# HG changeset patch # User Christian Urban # Date 1494329515 -3600 # Node ID 4fee50f3830572d469154c8cfb50aa9dbcc8f92f # Parent e28d7a327870af644be9f8855f96c591504c0961 updated diff -r e28d7a327870 -r 4fee50f38305 handouts/ho03.pdf Binary file handouts/ho03.pdf has changed diff -r e28d7a327870 -r 4fee50f38305 handouts/ho03.tex --- a/handouts/ho03.tex Sun May 07 03:01:29 2017 +0100 +++ b/handouts/ho03.tex Tue May 09 12:31:55 2017 +0100 @@ -40,7 +40,7 @@ \item $\delta$ is the transition function. \end{itemize} -\noindent I am sure you have seen this defininition already +\noindent I am sure you have seen this definition already before. The transition function determines how to ``transition'' from one state to the next state with respect to a character. We have the assumption that these transition functions do not need to be defined @@ -122,13 +122,11 @@ \] \noindent I let you think about a definition that describes the set of -all strings accepted by a determinsitic finite automaton. +all strings accepted by a deterministic finite automaton. \begin{figure}[p] \small -\lstinputlisting[numbers=left,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] - {../progs/display/dfa.scala} +\lstinputlisting[numbers=left]{../progs/display/dfa.scala} \caption{A Scala implementation of DFAs using partial functions. Note some subtleties: \texttt{deltas} implements the delta-hat construction by lifting the (partial) transition function to lists @@ -171,8 +169,7 @@ sink state (catch-all-state), you can use Scala's pattern matching and write something like -{\small\begin{lstlisting}[language=Scala,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] +{\small\begin{lstlisting}[language=Scala] abstract class State ... case object Sink extends State @@ -267,8 +264,7 @@ mathematical definition of NFAs. An example of a transition function in Scala for the NFA shown above is -{\small\begin{lstlisting}[language=Scala,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] +{\small\begin{lstlisting}[language=Scala] val nfa_delta : (State, Char) :=> Set[State] = { case (Q0, 'a') => Set(Q1, Q2) case (Q0, 'b') => Set(Q0) @@ -276,7 +272,7 @@ case (Q1, 'b') => Set(Q0, Q1) } \end{lstlisting}} -\noindent Like in the mathematical definition, \texttt{starts} is in +Like in the mathematical definition, \texttt{starts} is in NFAs a set of states; \texttt{fins} is again a function from states to booleans. The \texttt{next} function calculates the set of next states reachable from a single state \texttt{q} by a character~\texttt{c}. In @@ -287,9 +283,7 @@ \begin{figure}[p] \small -\lstinputlisting[numbers=left,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] - {../progs/display/nfa.scala} +\lstinputlisting[numbers=left]{../progs/display/nfa.scala} \caption{A Scala implementation of NFAs using partial functions. Notice that the function \texttt{accepts} implements the acceptance of a string in a breath-first search fashion. This can be a costly @@ -311,8 +305,7 @@ \texttt{accepts} using Scala's \texttt{exists}-function as follows: -{\small\begin{lstlisting}[language=Scala,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] +{\small\begin{lstlisting}[language=Scala] def search(q: A, s: List[C]) : Boolean = s match { case Nil => fins(q) case c::cs => next(q, c).exists(search(_, cs)) @@ -332,7 +325,7 @@ Ruby and Python. I like to show you this in the next two sections. -\subsubsection*{Epsilon NFAs} +\subsection*{Epsilon NFAs} In order to get an idea what calculations are performed by Java \& friends, we need a method for transforming a regular expression into @@ -389,7 +382,7 @@ this example, if you are in the starting state $Q_0$, you can silently move either to $Q_1$ or $Q_2$. You can see that once you are in $Q_1$, respectively $Q_2$, you cannot ``go back'' to the other states. So it -seems allowing $\epsilon$-transitions is a rather substancial +seems allowing $\epsilon$-transitions is a rather substantial extension to NFAs. On first appearances, $\epsilon$-transitions might even look rather strange, or even silly. To start with, silent transitions make the decision whether a string is accepted by an @@ -407,7 +400,7 @@ unfortunately. If we were to follow the many textbooks on the subject, we would now start with defining what $\epsilon$NFAs are---that would require extending the transition relation of -NFAs. Next, we woudl show that the $\epsilon$NFAs are equivalent to +NFAs. Next, we would show that the $\epsilon$NFAs are equivalent to NFAs and so on. Once we have done all this on paper, we would need to implement $\epsilon$NFAs. Lets try to take a shortcut instead. We are not really interested in $\epsilon$NFAs; they are only a convenient @@ -464,12 +457,10 @@ \begin{figure}[p] \small -\lstinputlisting[numbers=left,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] - {../progs/display/enfa.scala} +\lstinputlisting[numbers=left]{../progs/display/enfa.scala} \caption{A Scala function that translates $\epsilon$NFA into NFAs. The - transtion function of $\epsilon$NFA takes as input an \texttt{Option[C]}. + transition function of $\epsilon$NFA takes as input an \texttt{Option[C]}. \texttt{None} stands for an $\epsilon$-transition; \texttt{Some(c)} for a ``proper'' transition consuming a character. The functions in Lines 18--26 calculate @@ -480,13 +471,12 @@ Also look carefully how the transitions of $\epsilon$NFAs are implemented. The additional possibility of performing silent transitions is encoded by using \texttt{Option[C]} as the type for the -``input''. The \texttt{Some}s stand for ``propper'' transitions where +``input''. The \texttt{Some}s stand for ``proper'' transitions where a character is consumed; \texttt{None} stands for $\epsilon$-transitions. The transition functions for the two $\epsilon$NFAs from the beginning of this section can be defined as -{\small\begin{lstlisting}[language=Scala,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] +{\small\begin{lstlisting}[language=Scala] val enfa_trans1 : (State, Option[Char]) :=> Set[State] = { case (Q0, Some('a')) => Set(Q0) case (Q0, None) => Set(Q1, Q2) @@ -503,7 +493,7 @@ I hope you agree now with my earlier statement that the $\epsilon$NFAs are just an API for NFAs. -\subsubsection*{Thompson Construction} +\subsection*{Thompson Construction} Having the translation of $\epsilon$NFAs to NFAs in place, we can finally return to the problem of translating regular expressions into @@ -543,9 +533,7 @@ \begin{figure}[p] \small -\lstinputlisting[numbers=left,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] - {../progs/display/thompson1.scala} +\lstinputlisting[numbers=left]{../progs/display/thompson1.scala} \caption{The first part of the Thompson Construction. Lines 7--16 implement a way of how to create new states that are all distinct by virtue of a counter. This counter is @@ -558,11 +546,9 @@ \begin{figure}[p] \small -\lstinputlisting[numbers=left,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] - {../progs/display/thompson2.scala} +\lstinputlisting[numbers=left]{../progs/display/thompson2.scala} \caption{The second part of the Thompson Construction implementing - the composition of NFAs according to $\cdot$, $+$ and $\_^*$. + the composition of NFAs according to $\cdot$, $+$ and ${}^*$. The implicit class about rich partial functions implements the infix operation \texttt{+++} which combines an $\epsilon$NFA transition with a NFA transition @@ -650,25 +636,31 @@ string is first recognised by the first NFA, then we silently change to the second NFA; the ending of the string is recognised by the second NFA...just like matching of a string by the regular expression -$r_1\cdot r_2$. The Scala code for this constrction is given in +$r_1\cdot r_2$. The Scala code for this construction is given in Figure~\ref{thompson2} in Lines 16--23. The starting states of the $\epsilon$NFA are the starting states of the first NFA (corresponding to $r_1$); the accepting function is the accepting function of the second NFA (corresponding to $r_2$). The new transition function is all the ``old'' transitions plus the $\epsilon$-transitions connecting the accepting states of the first NFA to the starting states of the -first NFA (Lines 18 and 19). The $\epsilon$NFA is then immedately +first NFA (Lines 18 and 19). The $\epsilon$NFA is then immediately translated in a NFA. -The case for the choice regular expression $r_1 + r_2$ is slightly -different: We are given by recursion two NFAs representing $r_1$ and -$r_2$ respectively. +The case for the alternative regular expression $r_1 + r_2$ is +slightly different: We are given by recursion two NFAs representing +$r_1$ and $r_2$ respectively. Each NFA has some starting states and +some accepting states. We obtain a NFA for the regular expression $r_1 ++ r_2$ by composing the transition functions (this crucially depends +on knowing that the states of each component NFA are distinct); and +also combine the starting states and accepting functions. \begin{center} +\begin{tabular}[t]{ccc} \begin{tikzpicture}[node distance=3mm, >=stealth',very thick, - every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},] + every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20}, + baseline=(current bounding box.center)] \node at (0,0) (1) {$\mbox{}$}; \node (2) [above=10mm of 1] {}; \node[state, initial] (4) [above=1mm of 2] {$\mbox{}$}; @@ -691,18 +683,13 @@ \node [yshift=3mm] at (2.north) {$r_2$}; \end{pgfonlayer} \end{tikzpicture} -\end{center} - -\noindent Each NFA has some starting states and some accepting -states. We obtain a NFA for the regular expression $r_1 + r_2$ -by composing the transition functions (this crucially depends -on knowing that the states of each component NFA are distinct); -and also combine the starting states and accepting functions: - -\begin{center} +& +\mbox{}\qquad\tikz{\draw[>=stealth,line width=2mm,->] (0,0) -- (1, 0)}\quad\mbox{} +& \begin{tikzpicture}[node distance=3mm, >=stealth',very thick, - every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},] + every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20}, + baseline=(current bounding box.center)] \node at (0,0) (1) {$\mbox{}$}; \node (2) [above=10mm of 1] {$$}; \node[state, initial] (4) [above=1mm of 2] {$\mbox{}$}; @@ -727,14 +714,23 @@ \node [yshift=3mm] at (3.north) {$r_1+ r_2$}; \end{pgfonlayer} \end{tikzpicture} +\end{tabular} \end{center} \noindent The code for this construction is in Figure~\ref{thompson2} -in Lines 25--33. Finally for the $*$-case we have a NFA for $r$ +in Lines 25--33. + +Finally for the $*$-case we have a NFA for $r$ and connect its +accepting states to a new starting state via +$\epsilon$-transitions. This new starting state is also an accepting +state, because $r^*$ can recognise the empty string. \begin{center} +\begin{tabular}[b]{@{\hspace{-4mm}}ccc@{}} \begin{tikzpicture}[node distance=3mm, - >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},] + >=stealth',very thick, + every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20}, + baseline=(current bounding box.north)] \node at (0,0) (1) {$\mbox{}$}; \node[state, initial] (2) [right=16mm of 1] {$\mbox{}$}; \node (a) [right=of 2] {$\ldots$}; @@ -746,18 +742,13 @@ \node [yshift=3mm] at (1.north) {$r$}; \end{pgfonlayer} \end{tikzpicture} -\end{center} - -\noindent and connect its accepting states to a new starting state via -$\epsilon$-transitions. This new starting state is also an accepting -state, because $r^*$ can recognise the empty string. This gives the -following $\epsilon$NFA for $r^*$ (the corresponding code is in -Figure~\ref{thompson2} in Lines 35--43: - -\begin{center} +& +\raisebox{-16mm}{\;\tikz{\draw[>=stealth,line width=2mm,->] (0,0) -- (1, 0)}} +& \begin{tikzpicture}[node distance=3mm, >=stealth',very thick, - every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},] + every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20}, + baseline=(current bounding box.north)] \node at (0,0) [state, initial,accepting] (1) {$\mbox{}$}; \node[state] (2) [right=16mm of 1] {$\mbox{}$}; \node (a) [right=of 2] {$\ldots$}; @@ -772,21 +763,23 @@ \node (2) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (1) (a2) (a3)] {}; \node [yshift=3mm] at (2.north) {$r^*$}; \end{pgfonlayer} -\end{tikzpicture} +\end{tikzpicture} +\end{tabular} \end{center} +\noindent +The corresponding code is in Figure~\ref{thompson2} in Lines 35--43) -To sum ap, you can see in the sequence and star cases the need of +To sum up, you can see in the sequence and star cases the need of having silent $\epsilon$-transitions. Similarly the alternative case -shows the need of the NFA-nondeterminsim. It seems awkward to form the +shows the need of the NFA-nondeterminism. It seems awkward to form the `alternative' composition of two DFAs, because DFA do not allow several starting and successor states. All these constructions can now be put together in the following recursive function: -{\small\begin{lstlisting}[language=Scala,linebackgroundcolor= - {\ifodd\value{lstnumber}\color{capri!3}\fi}] -def thompson (r: Rexp) : NFAt = r match { +{\small\begin{lstlisting}[language=Scala] +def thompson(r: Rexp) : NFAt = r match { case ZERO => NFA_ZERO() case ONE => NFA_ONE() case CHAR(c) => NFA_CHAR(c) @@ -797,16 +790,25 @@ \end{lstlisting}} \noindent -It calculates a NFA from a regular expressions. At last we can run a -NFA for the our evil regular expression examples. +It calculates a NFA from a regular expressions. At last we can run +NFAs for the our evil regular expression examples. The graph on the +left shows that when translating a regular expression such as +$a^{\{n\}}$ into a NFA, the size can blow up and then even the +relative fast (on small examples) breadth-first search can be +slow. The graph on the right shows that with `evil' regular +expressions the depth-first search can be abysmally slow. Even if the +graphs not completely overlap with the curves of Python, Ruby and +Java, they are similar enough. OK\ldots now you know why regular +expression matchers in those languages are so slow. \begin{center} \begin{tabular}{@{\hspace{-1mm}}c@{\hspace{1mm}}c@{}} \begin{tikzpicture} \begin{axis}[ - title={Graph: $\texttt{a?\{n\}\,a{\{n\}}}$ and strings + title={Graph: $a^{?\{n\}} \cdot a^{\{n\}}$ and strings $\underbrace{\texttt{a}\ldots \texttt{a}}_{n}$}, + title style={yshift=-2ex}, xlabel={$n$}, x label style={at={(1.05,0.0)}}, ylabel={time in secs}, @@ -818,9 +820,9 @@ scaled ticks=false, axis lines=left, width=5.5cm, - height=4.5cm, - legend entries={Python,Ruby}, - legend pos=south east, + height=4cm, + legend entries={Python,Ruby, breadth-first NFA}, + legend style={at={(0.5,-0.25)},anchor=north,font=\small}, legend cell align=left] \addplot[blue,mark=*, mark options={fill=white}] table {re-python.data}; \addplot[brown,mark=triangle*, mark options={fill=white}] table {re-ruby.data}; @@ -845,8 +847,9 @@ & \begin{tikzpicture} \begin{axis}[ - title={Graph: $\texttt{(a*)*\,b}$ and strings + title={Graph: $(a^*)^* \cdot b$ and strings $\underbrace{\texttt{a}\ldots \texttt{a}}_{n}$}, + title style={yshift=-2ex}, xlabel={$n$}, x label style={at={(1.05,0.0)}}, ylabel={time in secs}, @@ -858,9 +861,9 @@ scaled ticks=false, axis lines=left, width=5.5cm, - height=4.5cm, - legend entries={Python, Java}, - legend pos=outer north east, + height=4cm, + legend entries={Python, Java, depth-first NFA}, + legend style={at={(0.5,-0.25)},anchor=north,font=\small}, legend cell align=left] \addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data}; \addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data}; @@ -881,166 +884,318 @@ -\subsubsection*{Subset Construction} +\subsection*{Subset Construction} + +Of course, some developers of regular expression matchers are aware +of these problems with sluggish NFAs and try to address them. One +common technique for this I like to show you in this section. It will +also explain why we insisted on polymorphic types in our DFA code +(remember I used \texttt{A} and \texttt{C} for the types of states and +the input, see Figure~\ref{dfa} on Page~\pageref{dfa}).\bigskip -Remember that we did not bother with defining and implementing -$\epsilon$NFA; we immediately translated them into equivalent -NFAs. Equivalent in the sense of accepting the same language (though -we only claimed this and did not prove it rigorously). Remember also -that NFAs have a non-deterministic transitions, given as a relation. -This non-determinism makes it harder to decide when a string is -accepted or not; such a decision is rather straightforward with DFAs -(remember their transition function). +\noindent +To start, remember that we did not bother with defining and +implementing $\epsilon$NFA; we immediately translated them into +equivalent NFAs. Equivalent in the sense of accepting the same +language (though we only claimed this and did not prove it +rigorously). Remember also that NFAs have non-deterministic +transitions defined as a relation or implemented as function returning +sets of states. This non-determinism is crucial for the Thompson +Construction to work (recall the cases for $\cdot$, $+$ and +${}^*$). But this non-determinism makes it harder with NFAs to decide +when a string is accepted or not; such a decision is rather +straightforward with DFAs: recall their transition function is a +\emph{function} that returns a single state. So we do not have to +search at all. What is perhaps interesting is the fact that for every +NFA we can find a DFA that also recognises the same language. This +might sound a bit paradoxical: NFA $\rightarrow$ decision of +acceptance hard; DFA $\rightarrow$ decision easy. But this \emph{is} +true\ldots but of course there is always a caveat---nothing is ever +for free in life. -What is interesting is that for every NFA we can find a DFA that also -recognises the same language. This might sound like a bit paradoxical, -but I litke to show you this next. There are a number of ways of -transforming a NFA into an equivalent DFA, but the most famous is -\emph{subset construction}. Consider again the NFA below on the left, -consisting of nodes labelled, say, with $0$, $1$ and $2$. +There are a number of techniques for transforming a NFA into an +equivalent DFA, but the most famous one is the \emph{subset + construction}. Consider the following NFA where the states are +labelled with, say, $0$, $1$ and $2$. \begin{center} \begin{tabular}{c@{\hspace{10mm}}c} \begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=0pt, draw=blue!50,very thick,fill=blue!20}, - baseline=0mm] + baseline=(current bounding box.center)] \node[state,initial] (Q_0) {$0$}; -\node[state] (Q_1) [above=of Q_0] {$1$}; -\node[state, accepting] (Q_2) [below=of Q_0] {$2$}; -\path[->] (Q_0) edge node [left] {$\epsilon$} (Q_1); -\path[->] (Q_0) edge node [left] {$\epsilon$} (Q_2); -\path[->] (Q_0) edge [loop right] node {$a$} (); -\path[->] (Q_1) edge [loop above] node {$a$} (); -\path[->] (Q_2) edge [loop below] node {$b$} (); +\node[state] (Q_1) [below=of Q_0] {$1$}; +\node[state, accepting] (Q_2) [below=of Q_1] {$2$}; + +\path[->] (Q_0) edge node [right] {$b$} (Q_1); +\path[->] (Q_1) edge node [right] {$a,b$} (Q_2); +\path[->] (Q_0) edge [loop above] node {$a, b$} (); \end{tikzpicture} & -\begin{tabular}{r|cl} -nodes & $a$ & $b$\\ +\begin{tabular}{r|ll} +states & $a$ & $b$\\ \hline $\{\}\phantom{\star}$ & $\{\}$ & $\{\}$\\ -$\{0\}\phantom{\star}$ & $\{0,1,2\}$ & $\{2\}$\\ -$\{1\}\phantom{\star}$ & $\{1\}$ & $\{\}$\\ -$\{2\}\star$ & $\{\}$ & $\{2\}$\\ -$\{0,1\}\phantom{\star}$ & $\{0,1,2\}$ & $\{2\}$\\ -$\{0,2\}\star$ & $\{0,1,2\}$ & $\{2\}$\\ -$\{1,2\}\star$ & $\{1\}$ & $\{2\}$\\ -s: $\{0,1,2\}\star$ & $\{0,1,2\}$ & $\{2\}$\\ +start: $\{0\}\phantom{\star}$ & $\{0\}$ & $\{0,1\}$\\ +$\{1\}\phantom{\star}$ & $\{2\}$ & $\{2\}$\\ +$\{2\}\star$ & $\{\}$ & $\{\}$\\ +$\{0,1\}\phantom{\star}$ & $\{0,2\}$ & $\{0,1,2\}$\\ +$\{0,2\}\star$ & $\{0\}$ & $\{0,1\}$\\ +$\{1,2\}\star$ & $\{2\}$ & $\{2\}$\\ +$\{0,1,2\}\star$ & $\{0,2\}$ & $\{0,1,2\}$\\ \end{tabular} \end{tabular} \end{center} -\noindent The nodes of the DFA are given by calculating all -subsets of the set of nodes of the NFA (seen in the nodes -column on the right). The table shows the transition function -for the DFA. The first row states that $\{\}$ is the -sink node which has transitions for $a$ and $b$ to itself. -The next three lines are calculated as follows: +\noindent The states of the corresponding DFA are given by generating +all subsets of the set of states of the NFA (seen in the states column +in the table on the right). The other columns define the transition +function for the DFA for input $a$ and $b$. The first row states that +$\{\}$ is the sink state which has transitions for $a$ and $b$ to +itself. The next three lines are calculated as follows: \begin{itemize} -\item suppose you calculate the entry for the transition for - $a$ and the node $\{0\}$ -\item start from the node $0$ in the NFA -\item do as many $\epsilon$-transition as you can obtaining a - set of nodes, in this case $\{0,1,2\}$ -\item filter out all notes that do not allow an $a$-transition - from this set, this excludes $2$ which does not permit a - $a$-transition -\item from the remaining set, do as many $\epsilon$-transition - as you can, this yields again $\{0,1,2\}$ -\item the resulting set specifies the transition from $\{0\}$ - when given an $a$ +\item Suppose you calculate the entry for the $a$-transition for state + $\{0\}$. Look for all states in the NFA that can be reached by such + a transition from this state; this is only state $0$; therefore from + state $\{0\}$ we can go to state $\{0\}$ via an $a$-transition. +\item Do the same for the $b$-transition; you can reach states $0$ and + $1$ in the NFA; therefore in the DFA we can go from state $\{0\}$ to + state $\{0,1\}$ via an $b$-transition. +\item Continue with the states $\{1\}$ and $\{2\}$. +\item Once you filled in the transitions for `simple' state, you only + have to build the union for the compound states $\{0,1\}$, $\{0,2\}$ + and so on. For example for $\{0,1\}$ you take the union of line + $\{0\}$ and line $\{1\}$, which gives $\{0,2\}$ for $a$, and + $\{0,1,2\}$ for $b$. And so on. +\item The starting state of the DFA can be calculated from the + starting states of the NFA, that is in this case $0$. But in general + there can be many starting states in the NFA and you would take the + corresponding subset as \emph{the} starting state of the DFA. +\item The accepting states in the DFA are given by all sets that + contain a $2$, which is the only accpting state in this NFA. But + again in general if the subset contains an accepting state from the + NFA, then the corresponding state in the DFA is accepting as well. \end{itemize} -\noindent So the transition from the state $\{0\}$ reading an -$a$ goes to the state $\{0,1,2\}$. Similarly for the other -entries in the rows for $\{0\}$, $\{1\}$ and $\{2\}$. The -other rows are calculated by just taking the union of the -single node entries. For example for $a$ and $\{0,1\}$ we need -to take the union of $\{0,1,2\}$ (for $0$) and $\{1\}$ (for -$1$). The starting state of the DFA can be calculated from the -starting state of the NFA, that is $0$, and then do as many -$\epsilon$-transitions as possible. This gives $\{0,1,2\}$ -which is the starting state of the DFA. The terminal states in -the DFA are given by all sets that contain a $2$, which is the -terminal state of the NFA. This completes the subset -construction. So the corresponding DFA to the NFA from -above is +\noindent This completes the subset construction. The corresponding +DFA for the NFA shown above is: \begin{center} -\begin{tikzpicture}[scale=0.7,>=stealth',very thick, +\begin{tikzpicture}[scale=0.8,>=stealth',very thick, every state/.style={minimum size=0pt, draw=blue!50,very thick,fill=blue!20}, baseline=0mm] -\node[state,initial,accepting] (q012) {$0,1,2$}; -\node[state,accepting] (q02) [right=of q012] {$0,2$}; -\node[state] (q01) [above=of q02] {$0,1$}; -\node[state,accepting] (q12) [below=of q02] {$1,2$}; -\node[state] (q0) [right=2cm of q01] {$0$}; -\node[state] (q1) [right=2.5cm of q02] {$1$}; -\node[state,accepting] (q2) [right=1.5cm of q12] {$2$}; -\node[state] (qn) [right=of q1] {$\{\}$}; +\node[state,initial] (q0) {$0$}; +\node[state] (q01) [right=of q0] {$0,1$}; +\node[state,accepting] (q02) [below=of q01] {$0,2$}; +\node[state,accepting] (q012) [right=of q02] {$0,1,2$}; +\node[state] (q1) [below=0.5cm of q0] {$1$}; +\node[state,accepting] (q2) [below=1cm of q1] {$2$}; +\node[state] (qn) [below left=1cm of q2] {$\{\}$}; +\node[state,accepting] (q12) [below right=1cm of q2] {$1,2$}; + +\path[->] (q0) edge node [above] {$b$} (q01); +\path[->] (q01) edge node [above] {$b$} (q012); +\path[->] (q0) edge [loop above] node {$a$} (); +\path[->] (q012) edge [loop right] node {$b$} (); +\path[->] (q012) edge node [below] {$a$} (q02); +\path[->] (q02) edge node [below] {$a$} (q0); +\path[->] (q01) edge [bend left] node [left] {$a$} (q02); +\path[->] (q02) edge [bend left] node [right] {$b$} (q01); +\path[->] (q1) edge node [left] {$a,b$} (q2); +\path[->] (q12) edge node [right] {$a, b$} (q2); +\path[->] (q2) edge node [right] {$a, b$} (qn); +\path[->] (qn) edge [loop left] node {$a,b$} (); +\end{tikzpicture} +\end{center} + +\noindent +Please check that this is indeed a DFA. The big question is whether +this DFA can recognise the same language as the NFA we started with. +I let you ponder about this question. + + +There are also two points to note: One is that very often the +resulting DFA contains a number of ``dead'' states that are never +reachable from the starting state. This is obvious in this case, where +state $\{1\}$, $\{2\}$, $\{1,2\}$ and $\{\}$ can never be reached from +the starting state. In effect the DFA in this example is not a +\emph{minimal} DFA (more about this in a minute). Such dead states can +be safely removed without changing the language that is recognised by +the DFA. Another point is that in some cases, however, the subset +construction produces a DFA that does \emph{not} contain any dead +states\ldots{}and further calculates a minimal DFA. Which in turn +means that in some cases the number of states can by going from NFAs +to DFAs exponentially increase, namely by $2^n$ (which is the number +of subsets you can form for $n$ states). This blow up the number of +states in the DFA is again bad news for how quickly you can decide +whether a string is accepted by a DFA or not. So the caveat with DFAs +is that they might make the task of finding the next state trival, but +might require $2^n$ times as many states as a NFA.\bigskip + +Lastly, can we + +{\small\begin{lstlisting}[language=Scala] +def subset[A, C](nfa: NFA[A, C]) : DFA[Set[A], C] = { + DFA(nfa.starts, + { case (qs, c) => nfa.nexts(qs, c) }, + _.exists(nfa.fins)) +} +\end{lstlisting}} + + + +\subsection*{DFA Minimisation} + +As seen in the subset construction, the translation +of a NFA to a DFA can result in a rather ``inefficient'' +DFA. Meaning there are states that are not needed. A +DFA can be \emph{minimised} by the following algorithm: + +\begin{enumerate} +\item Take all pairs $(q, p)$ with $q \not= p$ +\item Mark all pairs that accepting and non-accepting states +\item For all unmarked pairs $(q, p)$ and all characters $c$ + test whether + + \begin{center} + $(\delta(q, c), \delta(p,c))$ + \end{center} + + are marked. If there is one, then also mark $(q, p)$. +\item Repeat last step until no change. +\item All unmarked pairs can be merged. +\end{enumerate} + +\noindent To illustrate this algorithm, consider the following +DFA. -\path[->] (q012) edge [loop below] node {$a$} (); -\path[->] (q012) edge node [above] {$b$} (q2); -\path[->] (q12) edge [bend left] node [below,pos=0.4] {$a$} (q1); -\path[->] (q12) edge node [below] {$b$} (q2); -\path[->] (q02) edge node [above] {$a$} (q012); -\path[->] (q02) edge [bend left] node [above, pos=0.8] {$b$} (q2); -\path[->] (q01) edge node [below] {$a$} (q012); -\path[->] (q01) edge [bend left] node [above] {$b$} (q2); -\path[->] (q0) edge node [below] {$a$} (q012); -\path[->] (q0) edge node [right, pos=0.2] {$b$} (q2); -\path[->] (q1) edge [loop above] node {$a$} (); -\path[->] (q1) edge node [above] {$b$} (qn); -\path[->] (q2) edge [loop right] node {$b$} (); -\path[->] (q2) edge node [below] {$a$} (qn); -\path[->] (qn) edge [loop above] node {$a,b$} (); +\begin{center} +\begin{tikzpicture}[>=stealth',very thick,auto, + every state/.style={minimum size=0pt, + inner sep=2pt,draw=blue!50,very thick, + fill=blue!20}] +\node[state,initial] (Q_0) {$Q_0$}; +\node[state] (Q_1) [right=of Q_0] {$Q_1$}; +\node[state] (Q_2) [below right=of Q_0] {$Q_2$}; +\node[state] (Q_3) [right=of Q_2] {$Q_3$}; +\node[state, accepting] (Q_4) [right=of Q_1] {$Q_4$}; +\path[->] (Q_0) edge node [above] {$a$} (Q_1); +\path[->] (Q_1) edge node [above] {$a$} (Q_4); +\path[->] (Q_4) edge [loop right] node {$a, b$} (); +\path[->] (Q_3) edge node [right] {$a$} (Q_4); +\path[->] (Q_2) edge node [above] {$a$} (Q_3); +\path[->] (Q_1) edge node [right] {$b$} (Q_2); +\path[->] (Q_0) edge node [above] {$b$} (Q_2); +\path[->] (Q_2) edge [loop left] node {$b$} (); +\path[->] (Q_3) edge [bend left=95, looseness=1.3] node + [below] {$b$} (Q_0); +\end{tikzpicture} +\end{center} + +\noindent In Step 1 and 2 we consider essentially a triangle +of the form + +\begin{center} +\begin{tikzpicture}[scale=0.6,line width=0.8mm] +\draw (0,0) -- (4,0); +\draw (0,1) -- (4,1); +\draw (0,2) -- (3,2); +\draw (0,3) -- (2,3); +\draw (0,4) -- (1,4); + +\draw (0,0) -- (0, 4); +\draw (1,0) -- (1, 4); +\draw (2,0) -- (2, 3); +\draw (3,0) -- (3, 2); +\draw (4,0) -- (4, 1); + +\draw (0.5,-0.5) node {$Q_0$}; +\draw (1.5,-0.5) node {$Q_1$}; +\draw (2.5,-0.5) node {$Q_2$}; +\draw (3.5,-0.5) node {$Q_3$}; + +\draw (-0.5, 3.5) node {$Q_1$}; +\draw (-0.5, 2.5) node {$Q_2$}; +\draw (-0.5, 1.5) node {$Q_3$}; +\draw (-0.5, 0.5) node {$Q_4$}; + +\draw (0.5,0.5) node {\large$\star$}; +\draw (1.5,0.5) node {\large$\star$}; +\draw (2.5,0.5) node {\large$\star$}; +\draw (3.5,0.5) node {\large$\star$}; +\end{tikzpicture} +\end{center} + +\noindent where the lower row is filled with stars, because in +the corresponding pairs there is always one state that is +accepting ($Q_4$) and a state that is non-accepting (the other +states). + +Now in Step 3 we need to fill in more stars according whether +one of the next-state pairs are marked. We have to do this +for every unmarked field until there is no change anymore. +This gives the triangle + +\begin{center} +\begin{tikzpicture}[scale=0.6,line width=0.8mm] +\draw (0,0) -- (4,0); +\draw (0,1) -- (4,1); +\draw (0,2) -- (3,2); +\draw (0,3) -- (2,3); +\draw (0,4) -- (1,4); + +\draw (0,0) -- (0, 4); +\draw (1,0) -- (1, 4); +\draw (2,0) -- (2, 3); +\draw (3,0) -- (3, 2); +\draw (4,0) -- (4, 1); + +\draw (0.5,-0.5) node {$Q_0$}; +\draw (1.5,-0.5) node {$Q_1$}; +\draw (2.5,-0.5) node {$Q_2$}; +\draw (3.5,-0.5) node {$Q_3$}; + +\draw (-0.5, 3.5) node {$Q_1$}; +\draw (-0.5, 2.5) node {$Q_2$}; +\draw (-0.5, 1.5) node {$Q_3$}; +\draw (-0.5, 0.5) node {$Q_4$}; + +\draw (0.5,0.5) node {\large$\star$}; +\draw (1.5,0.5) node {\large$\star$}; +\draw (2.5,0.5) node {\large$\star$}; +\draw (3.5,0.5) node {\large$\star$}; +\draw (0.5,1.5) node {\large$\star$}; +\draw (2.5,1.5) node {\large$\star$}; +\draw (0.5,3.5) node {\large$\star$}; +\draw (1.5,2.5) node {\large$\star$}; +\end{tikzpicture} +\end{center} + +\noindent which means states $Q_0$ and $Q_2$, as well as $Q_1$ +and $Q_3$ can be merged. This gives the following minimal DFA + +\begin{center} +\begin{tikzpicture}[>=stealth',very thick,auto, + every state/.style={minimum size=0pt, + inner sep=2pt,draw=blue!50,very thick, + fill=blue!20}] +\node[state,initial] (Q_02) {$Q_{0, 2}$}; +\node[state] (Q_13) [right=of Q_02] {$Q_{1, 3}$}; +\node[state, accepting] (Q_4) [right=of Q_13] + {$Q_{4\phantom{,0}}$}; +\path[->] (Q_02) edge [bend left] node [above] {$a$} (Q_13); +\path[->] (Q_13) edge [bend left] node [below] {$b$} (Q_02); +\path[->] (Q_02) edge [loop below] node {$b$} (); +\path[->] (Q_13) edge node [above] {$a$} (Q_4); +\path[->] (Q_4) edge [loop above] node {$a, b$} (); \end{tikzpicture} \end{center} - -There are two points to note: One is that very often the -resulting DFA contains a number of ``dead'' nodes that are -never reachable from the starting state. For example -there is no way to reach node $\{0,2\}$ from the starting -state $\{0,1,2\}$. I let you find the other dead states. -In effect the DFA in this example is not a minimal DFA. Such -dead nodes can be safely removed without changing the language -that is recognised by the DFA. Another point is that in some -cases, however, the subset construction produces a DFA that -does \emph{not} contain any dead nodes\ldots{}that means it -calculates a minimal DFA. Which in turn means that in some -cases the number of nodes by going from NFAs to DFAs -exponentially increases, namely by $2^n$ (which is the number -of subsets you can form for $n$ nodes). - -Removing all the dead states in the automaton above, -gives a much more legible automaton, namely - -\begin{center} -\begin{tikzpicture}[scale=0.7,>=stealth',very thick, - every state/.style={minimum size=0pt, - draw=blue!50,very thick,fill=blue!20}, - baseline=0mm] -\node[state,initial,accepting] (q012) {$0,1,2$}; -\node[state,accepting] (q2) [right=of q012] {$2$}; -\node[state] (qn) [right=of q2] {$\{\}$}; - -\path[->] (q012) edge [loop below] node {$a$} (); -\path[->] (q012) edge node [above] {$b$} (q2); -\path[->] (q2) edge [loop below] node {$b$} (); -\path[->] (q2) edge node [below] {$a$} (qn); -\path[->] (qn) edge [loop above] node {$a,b$} (); -\end{tikzpicture} -\end{center} - -\noindent Now the big question is whether this DFA -can recognise the same language as the NFA we started with. -I let you ponder about this question. - -\subsubsection*{Brzozowski's Method} +\subsection*{Brzozowski's Method} As said before, we can also go into the other direction---from DFAs to regular expressions. Brzozowski's method calculates @@ -1190,154 +1345,8 @@ an equivalent regular expression for the automaton. But for the purposes of this module, we omit this. -\subsubsection*{Automata Minimization} -As seen in the subset construction, the translation -of a NFA to a DFA can result in a rather ``inefficient'' -DFA. Meaning there are states that are not needed. A -DFA can be \emph{minimised} by the following algorithm: - -\begin{enumerate} -\item Take all pairs $(q, p)$ with $q \not= p$ -\item Mark all pairs that accepting and non-accepting states -\item For all unmarked pairs $(q, p)$ and all characters $c$ - test whether - - \begin{center} - $(\delta(q, c), \delta(p,c))$ - \end{center} - - are marked. If there is one, then also mark $(q, p)$. -\item Repeat last step until no change. -\item All unmarked pairs can be merged. -\end{enumerate} - -\noindent To illustrate this algorithm, consider the following -DFA. - -\begin{center} -\begin{tikzpicture}[>=stealth',very thick,auto, - every state/.style={minimum size=0pt, - inner sep=2pt,draw=blue!50,very thick, - fill=blue!20}] -\node[state,initial] (Q_0) {$Q_0$}; -\node[state] (Q_1) [right=of Q_0] {$Q_1$}; -\node[state] (Q_2) [below right=of Q_0] {$Q_2$}; -\node[state] (Q_3) [right=of Q_2] {$Q_3$}; -\node[state, accepting] (Q_4) [right=of Q_1] {$Q_4$}; -\path[->] (Q_0) edge node [above] {$a$} (Q_1); -\path[->] (Q_1) edge node [above] {$a$} (Q_4); -\path[->] (Q_4) edge [loop right] node {$a, b$} (); -\path[->] (Q_3) edge node [right] {$a$} (Q_4); -\path[->] (Q_2) edge node [above] {$a$} (Q_3); -\path[->] (Q_1) edge node [right] {$b$} (Q_2); -\path[->] (Q_0) edge node [above] {$b$} (Q_2); -\path[->] (Q_2) edge [loop left] node {$b$} (); -\path[->] (Q_3) edge [bend left=95, looseness=1.3] node - [below] {$b$} (Q_0); -\end{tikzpicture} -\end{center} - -\noindent In Step 1 and 2 we consider essentially a triangle -of the form - -\begin{center} -\begin{tikzpicture}[scale=0.6,line width=0.8mm] -\draw (0,0) -- (4,0); -\draw (0,1) -- (4,1); -\draw (0,2) -- (3,2); -\draw (0,3) -- (2,3); -\draw (0,4) -- (1,4); - -\draw (0,0) -- (0, 4); -\draw (1,0) -- (1, 4); -\draw (2,0) -- (2, 3); -\draw (3,0) -- (3, 2); -\draw (4,0) -- (4, 1); - -\draw (0.5,-0.5) node {$Q_0$}; -\draw (1.5,-0.5) node {$Q_1$}; -\draw (2.5,-0.5) node {$Q_2$}; -\draw (3.5,-0.5) node {$Q_3$}; - -\draw (-0.5, 3.5) node {$Q_1$}; -\draw (-0.5, 2.5) node {$Q_2$}; -\draw (-0.5, 1.5) node {$Q_3$}; -\draw (-0.5, 0.5) node {$Q_4$}; - -\draw (0.5,0.5) node {\large$\star$}; -\draw (1.5,0.5) node {\large$\star$}; -\draw (2.5,0.5) node {\large$\star$}; -\draw (3.5,0.5) node {\large$\star$}; -\end{tikzpicture} -\end{center} - -\noindent where the lower row is filled with stars, because in -the corresponding pairs there is always one state that is -accepting ($Q_4$) and a state that is non-accepting (the other -states). - -Now in Step 3 we need to fill in more stars according whether -one of the next-state pairs are marked. We have to do this -for every unmarked field until there is no change anymore. -This gives the triangle - -\begin{center} -\begin{tikzpicture}[scale=0.6,line width=0.8mm] -\draw (0,0) -- (4,0); -\draw (0,1) -- (4,1); -\draw (0,2) -- (3,2); -\draw (0,3) -- (2,3); -\draw (0,4) -- (1,4); - -\draw (0,0) -- (0, 4); -\draw (1,0) -- (1, 4); -\draw (2,0) -- (2, 3); -\draw (3,0) -- (3, 2); -\draw (4,0) -- (4, 1); - -\draw (0.5,-0.5) node {$Q_0$}; -\draw (1.5,-0.5) node {$Q_1$}; -\draw (2.5,-0.5) node {$Q_2$}; -\draw (3.5,-0.5) node {$Q_3$}; - -\draw (-0.5, 3.5) node {$Q_1$}; -\draw (-0.5, 2.5) node {$Q_2$}; -\draw (-0.5, 1.5) node {$Q_3$}; -\draw (-0.5, 0.5) node {$Q_4$}; - -\draw (0.5,0.5) node {\large$\star$}; -\draw (1.5,0.5) node {\large$\star$}; -\draw (2.5,0.5) node {\large$\star$}; -\draw (3.5,0.5) node {\large$\star$}; -\draw (0.5,1.5) node {\large$\star$}; -\draw (2.5,1.5) node {\large$\star$}; -\draw (0.5,3.5) node {\large$\star$}; -\draw (1.5,2.5) node {\large$\star$}; -\end{tikzpicture} -\end{center} - -\noindent which means states $Q_0$ and $Q_2$, as well as $Q_1$ -and $Q_3$ can be merged. This gives the following minimal DFA - -\begin{center} -\begin{tikzpicture}[>=stealth',very thick,auto, - every state/.style={minimum size=0pt, - inner sep=2pt,draw=blue!50,very thick, - fill=blue!20}] -\node[state,initial] (Q_02) {$Q_{0, 2}$}; -\node[state] (Q_13) [right=of Q_02] {$Q_{1, 3}$}; -\node[state, accepting] (Q_4) [right=of Q_13] - {$Q_{4\phantom{,0}}$}; -\path[->] (Q_02) edge [bend left] node [above] {$a$} (Q_13); -\path[->] (Q_13) edge [bend left] node [below] {$b$} (Q_02); -\path[->] (Q_02) edge [loop below] node {$b$} (); -\path[->] (Q_13) edge node [above] {$a$} (Q_4); -\path[->] (Q_4) edge [loop above] node {$a, b$} (); -\end{tikzpicture} -\end{center} - -\subsubsection*{Regular Languages} +\subsection*{Regular Languages} Given the constructions in the previous sections we obtain the following overall picture: @@ -1381,7 +1390,7 @@ derivatives or NFAs or DFAs. But let us quickly look at what the differences mean in computational terms. Translating a regular expression into a NFA gives us an automaton that has -$O(n)$ nodes---that means the size of the NFA grows linearly +$O(n)$ states---that means the size of the NFA grows linearly with the size of the regular expression. The problem with NFAs is that the problem of deciding whether a string is accepted or not is computationally not cheap. Remember with NFAs we @@ -1441,11 +1450,11 @@ automaton for this language, but again that would lead us too far afield for what we want to do in this module. -\section*{Further Reading} +%\section*{Further Reading} -Compare what a ``human expert'' would create as an automaton for the -regular expression $a (b + c)^*$ and what the Thomson -algorithm generates. +%Compare what a ``human expert'' would create as an automaton for the +%regular expression $a\cdot (b + c)^*$ and what the Thomson +%algorithm generates. %http://www.inf.ed.ac.uk/teaching/courses/ct/ \end{document} diff -r e28d7a327870 -r 4fee50f38305 handouts/ho04.pdf Binary file handouts/ho04.pdf has changed diff -r e28d7a327870 -r 4fee50f38305 handouts/ho04.tex --- a/handouts/ho04.tex Sun May 07 03:01:29 2017 +0100 +++ b/handouts/ho04.tex Tue May 09 12:31:55 2017 +0100 @@ -91,11 +91,13 @@ letters, while values just start with a single upper-case character and the rest are lower-case letters. -{\small\lstinputlisting[language=Scala,numbers=none] +{\small\lstinputlisting[language=Scala,numbers=none,linebackgroundcolor= + {\ifodd\value{lstnumber}\color{capri!3}\fi}] {../progs/app01.scala}} -{\small\lstinputlisting[language=Scala,numbers=none] +{\small\lstinputlisting[language=Scala,numbers=none,linebackgroundcolor= + {\ifodd\value{lstnumber}\color{capri!3}\fi}] {../progs/app02.scala}} @@ -580,7 +582,8 @@ \end{figure} \begin{figure}[p] -\lstinputlisting{../progs/app61.scala} +\lstinputlisting[numbers=left,linebackgroundcolor= + {\ifodd\value{lstnumber}\color{capri!3}\fi}]{../progs/app61.scala} \caption{The Scala code for the simplification function. The first part defines some auxillary functions for the rectification. @@ -639,7 +642,8 @@ be regarded as a regular expression. The extended definition in Scala therefore looks as follows: -{\small\lstinputlisting[language=Scala] +{\small\lstinputlisting[language=Scala, numbers=none,linebackgroundcolor= + {\ifodd\value{lstnumber}\color{capri!3}\fi}] {../progs/app03.scala}} \noindent Since we regard records as regular expressions we @@ -648,7 +652,8 @@ to extend the definition of values, which in Scala looks as follows: -{\small\lstinputlisting[language=Scala] +{\small\lstinputlisting[language=Scala, numbers=none,linebackgroundcolor= + {\ifodd\value{lstnumber}\color{capri!3}\fi}] {../progs/app04.scala}} \noindent Let us now look at the purpose of records more diff -r e28d7a327870 -r 4fee50f38305 langs.sty --- a/langs.sty Sun May 07 03:01:29 2017 +0100 +++ b/langs.sty Tue May 09 12:31:55 2017 +0100 @@ -39,27 +39,6 @@ otherkeywords={=,!=,:=,<,>,\%;*,/}, }[keywords,comments,strings] -\lstdefinestyle{mystyle} - {basicstyle=\ttfamily, - keywordstyle=\color{codepurple}\bfseries, - stringstyle=\color{codegreen}, - commentstyle=\color{codegreen}, - morecomment=[s][\color{codedocblue}]{/**}{*/}, - numbers=left, - numberstyle=\tiny\color{black}, - stepnumber=1, - numbersep=10pt, - tabsize=2, - showspaces=false, - showstringspaces=false, - xleftmargin=8mm, - emphstyle=\color{codeblue}\bfseries, - keepspaces -} - -\lstset{language=Scala, - style=mystyle} - \newcommand{\code}[1]{{\lstinline{#1}}} \newcommand{\pcode}[1]{\mbox{\lstset{language={},keywordstyle=\color{black}}\lstinline!#1!}} @@ -69,8 +48,29 @@ %%\lstset{escapeinside={(*@}{@*)}} \lstset{escapeinside={/*@}{@*/}} - - %% stripy code \usepackage{lstlinebgrd} \definecolor{capri}{rgb}{0.0, 0.75, 1.0} + + +\lstdefinestyle{mystyle} + {basicstyle=\ttfamily, + keywordstyle=\color{codepurple}\bfseries, + stringstyle=\color{codegreen}, + commentstyle=\color{codegreen}, + morecomment=[s][\color{codedocblue}]{/**}{*/}, + numbers=none, + numberstyle=\tiny\color{black}, + stepnumber=1, + numbersep=10pt, + tabsize=2, + showspaces=false, + showstringspaces=false, + xleftmargin=8mm, + emphstyle=\color{codeblue}\bfseries, + keepspaces, + linebackgroundcolor={\ifodd\value{lstnumber}\color{capri!3}\fi} +} + +\lstset{language=Scala, + style=mystyle} diff -r e28d7a327870 -r 4fee50f38305 progs/nfa.scala --- a/progs/nfa.scala Sun May 07 03:01:29 2017 +0100 +++ b/progs/nfa.scala Tue May 09 12:31:55 2017 +0100 @@ -85,4 +85,15 @@ // subset constructions -//def subset(nfa: NFA[A, C]) : DFA[Set[A], C] = +def subset[A, C](nfa: NFA[A, C]) : DFA[Set[A], C] = { + DFA(nfa.starts, + { case (qs, c) => nfa.nexts(qs, c) }, + _.exists(nfa.fins)) +} + +subset(nfa1).accepts("aa".toList) // false +subset(nfa1).accepts("aaaaa".toList) // false +subset(nfa1).accepts("aaaaab".toList) // true +subset(nfa1).accepts("aaaaabbb".toList) // true +subset(nfa1).accepts("aaaaabbbaaa".toList) // false +subset(nfa1).accepts("ac".toList) // false diff -r e28d7a327870 -r 4fee50f38305 progs/re0.scala --- a/progs/re0.scala Sun May 07 03:01:29 2017 +0100 +++ b/progs/re0.scala Tue May 09 12:31:55 2017 +0100 @@ -1,4 +1,5 @@ import scala.annotation.tailrec +import scala.language.implicitConversions abstract class Rexp @@ -72,6 +73,7 @@ if (i == 0) NULL else SEQ(der(c, r), REP(r, i - 1)) } + // derivative w.r.t. a string (iterates der) @tailrec def ders (s: List[Char], r: Rexp) : Rexp = s match {