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\section*{Replacement Coursework 2 (Automata)}

This coursework is worth 10\%. It is about deterministic and
non-deterministic finite automata.  The coursework is due on ??? March
at 5pm.  Make sure the files you submit can be processed by just
calling \texttt{scala <<filename.scala>>}.\bigskip

\noindent
\textbf{Important:} Do not use any mutable data structures in your
submission! They are not needed. This means you cannot use 
\texttt{ListBuffer}s, for example. Do not use \texttt{return} in your
code! It has a different meaning in Scala, than in Java.  Do not use
\texttt{var}! This declares a mutable variable.  Make sure the
functions you submit are defined on the ``top-level'' of Scala, not
inside a class or object. Also note that the running time will be
restricted to a maximum of 360 seconds on my laptop.


\subsection*{Disclaimer}

It should be understood that the work you submit represents your own
effort! You have not copied from anyone else. An exception is the
Scala code I showed during the lectures or uploaded to KEATS, which
you can freely use.\bigskip


\subsection*{Part 1 (Deterministic Finite Automata)}

\noindent
There are many uses for Deterministic Finite Automata (DFAs), for
example testing whether a string should be accepted or not. The main
idea is that DFAs consist of some states (circles) and transitions
(edges) between states. For example consider the DFA

\begin{center}
\begin{tikzpicture}[scale=1.5,>=stealth',very thick,auto,
                    every state/.style={minimum size=4pt,
                    inner sep=4pt,draw=blue!50,very thick,
                    fill=blue!20}]
  \node[state, initial]        (q0) at ( 0,1) {$Q_0$};
  \node[state]                    (q1) at ( 1,1) {$Q_1$};
  \node[state, accepting] (q2) at ( 2,1) {$Q_2$};
  \path[->] (q0) edge[bend left] node[above] {$a$} (q1)
            (q1) edge[bend left] node[above] {$b$} (q0)
            (q2) edge[bend left=50] node[below] {$b$} (q0)
            (q1) edge node[above] {$a$} (q2)
            (q2) edge [loop right] node {$a$} ()
            (q0) edge [loop below] node {$b$} ();
\end{tikzpicture}
\end{center}

\noindent
where there are three states ($Q_0$, $Q_1$ and $Q_2$). The DFA has a
starting state ($Q_0$) and an accepting state ($Q_2$), the latter
indicated by double lines. In general, a DFA can have any number of
accepting states, but only a single starting state (in this example
only $a$ and $b$).

Transitions are edges between states labelled with a character. The
idea is that if I am in state $Q_0$, say, and get an $a$, I can go to
state $Q_1$. If I am in state $Q_2$ and get an $a$, I can stay in
state $Q_2$; if I get a $b$ in $Q_2$, then I have to go to state
$Q_0$. The main point of DFAs is that if I am in a state and get a
character, it is always clear which is the next state---there can only
be at most one. The task of Part 1 is to implement such DFAs in Scala
using partial functions for the transitions.

\subsubsection*{Tasks}

\begin{itemize}
\item[(1)] Write a polymorphic function, called \texttt{share}, that
  decides whether two sets share some elements (i.e.~the intersection
  is not empty).\hfill[1 Mark]
 
\item[(2)] The transitions of DFAs are given by partial functions,
  with the type of (state, character)-pair to state. For example
  the transitions of the DFA given above can be defined as

\begin{lstlisting}[language=Scala,numbers=none]
val dfa_trans : PartialFunction[(State,Char), State] = 
  { case (Q0, 'a') => Q1 
    case (Q0, 'b') => Q0
    case (Q1, 'a') => Q2 
    case (Q1, 'b') => Q0
    case (Q2, 'a') => Q2 
    case (Q2, 'b') => Q0 
  }
\end{lstlisting}

  The main idea of partial functions (as opposed to functions) is that they
  do not have to be defined everywhere. For example the transitions
  above only mention characters $a$ and $b$, but leave out any other
  characters. Partial functions come with a method \texttt{isDefinedAt}
  that can be used to check whether an input produces a result
  or not. For example

  \begin{lstlisting}[language=Scala,numbers=none]
    dfa_trans.isDefinedAt((Q0, 'a'))
    dfa_trans.isDefinedAt((Q0, 'c'))
  \end{lstlisting}   

  \noindent
  gives \texttt{true} in the first case and \texttt{false} in the second.

  Write a function that takes transition and a (state, character)-pair as arguments
  and produces an optional state (the state specified by the partial transition
  function whenever it is defined; if the transition function is undefined,
  return None).\hfill\mbox{[1 Mark]}

\item[(3)] 
  Write a function that ``lifts'' the function in (2) from characters to strings. That
  is, write a function that takes a transition, a state and a list of characters
  as arguments and produces the state generated by following the transitions for
  each character in the list. For example you are in state $Q_0$ in the DFA above
  and have the list \texttt{List(a,a,a,b,b,a)}, then you need to generate the
  state $Q_1$ (as option since there might not be such a state).\\
  \mbox{}\hfill\mbox{[1 Mark]}
  
\item[(4)] DFAs are defined as a triple: (staring state, transitions, final states).
  Write a function \texttt{accepts} that tests whether a string is accepted
  by an DFA or not. For this start in the starting state of the DFA,
  use the function under (3) to calculate the state after following all transitions
  according to the characters in the string. If the state is a final state, return
  true; otherwise false.\\\mbox{}\hfill\mbox{[1 Mark]}
\end{itemize}


\subsection*{Part 2 (Non-Deterministic Finite Automata)}

The main point of DFAs is that for every given state and character
there is at most one next state (one if the transition is defined;
none otherwise). However, this restriction to at most one state can be
quite limiting for some applications.\footnote{Though there is a
  curious fact that every NFA can be translated into an ``equivalent''
  DFA, that is accepting the same set of strings. However this might
  increase drastically the number of states in the DFA.}  
Non-Deterministic Automata (NFAs) remove this restriction: there can
be more than one starting state, and given a state and a character
there can be more than one next state. Consider for example

\begin{center}
\begin{tikzpicture}[scale=0.7,>=stealth',very thick,
    every state/.style={minimum size=0pt,
      draw=blue!50,very thick,fill=blue!20},]
\node[state,initial]  (R_1)  {$R_1$};
\node[state,initial] (R_2) [above=of R_1] {$R_2$};
\node[state, accepting] (R_3) [right=of R_1] {$R_3$};
\path[->] (R_1) edge node [below]  {$b$} (R_3);
\path[->] (R_2) edge [bend left] node [above]  {$a$} (R_3);
\path[->] (R_1) edge [bend left] node  [left] {$c$} (R_2);
\path[->] (R_2) edge [bend left] node  [right] {$a$} (R_1);
\end{tikzpicture}
\end{center}

\noindent
where in state $R_2$ if you get an $a$, you can go to state $R_1$
\emph{or} $R_3$. If we want to find out whether a NFA accepts a
string, then we need to explore both possibilities. We will do this
``exploration'' in the tasks below in a breath-first manner.
The possibility of having more than one next state in NFAs will
be implemented by having a \emph{set} of partial transition
functions. For example the NFA shown above will be represented by the
set of partial functions

\begin{lstlisting}[language=Scala,numbers=none]
val nfa_trans : NTrans = Set(
  { case (R1, 'c') => R2 },
  { case (R1, 'b') => R3 },
  { case (R2, 'a') => R1 },
  { case (R2, 'a') => R3 }
)
\end{lstlisting}

\noindent
The point is that the 3rd element in this set states that
in $R_2$ and given an $a$, I can go to state $R_1$; and the
4th element, in $R_2$, given an $a$, I can go to state $R_3$.
When following transitions from a state, we have to look at all
partial functions in the set and generate the set of all possible
next states.

\subsubsection*{Tasks}

\begin{itemize}
\item[(5)]
  Write a function \texttt{nnext} which takes a transition set, a state
  and a character as arguments, and calculates all possible next states
  (returned as set).\\
  \mbox{}\hfill\mbox{[1 Mark]}

\item[(6)] Write a function \texttt{nnexts} which takes a transition
  set, a \emph{set} of states and a character as arguments, and
  calculates \emph{all} possible next states that can be reached from
  any state in the set.\\ \mbox{}\hfill\mbox{[1 Mark]}

\item[(7)] Like in (3), write a function \texttt{nnextss} that lifts
  \texttt{nnexts} from (6) from single characters to lists of characters.
  \mbox{}\hfill\mbox{[1 Mark]}
  
\item[(8)] NFAs are also defined as a triple: (set of staring states,
  set of transitions, final states).  Write a function
  \texttt{naccepts} that tests whether a string is accepted by a NFA
  or not. For this start in all starting states of the NFA, use the
  function under (7) to calculate the set of states following all
  transitions according to the characters in the string. If the set of
  states shares and state with the set of final states, return true;
  otherwise false.  Use the function under (1) in order to test
  whether these two sets of states share any
  states\mbox{}\hfill\mbox{[1 Mark]}

\item[(9)] Since we explore in functions under (6) and (7) all
  possible next states, we decide whether a string is accepted in a
  breath-first manner. (Depth-first would be to choose one state,
  follow all next states of this single state; check whether it leads
  to a accepting state. If not, we backtrack and choose another
  state). The disadvantage of breath-first search is that at every
  step a non-empty set of states are ``active''\ldots that need to be
  followed at the same time.  Write similar functions as in (7) and
  (8), but instead of returning states or a boolean, these functions
  return the number of states that need to be followed in each
  step. The function \texttt{max\_accept} should return the maximum
  of all these numbers.

  Consider again the NFA shown above. At the beginning the number of
  active states will be 2 (since there are two starting states, namely
  $R_1$ and $R_2$). If we get an $a$, there will be still 2 active
  states, namely $R_1$ and $R_3$ both reachable from $R_2$. There is
  no transition for $a$ and $R_1$. So for a string, say, $ab$ which is
  accepted by the NFA, the maximum number of active states is 2 (it is
  not possible that all states are active with this NFA; is it possible
  that no state is active?).
  \hfill\mbox{[2 Marks]}

  
\end{itemize}
  

\end{document}


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