Past exam of the computer science course of the University of Oxford / 2015 / Models of computation / 1

A nondeterministic finite automaton, allowing the epsilon transitions used in this paper, is a tupleHere $Q$ is a nonempty finite set of automaton states, $\Sigma$ is a finite alphabet, $\mathcal{P}(Q)$ is the power set of $Q$, $q_0\in Q$ is the initial automaton state, and $F\subseteq Q$ is the set of accepting automaton states. The symbol $\epsilon \notin \Sigma$ denotes the empty word. Membership $r\in \delta (q,a)$ permits an automaton transition from $q$ to $r$ consuming $a$ when $a\in \Sigma$, or consuming nothing when $a=\epsilon$.

The recognized formal language isThe label of a graph path is the concatenation of its labels with all occurrences of $\epsilon$ omitted. Thus acceptance requires consuming the entire input word; a single successful choice of automaton transitions suffices. A path of length zero is allowed, so the empty word is accepted whenever $q_0\in F$, or an accepting automaton state is reachable by epsilon transitions. If epsilon transitions are excluded from the definition of a nondeterministic finite automaton, the displayed version is called an Epsilon-NFA.

For $x$, there is an accepting graph pathIts nonempty labels spell $1100110$, and its last automaton state is accepting. Therefore $x$ is accepted. The last epsilon transition matters: finishing the consumed input at $q_0$ does not prevent a subsequent move to $q_2$.

For $y$, track all possible automaton states, including their epsilon closure, after each input prefix:

The final set contains no accepting automaton state, and $q_1$ has no outgoing epsilon transition. Therefore $y$ is rejected.

Use state elimination for finite automata, with regular expression labels on a generalized finite automaton. Add a fresh initial automaton state $s$ with $s\stackrel{\epsilon }{\to }{q}_0$, and a fresh final automaton state $f$ with $q_2\stackrel{\epsilon }{\to }f$. Only $f$ is now accepting. Missing labels represent the empty language, and parallel alternatives are combined by union.

When eliminating an internal automaton state $k$, update every surviving label byThe added regular expression describes entering $k$, traversing its loop zero or more times by Kleene star, and leaving it. This accounts for exactly the graph paths newly bypassing $k$.

First eliminate $q_1$. Its only incoming and outgoing labels are both $1$, and it has no loop. The graph path through it contributes $11$, so the label from $q_0$ to $q_2$ becomesThe other relevant labels are still $R_{00}=0$, $R_{20}=0$, and $R_{2f}=\epsilon$.

Next eliminate $q_2$. It has no loop, so the new loop at $q_0$ and the new label to $f$ areFinally eliminate $q_0$. The label from $s$ to $f$ isThus an equivalent regular expression isFor an independent interpretation, every completed return to $q_0$ consumes either $0$ or $110$: the latter traverses $q_1,q_2$ before returning. The last visit to $q_2$ consumes either nothing or $11$. Consequently every run of ones has length exactly two, and successive such runs require at least one zero between them. In particular, consecutive copies of $11$ without a separating zero are not accepted.

Let $A=(Q,\{0,1\},\delta ,q_0,F)$ be a deterministic finite automaton recognizing $L$. Construct a nondeterministic finite automaton $B$ with automaton state set $Q\times \{0,1\}$. Its second coordinate records whether the extra zero has already been consumed. Start at $(q_0,0)$ and make exactly the automaton states in $F\times \{1\}$ accepting.

For each $q\in Q$, $i\in \{0,1\}$, and $a\in \{0,1\}$, include the ordinary automaton transitionAlso include the extra automaton transitionThis consumes the inserted zero without changing the simulated automaton state of $A$. At a zero in layer zero, both the ordinary and extra choices are available. There are no automaton transitions from layer one back to layer zero.

If $xy\in L$, run $A$ on $x$ in layer zero, consume the extra zero by the displayed automaton transition, and run $A$ on $y$ in layer one. The resulting run accepts $x0y$ in $B$.

Conversely, every accepting run of $B$ starts in layer zero and finishes in layer one, so it uses exactly one extra automaton transition. Split its input as $x0y$ at that move. Removing that consumed zero leaves precisely a run of $A$ on $xy$, ending in $F$. Hence $xy\in L$. The construction also permits an insertion at either end, including insertion into the empty word.

Therefore $L(B)=L^{\prime }$. Since $B$ has $2|Q|$ automaton states, $L^{\prime }$ is a regular language. This is a single-symbol insertion into a regular language.

No: regular languages are not closed under countably infinite intersection. Let $\Sigma =\{0,1\}$ and putThe first term in $L_i$ has regular expression $(0\cup 1{)}^i(0\cup 1{)}^{\ast }$, where the fixed power $i$ means $i$ copies. The second term is a finite language, a finite union of single words. Their union is therefore a regular language.

If $w\in K$, then for each $i$ either $|w|\ge i$ or $w$ belongs to the second term, so $w\in L_i$. If $w\notin K$, choose $i>|w|$. It belongs to neither term of $L_i$. ConsequentlyThe formal language $K$ is not a regular language. Indeed, for a proposed pumping length $p\ge 1$, the pumping lemma for regular languages applied to $0^p{1}^p$ must pump a nonempty block consisting only of zeros. Repeating that block increases the number of zeros without increasing the number of ones, leaving $K$. This contradiction proves the counterexample. Finite intersection closure does not extend to this infinite intersection of regular languages.