afl-material: comparison hws/hw03.tex

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 \documentclass{article}
 \usepackage{../style}
 \usepackage{../graphics}
+\newcommand{\solution}[1]{%
+\begin{quote}\sf%
+#1%
+\end{quote}}
+\renewcommand{\solution}[1]{}
 \begin{document}
 \section*{Homework 3}
 \begin{enumerate}
 \item The regular expression matchers in Java, Python and Ruby can be
 very slow with some (basic) regular expressions. What is the main
 reason for this inefficient computation?
+\solution{Many matchers employ DFS type of algorithms to check
+if a string is matched by the regex or not. Such algorithms
+require backtracking if have gone down the wrong path which
+can be very slow. There are also problems with bounded regular
+expressions and backreferences.}
 \item What is a regular language? Are there alternative ways
 to define this notion? If yes, give an explanation why
 they define the same notion.
+\solution{A regular language is a language for which every string
+can be recognized by some regular expression. Another definition is
+that it is a language for which a finite automaton can be
+constructed. Both define the same set of languages.}
 \item Why is every finite set of strings a regular language?
+\solution{Take a regex composed of all strings (works for finite languages)}
 \item Assume you have an alphabet consisting of the letters
 $a$, $b$ and $c$ only. (1) Find a regular expression
 that recognises the two strings $ab$ and $ac$. (2) Find
 a regular expression that matches all strings
 \emph{except} these two strings. Note, you can only use
 %  \begin{center}
 %    $\textit{zeroable(r)} \;\text{if and only if}\;
 %    L(r) = \{\}$
 %  \end{center}
+\solution{Done in the video but there I forgot to include the empty string.}
 \item Given the alphabet $\{a,b\}$. Draw the automaton that has two
 states, say $Q_0$ and $Q_1$.  The starting state is $Q_0$ and the
 final state is $Q_1$. The transition function is given by
 \begin{center}
 \item Given a deterministic finite automaton $A(\varSigma, Q, Q_0, F,
 \delta)$, define which language is recognised by this
 automaton. Can you define also the language defined by a
 non-deterministic automaton?
+\solution{
+A formula for DFAs is
+\[L(A) \dn \{s \;|\; \hat{\delta}(start_q, s) \in F\}\]
+For NFAs you need to first define what $\hat{\rho}$ means. If
+$\rho$ is given as a relation, you can define:
+\[
+\hat{\rho}(qs, []) \dn qs \qquad
+\hat{\rho}(qs, c::s) \dn \bigcup_{q\in qs} \{ q' \; | \; \rho(q, c, q')\}
+\]
+This ``collects'' all the states reachable in a breadth-first
+manner. Once you have all the states reachable by an NFA, you can define
+the language as
+\[
+L(N) \dn \{s \;|\; \hat{\rho}(qs_{start}, s) \cap F \not= \emptyset\}
+\]
+Here you test whether the all states reachable (for $s$) contain at least
+a single accepting state.
+}
 \item Given the following deterministic finite automaton over
 the alphabet $\{a, b\}$, find an automaton that
 recognises the complement language. (Hint: Recall that
 for the algorithm from the lectures, the automaton needs
 to be in completed form, that is have a transition for
 every letter from the alphabet.)
+\solution{
+Before exchanging accepting and non-accepting states, it is important that
+the automaton is completed (meamning has a transition for every letter
+of the alphabet). If not completed, you have to introduce a sink state.
+For fun you can try out the example with
+out completion: Then the original automaton can recognise
+strings of the form $a$, $ab...b$; but the ``uncompleted'' automaton would
+recognise only the empty string.
+}
 \begin{center}
 \begin{tikzpicture}[>=stealth',very thick,auto,
 every state/.style={minimum size=0pt,
 inner sep=2pt,draw=blue!50,very thick,
 (q3) edge [bend left=95, looseness = 2.2] node [left=2mm] {$1$} (q0)
 (q4) edge [loop right] node {$0, 1$} ();
 \end{tikzpicture}
 \end{center}
+\solution{Q0 and Q2 can be merged; and Q1 and Q3 as well}
 \item Given the following finite deterministic automaton over the alphabet $\{a, b\}$:
 \begin{center}
 \begin{tikzpicture}[scale=2,>=stealth',very thick,auto,
 every state/.style={minimum size=0pt,
 \item If a non-deterministic finite automaton (NFA) has
 $n$ states. How many states does a deterministic
 automaton (DFA) that can recognise the same language
 as the NFA maximal need?
+\solution{ $2^n$ in the worst-case and for some regexes the worst case
+cannot be avoided.
+Other comments: $r^{\{n\}}$ can only be represented as $n$
+copies of the automaton for $r$, which can explode the automaton for bounded
+regular expressions. Similarly, we have no idea how backreferences can be
+represented as automaton.
+}
 \item Prove that for all regular expressions $r$ we have
 \begin{center}
 $\textit{nullable}(r) \quad \text{if and only if}
 \quad [] \in L(r)$

changeset 892	f4df090a84d0
parent 778	3e5f5d19f514
child 916	10f834eb0a9e