Skip to content
Snippets Groups Projects
Commit 3cbe13d6 authored by Sankalp Gambhir's avatar Sankalp Gambhir
Browse files

Merge branch 'ex01' into 'main' 

Add exercise set 1

See merge request lara/cs320!5
parents 7e0c4dd1 179de5da
No related branches found
No related tags found
No related merge requests found
Hide whitespace changes
Inline Side-by-side
Showing
with 786 additions and 0 deletions
info/exercises/ex-01-sol.pdf 0 → 100644
+ 0
− 0
View file @ 3cbe13d6
File added
info/exercises/ex-01.pdf 0 → 100644
+ 0
− 0
View file @ 3cbe13d6
File added
info/exercises/past-exercises/README.md 0 → 100644
+ 5
− 0
View file @ 3cbe13d6
# Past Exercises
This is the collection of the exercises used in previous iterations of the
course. They are here if you wish to try additional exercises, but note that not
everything may necessarily match the current content of the course.
info/exercises/src/.gitignore 0 → 100644
+ 7
− 0
View file @ 3cbe13d6
*.aux
*.fdb_latexmk
*.fls
*.log
*.out
*.synctex.gz
*.pdf
info/exercises/src/ex-01/ex/dfa.tex 0 → 100644
+ 774
− 0
View file @ 3cbe13d6
\begin{exercise}{}
For each the following languages, construct an NFA \(\mathcal{A}\) that
recognizes them, i.e. \(L(\mathcal{A})\) = \(L_i\):
\begin{enumerate}
\item \(L_1\): binary strings divisible by 3
\item \(L_2\): binary strings divisible by 4
\item \(L_3\): \(\{(w_1 \oplus w_2) \mid w_1 \in L_1 \land w_2 \in L_2 \land |w_1| = |w_2|\}\)
\end{enumerate}
where \(\oplus\) is the bitwise-xor operation on binary strings.
\begin{solution}
\begin{enumerate}
\item The language of binary strings divisible by 3. We need two
observations to construct this automaton:
\begin{enumerate}
\item If the automaton has consumed a binary string \(s\) with decimal
value, say, \(val(s) = n\), then we can determine the decimal value of
the string after reading one more character as either \(val(s0) = 2n\)
or \(val(s1) = 2n + 1\).
\item The set of strings is finite, but it is sufficient to know only
the value of the string \emph{modulo 3} to determine if it is divisible.
\end{enumerate}
We construct the automaton \(\mathcal{A}_1 = (Q, \Sigma, \delta, q_{init}, F)\)
where:
\begin{itemize}
\item \(Q = \{q_{init}, q_0, q_1, q_2\}\), representing the initial
state (empty word has no value), and the states corresponding to the
values \(0, 1, 2\) modulo 3.
\item \(\Sigma = \{0, 1\}\), as required.
\item \(\delta = \{(q_i, 0, q_j) \mid 2i \mod 3 = j\} \cup \{(q_i, 1,
q_j) \mid (2i + 1) mod 3 = j\} \cup \{(q_{init}, 0, q_0), (q_{init}, 1,
q_1)\}\), i.e., there is a transition from \(q_i\) to \(q_j\) if, as the
currently known value modulo 3 is \(i\), on reading \(0\) the next value
is \(j = 2i \mod 3\). We use the fact that \(2n \mod 3 = 2 (n \mod 3)
\mod 3\). The case for reading \(1\) is similar. The translations from
the initial state are to the states corresponding to the values \(0, 1\)
modulo 3.
For example, if we have read ``1101'', with decimal value \(13\), we
must be in state \(q_1\), as \(13 \mod 3 = 1\). On reading a \(0\), we
have the string ``11010'' with decimal value \(26\), and \(26 \mod 3 =
2\), so we transition to \(q_2\).
The full automaton is below.
\item \(F = \{q_0\}\) as we accept that words that are divisible by 3,
and are hence equal to 0 modulo 3.
\end{itemize}
The automaton is:
\begin{center}
\begin{tikzpicture}[node distance = 2cm, on grid]
\node[state, initial] (qi) {\(q_{init}\)};
\node[state, accepting, right of = qi] (q0) {\(q_0\)};
\node[state, above right of = q0] (q1) {\(q_1\)};
\node[state, below right of = q1] (q2) {\(q_2\)};
\draw[->]
(qi) edge node[above] {\(0\)} (q0)
(qi) edge node[above] {\(1\)} (q1)
%
(q0) edge[loop below] node[below] {\(0\)} (q0)
(q0) edge node[above] {\(1\)} (q1)
%
(q1) edge[bend right] node[below] {\(0\)} (q2)
(q1) edge[bend left] node[below] {\(1\)} (q0)
%
(q2) edge node[above] {\(0\)} (q1)
(q2) edge[loop below] node[below] {\(1\)} (q2)
;
\end{tikzpicture}
\end{center}
\item The language of binary strings divisible by 4. The construction is
similar to the one above, now with 5 states.
\begin{center}
\begin{tikzpicture}[node distance = 2cm, on grid]
\node[state, initial] (qi) {\(q_{init}\)};
\node[state, accepting, right of = qi] (q0) {\(q_0\)};
\node[state, above of = q0] (q1) {\(q_1\)};
\node[state, right of = q1] (q2) {\(q_2\)};
\node[state, right of = q0] (q3) {\(q_3\)};
\draw[->]
(qi) edge node[above] {\(0\)} (q0)
(qi) edge node[above] {\(1\)} (q1)
%
(q0) edge[loop below] node[below] {\(0\)} (q0)
(q0) edge node[left] {\(1\)} (q1)
%
(q1) edge node[above] {\(0\)} (q2)
(q1) edge[bend right] node[above] {\(1\)} (q3)
%
(q2) edge node[above] {\(0\)} (q0)
(q2) edge[bend left] node[above] {\(1\)} (q1)
%
(q3) edge node[right] {\(0\)} (q2)
(q3) edge[loop below] node[below] {\(1\)} (q3)
;
\end{tikzpicture}
\end{center}
\item To compute the bitwise-xor of two strings, we must compute a product
automaton. To accept a word \(w\), there must exist \(w_1, w_2\) such that
\(w_1 \in L_1\), and \(w_2 \in L_2\).
We do not explicitly construct the automaton, but present an argument.
First, consider the truth table for xor:
\begin{center}
\begin{tabular}{c c | c}
\(b_1\) & \(b_2\) & \(b_1 \oplus b_2\) \\
\hline
0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{tabular}
\end{center}
Notably, given a xor result, we cannot exactly determine the input bits.
In essence, we construct an automaton that, given a string, tries to
simulate the two input automata in parallel non-deterministically on all
possible pairs of input strings. If any of them are accepted, that means
we found a pair of strings that, one, are accepted by the two original
automata, and two, have the input string as their bitwise-xor.
Formally, the automaton \(\mathcal{A}_3 = (Q, \Sigma, \delta, q_{init},
F)\) has:
\begin{itemize}
\item \(Q = Q_1 \times Q_2\), where \(Q_1\) and \(Q_2\) are the state
sets of \(\mathcal{A}_1\) and \(\mathcal{A}_2\).
\item \(\Sigma = \{0, 1\}\) as before.
\item \(q_{init} = (q_{1, init}, q_{2, init})\) where \(q_{1, init}\)
and \(q_{2, init}\) are the initial states of \(\mathcal{A}_1\) and
\(\mathcal{A}_2\).
\item \(F = F_1 \times F_2\) similarly.
\item \(\delta\) is constructed as follows: for a pair of states \((q_1,
q_2)\), on reading a \(0\), we look at the truth table of xor; two input
pairs \((0, 0)\) and \((1, 1)\) could have produced this result bit.
Hence, we add transitions for both automata simultaneously, \((((q_1,
q_2), 0, (q_1', q_2')))\) corresponding to possible inputs \((0, 0)\) if
\(\delta_1(q_1, 0, q_1')\) and \(\delta_2(q_2, 0, q_2')\), and similarly
\(((q_1, q_2), 0, (q_1', q_2'))\) corresponding to possible inputs \((1,
1)\) if \(\delta_1(q_1, 1, q_1'')\) and \(\delta_2(q_2, 1, q_2'')\).
The case for reading a \(1\) is similar, with possible input pairs \((0,
1)\) and \((1, 0)\).
\end{itemize}
\end{enumerate}
\end{solution}
\end{exercise}
% \begin{exercise}{}
% \todo{Suggestion: From Kozen H2.3, recognizing with errors; regular language modulo n Hamming distance is regular}
% \end{exercise}
\begin{exercise}{}
Give a verbal and a set-notational description of the language accepted by
each of the following automata. You can assume that the alphabet is \(\Sigma =
\{a, b\}\).
\begin{enumerate}
\item \(\mathcal{A}_1\)
\begin{center}
\begin{tikzpicture}[node distance = 2cm, on grid]
\node[state, initial] (q0) {\(q_0\)};
\node[state, accepting, right of = q0] (q1) {\(q_1\)};
\node[state, right of = q1] (q2) {\(q_2\)};
\draw[->] (q0) edge node[above] {\(a\)} (q1)
(q1) edge node[above] {\(a\)} (q2)
(q2) edge[loop above] node[above] {\(a, b\)} (q2)
(q0) edge[loop above] node[above] {\(b\)} (q0)
(q1) edge[loop above] node[above] {\(b\)} (q1)
;
\end{tikzpicture}
\end{center}
\item \(\mathcal{A}_2\)
\begin{center}
\begin{tikzpicture}[node distance = 2cm, on grid]
\node[state, accepting, initial] (q0) {\(q_0\)};
\node[state, accepting, right of = q0] (q1) {\(q_1\)};
\node[state, right of = q1] (q2) {\(q_2\)};
\draw[->] (q0) edge node[below] {\(a\)} (q1)
(q1) edge[bend right] node[above] {\(b\)} (q0)
(q2) edge[loop above] node[above] {\(a, b\)} (q2)
(q0) edge[loop above] node[above] {\(b\)} (q0)
(q1) edge node[above] {\(a\)} (q2)
;
\end{tikzpicture}
\end{center}
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item
As regular expression: \(b^*ab^*\), this is the language of words that
contain exactly one \(a\). In set-notation:
\begin{equation*}
\{w \mid \exists! i.\; 0 \leq i \leq |w| \land w_{(i)} = a\}
\end{equation*}
\item
As generalized regular expression (with complement): \((\Sigma^* aa
\Sigma^*)^c\). Without complement: \((b^*(ab^*)^*)^*\). This is the language
of words that contain no consecutive pair of \(a\)'s. In set-notation:
\begin{equation*}
\{w \mid \forall i.\; 0 \leq i < |w| \land w_{(i)} = a \implies (i + 1 \geq |w| \lor w_{(i + 1)} \neq a)\}
\end{equation*}
\end{enumerate}
\end{solution}
\end{exercise}
% \begin{exercise}{}
% Construct a DFA for the following languages:
% \begin{enumerate}
% \item the set of strings over \(\Sigma = \{0, 1\}\) such that the number
% of 0's is a multiple of 2, and the number of 1's is a multiple of 3.
% \note{\cite[HW 1.1.c]{kozen2007automata}}
% \item the set of strings over \(\Sigma = \{0, 1\}\) that, as binary
% numerals, have decimal value 42. \note{\cite[Ex 2.1.a]{mogensen2010basics}}
% \end{enumerate}
% For the cases above:
% \begin{enumerate}
% \item is the complement of each language regular as well?
% \item is their union regular?
% \item is their intersection regular?
% \item is their difference regular?
% \end{enumerate}
% \begin{solution}
% The DFA constructions:
% \paragraph*{The set of strings over \(\Sigma = \{0, 1\}\) such that the
% number of 0's is a multiple of 2, and the number of 1's is a
% multiple of 3.} The DFA states count the number of 0's and 1's modulo
% 2 and 3 respectively. The transitions correspond to incrementing
% modulo 2 and 3. A state \(q_{n, m}\) represents the set of words
% such that the number of 0's \(= n \mod 2\) and the number of 1's \(=
% m \mod 3\).
% The state \(q_{0, 0}\) is the desired initial \emph{and}
% final state (why?).
% The transitions are:
% \begin{align*}
% q_{0, 0} \xrightarrow{0} q_{1, 0} & & q_{0, 0} \xrightarrow{1} q_{0, 1} \\
% q_{0, 1} \xrightarrow{0} q_{1, 1} & & q_{0, 1} \xrightarrow{1} q_{0, 2} \\
% q_{0, 2} \xrightarrow{0} q_{1, 2} & & q_{0, 2} \xrightarrow{1} q_{0, 0} \\
% q_{1, 0} \xrightarrow{0} q_{0, 0} & & q_{1, 0} \xrightarrow{1} q_{1, 1} \\
% q_{1, 1} \xrightarrow{0} q_{0, 1} & & q_{1, 1} \xrightarrow{1} q_{1, 2} \\
% q_{1, 2} \xrightarrow{0} q_{0, 2} & & q_{1, 2} \xrightarrow{1} q_{1, 0}
% \end{align*}
% In this manner, DFA states represent a \emph{finite partitioning} of the
% set of words \(\Sigma^*\), and the transitions define how these
% partitions correspond to each other.
% \textbf{Extra}: This is made more apparent in the algorithms for DFA
% minimization and equivalence checking and makes DFAs as interesting as
% they are theoretically. See
% \href{https://en.wikipedia.org/wiki/Myhill%E2%80%93Nerode_theorem}{Myhill-Nerode Theorem}
% or
% \href{https://en.wikipedia.org/wiki/Brzozowski_derivative}{Brzozowski derivatives}
% on Wikipedia.
% \paragraph*{The set of binary strings whose value is less than 42}.
% First, let's attempt to describe this set as a regular expression. The
% binary representation of 42 is \(101010_2\). We write the expression in
% a natural way, decomposing case by case looking at the bits:
% \begin{tikzpicture}[node distance = 3cm, on grid]
% \node (leading) {\(0^*\)};
% \node[below left = 1cm of leading] (leadinglabel) {Leading zeroes};
% \node[above right of = leading] (b50) {\(0 \,\Sigma^5\)};
% \node[below = 0.5cm of b50] (b50label) {\(< 32\)};
% \node[below right of = leading] (b51) {\(1\)};
% \node[below = 0.5cm of b51] (b51label) {\(\geq 32\)};
% \node[above right of = b51] (b40) {\(1 \,\Sigma^4\)};
% \node[below = 0.5cm of b40] (b40label) {\(\geq 48\)};
% \node[below right of = b51] (b41) {\(0\)};
% \node[below = 0.5cm of b41] (b41label) {\(< 48\)};
% \node[right of = b41] (dots) {\(\ldots\)};
% \path[->] (leading) edge node[above left] {read 5th bit as 0} ( b50)
% (leading) edge (b51)
% (b51) edge (b40)
% (b51) edge node[above right] {read 4th bit as 0, continue} (b41)
% (b41) edge (dots)
% ;
% \end{tikzpicture}
% And finally collapse this diagram into an expression:
% %
% \begin{gather*}
% L_{\leq 42} = 0^*(0\,\Sigma^5 \mid 1(0(0\,\Sigma^3 \mid 1(0(0\,\Sigma^1 \mid 10)))))
% \end{gather*}
% The `quick terminating' branches read \(010101\) (one's complement of
% \(42_{10}\)) and the `continuing' branches read \(101010\), i.e.
% \(42_{10}\). This is just a binary search for the number 42!
% The DFA for this regular expression (not provided) looks nearly
% identical to the diagram above. \(L_{< 42}\) requires only a minor
% change at the end.
% This can be done more generally, given any binary string, constructing a
% DFA recognizing all binary strings numerically smaller than it.
% \vspace{1em}
% For the regularity on applying operations:
% \begin{enumerate}
% \item the complements are regular. In the DFA, change all final
% states to non-final, and all previously non-final states to final.
% This is an NFA accepting the complement of the original language.
% \item the union is regular. Take the new NFA \((Q_1 \cup Q_2,
% \Sigma, \delta_1 \cup \delta_2, S_1 \cup S_2, F_1 \cup F_2)\) where
% \(A_1 = (Q_1, \Sigma, \delta_1, S_1, F_1)\) and \(A_2 = (Q_2,
% \Sigma, \delta_2, S_2, F_2)\) are the individual DFAs for the
% languages above. This is the `trivial' union of the automata, and
% accepts exactly the words that are in one or both of the languages.
% \item the intersection is regular. The construction is a bit more
% involved. Intuitively, in the way that the union NFA allows
% accepting runs from either automaton, the intersection NFA must
% simulate the runs of both automata together. Given two automata
% \(A_1 = (Q_1, \Sigma, \delta_1, S_1, F_1)\) and \(A_2 = (Q_2,
% \Sigma, \delta_2, S_2, F_2)\) representing the two languages, define
% the intersection NFA \(A_\cap = (Q_1 \times Q_2, \Sigma \times
% \Sigma, \delta, S_1 \times S_2, F_1 \times F_2)\), where \(\delta\)
% is defined as
% \begin{gather*}
% \delta \subseteq (Q_1 \times Q_2) \times (\Sigma \times \Sigma) \times (Q_1 \times Q_2) \\
% \delta((q_1, q_2), (a_1, a_2), (q_1', q_2')) \iff \delta_1(q_1, a_1, q_1') \land \delta_2(q_2, a_2, q_2')~.
% \end{gather*}
% Thus, \(A_\cap\) operates simultaneously on a state from \(A_1\),
% and a state from \(A_2\). There is a transition from \((q_1, q_2)\)
% to \((q_1', q_2')\) when reading the letters \((a_1, a_2)\)
% simultaneously if the transitions on \emph{both} sides are defined.
% Discuss and convince yourself that the language of \(A_\cap\) is
% precisely the intersection of the languages of \(A_1\) and \(A_2\).
% \item the difference is regular. It follows from the complement and
% intersection regularities as \(L_1 \setminus L_2 = L_1 \cap \bar
% L_2\).
% \end{enumerate}
% This has some interesting consequences! Consider the predicate \(x \geq
% 42 \land x > 14\). We can construct DFAs representing the individual
% conditions, and to represent the logical-and, take their intersection.
% This gives us a way to solve some arithmetic constraints using automata!
% \textbf{Extra}: Automata (of different kinds) have been extensively
% studied as ways to solve many classes of constraints. Regarding
% arithmetic, there is the seminal result by B\"uchi, Weak Second-order
% Arithmetic and Finite Automata \cite{buchi1960weak}.
% \end{solution}
% \end{exercise}
% \begin{exercise}{}
% \todo{Optionally, ask them to construct the NFA? But it's too simple}
% Consider the following NFA, \(A\):
% %
% \begin{center}
% \begin{tikzpicture}[node distance = 2cm, on grid]
% \node[state, initial] (q0) {\(q_0\)};
% \node[state, right of = q0] (q1) {\(q_1\)};
% \node[state, right of = q1] (q2) {\(q_2\)};
% \node[state, accepting, right of = q2] (q3) {\(q_3\)};
% \draw[->, loop] (q0) edge node[above] {\(a, b\)} (q0);
% \path[->] (q0) edge node[above] {\(a\)} (q1)
% (q1) edge node[above] {\(a\)} (q2)
% (q2) edge node[above] {\(a\)} (q3)
% ;
% \end{tikzpicture}
% \end{center}
% Describe the language \(L(A)\). Is \(A\) a DFA (barring missing
% transitions)? If not, determinize it using the subset construction seen in
% class
% %
% \footnote{This is called subset / product / powerset construction by
% different authors. Wikipedia for quick reference:
% \url{https://en.wikipedia.org/wiki/Powerset_construction}}.
% \begin{solution}
% \(L(A)\) is the set of strings over \(\{a, b\}\) ending in three a's. As
% a regular expression, \(L(A) = (a \mid b)^*aaa\).
% It is not deterministic due to the two outgoing transitions from \(q_0\)
% labelled \(a\).
% For the subset construction, we define the power set automaton \(A' =
% (2^Q, \Sigma, \delta', S', F')\), where \(Q = \{q_0, q_1, q_2, q_3\}\).
% The set of initial states \(S'\) only contains the singleton
% \(\{q_0\}\), as we start in solely the initial state \(q_0\). So, \(S' =
% \{\{q_0\}\}\).
% The set of final states \(F'\) is any set that contains a final state,
% here only \(q_3\). Thus, \(F' = \{\{s \in 2^Q \mid q_3 \in s\}\}\). This
% corresponds to `angelic' executions, i.e., a word is accepted if there
% is \emph{a} run of it that leads to a final state.
% We draw the set of states \(2^Q\). The states are represented compactly
% due to space constraints: \(s_{23}\) represents the set \(\{q_2,
% q_3\}\), for example. The empty set is marked by \(s_\emptyset\).
% %
% \begin{center}
% \begin{tikzpicture}[node distance = 2cm, on grid]
% \node[state] (q0000) {\(s_\emptyset\)};
% \node[state, right of = q0000] (q0001) {\(s_0\)};
% \node[state, right of = q0001] (q0011) {\(s_{01}\)};
% \node[state, right of = q0011] (q0010) {\(s_1\)};
% \node[state, below of = q0000] (q0100) {\(s_2\)};
% \node[state, right of = q0100] (q0101) {\(s_{02}\)};
% \node[state, right of = q0101] (q0111) {\(s_{012}\)};
% \node[state, right of = q0111] (q0110) {\(s_{01}\)};
% \node[state, accepting, below of = q0100] (q1100) {\(s_{23}\)};
% \node[state, accepting, right of = q1100] (q1101) {\(s_{023}\)};
% \node[state, accepting, right of = q1101] (q1111) {\(s_{0123}\)};
% \node[state, accepting, right of = q1111] (q1110) {\(s_{123}\)};
% \node[state, accepting, below of = q1100] (q1000) {\(s_3\)};
% \node[state, accepting, right of = q1000] (q1001) {\(s_{03}\)};
% \node[state, accepting, right of = q1001] (q1011) {\(s_{013}\)};
% \node[state, accepting, right of = q1011] (q1010) {\(s_{13}\)};
% \node[above = 1.5cm of q0001] (start) {start};
% \draw[->] (start) -- (q0001);
% \end{tikzpicture}
% \end{center}
% % not relevant anymore
% % The states are drawn as a
% % \href{https://en.wikipedia.org/wiki/Karnaugh_map}{Karnaugh Map} on 4
% % bits for consistency. Each adjacent cell differs by only `one bit'.
% %
% Finally, we add the transitions corresponding to the outgoing
% transitions \(t_0 - t_5\). We define the new transition relation as
% %
% \begin{gather*}
% \forall\; s, s' \in 2^Q.\; \delta'(s, a, s') \iff s' = \cup \{p \mid \exists\; q \in s.\; \delta(q, a, p)\}
% \end{gather*}
% Example: what is the transition from the set \(\{q_0, q_2\}\) on \(a\)?
% From \(q_0\) we can transition to \(q_0\) (by \(t_0\)) or to \(q_1\) (by
% \(t_2\)), and from \(q_2\), we can only transition to \(q_3\) (by
% \(t_4\)). Thus, we have the transition \(\{q_0, q_2\} \xrightarrow{a}
% \{q_0, q_1\} \cup \{q_3\}\), or in the notation as above, \(s_{02}
% \xrightarrow{a} s_{013}\). We do this for every state and every letter
% (\(16 \times 2\)).
% The resulting automaton is too complicated to draw fully while
% maintaining any readability! Below, we draw all the relevant
% transitions. Any missing transitions go to the dead state
% \(s_\emptyset\) (marked by the input `several').
% %
% % \begin{center}
% % \begin{tikzpicture}[node distance = 2cm, on grid]
% % \node[state] (q0000) {\(s_\emptyset\)};
% % \node[state, right of = q0000] (q0001) {\(s_0\)};
% % \node[state, right of = q0001] (q0011) {\(s_{01}\)};
% % \node[state, right of = q0011] (q0010) {\(s_1\)};
% % \node[state, below of = q0000] (q0100) {\(s_2\)};
% % \node[state, right of = q0100] (q0101) {\(s_{02}\)};
% % \node[state, right of = q0101] (q0111) {\(s_{012}\)};
% % \node[state, right of = q0111] (q0110) {\(s_{12}\)};
% % \node[state, accepting, below of = q0100] (q1100) {\(s_{23}\)};
% % \node[state, accepting, right of = q1100] (q1101) {\(s_{023}\)};
% % \node[state, accepting, right of = q1101] (q1111) {\(s_{0123}\)};
% % \node[state, accepting, right of = q1111] (q1110) {\(s_{123}\)};
% % \node[state, accepting, below of = q1100] (q1000) {\(s_3\)};
% % \node[state, accepting, right of = q1000] (q1001) {\(s_{03}\)};
% % \node[state, accepting, right of = q1001] (q1011) {\(s_{013}\)};
% % \node[state, accepting, right of = q1011] (q1010) {\(s_{13}\)};
% % \node[above = 1.5cm of q0001] (start) {start};
% % \draw[->] (start) -- (q0001);
% % \path[->, color = red]
% % (q0001.45) edge[out=0, in=90, loop] node[above] {\calpha} (q0001.45)
% % (q0011.45) edge[out=0, in=90, loop] node[above] {\calpha} (q0011.45)
% % (q0101.45) edge[out=0, in=90, loop] node[above] {\calpha} (q0101.45)
% % (q0111.45) edge[out=0, in=90, loop] node[above] {\calpha} (q0111.45)
% % (q1111.45) edge[out=0, in=90, loop] node[above] {\calpha} (q1111.45)
% % (q1101.45) edge[out=0, in=90, loop] node[above] {\calpha} (q1101.45)
% % (q1001.45) edge[out=0, in=90, loop] node[above] {\calpha} (q1001.45)
% % (q1011.45) edge[out=0, in=90, loop] node[above] {\calpha} (q1011.45)
% % ;
% % \path[->, color = blue]
% % (q0001.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q0001.225)
% % (q0011.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q0011.225)
% % (q0101.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q0101.225)
% % (q0111.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q0111.225)
% % (q1111.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q1111.225)
% % (q1101.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q1101.225)
% % (q1001.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q1001.225)
% % (q1011.225) edge[out=180, in=270, loop] node[below left] {\cbeta} (q1011.225)
% % ;
% % \path[->, color = OliveGreen]
% % (q0001) edge[] node[below] {\cgamma} (q0011)
% % (q0011.315) edge[in=-90, out=0, loop] node[below] {\cgamma} (q0011.315)
% % (q0101) edge[] node[below] {\cgamma} (q0111)
% % (q0111.315) edge[in=-90, out=0, loop] node[below] {\cgamma} (q0111.315)
% % (q1111.315) edge[in=-90, out=0, loop] node[below] {\cgamma} (q1111.315)
% % (q1101) edge[] node[below] {\cgamma} (q1111)
% % (q1001) edge[] node[below] {\cgamma} (q1011)
% % (q1011.315) edge[in=-90, out=0, loop] node[below] {\cgamma} (q1011.315)
% % ;
% % \path[->, color = purple]
% % (q0010) edge[] node[right] {\ceta} (q0110)
% % (q0011) edge[] node[right] {\ceta} (q0111)
% % (q0110.135) edge[out=90, in=180, loop] node[above] {\ceta} (q0110.135)
% % (q0111.135) edge[out=90, in=180, loop] node[above] {\ceta} (q0111.135)
% % (q1010) edge[] node[right] {\ceta} (q1110)
% % (q1011) edge[] node[right] {\ceta} (q1111)
% % (q1110.135) edge[out=90, in=180, loop] node[above] {\ceta} (q1110.135)
% % (q1111.135) edge[out=90, in=180, loop] node[above] {\ceta} (q1111.135)
% % ;
% % \path[->, color = orange]
% % (q0100) edge[] node[left] {\cpi} (q1100)
% % (q0101) edge[] node[left] {\cpi} (q1101)
% % (q0110) edge[] node[left] {\cpi} (q1110)
% % (q0111) edge[] node[left] {\cpi} (q1111)
% % (q1100.0) edge[in=0, out=90, loop] node[above right] {\cpi} (q1100.0)
% % (q1101.0) edge[in=0, out=90, loop] node[above right] {\cpi} (q1101.0)
% % (q1110.0) edge[in=0, out=90, loop] node[above right] {\cpi} (q1110.0)
% % (q1111.0) edge[in=0, out=90, loop] node[above right] {\cpi} (q1111.0)
% % ;
% % \end{tikzpicture}
% % \end{center}
% % \begin{center}
% % \begin{tikzpicture}[node distance = 2cm, on grid]
% % \node[state] (q0000) {\(s_\emptyset\)};
% % \node[state, right of = q0000] (q0001) {\(s_0\)};
% % \node[state, right of = q0001] (q0011) {\(s_{01}\)};
% % \node[state, right of = q0011] (q0010) {\(s_1\)};
% % \node[state, below of = q0000] (q0100) {\(s_2\)};
% % \node[state, right of = q0100] (q0101) {\(s_{02}\)};
% % \node[state, right of = q0101] (q0111) {\(s_{012}\)};
% % \node[state, right of = q0111] (q0110) {\(s_{12}\)};
% % \node[state, accepting, below of = q0100] (q1100) {\(s_{23}\)};
% % \node[state, accepting, right of = q1100] (q1101) {\(s_{023}\)};
% % \node[state, accepting, right of = q1101] (q1111) {\(s_{0123}\)};
% % \node[state, accepting, right of = q1111] (q1110) {\(s_{123}\)};
% % \node[state, accepting, below of = q1100] (q1000) {\(s_3\)};
% % \node[state, accepting, right of = q1000] (q1001) {\(s_{03}\)};
% % \node[state, accepting, right of = q1001] (q1011) {\(s_{013}\)};
% % \node[state, accepting, right of = q1011] (q1010) {\(s_{13}\)};
% % \node[above = 1.5cm of q0001] (start) {start};
% % \draw[->] (start) -- (q0001);
% % % (q0000)
% % % (q0001)
% % % (q0010)
% % % (q0011)
% % % (q0100)
% % % (q0101)
% % % (q0110)
% % % (q0111)
% % % (q1000)
% % % (q1001)
% % % (q1010)
% % % (q1011)
% % % (q1100)
% % % (q1101)
% % % (q1110)
% % % (q1111)
% % \path[->]
% % (q0000) edge[loop] node[above] {\(a\)} (q0000)
% % (q0001) edge[] node[above] {\(a\)} (q0011)
% % (q0010) edge[] node[above] {\(a\)} (q0100)
% % (q0011) edge[] node[above] {\(a\)} (q0111)
% % (q0100) edge[] node[above] {\(a\)} (q1000)
% % (q0101) edge[] node[above] {\(a\)} (q1011)
% % (q0110) edge[] node[above] {\(a\)} (q1100)
% % (q0111) edge[] node[above] {\(a\)} (q1111)
% % (q1000) edge[] node[above] {\(a\)} (q0000)
% % (q1001) edge[] node[above] {\(a\)} (q0011)
% % (q1010) edge[] node[above] {\(a\)} (q0100)
% % (q1011) edge[] node[above] {\(a\)} (q0111)
% % (q1100) edge[] node[above] {\(a\)} (q1000)
% % (q1101) edge[] node[above] {\(a\)} (q1011)
% % (q1110) edge[] node[above] {\(a\)} (q1100)
% % (q1111) edge[loop] node[above] {\(a\)} (q1111)
% % ;
% % \path[->]
% % (q0000) edge[loop] node[right] {\(b\)} (q0000)
% % (q0001) edge[] node[right] {\(b\)} (q0001)
% % (q0010) edge[] node[right] {\(b\)} (q0000)
% % (q0011) edge[] node[right] {\(b\)} (q0001)
% % (q0100) edge[] node[right] {\(b\)} (q0000)
% % (q0101) edge[] node[right] {\(b\)} (q0001)
% % (q0110) edge[] node[right] {\(b\)} (q0000)
% % (q0111) edge[] node[right] {\(b\)} (q0001)
% % (q1000) edge[] node[right] {\(b\)} (q0000)
% % (q1001) edge[] node[right] {\(b\)} (q0001)
% % (q1010) edge[] node[right] {\(b\)} (q0000)
% % (q1011) edge[] node[right] {\(b\)} (q0001)
% % (q1100) edge[] node[right] {\(b\)} (q0000)
% % (q1101) edge[] node[right] {\(b\)} (q0001)
% % (q1110) edge[] node[right] {\(b\)} (q0000)
% % (q1111) edge[loop] node[right] {\(b\)} (q0001)
% % ;
% % \end{tikzpicture}
% % \end{center}
% \begin{center}
% \begin{tikzpicture}[node distance = 2cm, on grid]
% \node[state] (q0000) {\(s_\emptyset\)};
% \node[state, right of = q0000] (q0001) {\(s_0\)};
% \node[state, right of = q0001] (q0011) {\(s_{01}\)};
% \node[state, right of = q0011] (q0010) {\(s_1\)};
% \node[state, below of = q0000] (q0100) {\(s_2\)};
% \node[state, right of = q0100] (q0101) {\(s_{02}\)};
% \node[state, right of = q0101] (q0111) {\(s_{012}\)};
% \node[state, right of = q0111] (q0110) {\(s_{12}\)};
% \node[state, accepting, below of = q0100] (q1100) {\(s_{23}\)};
% \node[state, accepting, right of = q1100] (q1101) {\(s_{023}\)};
% \node[state, accepting, right of = q1101] (q1111) {\(s_{0123}\)};
% \node[state, accepting, right of = q1111] (q1110) {\(s_{123}\)};
% \node[state, accepting, below of = q1100] (q1000) {\(s_3\)};
% \node[state, accepting, right of = q1000] (q1001) {\(s_{03}\)};
% \node[state, accepting, right of = q1001] (q1011) {\(s_{013}\)};
% \node[state, accepting, right of = q1011] (q1010) {\(s_{13}\)};
% \node[above = 1.5cm of q0001] (start) {start};
% \draw[->] (start) -- (q0001);
% \node[left = 1.5cm of q0000] (several) {several};
% \draw[->] (several) -- (q0000);
% % (q0000)
% % (q0001)
% % (q0010)
% % (q0011)
% % (q0100)
% % (q0101)
% % (q0110)
% % (q0111)
% % (q1000)
% % (q1001)
% % (q1010)
% % (q1011)
% % (q1100)
% % (q1101)
% % (q1110)
% % (q1111)
% \path[->]
% (q0000.north) edge[out=45, in=135, loop] node[above] {\(a, b\)} (q0000.north)
% (q0001) edge[] node[above] {\(a\)} (q0011)
% (q0010) edge[] node[above] {\(a\)} (q0100)
% (q0011) edge[] node[right] {\(a\)} (q0111)
% (q0100) edge[bend right] node[left] {\(a\)} (q1000)
% (q0101) edge[] node[above] {\(a\)} (q1011)
% (q0110) edge[] node[above] {\(a\)} (q1100)
% (q0111) edge[] node[right] {\(a\)} (q1111)
% (q1001) edge[] node[above] {\(a\)} (q0011)
% (q1010) edge[] node[above] {\(a\)} (q0100)
% (q1011) edge[bend right] node[right] {\(a\)} (q0111)
% (q1100) edge[] node[right] {\(a\)} (q1000)
% (q1101) edge[] node[above] {\(a\)} (q1011)
% (q1110) edge[bend left] node[below] {\(a\)} (q1100)
% (q1111.315) edge[out=270, in=0, loop] node[right] {\(a\)} (q1111.315)
% ;
% \path[->]
% (q0001.45) edge[out=0, in=90, loop] node[right] {\(b\)} (q0001.45)
% (q0011.north) edge[bend right] node[above] {\(b\)} (q0001.north)
% (q0101) edge[] node[right] {\(b\)} (q0001)
% (q0111) edge[] node[right] {\(b\)} (q0001)
% (q1001) edge[bend left] node[left] {\(b\)} (q0001)
% (q1011) edge[] node[right] {\(b\)} (q0001)
% (q1101) edge[bend left = 45] node[above left] {\(b\)} (q0001)
% (q1111) edge[] node[right] {\(b\)} (q0001)
% ;
% \end{tikzpicture}
% \end{center}
% After removing the remaining dead states and transitions for visibility,
% we can see the DFA:
% \begin{center}
% \begin{tikzpicture}[node distance = 2cm, on grid]
% \node[state, initial] (q0) {\(s_0\)};
% \node[state, right of = q0] (q1) {\(s_{01}\)};
% \node[state, right of = q1] (q2) {\(s_{012}\)};
% \node[state, accepting, right of = q2] (q3) {\(s_{0123}\)};
% \draw[->, in=180, out=90, loop] (q0.135) edge node[above] {\(b\)} (q0.135);
% \draw[->, in=45, out=-45, loop] (q3.0) edge node[right] {\(a\)} (q3.0);
% \path[->] (q0) edge node[above] {\(a\)} (q1)
% (q1) edge node[above] {\(a\)} (q2)
% (q2) edge node[above] {\(a\)} (q3)
% ;
% \path[->] (q1) edge[bend left] node[below left] {\(b\)} (q0)
% (q2) edge[bend right] node[above] {\(b\)} (q0)
% (q3) edge[bend left] node[below] {\(b\)} (q0)
% ;
% \end{tikzpicture}
% \end{center}
% This is an `eager' version of our original NFA. It accepts the same
% language. However, since it cannot be `non-deterministic' by
% definition, every time it comes across an \(a\), it must `guess' that it
% is the first of three ending \(a\)'s, and backtrack if it is not the
% case.
% % Although equally powerful, the DFA more accurately captures how a
% % human-written parser would operate in practice on this language.
% \end{solution}
% \end{exercise}
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment