192 | meant to do with DFAs. Which fishy things could that be? | |
| 480 | 193 | |
| 482 | 194 | |
| 195 | ||
| 196 | \subsection*{Non-Deterministic Finite Automata}
 | |
| 197 | ||
| 485 | 198 | Remember we want to find out what the relation is between regular | 
| 199 | expressions and automata. To do this with DFAs is a bit unwieldy. | |
| 200 | While with DFAs it is always clear that given a state and a character | |
| 201 | what the next state is (potentially none), it will be convenient to | |
| 202 | relax this restriction. That means we allow states to have several | |
| 482 | 203 | potential successor states. We even allow more than one starting | 
| 484 | 204 | state. The resulting construction is called a \emph{Non-Deterministic
 | 
| 485 | 205 |   Finite Automaton} (NFA) given also as a five-tuple ${\cal
 | 
| 206 |   A}(\varSigma, Qs, Q_{0s}, F, \rho)$ where
 | |
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changeset | 208 | \begin{itemize}
 | 
| 484 | 209 | \item $\varSigma$ is an alphabet, | 
| 482 | 210 | \item $Qs$ is a finite set of states | 
| 211 | \item $Q_{0s}$ is a set of start states ($Q_{0s} \subseteq Qs$)
 | |
| 212 | \item $F$ are some accepting states with $F \subseteq Qs$, and | |
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changeset | 213 | \item $\rho$ is a transition relation. | 
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changeset | 214 | \end{itemize}
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changeset | 216 | \noindent | 
| 483 | 217 | A typical example of a NFA is | 
| 482 | 218 | |
| 219 | % A NFA for (ab* + b)*a | |
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changeset | 220 | \begin{center}
 | 
| 483 | 221 | \begin{tikzpicture}[>=stealth',very thick, auto,
 | 
| 222 |     every state/.style={minimum size=0pt,inner sep=3pt,
 | |
| 223 | draw=blue!50,very thick,fill=blue!20},scale=2] | |
| 482 | 224 | \node[state,initial]  (Q_0)  {$Q_0$};
 | 
| 225 | \node[state] (Q_1) [right=of Q_0] {$Q_1$};
 | |
| 226 | \node[state, accepting] (Q_2) [right=of Q_1] {$Q_2$};
 | |
| 227 | \path[->] (Q_0) edge [loop above] node  {$b$} ();
 | |
| 228 | \path[<-] (Q_0) edge node [below]  {$b$} (Q_1);
 | |
| 229 | \path[->] (Q_0) edge [bend left] node [above]  {$a$} (Q_1);
 | |
| 230 | \path[->] (Q_0) edge [bend right] node [below]  {$a$} (Q_2);
 | |
| 231 | \path[->] (Q_1) edge [loop above] node  {$a,b$} ();
 | |
| 232 | \path[->] (Q_1) edge node  [above] {$a$} (Q_2);
 | |
| 233 | \end{tikzpicture}
 | |
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changeset | 234 | \end{center}
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changeset | 235 | |
| 482 | 236 | \noindent | 
| 237 | This NFA happens to have only one starting state, but in general there | |
| 238 | could be more. Notice that in state $Q_0$ we might go to state $Q_1$ | |
| 483 | 239 | \emph{or} to state $Q_2$ when receiving an $a$. Similarly in state
 | 
| 484 | 240 | $Q_1$ and receiving an $a$, we can stay in state $Q_1$ \emph{or} go to
 | 
| 241 | $Q_2$. This kind of choice is not allowed with DFAs. The downside of | |
| 488 | 242 | this choice in NFAs is that when it comes to deciding whether a string is | 
| 484 | 243 | accepted by a NFA we potentially have to explore all possibilities. I | 
| 488 | 244 | let you think which strings the above NFA accepts. | 
| 482 | 245 | |
| 246 | ||
| 485 | 247 | There are a number of additional points you should note about | 
| 483 | 248 | NFAs. Every DFA is a NFA, but not vice versa. The $\rho$ in NFAs is a | 
| 249 | transition \emph{relation} (DFAs have a transition function). The
 | |
| 250 | difference between a function and a relation is that a function has | |
| 251 | always a single output, while a relation gives, roughly speaking, | |
| 252 | several outputs. Look again at the NFA above: if you are currently in | |
| 253 | the state $Q_1$ and you read a character $b$, then you can transition | |
| 254 | to either $Q_0$ \emph{or} $Q_2$. Which route, or output, you take is
 | |
| 255 | not determined. This non-determinism can be represented by a | |
| 256 | relation. | |
| 482 | 257 | |
| 483 | 258 | My implementation of NFAs in Scala is shown in Figure~\ref{nfa}.
 | 
| 259 | Perhaps interestingly, I do not actually use relations for my NFAs, | |
| 485 | 260 | but use transition functions that return sets of states. DFAs have | 
| 261 | partial transition functions that return a single state; my NFAs | |
| 488 | 262 | return a set of states. I let you think about this representation for | 
| 485 | 263 | NFA-transitions and how it corresponds to the relations used in the | 
| 487 | 264 | mathematical definition of NFAs. An example of a transition function | 
| 488 | 265 | in Scala for the NFA shown above is | 
| 482 | 266 | |
| 490 | 267 | {\small\begin{lstlisting}[language=Scala]
 | 
| 487 | 268 | val nfa_delta : (State, Char) :=> Set[State] = | 
| 269 |   { case (Q0, 'a') => Set(Q1, Q2)
 | |
| 270 | case (Q0, 'b') => Set(Q0) | |
| 271 | case (Q1, 'a') => Set(Q1, Q2) | |
| 272 | case (Q1, 'b') => Set(Q0, Q1) } | |
| 273 | \end{lstlisting}}  
 | |
| 274 | ||
| 490 | 275 | Like in the mathematical definition, \texttt{starts} is in
 | 
| 487 | 276 | NFAs a set of states; \texttt{fins} is again a function from states to
 | 
| 485 | 277 | booleans. The \texttt{next} function calculates the set of next states
 | 
| 278 | reachable from a single state \texttt{q} by a character~\texttt{c}. In
 | |
| 279 | case there is no such state---the partial transition function is | |
| 280 | undefined---the empty set is returned (see function | |
| 281 | \texttt{applyOrElse} in Lines 9 and 10). The function \texttt{nexts}
 | |
| 282 | just lifts this function to sets of states. | |
| 283 | ||
| 484 | 284 | \begin{figure}[p]
 | 
| 482 | 285 | \small | 
| 490 | 286 | \lstinputlisting[numbers=left]{../progs/display/nfa.scala}
 | 
| 485 | 287 | \caption{A Scala implementation of NFAs using partial functions.
 | 
| 288 |   Notice that the function \texttt{accepts} implements the
 | |
| 289 | acceptance of a string in a breath-first search fashion. This can be a costly | |
| 290 | way of deciding whether a string is accepted or not in applications that need to handle | |
| 291 |   large NFAs and large inputs.\label{nfa}}
 | |
| 482 | 292 | \end{figure}
 | 
| 293 | ||
| 485 | 294 | Look very careful at the \texttt{accepts} and \texttt{deltas}
 | 
| 295 | functions in NFAs and remember that when accepting a string by a NFA | |
| 484 | 296 | we might have to explore all possible transitions (recall which state | 
| 485 | 297 | to go to is not unique anymore with NFAs\ldots{}we need to explore
 | 
| 298 | potentially all next states). The implementation achieves this | |
| 487 | 299 | exploration through a \emph{breadth-first search}. This is fine for
 | 
| 485 | 300 | small NFAs, but can lead to real memory problems when the NFAs are | 
| 301 | bigger and larger strings need to be processed. As result, some | |
| 302 | regular expression matching engines resort to a \emph{depth-first
 | |
| 303 |   search} with \emph{backtracking} in unsuccessful cases. In our
 | |
| 304 | implementation we can implement a depth-first version of | |
| 305 | \texttt{accepts} using Scala's \texttt{exists}-function as follows:
 | |
| 483 | 306 | |
| 307 | ||
| 490 | 308 | {\small\begin{lstlisting}[language=Scala]
 | 
| 483 | 309 | def search(q: A, s: List[C]) : Boolean = s match {
 | 
| 310 | case Nil => fins(q) | |
| 485 | 311 | case c::cs => next(q, c).exists(search(_, cs)) | 
| 483 | 312 | } | 
| 313 | ||
| 485 | 314 | def accepts2(s: List[C]) : Boolean = | 
| 483 | 315 | starts.exists(search(_, s)) | 
| 316 | \end{lstlisting}}
 | |
| 317 | ||
| 318 | \noindent | |
| 487 | 319 | This depth-first way of exploration seems to work quite efficiently in | 
| 320 | many examples and is much less of a strain on memory. The problem is | |
| 321 | that the backtracking can get ``catastrophic'' in some | |
| 322 | examples---remember the catastrophic backtracking from earlier | |
| 323 | lectures. This depth-first search with backtracking is the reason for | |
| 324 | the abysmal performance of some regular expression matchings in Java, | |
| 325 | Ruby and Python. I like to show you this in the next two sections. | |
| 482 | 326 | |
| 268 
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| 490 | 328 | \subsection*{Epsilon NFAs}
 | 
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| 485 | 330 | In order to get an idea what calculations are performed by Java \& | 
| 331 | friends, we need a method for transforming a regular expression into | |
| 332 | an automaton. This automaton should accept exactly those strings that | |
| 333 | are accepted by the regular expression. The simplest and most | |
| 334 | well-known method for this is called \emph{Thompson Construction},
 | |
| 335 | after the Turing Award winner Ken Thompson. This method is by | |
| 487 | 336 | recursion over regular expressions and depends on the non-determinism | 
| 488 | 337 | in NFAs described in the previous section. You will see shortly why | 
| 487 | 338 | this construction works well with NFAs, but is not so straightforward | 
| 339 | with DFAs. | |
| 340 | ||
| 341 | Unfortunately we are still one step away from our intended target | |
| 342 | though---because this construction uses a version of NFAs that allows | |
| 343 | ``silent transitions''. The idea behind silent transitions is that | |
| 344 | they allow us to go from one state to the next without having to | |
| 345 | consume a character. We label such silent transition with the letter | |
| 346 | $\epsilon$ and call the automata $\epsilon$NFAs. Two typical examples | |
| 347 | of $\epsilon$NFAs are: | |
| 484 | 348 | |
| 349 | ||
| 485 | 350 | \begin{center}
 | 
| 351 | \begin{tabular}[t]{c@{\hspace{9mm}}c}
 | |
| 352 | \begin{tikzpicture}[>=stealth',very thick,
 | |
| 353 |     every state/.style={minimum size=0pt,draw=blue!50,very thick,fill=blue!20},]
 | |
| 354 | \node[state,initial]  (Q_0)  {$Q_0$};
 | |
| 355 | \node[state] (Q_1) [above=of Q_0] {$Q_1$};
 | |
| 356 | \node[state, accepting] (Q_2) [below=of Q_0] {$Q_2$};
 | |
| 357 | \path[->] (Q_0) edge node [left]  {$\epsilon$} (Q_1);
 | |
| 358 | \path[->] (Q_0) edge node [left]  {$\epsilon$} (Q_2);
 | |
| 359 | \path[->] (Q_0) edge [loop right] node  {$a$} ();
 | |
| 360 | \path[->] (Q_1) edge [loop right] node  {$a$} ();
 | |
| 361 | \path[->] (Q_2) edge [loop right] node  {$b$} ();
 | |
| 362 | \end{tikzpicture} &
 | |
| 363 | ||
| 364 | \raisebox{20mm}{
 | |
| 365 | \begin{tikzpicture}[>=stealth',very thick,
 | |
| 366 |     every state/.style={minimum size=0pt,draw=blue!50,very thick,fill=blue!20},]
 | |
| 367 | \node[state,initial]  (r_1)  {$R_1$};
 | |
| 368 | \node[state] (r_2) [above=of r_1] {$R_2$};
 | |
| 369 | \node[state, accepting] (r_3) [right=of r_1] {$R_3$};
 | |
| 370 | \path[->] (r_1) edge node [below]  {$b$} (r_3);
 | |
| 371 | \path[->] (r_2) edge [bend left] node [above]  {$a$} (r_3);
 | |
| 372 | \path[->] (r_1) edge [bend left] node  [left] {$\epsilon$} (r_2);
 | |
| 373 | \path[->] (r_2) edge [bend left] node  [right] {$a$} (r_1);
 | |
| 374 | \end{tikzpicture}}
 | |
| 375 | \end{tabular}
 | |
| 376 | \end{center}
 | |
| 377 | ||
| 378 | \noindent | |
| 487 | 379 | Consider the $\epsilon$NFA on the left-hand side: the | 
| 380 | $\epsilon$-transitions mean you do not have to ``consume'' any part of | |
| 381 | the input string, but ``silently'' change to a different state. In | |
| 382 | this example, if you are in the starting state $Q_0$, you can silently | |
| 383 | move either to $Q_1$ or $Q_2$. You can see that once you are in $Q_1$, | |
| 384 | respectively $Q_2$, you cannot ``go back'' to the other states. So it | |
| 490 | 385 | seems allowing $\epsilon$-transitions is a rather substantial | 
| 487 | 386 | extension to NFAs. On first appearances, $\epsilon$-transitions might | 
| 387 | even look rather strange, or even silly. To start with, silent | |
| 388 | transitions make the decision whether a string is accepted by an | |
| 389 | automaton even harder: with $\epsilon$NFAs we have to look whether we | |
| 390 | can do first some $\epsilon$-transitions and then do a | |
| 391 | ``proper''-transition; and after any ``proper''-transition we again | |
| 392 | have to check whether we can do again some silent transitions. Even | |
| 393 | worse, if there is a silent transition pointing back to the same | |
| 394 | state, then we have to be careful our decision procedure for strings | |
| 395 | does not loop (remember the depth-first search for exploring all | |
| 396 | states). | |
| 485 | 397 | |
| 398 | The obvious question is: Do we get anything in return for this hassle | |
| 399 | with silent transitions? Well, we still have to work for it\ldots | |
| 400 | unfortunately. If we were to follow the many textbooks on the | |
| 401 | subject, we would now start with defining what $\epsilon$NFAs | |
| 402 | are---that would require extending the transition relation of | |
| 490 | 403 | NFAs. Next, we would show that the $\epsilon$NFAs are equivalent to | 
| 488 | 404 | NFAs and so on. Once we have done all this on paper, we would need to | 
| 405 | implement $\epsilon$NFAs. Lets try to take a shortcut instead. We are | |
| 406 | not really interested in $\epsilon$NFAs; they are only a convenient | |
| 407 | tool for translating regular expressions into automata. So we are not | |
| 408 | going to implementing them explicitly, but translate them immediately | |
| 409 | into NFAs (in a sense $\epsilon$NFAs are just a convenient API for | |
| 410 | lazy people ;o). How does the translation work? Well we have to find | |
| 411 | all transitions of the form | |
| 485 | 412 | |
| 413 | \[ | |
| 414 | q\stackrel{\epsilon}{\longrightarrow}\ldots\stackrel{\epsilon}{\longrightarrow}
 | |
| 415 | \;\stackrel{a}{\longrightarrow}\;
 | |
| 416 | \stackrel{\epsilon}{\longrightarrow}\ldots\stackrel{\epsilon}{\longrightarrow} q'
 | |
| 417 | \] | |
| 418 | ||
| 492 | 419 | \noindent where somewhere in the ``middle'' is an $a$-transition. We | 
| 420 | replace them with $q \stackrel{a}{\longrightarrow} q'$. Doing this to
 | |
| 421 | the $\epsilon$NFA on the right-hand side above gives the NFA | |
| 485 | 422 | |
| 423 | \begin{center}
 | |
| 424 | \begin{tikzpicture}[>=stealth',very thick,
 | |
| 425 |     every state/.style={minimum size=0pt,draw=blue!50,very thick,fill=blue!20},]
 | |
| 426 | \node[state,initial]  (r_1)  {$R_1$};
 | |
| 427 | \node[state] (r_2) [above=of r_1] {$R_2$};
 | |
| 428 | \node[state, accepting] (r_3) [right=of r_1] {$R_3$};
 | |
| 429 | \path[->] (r_1) edge node [above]  {$b$} (r_3);
 | |
| 430 | \path[->] (r_2) edge [bend left] node [above]  {$a$} (r_3);
 | |
| 431 | \path[->] (r_1) edge [bend left] node  [left] {$a$} (r_2);
 | |
| 432 | \path[->] (r_2) edge [bend left] node  [right] {$a$} (r_1);
 | |
| 433 | \path[->] (r_1) edge [loop below] node  {$a$} ();
 | |
| 434 | \path[->] (r_1) edge [bend right] node [below]  {$a$} (r_3);
 | |
| 435 | \end{tikzpicture}
 | |
| 436 | \end{center}
 | |
| 437 | ||
| 487 | 438 | \noindent where the single $\epsilon$-transition is replaced by | 
| 439 | three additional $a$-transitions. Please do the calculations yourself | |
| 440 | and verify that I did not forget any transition. | |
| 441 | ||
| 442 | So in what follows, whenever we are given an $\epsilon$NFA we will | |
| 488 | 443 | replace it by an equivalent NFA. The Scala code for this translation | 
| 444 | is given in Figure~\ref{enfa}. The main workhorse in this code is a
 | |
| 445 | function that calculates a fixpoint of function (Lines 5--10). This | |
| 446 | function is used for ``discovering'' which states are reachable by | |
| 487 | 447 | $\epsilon$-transitions. Once no new state is discovered, a fixpoint is | 
| 448 | reached. This is used for example when calculating the starting states | |
| 449 | of an equivalent NFA (see Line 36): we start with all starting states | |
| 450 | of the $\epsilon$NFA and then look for all additional states that can | |
| 451 | be reached by $\epsilon$-transitions. We keep on doing this until no | |
| 452 | new state can be reached. This is what the $\epsilon$-closure, named | |
| 453 | in the code \texttt{ecl}, calculates. Similarly, an accepting state of
 | |
| 454 | the NFA is when we can reach an accepting state of the $\epsilon$NFA | |
| 455 | by $\epsilon$-transitions. | |
| 456 | ||
| 485 | 457 | |
| 458 | \begin{figure}[p]
 | |
| 459 | \small | |
| 490 | 460 | \lstinputlisting[numbers=left]{../progs/display/enfa.scala}
 | 
| 485 | 461 | |
| 462 | \caption{A Scala function that translates $\epsilon$NFA into NFAs. The
 | |
| 490 | 463 |   transition function of $\epsilon$NFA takes as input an \texttt{Option[C]}.
 | 
| 485 | 464 |   \texttt{None} stands for an $\epsilon$-transition; \texttt{Some(c)}
 | 
| 488 | 465 | for a ``proper'' transition consuming a character. The functions in | 
| 466 | Lines 18--26 calculate | |
| 485 | 467 | all states reachable by one or more $\epsilon$-transition for a given | 
| 491 | 468 | set of states. The NFA is constructed in Lines 36--38. | 
| 469 | Note the interesting commands in Lines 5 and 6: their purpose is | |
| 470 |   to ensure that \texttt{fixpT} is the tail-recursive version of
 | |
| 471 | the fixpoint construction; otherwise we would quickly get a | |
| 472 | stack-overflow exception, even on small examples, due to limitations | |
| 473 | of the JVM. | |
| 474 |   \label{enfa}}
 | |
| 485 | 475 | \end{figure}
 | 
| 476 | ||
| 487 | 477 | Also look carefully how the transitions of $\epsilon$NFAs are | 
| 478 | implemented. The additional possibility of performing silent | |
| 479 | transitions is encoded by using \texttt{Option[C]} as the type for the
 | |
| 490 | 480 | ``input''. The \texttt{Some}s stand for ``proper'' transitions where
 | 
| 487 | 481 | a character is consumed; \texttt{None} stands for
 | 
| 482 | $\epsilon$-transitions. The transition functions for the two | |
| 483 | $\epsilon$NFAs from the beginning of this section can be defined as | |
| 485 | 484 | |
| 490 | 485 | {\small\begin{lstlisting}[language=Scala]
 | 
| 487 | 486 | val enfa_trans1 : (State, Option[Char]) :=> Set[State] = | 
| 487 |   { case (Q0, Some('a')) => Set(Q0)
 | |
| 488 | case (Q0, None) => Set(Q1, Q2) | |
| 489 |     case (Q1, Some('a')) => Set(Q1)
 | |
| 490 |     case (Q2, Some('b')) => Set(Q2) }
 | |
| 491 | ||
| 492 | val enfa_trans2 : (State, Option[Char]) :=> Set[State] = | |
| 493 |   { case (R1, Some('b')) => Set(R3)
 | |
| 494 | case (R1, None) => Set(R2) | |
| 495 |     case (R2, Some('a')) => Set(R1, R3) }
 | |
| 496 | \end{lstlisting}}
 | |
| 497 | ||
| 498 | \noindent | |
| 499 | I hope you agree now with my earlier statement that the $\epsilon$NFAs | |
| 500 | are just an API for NFAs. | |
| 501 | ||
| 490 | 502 | \subsection*{Thompson Construction}
 | 
| 487 | 503 | |
| 504 | Having the translation of $\epsilon$NFAs to NFAs in place, we can | |
| 505 | finally return to the problem of translating regular expressions into | |
| 506 | equivalent NFAs. Recall that by equivalent we mean that the NFAs | |
| 485 | 507 | recognise the same language. Consider the simple regular expressions | 
| 508 | $\ZERO$, $\ONE$ and $c$. They can be translated into equivalent NFAs | |
| 509 | as follows: | |
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| 488 | 511 | \begin{equation}\mbox{
 | 
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 | 
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changeset | 513 | \raisebox{1mm}{$\ZERO$} & 
 | 
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changeset | 514 | \begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]
 | 
| 482 | 515 | \node[state, initial]  (Q_0)  {$\mbox{}$};
 | 
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changeset | 516 | \end{tikzpicture}\\\\
 | 
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changeset | 517 | \raisebox{1mm}{$\ONE$} & 
 | 
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changeset | 518 | \begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]
 | 
| 482 | 519 | \node[state, initial, accepting]  (Q_0)  {$\mbox{}$};
 | 
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changeset | 520 | \end{tikzpicture}\\\\
 | 
| 487 | 521 | \raisebox{3mm}{$c$} & 
 | 
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changeset | 522 | \begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]
 | 
| 482 | 523 | \node[state, initial]  (Q_0)  {$\mbox{}$};
 | 
| 524 | \node[state, accepting]  (Q_1)  [right=of Q_0] {$\mbox{}$};
 | |
| 525 | \path[->] (Q_0) edge node [below]  {$c$} (Q_1);
 | |
| 487 | 526 | \end{tikzpicture}\\
 | 
| 488 | 527 | \end{tabular}}\label{simplecases}
 | 
| 528 | \end{equation}
 | |
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| 487 | 530 | \noindent | 
| 531 | I let you think whether the NFAs can match exactly those strings the | |
| 532 | regular expressions can match. To do this translation in code we need | |
| 533 | a way to construct states programatically...and as an additional | |
| 488 | 534 | constrain Scala needs to recognise that these states are being distinct. | 
| 487 | 535 | For this I implemented in Figure~\ref{thompson1} a class
 | 
| 536 | \texttt{TState} that includes a counter and a companion object that
 | |
| 488 | 537 | increases this counter whenever a new state is created.\footnote{You might
 | 
| 538 |   have to read up what \emph{companion objects} do in Scala.}
 | |
| 487 | 539 | |
| 485 | 540 | \begin{figure}[p]
 | 
| 541 | \small | |
| 490 | 542 | \lstinputlisting[numbers=left]{../progs/display/thompson1.scala}
 | 
| 487 | 543 | \caption{The first part of the Thompson Construction. Lines 7--16
 | 
| 488 | 544 | implement a way of how to create new states that are all | 
| 487 | 545 | distinct by virtue of a counter. This counter is | 
| 546 |   increased in the companion object of \texttt{TState}
 | |
| 547 | whenever a new state is created. The code in Lines 24--40 | |
| 488 | 548 | constructs NFAs for the simple regular expressions $\ZERO$, $\ONE$ and $c$. | 
| 549 |   Compare the pictures given in \eqref{simplecases}.
 | |
| 487 | 550 |   \label{thompson1}}
 | 
| 485 | 551 | \end{figure}
 | 
| 552 | ||
| 487 | 553 | \begin{figure}[p]
 | 
| 554 | \small | |
| 490 | 555 | \lstinputlisting[numbers=left]{../progs/display/thompson2.scala}
 | 
| 487 | 556 | \caption{The second part of the Thompson Construction implementing
 | 
| 490 | 557 |   the composition of NFAs according to $\cdot$, $+$ and ${}^*$.
 | 
| 487 | 558 | The implicit class about rich partial functions | 
| 559 |   implements the infix operation \texttt{+++} which
 | |
| 560 | combines an $\epsilon$NFA transition with a NFA transition | |
| 561 |   (both given as partial functions).\label{thompson2}}
 | |
| 562 | \end{figure}
 | |
| 485 | 563 | |
| 488 | 564 | The case for the sequence regular expression $r_1 \cdot r_2$ is a bit more | 
| 489 | 565 | complicated: Say, we are given by recursion two NFAs representing the regular | 
| 488 | 566 | expressions $r_1$ and $r_2$ respectively. | 
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 | 
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 | 
| 488 | 570 | >=stealth',very thick, | 
| 571 |     every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]
 | |
| 482 | 572 | \node[state, initial]  (Q_0)  {$\mbox{}$};
 | 
| 488 | 573 | \node[state, initial]  (Q_01) [below=1mm of Q_0] {$\mbox{}$};  
 | 
| 574 | \node[state, initial]  (Q_02) [above=1mm of Q_0] {$\mbox{}$};  
 | |
| 575 | \node (R_1)  [right=of Q_0] {$\ldots$};
 | |
| 576 | \node[state, accepting]  (T_1)  [right=of R_1] {$\mbox{}$};
 | |
| 577 | \node[state, accepting]  (T_2)  [above=of T_1] {$\mbox{}$};
 | |
| 578 | \node[state, accepting]  (T_3)  [below=of T_1] {$\mbox{}$};
 | |
| 579 | ||
| 580 | \node (A_0)  [right=2.5cm of T_1] {$\mbox{}$};
 | |
| 581 | \node[state, initial]  (A_01)  [above=1mm of A_0] {$\mbox{}$};
 | |
| 582 | \node[state, initial]  (A_02)  [below=1mm of A_0] {$\mbox{}$};
 | |
| 583 | ||
| 584 | \node (b_1)  [right=of A_0] {$\ldots$};
 | |
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 | 
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changeset | 586 | \node[state, accepting]  (c_2)  [above=of c_1] {$\mbox{}$};
 | 
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 | 
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changeset | 588 | \begin{pgfonlayer}{background}
 | 
| 488 | 589 | \node (1) [rounded corners, inner sep=1mm, thick, | 
| 590 |     draw=black!60, fill=black!20, fit= (Q_0) (R_1) (T_1) (T_2) (T_3)] {};
 | |
| 591 | \node (2) [rounded corners, inner sep=1mm, thick, | |
| 592 |     draw=black!60, fill=black!20, fit= (A_0) (b_1) (c_1) (c_2) (c_3)] {};
 | |
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changeset | 593 | \node [yshift=2mm] at (1.north) {$r_1$};
 | 
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 | 
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changeset | 595 | \end{pgfonlayer}
 | 
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changeset | 596 | \end{tikzpicture}
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changeset | 597 | \end{center}
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changeset | 598 | |
| 488 | 599 | \noindent The first NFA has some accepting states and the second some | 
| 489 | 600 | starting states. We obtain an $\epsilon$NFA for $r_1\cdot r_2$ by | 
| 601 | connecting the accepting states of the first NFA with | |
| 602 | $\epsilon$-transitions to the starting states of the second | |
| 603 | automaton. By doing so we make the accepting states of the first NFA | |
| 604 | to be non-accepting like so: | |
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changeset | 605 | |
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changeset | 606 | \begin{center}
 | 
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changeset | 607 | \begin{tikzpicture}[node distance=3mm,
 | 
| 488 | 608 | >=stealth',very thick, | 
| 609 |     every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]
 | |
| 482 | 610 | \node[state, initial]  (Q_0)  {$\mbox{}$};
 | 
| 488 | 611 | \node[state, initial]  (Q_01) [below=1mm of Q_0] {$\mbox{}$};  
 | 
| 612 | \node[state, initial]  (Q_02) [above=1mm of Q_0] {$\mbox{}$};  
 | |
| 482 | 613 | \node (r_1)  [right=of Q_0] {$\ldots$};
 | 
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changeset | 614 | \node[state]  (t_1)  [right=of r_1] {$\mbox{}$};
 | 
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changeset | 615 | \node[state]  (t_2)  [above=of t_1] {$\mbox{}$};
 | 
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changeset | 616 | \node[state]  (t_3)  [below=of t_1] {$\mbox{}$};
 | 
| 488 | 617 | |
| 618 | \node  (A_0)  [right=2.5cm of t_1] {$\mbox{}$};
 | |
| 619 | \node[state]  (A_01)  [above=1mm of A_0] {$\mbox{}$};
 | |
| 620 | \node[state]  (A_02)  [below=1mm of A_0] {$\mbox{}$};
 | |
| 621 | ||
| 622 | \node (b_1)  [right=of A_0] {$\ldots$};
 | |
| 143 
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changeset | 623 | \node[state, accepting]  (c_1)  [right=of b_1] {$\mbox{}$};
 | 
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changeset | 624 | \node[state, accepting]  (c_2)  [above=of c_1] {$\mbox{}$};
 | 
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changeset | 625 | \node[state, accepting]  (c_3)  [below=of c_1] {$\mbox{}$};
 | 
| 488 | 626 | \path[->] (t_1) edge (A_01); | 
| 492 | 627 | \path[->] (t_2) edge node [above]  {$\epsilon$s} (A_01);
 | 
| 488 | 628 | \path[->] (t_3) edge (A_01); | 
| 629 | \path[->] (t_1) edge (A_02); | |
| 630 | \path[->] (t_2) edge (A_02); | |
| 492 | 631 | \path[->] (t_3) edge node [below]  {$\epsilon$s} (A_02);
 | 
| 488 | 632 | |
| 143 
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changeset | 633 | \begin{pgfonlayer}{background}
 | 
| 488 | 634 | \node (3) [rounded corners, inner sep=1mm, thick, | 
| 635 |     draw=black!60, fill=black!20, fit= (Q_0) (c_1) (c_2) (c_3)] {};
 | |
| 143 
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changeset | 636 | \node [yshift=2mm] at (3.north) {$r_1\cdot r_2$};
 | 
| 
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changeset | 637 | \end{pgfonlayer}
 | 
| 
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changeset | 638 | \end{tikzpicture}
 | 
| 
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changeset | 639 | \end{center}
 | 
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changeset | 640 | |
| 489 | 641 | \noindent The idea behind this construction is that the start of any | 
| 642 | string is first recognised by the first NFA, then we silently change | |
| 643 | to the second NFA; the ending of the string is recognised by the | |
| 644 | second NFA...just like matching of a string by the regular expression | |
| 490 | 645 | $r_1\cdot r_2$. The Scala code for this construction is given in | 
| 489 | 646 | Figure~\ref{thompson2} in Lines 16--23. The starting states of the
 | 
| 647 | $\epsilon$NFA are the starting states of the first NFA (corresponding | |
| 648 | to $r_1$); the accepting function is the accepting function of the | |
| 649 | second NFA (corresponding to $r_2$). The new transition function is | |
| 650 | all the ``old'' transitions plus the $\epsilon$-transitions connecting | |
| 651 | the accepting states of the first NFA to the starting states of the | |
| 490 | 652 | first NFA (Lines 18 and 19). The $\epsilon$NFA is then immediately | 
| 489 | 653 | translated in a NFA. | 
| 654 | ||
| 655 | ||
| 490 | 656 | The case for the alternative regular expression $r_1 + r_2$ is | 
| 657 | slightly different: We are given by recursion two NFAs representing | |
| 658 | $r_1$ and $r_2$ respectively. Each NFA has some starting states and | |
| 659 | some accepting states. We obtain a NFA for the regular expression $r_1 | |
| 660 | + r_2$ by composing the transition functions (this crucially depends | |
| 661 | on knowing that the states of each component NFA are distinct); and | |
| 662 | also combine the starting states and accepting functions. | |
| 143 
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changeset | 663 | |
| 
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changeset | 664 | \begin{center}
 | 
| 490 | 665 | \begin{tabular}[t]{ccc}
 | 
| 143 
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changeset | 666 | \begin{tikzpicture}[node distance=3mm,
 | 
| 488 | 667 | >=stealth',very thick, | 
| 490 | 668 |     every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},
 | 
| 669 | baseline=(current bounding box.center)] | |
| 143 
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changeset | 670 | \node at (0,0)  (1)  {$\mbox{}$};
 | 
| 489 | 671 | \node  (2)  [above=10mm of 1] {};
 | 
| 672 | \node[state, initial]  (4)  [above=1mm of 2] {$\mbox{}$};
 | |
| 673 | \node[state, initial]  (5)  [below=1mm of 2] {$\mbox{}$};
 | |
| 674 | \node[state, initial]  (3)  [below=10mm of 1] {$\mbox{}$};
 | |
| 143 
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changeset | 675 | |
| 489 | 676 | \node (a)  [right=of 2] {$\ldots\,$};
 | 
| 677 | \node  (a1)  [right=of a] {$$};
 | |
| 143 
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changeset | 678 | \node[state, accepting]  (a2)  [above=of a1] {$\mbox{}$};
 | 
| 
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changeset | 679 | \node[state, accepting]  (a3)  [below=of a1] {$\mbox{}$};
 | 
| 
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changeset | 680 | |
| 
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changeset | 681 | \node (b)  [right=of 3] {$\ldots$};
 | 
| 
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changeset | 682 | \node[state, accepting]  (b1)  [right=of b] {$\mbox{}$};
 | 
| 
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changeset | 683 | \node[state, accepting]  (b2)  [above=of b1] {$\mbox{}$};
 | 
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changeset | 684 | \node[state, accepting]  (b3)  [below=of b1] {$\mbox{}$};
 | 
| 
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changeset | 685 | \begin{pgfonlayer}{background}
 | 
| 
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changeset | 686 | \node (1) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (2) (a1) (a2) (a3)] {};
 | 
| 
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changeset | 687 | \node (2) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (3) (b1) (b2) (b3)] {};
 | 
| 
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changeset | 688 | \node [yshift=3mm] at (1.north) {$r_1$};
 | 
| 
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changeset | 689 | \node [yshift=3mm] at (2.north) {$r_2$};
 | 
| 
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changeset | 690 | \end{pgfonlayer}
 | 
| 
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changeset | 691 | \end{tikzpicture}
 | 
| 490 | 692 | & | 
| 693 | \mbox{}\qquad\tikz{\draw[>=stealth,line width=2mm,->] (0,0) -- (1, 0)}\quad\mbox{}
 | |
| 694 | & | |
| 489 | 695 | \begin{tikzpicture}[node distance=3mm,
 | 
| 696 | >=stealth',very thick, | |
| 490 | 697 |     every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},
 | 
| 698 | baseline=(current bounding box.center)] | |
| 489 | 699 | \node at (0,0) (1)  {$\mbox{}$};
 | 
| 700 | \node (2)  [above=10mm of 1] {$$};
 | |
| 701 | \node[state, initial]  (4)  [above=1mm of 2] {$\mbox{}$};
 | |
| 702 | \node[state, initial]  (5)  [below=1mm of 2] {$\mbox{}$};
 | |
| 703 | \node[state, initial]  (3)  [below=10mm of 1] {$\mbox{}$};
 | |
| 143 
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changeset | 704 | |
| 489 | 705 | \node (a)  [right=of 2] {$\ldots\,$};
 | 
| 706 | \node (a1)  [right=of a] {$$};
 | |
| 143 
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changeset | 707 | \node[state, accepting]  (a2)  [above=of a1] {$\mbox{}$};
 | 
| 
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changeset | 708 | \node[state, accepting]  (a3)  [below=of a1] {$\mbox{}$};
 | 
| 
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changeset | 709 | |
| 
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changeset | 710 | \node (b)  [right=of 3] {$\ldots$};
 | 
| 
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changeset | 711 | \node[state, accepting]  (b1)  [right=of b] {$\mbox{}$};
 | 
| 
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changeset | 712 | \node[state, accepting]  (b2)  [above=of b1] {$\mbox{}$};
 | 
| 
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changeset | 713 | \node[state, accepting]  (b3)  [below=of b1] {$\mbox{}$};
 | 
| 
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142diff
changeset | 714 | |
| 489 | 715 | %\path[->] (1) edge node [above]  {$\epsilon$} (2);
 | 
| 716 | %\path[->] (1) edge node [below]  {$\epsilon$} (3);
 | |
| 143 
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changeset | 717 | |
| 
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changeset | 718 | \begin{pgfonlayer}{background}
 | 
| 
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changeset | 719 | \node (3) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (1) (a2) (a3) (b2) (b3)] {};
 | 
| 
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changeset | 720 | \node [yshift=3mm] at (3.north) {$r_1+ r_2$};
 | 
| 
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142diff
changeset | 721 | \end{pgfonlayer}
 | 
| 
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changeset | 722 | \end{tikzpicture}
 | 
| 490 | 723 | \end{tabular}
 | 
| 143 
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142diff
changeset | 724 | \end{center}
 | 
| 
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changeset | 725 | |
| 489 | 726 | \noindent The code for this construction is in Figure~\ref{thompson2}
 | 
| 490 | 727 | in Lines 25--33. | 
| 728 | ||
| 729 | Finally for the $*$-case we have a NFA for $r$ and connect its | |
| 730 | accepting states to a new starting state via | |
| 731 | $\epsilon$-transitions. This new starting state is also an accepting | |
| 732 | state, because $r^*$ can recognise the empty string. | |
| 143 
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changeset | 733 | |
| 
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changeset | 734 | \begin{center}
 | 
| 490 | 735 | \begin{tabular}[b]{@{\hspace{-4mm}}ccc@{}}  
 | 
| 143 
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changeset | 736 | \begin{tikzpicture}[node distance=3mm,
 | 
| 490 | 737 | >=stealth',very thick, | 
| 738 |     every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},
 | |
| 739 | baseline=(current bounding box.north)] | |
| 143 
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changeset | 740 | \node at (0,0)  (1)  {$\mbox{}$};
 | 
| 
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changeset | 741 | \node[state, initial]  (2)  [right=16mm of 1] {$\mbox{}$};
 | 
| 
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changeset | 742 | \node (a)  [right=of 2] {$\ldots$};
 | 
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changeset | 743 | \node[state, accepting]  (a1)  [right=of a] {$\mbox{}$};
 | 
| 
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changeset | 744 | \node[state, accepting]  (a2)  [above=of a1] {$\mbox{}$};
 | 
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changeset | 745 | \node[state, accepting]  (a3)  [below=of a1] {$\mbox{}$};
 | 
| 
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changeset | 746 | \begin{pgfonlayer}{background}
 | 
| 
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changeset | 747 | \node (1) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (2) (a1) (a2) (a3)] {};
 | 
| 
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changeset | 748 | \node [yshift=3mm] at (1.north) {$r$};
 | 
| 
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changeset | 749 | \end{pgfonlayer}
 | 
| 
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changeset | 750 | \end{tikzpicture}
 | 
| 490 | 751 | & | 
| 752 | \raisebox{-16mm}{\;\tikz{\draw[>=stealth,line width=2mm,->] (0,0) -- (1, 0)}}
 | |
| 753 | & | |
| 143 
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changeset | 754 | \begin{tikzpicture}[node distance=3mm,
 | 
| 489 | 755 | >=stealth',very thick, | 
| 490 | 756 |     every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},
 | 
| 757 | baseline=(current bounding box.north)] | |
| 143 
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changeset | 758 | \node at (0,0) [state, initial,accepting]  (1)  {$\mbox{}$};
 | 
| 
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changeset | 759 | \node[state]  (2)  [right=16mm of 1] {$\mbox{}$};
 | 
| 
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changeset | 760 | \node (a)  [right=of 2] {$\ldots$};
 | 
| 
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changeset | 761 | \node[state]  (a1)  [right=of a] {$\mbox{}$};
 | 
| 
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changeset | 762 | \node[state]  (a2)  [above=of a1] {$\mbox{}$};
 | 
| 
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changeset | 763 | \node[state]  (a3)  [below=of a1] {$\mbox{}$};
 | 
| 
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changeset | 764 | \path[->] (1) edge node [above]  {$\epsilon$} (2);
 | 
| 
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changeset | 765 | \path[->] (a1) edge [bend left=45] node [above]  {$\epsilon$} (1);
 | 
| 
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changeset | 766 | \path[->] (a2) edge [bend right] node [below]  {$\epsilon$} (1);
 | 
| 
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changeset | 767 | \path[->] (a3) edge [bend left=45] node [below]  {$\epsilon$} (1);
 | 
| 
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changeset | 768 | \begin{pgfonlayer}{background}
 | 
| 
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changeset | 769 | \node (2) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (1) (a2) (a3)] {};
 | 
| 
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changeset | 770 | \node [yshift=3mm] at (2.north) {$r^*$};
 | 
| 
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changeset | 771 | \end{pgfonlayer}
 | 
| 490 | 772 | \end{tikzpicture}    
 | 
| 773 | \end{tabular}
 | |
| 143 
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changeset | 774 | \end{center}
 | 
| 
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changeset | 775 | |
| 490 | 776 | \noindent | 
| 777 | The corresponding code is in Figure~\ref{thompson2} in Lines 35--43)
 | |
| 489 | 778 | |
| 490 | 779 | To sum up, you can see in the sequence and star cases the need of | 
| 489 | 780 | having silent $\epsilon$-transitions. Similarly the alternative case | 
| 490 | 781 | shows the need of the NFA-nondeterminism. It seems awkward to form the | 
| 489 | 782 | `alternative' composition of two DFAs, because DFA do not allow | 
| 783 | several starting and successor states. All these constructions can now | |
| 784 | be put together in the following recursive function: | |
| 785 | ||
| 786 | ||
| 490 | 787 | {\small\begin{lstlisting}[language=Scala]
 | 
| 788 | def thompson(r: Rexp) : NFAt = r match {
 | |
| 488 | 789 | case ZERO => NFA_ZERO() | 
| 790 | case ONE => NFA_ONE() | |
| 791 | case CHAR(c) => NFA_CHAR(c) | |
| 792 | case ALT(r1, r2) => NFA_ALT(thompson(r1), thompson(r2)) | |
| 793 | case SEQ(r1, r2) => NFA_SEQ(thompson(r1), thompson(r2)) | |
| 794 | case STAR(r1) => NFA_STAR(thompson(r1)) | |
| 795 | } | |
| 796 | \end{lstlisting}}
 | |
| 797 | ||
| 489 | 798 | \noindent | 
| 490 | 799 | It calculates a NFA from a regular expressions. At last we can run | 
| 800 | NFAs for the our evil regular expression examples. The graph on the | |
| 801 | left shows that when translating a regular expression such as | |
| 802 | $a^{\{n\}}$ into a NFA, the size can blow up and then even the
 | |
| 803 | relative fast (on small examples) breadth-first search can be | |
| 804 | slow. The graph on the right shows that with `evil' regular | |
| 805 | expressions the depth-first search can be abysmally slow. Even if the | |
| 806 | graphs not completely overlap with the curves of Python, Ruby and | |
| 807 | Java, they are similar enough. OK\ldots now you know why regular | |
| 808 | expression matchers in those languages are so slow. | |
| 489 | 809 | |
| 488 | 810 | |
| 811 | \begin{center}
 | |
| 812 | \begin{tabular}{@{\hspace{-1mm}}c@{\hspace{1mm}}c@{}}  
 | |
| 813 | \begin{tikzpicture}
 | |
| 814 | \begin{axis}[
 | |
| 490 | 815 |     title={Graph: $a^{?\{n\}} \cdot a^{\{n\}}$ and strings 
 | 
| 489 | 816 |            $\underbrace{\texttt{a}\ldots \texttt{a}}_{n}$},
 | 
| 490 | 817 |     title style={yshift=-2ex},
 | 
| 489 | 818 |     xlabel={$n$},
 | 
| 819 |     x label style={at={(1.05,0.0)}},
 | |
| 820 |     ylabel={time in secs},
 | |
| 821 | enlargelimits=false, | |
| 822 |     xtick={0,5,...,30},
 | |
| 823 | xmax=33, | |
| 824 | ymax=35, | |
| 825 |     ytick={0,5,...,30},
 | |
| 826 | scaled ticks=false, | |
| 827 | axis lines=left, | |
| 828 | width=5.5cm, | |
| 490 | 829 | height=4cm, | 
| 830 |     legend entries={Python,Ruby, breadth-first NFA},
 | |
| 831 |     legend style={at={(0.5,-0.25)},anchor=north,font=\small},
 | |
| 489 | 832 | legend cell align=left] | 
| 833 |   \addplot[blue,mark=*, mark options={fill=white}] table {re-python.data};
 | |
| 834 |   \addplot[brown,mark=triangle*, mark options={fill=white}] table {re-ruby.data};
 | |
| 835 | % breath-first search in NFAs | |
| 836 |   \addplot[red,mark=*, mark options={fill=white}] table {
 | |
| 837 | 1 0.00586 | |
| 838 | 2 0.01209 | |
| 839 | 3 0.03076 | |
| 840 | 4 0.08269 | |
| 841 | 5 0.12881 | |
| 842 | 6 0.25146 | |
| 843 | 7 0.51377 | |
| 844 | 8 0.89079 | |
| 845 | 9 1.62802 | |
| 846 | 10 3.05326 | |
| 847 | 11 5.92437 | |
| 848 | 12 11.67863 | |
| 849 | 13 24.00568 | |
| 850 | }; | |
| 851 | \end{axis}
 | |
| 852 | \end{tikzpicture}
 | |
| 853 | & | |
| 854 | \begin{tikzpicture}
 | |
| 855 | \begin{axis}[
 | |
| 490 | 856 |     title={Graph: $(a^*)^* \cdot b$ and strings 
 | 
| 488 | 857 |            $\underbrace{\texttt{a}\ldots \texttt{a}}_{n}$},
 | 
| 490 | 858 |     title style={yshift=-2ex},
 | 
| 488 | 859 |     xlabel={$n$},
 | 
| 860 |     x label style={at={(1.05,0.0)}},
 | |
| 861 |     ylabel={time in secs},
 | |
| 862 | enlargelimits=false, | |
| 863 |     xtick={0,5,...,30},
 | |
| 864 | xmax=33, | |
| 865 | ymax=35, | |
| 866 |     ytick={0,5,...,30},
 | |
| 867 | scaled ticks=false, | |
| 868 | axis lines=left, | |
| 869 | width=5.5cm, | |
| 490 | 870 | height=4cm, | 
| 871 |     legend entries={Python, Java, depth-first NFA},
 | |
| 872 |     legend style={at={(0.5,-0.25)},anchor=north,font=\small},
 | |
| 488 | 873 | legend cell align=left] | 
| 874 |   \addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data};
 | |
| 875 |   \addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data};  
 | |
| 876 | % depth-first search in NFAs | |
| 877 |   \addplot[red,mark=*, mark options={fill=white}] table {
 | |
| 878 | 1 0.00605 | |
| 879 | 2 0.03086 | |
| 880 | 3 0.11994 | |
| 881 | 4 0.45389 | |
| 882 | 5 2.06192 | |
| 883 | 6 8.04894 | |
| 884 | 7 32.63549 | |
| 885 | }; | |
| 886 | \end{axis}
 | |
| 887 | \end{tikzpicture}
 | |
| 888 | \end{tabular}
 | |
| 889 | \end{center}
 | |
| 890 | ||
| 891 | ||
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changeset | 892 | |
| 490 | 893 | \subsection*{Subset Construction}
 | 
| 894 | ||
| 491 | 895 | Of course, some developers of regular expression matchers are aware of | 
| 896 | these problems with sluggish NFAs and try to address them. One common | |
| 897 | technique for alleviating the problem I like to show you in this | |
| 898 | section. This will also explain why we insisted on polymorphic types in | |
| 899 | our DFA code (remember I used \texttt{A} and \texttt{C} for the types
 | |
| 900 | of states and the input, see Figure~\ref{dfa} on
 | |
| 901 | Page~\pageref{dfa}).\bigskip
 | |
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| 490 | 903 | \noindent | 
| 491 | 904 | To start remember that we did not bother with defining and | 
| 905 | implementing $\epsilon$NFAs: we immediately translated them into | |
| 490 | 906 | equivalent NFAs. Equivalent in the sense of accepting the same | 
| 907 | language (though we only claimed this and did not prove it | |
| 908 | rigorously). Remember also that NFAs have non-deterministic | |
| 909 | transitions defined as a relation or implemented as function returning | |
| 910 | sets of states. This non-determinism is crucial for the Thompson | |
| 911 | Construction to work (recall the cases for $\cdot$, $+$ and | |
| 912 | ${}^*$). But this non-determinism makes it harder with NFAs to decide
 | |
| 491 | 913 | when a string is accepted or not; whereas such a decision is rather | 
| 490 | 914 | straightforward with DFAs: recall their transition function is a | 
| 491 | 915 | \emph{function} that returns a single state. So with DFAs we do not
 | 
| 916 | have to search at all. What is perhaps interesting is the fact that | |
| 917 | for every NFA we can find a DFA that also recognises the same | |
| 918 | language. This might sound a bit paradoxical: NFA $\rightarrow$ | |
| 919 | decision of acceptance hard; DFA $\rightarrow$ decision easy. But this | |
| 920 | \emph{is} true\ldots but of course there is always a caveat---nothing
 | |
| 921 | ever is for free in life. | |
| 488 | 922 | |
| 491 | 923 | There are actually a number of methods for transforming a NFA into | 
| 924 | an equivalent DFA, but the most famous one is the \emph{subset
 | |
| 490 | 925 | construction}. Consider the following NFA where the states are | 
| 491 | 926 | labelled with $0$, $1$ and $2$. | 
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changeset | 927 | |
| 
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changeset | 928 | \begin{center}
 | 
| 
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changeset | 929 | \begin{tabular}{c@{\hspace{10mm}}c}
 | 
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changeset | 930 | \begin{tikzpicture}[scale=0.7,>=stealth',very thick,
 | 
| 
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changeset | 931 |                     every state/.style={minimum size=0pt,
 | 
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changeset | 932 | draw=blue!50,very thick,fill=blue!20}, | 
| 490 | 933 | baseline=(current bounding box.center)] | 
| 482 | 934 | \node[state,initial]  (Q_0)  {$0$};
 | 
| 490 | 935 | \node[state] (Q_1) [below=of Q_0] {$1$};
 | 
| 936 | \node[state, accepting] (Q_2) [below=of Q_1] {$2$};
 | |
| 937 | ||
| 938 | \path[->] (Q_0) edge node [right]  {$b$} (Q_1);
 | |
| 939 | \path[->] (Q_1) edge node [right]  {$a,b$} (Q_2);
 | |
| 940 | \path[->] (Q_0) edge [loop above] node  {$a, b$} ();
 | |
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changeset | 941 | \end{tikzpicture}
 | 
| 
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changeset | 942 | & | 
| 490 | 943 | \begin{tabular}{r|ll}
 | 
| 944 | states & $a$ & $b$\\ | |
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changeset | 945 | \hline | 
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changeset | 946 | $\{\}\phantom{\star}$ & $\{\}$ & $\{\}$\\
 | 
| 490 | 947 | start: $\{0\}\phantom{\star}$       & $\{0\}$   & $\{0,1\}$\\
 | 
| 948 | $\{1\}\phantom{\star}$       & $\{2\}$       & $\{2\}$\\
 | |
| 949 | $\{2\}\star$  & $\{\}$ & $\{\}$\\
 | |
| 950 | $\{0,1\}\phantom{\star}$     & $\{0,2\}$   & $\{0,1,2\}$\\
 | |
| 951 | $\{0,2\}\star$ & $\{0\}$   & $\{0,1\}$\\
 | |
| 952 | $\{1,2\}\star$ & $\{2\}$       & $\{2\}$\\
 | |
| 953 | $\{0,1,2\}\star$ & $\{0,2\}$ & $\{0,1,2\}$\\
 | |
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changeset | 954 | \end{tabular}
 | 
| 
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changeset | 955 | \end{tabular}
 | 
| 
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changeset | 956 | \end{center}
 | 
| 
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changeset | 957 | |
| 490 | 958 | \noindent The states of the corresponding DFA are given by generating | 
| 491 | 959 | all subsets of the set $\{0,1,2\}$ (seen in the states column
 | 
| 490 | 960 | in the table on the right). The other columns define the transition | 
| 491 | 961 | function for the DFA for inputs $a$ and $b$. The first row states that | 
| 490 | 962 | $\{\}$ is the sink state which has transitions for $a$ and $b$ to
 | 
| 963 | itself. The next three lines are calculated as follows: | |
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changeset | 964 | |
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changeset | 965 | \begin{itemize}
 | 
| 490 | 966 | \item Suppose you calculate the entry for the $a$-transition for state | 
| 967 |   $\{0\}$. Look for all states in the NFA that can be reached by such
 | |
| 968 | a transition from this state; this is only state $0$; therefore from | |
| 969 |   state $\{0\}$ we can go to state $\{0\}$ via an $a$-transition.
 | |
| 970 | \item Do the same for the $b$-transition; you can reach states $0$ and | |
| 971 |   $1$ in the NFA; therefore in the DFA we can go from state $\{0\}$ to
 | |
| 972 |   state $\{0,1\}$ via an $b$-transition.
 | |
| 973 | \item Continue with the states $\{1\}$ and $\{2\}$.
 | |
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changeset | 974 | \end{itemize}
 | 
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changeset | 975 | |
| 491 | 976 | \noindent | 
| 977 | Once you filled in the transitions for `simple' states $\{0\}$
 | |
| 978 | .. $\{2\}$, you only have to build the union for the compound states
 | |
| 979 | $\{0,1\}$, $\{0,2\}$ and so on. For example for $\{0,1\}$ you take the
 | |
| 980 | union of Line $\{0\}$ and Line $\{1\}$, which gives $\{0,2\}$ for $a$,
 | |
| 981 | and $\{0,1,2\}$ for $b$. And so on.
 | |
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changeset | 982 | |
| 491 | 983 | The starting state of the DFA can be calculated from the starting | 
| 984 | states of the NFA, that is in this case $\{0\}$. But in general there
 | |
| 985 | can of course be many starting states in the NFA and you would take | |
| 986 | the corresponding subset as \emph{the} starting state of the DFA.
 | |
| 987 | ||
| 988 | The accepting states in the DFA are given by all sets that contain a | |
| 989 | $2$, which is the only accpting state in this NFA. But again in | |
| 990 | general if the subset contains any accepting state from the NFA, then | |
| 991 | the corresponding state in the DFA is accepting as well. This | |
| 992 | completes the subset construction. The corresponding DFA for the NFA | |
| 993 | shown above is: | |
| 994 | ||
| 995 | \begin{equation}
 | |
| 490 | 996 | \begin{tikzpicture}[scale=0.8,>=stealth',very thick,
 | 
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changeset | 997 |                     every state/.style={minimum size=0pt,
 | 
| 491 | 998 | draw=blue!50,very thick,fill=blue!20}, | 
| 999 | baseline=(current bounding box.center)] | |
| 490 | 1000 | \node[state,initial]  (q0)  {$0$};
 | 
| 1001 | \node[state] (q01) [right=of q0] {$0,1$};
 | |
| 1002 | \node[state,accepting] (q02) [below=of q01] {$0,2$};
 | |
| 1003 | \node[state,accepting] (q012) [right=of q02] {$0,1,2$};
 | |
| 1004 | \node[state] (q1) [below=0.5cm of q0] {$1$};
 | |
| 1005 | \node[state,accepting] (q2) [below=1cm of q1] {$2$};
 | |
| 1006 | \node[state] (qn) [below left=1cm of q2] {$\{\}$};
 | |
| 1007 | \node[state,accepting] (q12) [below right=1cm of q2] {$1,2$};
 | |
| 1008 | ||
| 1009 | \path[->] (q0) edge node [above] {$b$} (q01);
 | |
| 1010 | \path[->] (q01) edge node [above] {$b$} (q012);
 | |
| 1011 | \path[->] (q0) edge [loop above] node  {$a$} ();
 | |
| 1012 | \path[->] (q012) edge [loop right] node  {$b$} ();
 | |
| 1013 | \path[->] (q012) edge node [below] {$a$} (q02);
 | |
| 1014 | \path[->] (q02) edge node [below] {$a$} (q0);
 | |
| 1015 | \path[->] (q01) edge [bend left] node [left]  {$a$} (q02);
 | |
| 1016 | \path[->] (q02) edge [bend left] node [right]  {$b$} (q01);
 | |
| 1017 | \path[->] (q1) edge node [left] {$a,b$} (q2);
 | |
| 1018 | \path[->] (q12) edge node [right] {$a, b$} (q2);
 | |
| 1019 | \path[->] (q2) edge node [right] {$a, b$} (qn);
 | |
| 1020 | \path[->] (qn) edge [loop left] node  {$a,b$} ();
 | |
| 491 | 1021 | \end{tikzpicture}\label{subsetdfa}
 | 
| 1022 | \end{equation}
 | |
| 490 | 1023 | |
| 1024 | \noindent | |
| 1025 | Please check that this is indeed a DFA. The big question is whether | |
| 491 | 1026 | this DFA can recognise the same language as the NFA we started with? | 
| 490 | 1027 | I let you ponder about this question. | 
| 1028 | ||
| 1029 | ||
| 491 | 1030 | There are also two points to note: One is that very often in the | 
| 1031 | subset construction the resulting DFA contains a number of ``dead'' | |
| 1032 | states that are never reachable from the starting state. This is | |
| 1033 | obvious in the example, where state $\{1\}$, $\{2\}$, $\{1,2\}$ and
 | |
| 1034 | $\{\}$ can never be reached from the starting state. But this might
 | |
| 1035 | not always be as obvious as that. In effect the DFA in this example is | |
| 1036 | not a \emph{minimal} DFA (more about this in a minute). Such dead
 | |
| 1037 | states can be safely removed without changing the language that is | |
| 1038 | recognised by the DFA. Another point is that in some cases, however, | |
| 1039 | the subset construction produces a DFA that does \emph{not} contain
 | |
| 1040 | any dead states\ldots{}this means it calculates a minimal DFA. Which
 | |
| 1041 | in turn means that in some cases the number of states can by going | |
| 1042 | from NFAs to DFAs exponentially increase, namely by $2^n$ (which is | |
| 1043 | the number of subsets you can form for sets of $n$ states). This blow | |
| 1044 | up in the number of states in the DFA is again bad news for how | |
| 1045 | quickly you can decide whether a string is accepted by a DFA or | |
| 1046 | not. So the caveat with DFAs is that they might make the task of | |
| 1047 | finding the next state trival, but might require $2^n$ times as many | |
| 1048 | states then a NFA.\bigskip | |
| 490 | 1049 | |
| 491 | 1050 | \noindent | 
| 1051 | To conclude this section, how conveniently we can | |
| 1052 | implement the subset construction with our versions of NFAs and | |
| 1053 | DFAs? Very conveninetly. The code is just: | |
| 490 | 1054 | |
| 1055 | {\small\begin{lstlisting}[language=Scala]
 | |
| 1056 | def subset[A, C](nfa: NFA[A, C]) : DFA[Set[A], C] = {
 | |
| 1057 | DFA(nfa.starts, | |
| 1058 |       { case (qs, c) => nfa.nexts(qs, c) }, 
 | |
| 1059 | _.exists(nfa.fins)) | |
| 1060 | } | |
| 1061 | \end{lstlisting}}  
 | |
| 1062 | ||
| 491 | 1063 | \noindent | 
| 1064 | The interesting point in this code is that the state type of the | |
| 1065 | calculated DFA is \texttt{Set[A]}. Think carefully that this works out
 | |
| 1066 | correctly. | |
| 490 | 1067 | |
| 491 | 1068 | The DFA is then given by three components: the starting states, the | 
| 1069 | transition function and the accepting-states function. The starting | |
| 1070 | states are a set in the given NFA, but a single state in the DFA. The | |
| 1071 | transition function, given the state \texttt{qs} and input \texttt{c},
 | |
| 1072 | needs to produce the next state: this is the set of all NFA states | |
| 1073 | that are reachable from each state in \texttt{qs}. The function
 | |
| 1074 | \texttt{nexts} from the NFA class already calculates this for us. The
 | |
| 1075 | accepting-states function for the DFA is true henevner at least one | |
| 1076 | state in the subset is accepting (that is true) in the NFA.\medskip | |
| 1077 | ||
| 1078 | \noindent | |
| 1079 | You might be able to spend some quality tinkering with this code and | |
| 492 | 1080 | time to ponder about it. Then you will probably notice it is actually | 
| 1081 | a bit silly. The whole point of translating the NFA into a DFA via the | |
| 491 | 1082 | subset construction is to make the decision of whether a string is | 
| 1083 | accepted or not faster. Given the code above, the generated DFA will | |
| 1084 | be exactly as fast, or as slow, as the NFA we started with (actually | |
| 1085 | it will even be a tiny bit slower). The reason is that we just re-use | |
| 492 | 1086 | the \texttt{nexts} function from the NFA. This function implements the
 | 
| 1087 | non-deterministic breadth-first search. You might be thinking: This | |
| 491 | 1088 | is cheating! \ldots{} Well, not quite as you will see later, but in
 | 
| 1089 | terms of speed we still need to work a bit in order to get | |
| 1090 | sometimes(!) a faster DFA. Let's do this next. | |
| 490 | 1091 | |
| 1092 | \subsection*{DFA Minimisation}
 | |
| 1093 | ||
| 491 | 1094 | As seen in \eqref{subsetdfa}, the subset construction from NFA to a
 | 
| 1095 | DFA can result in a rather ``inefficient'' DFA. Meaning there are | |
| 1096 | states that are not needed. There are two kinds of such unneeded | |
| 1097 | states: \emph{unreachable} states and \emph{nondistinguishable}
 | |
| 1098 | states. The first kind of states can just be removed without affecting | |
| 1099 | the language that can be recognised (after all they are | |
| 1100 | unreachable). The second kind can also be recognised and thus a DFA | |
| 1101 | can be \emph{minimised} by the following algorithm:
 | |
| 490 | 1102 | |
| 1103 | \begin{enumerate}
 | |
| 1104 | \item Take all pairs $(q, p)$ with $q \not= p$ | |
| 1105 | \item Mark all pairs that accepting and non-accepting states | |
| 1106 | \item For all unmarked pairs $(q, p)$ and all characters $c$ | |
| 1107 | test whether | |
| 1108 | ||
| 1109 |       \begin{center} 
 | |
| 1110 | $(\delta(q, c), \delta(p,c))$ | |
| 1111 |       \end{center} 
 | |
| 1112 | ||
| 1113 | are marked. If there is one, then also mark $(q, p)$. | |
| 1114 | \item Repeat last step until no change. | |
| 1115 | \item All unmarked pairs can be merged. | |
| 1116 | \end{enumerate}
 | |
| 1117 | ||
| 491 | 1118 | \noindent Unfortunately, once we throw away all unreachable states in | 
| 1119 | \eqref{subsetdfa}, all remaining states are needed.  In order to
 | |
| 1120 | illustrate the minimisation algorithm, consider the following DFA. | |
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changeset | 1121 | |
| 490 | 1122 | \begin{center}
 | 
| 1123 | \begin{tikzpicture}[>=stealth',very thick,auto,
 | |
| 1124 |                     every state/.style={minimum size=0pt,
 | |
| 1125 | inner sep=2pt,draw=blue!50,very thick, | |
| 1126 | fill=blue!20}] | |
| 1127 | \node[state,initial]  (Q_0)  {$Q_0$};
 | |
| 1128 | \node[state] (Q_1) [right=of Q_0] {$Q_1$};
 | |
| 1129 | \node[state] (Q_2) [below right=of Q_0] {$Q_2$};
 | |
| 1130 | \node[state] (Q_3) [right=of Q_2] {$Q_3$};
 | |
| 1131 | \node[state, accepting] (Q_4) [right=of Q_1] {$Q_4$};
 | |
| 1132 | \path[->] (Q_0) edge node [above]  {$a$} (Q_1);
 | |
| 1133 | \path[->] (Q_1) edge node [above]  {$a$} (Q_4);
 | |
| 1134 | \path[->] (Q_4) edge [loop right] node  {$a, b$} ();
 | |
| 1135 | \path[->] (Q_3) edge node [right]  {$a$} (Q_4);
 | |
| 1136 | \path[->] (Q_2) edge node [above]  {$a$} (Q_3);
 | |
| 1137 | \path[->] (Q_1) edge node [right]  {$b$} (Q_2);
 | |
| 1138 | \path[->] (Q_0) edge node [above]  {$b$} (Q_2);
 | |
| 1139 | \path[->] (Q_2) edge [loop left] node  {$b$} ();
 | |
| 1140 | \path[->] (Q_3) edge [bend left=95, looseness=1.3] node | |
| 1141 |   [below]  {$b$} (Q_0);
 | |
| 1142 | \end{tikzpicture}
 | |
| 1143 | \end{center}
 | |
| 1144 | ||
| 1145 | \noindent In Step 1 and 2 we consider essentially a triangle | |
| 1146 | of the form | |
| 1147 | ||
| 1148 | \begin{center}
 | |
| 1149 | \begin{tikzpicture}[scale=0.6,line width=0.8mm]
 | |
| 1150 | \draw (0,0) -- (4,0); | |
| 1151 | \draw (0,1) -- (4,1); | |
| 1152 | \draw (0,2) -- (3,2); | |
| 1153 | \draw (0,3) -- (2,3); | |
| 1154 | \draw (0,4) -- (1,4); | |
| 1155 | ||
| 1156 | \draw (0,0) -- (0, 4); | |
| 1157 | \draw (1,0) -- (1, 4); | |
| 1158 | \draw (2,0) -- (2, 3); | |
| 1159 | \draw (3,0) -- (3, 2); | |
| 1160 | \draw (4,0) -- (4, 1); | |
| 1161 | ||
| 1162 | \draw (0.5,-0.5) node {$Q_0$}; 
 | |
| 1163 | \draw (1.5,-0.5) node {$Q_1$}; 
 | |
| 1164 | \draw (2.5,-0.5) node {$Q_2$}; 
 | |
| 1165 | \draw (3.5,-0.5) node {$Q_3$};
 | |
| 1166 | ||
| 1167 | \draw (-0.5, 3.5) node {$Q_1$}; 
 | |
| 1168 | \draw (-0.5, 2.5) node {$Q_2$}; 
 | |
| 1169 | \draw (-0.5, 1.5) node {$Q_3$}; 
 | |
| 1170 | \draw (-0.5, 0.5) node {$Q_4$}; 
 | |
| 1171 | ||
| 1172 | \draw (0.5,0.5) node {\large$\star$}; 
 | |
| 1173 | \draw (1.5,0.5) node {\large$\star$}; 
 | |
| 1174 | \draw (2.5,0.5) node {\large$\star$}; 
 | |
| 1175 | \draw (3.5,0.5) node {\large$\star$};
 | |
| 1176 | \end{tikzpicture}
 | |
| 1177 | \end{center}
 | |
| 1178 | ||
| 1179 | \noindent where the lower row is filled with stars, because in | |
| 1180 | the corresponding pairs there is always one state that is | |
| 1181 | accepting ($Q_4$) and a state that is non-accepting (the other | |
| 1182 | states). | |
| 1183 | ||
| 491 | 1184 | In Step 3 we need to fill in more stars according whether | 
| 490 | 1185 | one of the next-state pairs are marked. We have to do this | 
| 1186 | for every unmarked field until there is no change anymore. | |
| 1187 | This gives the triangle | |
| 1188 | ||
| 1189 | \begin{center}
 | |
| 1190 | \begin{tikzpicture}[scale=0.6,line width=0.8mm]
 | |
| 1191 | \draw (0,0) -- (4,0); | |
| 1192 | \draw (0,1) -- (4,1); | |
| 1193 | \draw (0,2) -- (3,2); | |
| 1194 | \draw (0,3) -- (2,3); | |
| 1195 | \draw (0,4) -- (1,4); | |
| 1196 | ||
| 1197 | \draw (0,0) -- (0, 4); | |
| 1198 | \draw (1,0) -- (1, 4); | |
| 1199 | \draw (2,0) -- (2, 3); | |
| 1200 | \draw (3,0) -- (3, 2); | |
| 1201 | \draw (4,0) -- (4, 1); | |
| 1202 | ||
| 1203 | \draw (0.5,-0.5) node {$Q_0$}; 
 | |
| 1204 | \draw (1.5,-0.5) node {$Q_1$}; 
 | |
| 1205 | \draw (2.5,-0.5) node {$Q_2$}; 
 | |
| 1206 | \draw (3.5,-0.5) node {$Q_3$};
 | |
| 1207 | ||
| 1208 | \draw (-0.5, 3.5) node {$Q_1$}; 
 | |
| 1209 | \draw (-0.5, 2.5) node {$Q_2$}; 
 | |
| 1210 | \draw (-0.5, 1.5) node {$Q_3$}; 
 | |
| 1211 | \draw (-0.5, 0.5) node {$Q_4$}; 
 | |
| 1212 | ||
| 1213 | \draw (0.5,0.5) node {\large$\star$}; 
 | |
| 1214 | \draw (1.5,0.5) node {\large$\star$}; 
 | |
| 1215 | \draw (2.5,0.5) node {\large$\star$}; 
 | |
| 1216 | \draw (3.5,0.5) node {\large$\star$};
 | |
| 1217 | \draw (0.5,1.5) node {\large$\star$}; 
 | |
| 1218 | \draw (2.5,1.5) node {\large$\star$}; 
 | |
| 1219 | \draw (0.5,3.5) node {\large$\star$}; 
 | |
| 1220 | \draw (1.5,2.5) node {\large$\star$}; 
 | |
| 1221 | \end{tikzpicture}
 | |
| 1222 | \end{center}
 | |
| 1223 | ||
| 1224 | \noindent which means states $Q_0$ and $Q_2$, as well as $Q_1$ | |
| 1225 | and $Q_3$ can be merged. This gives the following minimal DFA | |
| 1226 | ||
| 1227 | \begin{center}
 | |
| 1228 | \begin{tikzpicture}[>=stealth',very thick,auto,
 | |
| 1229 |                     every state/.style={minimum size=0pt,
 | |
| 1230 | inner sep=2pt,draw=blue!50,very thick, | |
| 1231 | fill=blue!20}] | |
| 1232 | \node[state,initial]  (Q_02)  {$Q_{0, 2}$};
 | |
| 1233 | \node[state] (Q_13) [right=of Q_02] {$Q_{1, 3}$};
 | |
| 1234 | \node[state, accepting] (Q_4) [right=of Q_13] | |
| 1235 |   {$Q_{4\phantom{,0}}$};
 | |
| 1236 | \path[->] (Q_02) edge [bend left] node [above]  {$a$} (Q_13);
 | |
| 1237 | \path[->] (Q_13) edge [bend left] node [below]  {$b$} (Q_02);
 | |
| 1238 | \path[->] (Q_02) edge [loop below] node  {$b$} ();
 | |
| 1239 | \path[->] (Q_13) edge node [above]  {$a$} (Q_4);
 | |
| 1240 | \path[->] (Q_4) edge [loop above] node  {$a, b$} ();
 | |
| 344 
408fd5994288
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
333diff
changeset | 1241 | \end{tikzpicture}
 | 
| 
408fd5994288
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
333diff
changeset | 1242 | \end{center}
 | 
| 
408fd5994288
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
333diff
changeset | 1243 | |
| 
408fd5994288
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
333diff
changeset | 1244 | |
| 492 | 1245 | By the way, we are not bothering with implementing the above | 
| 1246 | minimisation algorith: while up to now all the transformations used | |
| 1247 | some clever composition of functions, the minimisation algorithm | |
| 1248 | cannot be implemented by just composing some functions. For this we | |
| 1249 | would require a more concrete representation of the transition | |
| 1250 | function (like maps). If we did this, however, then many advantages of | |
| 1251 | the functions would be thrown away. So the compromise is to not being | |
| 1252 | able to minimise (easily) our DFAs. | |
| 1253 | ||
| 490 | 1254 | \subsection*{Brzozowski's Method}
 | 
| 269 
83e6cb90216d
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
268diff
changeset | 1255 | |
| 491 | 1256 | I know tyhis is already a long, long rant: but after all it is a topic | 
| 1257 | that has been researched for more than 60 years. If you reflect on | |
| 492 | 1258 | what you have read so far, the story is that you can take a regular | 
| 491 | 1259 | expression, translate it via the Thompson Construction into an | 
| 1260 | $\epsilon$NFA, then translate it into a NFA by removing all | |
| 1261 | $\epsilon$-transitions, and then via the subset construction obtain a | |
| 1262 | DFA. In all steps we made sure the language, or which strings can be | |
| 1263 | recognised, stays the same. After the last section, we can even | |
| 492 | 1264 | minimise the DFA (maybe not in code). But again we made sure the same | 
| 1265 | language is recognised. You might be wondering: Can we go into the | |
| 1266 | other direction? Can we go from a DFA and obtain a regular expression | |
| 1267 | that can recognise the same language as the DFA?\medskip | |
| 491 | 1268 | |
| 1269 | \noindent | |
| 1270 | The answer is yes. Again there are several methods for calculating a | |
| 1271 | regular expression for a DFA. I will show you Brzozowski's method | |
| 1272 | because it calculates a regular expression using quite familiar | |
| 1273 | transformations for solving equational systems. Consider the DFA: | |
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1274 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1275 | \begin{center}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1276 | \begin{tikzpicture}[scale=1.5,>=stealth',very thick,auto,
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1277 |                     every state/.style={minimum size=0pt,
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1278 | inner sep=2pt,draw=blue!50,very thick, | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1279 | fill=blue!20}] | 
| 482 | 1280 |   \node[state, initial]        (q0) at ( 0,1) {$Q_0$};
 | 
| 1281 |   \node[state]                    (q1) at ( 1,1) {$Q_1$};
 | |
| 1282 |   \node[state, accepting] (q2) at ( 2,1) {$Q_2$};
 | |
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1283 |   \path[->] (q0) edge[bend left] node[above] {$a$} (q1)
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1284 |             (q1) edge[bend left] node[above] {$b$} (q0)
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1285 |             (q2) edge[bend left=50] node[below] {$b$} (q0)
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1286 |             (q1) edge node[above] {$a$} (q2)
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1287 |             (q2) edge [loop right] node {$a$} ()
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1288 |             (q0) edge [loop below] node {$b$} ();
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1289 | \end{tikzpicture}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1290 | \end{center}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1291 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1292 | \noindent for which we can set up the following equational | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1293 | system | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1294 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1295 | \begin{eqnarray}
 | 
| 482 | 1296 | Q_0 & = & \ONE + Q_0\,b + Q_1\,b + Q_2\,b\\ | 
| 1297 | Q_1 & = & Q_0\,a\\ | |
| 1298 | Q_2 & = & Q_1\,a + Q_2\,a | |
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1299 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1300 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1301 | \noindent There is an equation for each node in the DFA. Let | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1302 | us have a look how the right-hand sides of the equations are | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1303 | constructed. First have a look at the second equation: the | 
| 482 | 1304 | left-hand side is $Q_1$ and the right-hand side $Q_0\,a$. The | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1305 | right-hand side is essentially all possible ways how to end up | 
| 482 | 1306 | in node $Q_1$. There is only one incoming edge from $Q_0$ consuming | 
| 322 
698ed1c96cd0
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
318diff
changeset | 1307 | an $a$. Therefore the right hand side is this | 
| 482 | 1308 | state followed by character---in this case $Q_0\,a$. Now lets | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1309 | have a look at the third equation: there are two incoming | 
| 482 | 1310 | edges for $Q_2$. Therefore we have two terms, namely $Q_1\,a$ and | 
| 1311 | $Q_2\,a$. These terms are separated by $+$. The first states | |
| 1312 | that if in state $Q_1$ consuming an $a$ will bring you to | |
| 485 | 1313 | $Q_2$, and the second that being in $Q_2$ and consuming an $a$ | 
| 482 | 1314 | will make you stay in $Q_2$. The right-hand side of the | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1315 | first equation is constructed similarly: there are three | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1316 | incoming edges, therefore there are three terms. There is | 
| 444 
3056a4c071b0
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
349diff
changeset | 1317 | one exception in that we also ``add'' $\ONE$ to the | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1318 | first equation, because it corresponds to the starting state | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1319 | in the DFA. | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1320 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1321 | Having constructed the equational system, the question is | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1322 | how to solve it? Remarkably the rules are very similar to | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1323 | solving usual linear equational systems. For example the | 
| 482 | 1324 | second equation does not contain the variable $Q_1$ on the | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1325 | right-hand side of the equation. We can therefore eliminate | 
| 482 | 1326 | $Q_1$ from the system by just substituting this equation | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1327 | into the other two. This gives | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1328 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1329 | \begin{eqnarray}
 | 
| 482 | 1330 | Q_0 & = & \ONE + Q_0\,b + Q_0\,a\,b + Q_2\,b\\ | 
| 1331 | Q_2 & = & Q_0\,a\,a + Q_2\,a | |
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1332 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1333 | |
| 485 | 1334 | \noindent where in Equation (4) we have two occurrences | 
| 482 | 1335 | of $Q_0$. Like the laws about $+$ and $\cdot$, we can simplify | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1336 | Equation (4) to obtain the following two equations: | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1337 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1338 | \begin{eqnarray}
 | 
| 482 | 1339 | Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_2\,b\\ | 
| 1340 | Q_2 & = & Q_0\,a\,a + Q_2\,a | |
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1341 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1342 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1343 | \noindent Unfortunately we cannot make any more progress with | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1344 | substituting equations, because both (6) and (7) contain the | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1345 | variable on the left-hand side also on the right-hand side. | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1346 | Here we need to now use a law that is different from the usual | 
| 349 
434891622131
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
344diff
changeset | 1347 | laws about linear equations. It is called \emph{Arden's rule}.
 | 
| 
434891622131
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
344diff
changeset | 1348 | It states that if an equation is of the form $q = q\,r + s$ | 
| 
434891622131
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
344diff
changeset | 1349 | then it can be transformed to $q = s\, r^*$. Since we can | 
| 
434891622131
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
344diff
changeset | 1350 | assume $+$ is symmetric, Equation (7) is of that form: $s$ is | 
| 482 | 1351 | $Q_0\,a\,a$ and $r$ is $a$. That means we can transform | 
| 349 
434891622131
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
344diff
changeset | 1352 | (7) to obtain the two new equations | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1353 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1354 | \begin{eqnarray}
 | 
| 482 | 1355 | Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_2\,b\\ | 
| 1356 | Q_2 & = & Q_0\,a\,a\,(a^*) | |
| 292 
7ed2a25dd115
updated
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270diff
changeset | 1357 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1358 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1359 | \noindent Now again we can substitute the second equation into | 
| 482 | 1360 | the first in order to eliminate the variable $Q_2$. | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1361 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1362 | \begin{eqnarray}
 | 
| 482 | 1363 | Q_0 & = & \ONE + Q_0\,(b + a\,b) + Q_0\,a\,a\,(a^*)\,b | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1364 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1365 | |
| 482 | 1366 | \noindent Pulling $Q_0$ out as a single factor gives: | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1367 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1368 | \begin{eqnarray}
 | 
| 482 | 1369 | Q_0 & = & \ONE + Q_0\,(b + a\,b + a\,a\,(a^*)\,b) | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1370 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1371 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1372 | \noindent This equation is again of the form so that we can | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1373 | apply Arden's rule ($r$ is $b + a\,b + a\,a\,(a^*)\,b$ and $s$ | 
| 482 | 1374 | is $\ONE$). This gives as solution for $Q_0$ the following | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1375 | regular expression: | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1376 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1377 | \begin{eqnarray}
 | 
| 482 | 1378 | Q_0 & = & \ONE\,(b + a\,b + a\,a\,(a^*)\,b)^* | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1379 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1380 | |
| 349 
434891622131
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
344diff
changeset | 1381 | \noindent Since this is a regular expression, we can simplify | 
| 444 
3056a4c071b0
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
349diff
changeset | 1382 | away the $\ONE$ to obtain the slightly simpler regular | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1383 | expression | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1384 | |
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1385 | \begin{eqnarray}
 | 
| 482 | 1386 | Q_0 & = & (b + a\,b + a\,a\,(a^*)\,b)^* | 
| 292 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1387 | \end{eqnarray}
 | 
| 
7ed2a25dd115
updated
 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
270diff
changeset | 1388 | |
| 
7ed2a25dd115
updated
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270diff
changeset | 1389 | \noindent | 
| 
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changeset | 1390 | Now we can unwind this process and obtain the solutions | 
| 
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changeset | 1391 | for the other equations. This gives: | 
| 
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changeset | 1392 | |
| 
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changeset | 1393 | \begin{eqnarray}
 | 
| 482 | 1394 | Q_0 & = & (b + a\,b + a\,a\,(a^*)\,b)^*\\ | 
| 1395 | Q_1 & = & (b + a\,b + a\,a\,(a^*)\,b)^*\,a\\ | |
| 1396 | Q_2 & = & (b + a\,b + a\,a\,(a^*)\,b)^*\,a\,a\,(a)^* | |
| 292 
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changeset | 1397 | \end{eqnarray}
 | 
| 
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270diff
changeset | 1398 | |
| 
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changeset | 1399 | \noindent Finally, we only need to ``add'' up the equations | 
| 
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changeset | 1400 | which correspond to a terminal state. In our running example, | 
| 482 | 1401 | this is just $Q_2$. Consequently, a regular expression | 
| 491 | 1402 | that recognises the same language as the DFA is | 
| 292 
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270diff
changeset | 1403 | |
| 
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270diff
changeset | 1404 | \[ | 
| 
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changeset | 1405 | (b + a\,b + a\,a\,(a^*)\,b)^*\,a\,a\,(a)^* | 
| 
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changeset | 1406 | \] | 
| 
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changeset | 1407 | |
| 491 | 1408 | \noindent You can somewhat crosscheck your solution by taking a string | 
| 1409 | the regular expression can match and and see whether it can be matched | |
| 1410 | by the DFA.  One string for example is $aaa$ and \emph{voila} this
 | |
| 292 
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270diff
changeset | 1411 | string is also matched by the automaton. | 
| 
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changeset | 1412 | |
| 491 | 1413 | We should prove that Brzozowski's method really produces an equivalent | 
| 1414 | regular expression. But for the purposes of this module, we omit | |
| 1415 | this. I guess you are relieved. | |
| 269 
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268diff
changeset | 1416 | |
| 
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268diff
changeset | 1417 | |
| 490 | 1418 | \subsection*{Regular Languages}
 | 
| 269 
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268diff
changeset | 1419 | |
| 
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changeset | 1420 | Given the constructions in the previous sections we obtain | 
| 349 
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changeset | 1421 | the following overall picture: | 
| 269 
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268diff
changeset | 1422 | |
| 
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268diff
changeset | 1423 | \begin{center}
 | 
| 
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268diff
changeset | 1424 | \begin{tikzpicture}
 | 
| 
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 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
268diff
changeset | 1425 | \node (rexp)  {\bf Regexps};
 | 
| 
83e6cb90216d
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268diff
changeset | 1426 | \node (nfa) [right=of rexp] {\bf NFAs};
 | 
| 
83e6cb90216d
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268diff
changeset | 1427 | \node (dfa) [right=of nfa] {\bf DFAs};
 | 
| 
83e6cb90216d
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268diff
changeset | 1428 | \node (mdfa) [right=of dfa] {\bf\begin{tabular}{c}minimal\\ DFAs\end{tabular}};
 | 
| 
83e6cb90216d
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268diff
changeset | 1429 | \path[->,line width=1mm] (rexp) edge node [above=4mm, black] {\begin{tabular}{c@{\hspace{9mm}}}Thompson's\\[-1mm] construction\end{tabular}} (nfa);
 | 
| 
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268diff
changeset | 1430 | \path[->,line width=1mm] (nfa) edge node [above=4mm, black] {\begin{tabular}{c}subset\\[-1mm] construction\end{tabular}}(dfa);
 | 
| 
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changeset | 1431 | \path[->,line width=1mm] (dfa) edge node [below=5mm, black] {minimisation} (mdfa);
 | 
| 344 
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changeset | 1432 | \path[->,line width=1mm] (dfa) edge [bend left=45] node [below] {\begin{tabular}{l}Brzozowski's\\ method\end{tabular}} (rexp);
 | 
| 269 
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changeset | 1433 | \end{tikzpicture}
 | 
| 
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268diff
changeset | 1434 | \end{center}
 | 
| 
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268diff
changeset | 1435 | |
| 
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changeset | 1436 | \noindent By going from regular expressions over NFAs to DFAs, | 
| 
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changeset | 1437 | we can always ensure that for every regular expression there | 
| 349 
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344diff
changeset | 1438 | exists a NFA and a DFA that can recognise the same language. | 
| 269 
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changeset | 1439 | Although we did not prove this fact. Similarly by going from | 
| 
83e6cb90216d
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268diff
changeset | 1440 | DFAs to regular expressions, we can make sure for every DFA | 
| 
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268diff
changeset | 1441 | there exists a regular expression that can recognise the same | 
| 
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changeset | 1442 | language. Again we did not prove this fact. | 
| 
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268diff
changeset | 1443 | |
| 491 | 1444 | The fundamental conclusion we can draw is that automata and regular | 
| 269 
83e6cb90216d
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268diff
changeset | 1445 | expressions can recognise the same set of languages: | 
| 
83e6cb90216d
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268diff
changeset | 1446 | |
| 
83e6cb90216d
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changeset | 1447 | \begin{quote} A language is \emph{regular} iff there exists a
 | 
| 
83e6cb90216d
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268diff
changeset | 1448 | regular expression that recognises all its strings. | 
| 
83e6cb90216d
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268diff
changeset | 1449 | \end{quote}
 | 
| 
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268diff
changeset | 1450 | |
| 
83e6cb90216d
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268diff
changeset | 1451 | \noindent or equivalently | 
| 
83e6cb90216d
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268diff
changeset | 1452 | |
| 
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 Christian Urban <christian dot urban at kcl dot ac dot uk> parents: 
268diff
changeset | 1453 | \begin{quote} A language is \emph{regular} iff there exists an
 | 
| 
83e6cb90216d
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268diff
changeset | 1454 | automaton that recognises all its strings. | 
| 
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268diff
changeset | 1455 | \end{quote}
 | 
| 268 
18bef085a7ca
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251diff
changeset | 1456 | |
| 491 | 1457 | \noindent Note that this is not a stement for a particular language | 
| 1458 | (that is a particular set of strings), but about a large class of | |
| 1459 | languages, namely the regular ones. | |
| 268 
18bef085a7ca
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251diff
changeset | 1460 | |
| 491 | 1461 | As a consequence for deciding whether a string is recognised by a | 
| 1462 | regular expression, we could use our algorithm based on derivatives or | |
| 1463 | NFAs or DFAs. But let us quickly look at what the differences mean in | |
| 1464 | computational terms. Translating a regular expression into a NFA gives | |
| 1465 | us an automaton that has $O(n)$ states---that means the size of the | |
| 1466 | NFA grows linearly with the size of the regular expression. The | |
| 1467 | problem with NFAs is that the problem of deciding whether a string is | |
| 1468 | accepted or not is computationally not cheap. Remember with NFAs we | |
| 1469 | have potentially many next states even for the same input and also | |
| 1470 | have the silent $\epsilon$-transitions. If we want to find a path from | |
| 1471 | the starting state of a NFA to an accepting state, we need to consider | |
| 1472 | all possibilities. In Ruby, Python and Java this is done by a | |
| 1473 | depth-first search, which in turn means that if a ``wrong'' choice is | |
| 1474 | made, the algorithm has to backtrack and thus explore all potential | |
| 1475 | candidates. This is exactly the reason why Ruby, Python and Java are | |
| 1476 | so slow for evil regular expressions. An alternative to the | |
| 1477 | potentially slow depth-first search is to explore the search space in | |
| 1478 | a breadth-first fashion, but this might incur a big memory penalty. | |
| 1479 | ||
| 1480 | To avoid the problems with NFAs, we can translate them into DFAs. With | |
| 1481 | DFAs the problem of deciding whether a string is recognised or not is | |
| 1482 | much simpler, because in each state it is completely determined what | |
| 1483 | the next state will be for a given input. So no search is needed. The | |
| 1484 | problem with this is that the translation to DFAs can explode | |
| 1485 | exponentially the number of states. Therefore when this route is | |
| 1486 | taken, we definitely need to minimise the resulting DFAs in order to | |
| 1487 | have an acceptable memory and runtime behaviour. But remember the | |
| 1488 | subset construction in the worst case explodes the number of states by | |
| 1489 | $2^n$. Effectively also the translation to DFAs can incur a big | |
| 349 
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344diff
changeset | 1490 | runtime penalty. | 
| 269 
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changeset | 1491 | |
| 491 | 1492 | But this does not mean that everything is bad with automata. Recall | 
| 1493 | the problem of finding a regular expressions for the language that is | |
| 1494 | \emph{not} recognised by a regular expression. In our implementation
 | |
| 1495 | we added explicitly such a regular expressions because they are useful | |
| 1496 | for recognising comments. But in principle we did not need to. The | |
| 1497 | argument for this is as follows: take a regular expression, translate | |
| 349 
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344diff
changeset | 1498 | it into a NFA and then a DFA that both recognise the same | 
| 491 | 1499 | language. Once you have the DFA it is very easy to construct the | 
| 1500 | automaton for the language not recognised by a DFA. If the DFA is | |
| 1501 | completed (this is important!), then you just need to exchange the | |
| 1502 | accepting and non-accepting states. You can then translate this DFA | |
| 1503 | back into a regular expression and that will be the regular expression | |
| 1504 | that can match all strings the original regular expression could | |
| 1505 | \emph{not} match.
 | |
| 268 
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changeset | 1506 | |
| 491 | 1507 | It is also interesting that not all languages are regular. The most | 
| 1508 | well-known example of a language that is not regular consists of all | |
| 1509 | the strings of the form | |
| 292 
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changeset | 1510 | |
| 
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changeset | 1511 | \[a^n\,b^n\] | 
| 
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270diff
changeset | 1512 | |
| 491 | 1513 | \noindent meaning strings that have the same number of $a$s and | 
| 1514 | $b$s. You can try, but you cannot find a regular expression for this | |
| 1515 | language and also not an automaton. One can actually prove that there | |
| 1516 | is no regular expression nor automaton for this language, but again | |
| 1517 | that would lead us too far afield for what we want to do in this | |
| 1518 | module. | |
| 270 
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269diff
changeset | 1519 | |
| 492 | 1520 | |
| 1521 | \subsection*{Where Have Derivatives Gone?}
 | |
| 1522 | ||
| 1523 | By now you are probably fed up by this text. It is now way too long | |
| 1524 | for one lecture, but there is still one aspect of the | |
| 1525 | automata-connection I like to highlight for you. Perhaps by now you | |
| 1526 | are asking yourself: Where have the derivatives gone? Did we just | |
| 1527 | forget them? Well, they have a place in the picture of calculating a | |
| 1528 | DFA from the regular expression. | |
| 1529 | ||
| 1530 | To be done | |
| 1531 | ||
| 1532 | ||
| 490 | 1533 | %\section*{Further Reading}
 | 
| 270 
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269diff
changeset | 1534 | |
| 490 | 1535 | %Compare what a ``human expert'' would create as an automaton for the | 
| 1536 | %regular expression $a\cdot (b + c)^*$ and what the Thomson | |
| 1537 | %algorithm generates. | |
| 325 
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324diff
changeset | 1538 | |
| 
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324diff
changeset | 1539 | %http://www.inf.ed.ac.uk/teaching/courses/ct/ | 
| 140 
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changeset | 1540 | \end{document}
 | 
| 
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changeset | 1541 | |
| 
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changeset | 1542 | %%% Local Variables: | 
| 
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changeset | 1543 | %%% mode: latex | 
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changeset | 1544 | %%% TeX-master: t | 
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changeset | 1545 | %%% End: | 
| 482 | 1546 | |
| 1547 |