Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 9

A regular language over a finite alphabet is a formal language recognized by a deterministic finite automaton; equivalently it is described by a regular expression. A nondeterministic finite automaton accepts an input string if at least one permitted computation consumes that string and reaches an accepting state. The connection is the equivalenceEvery deterministic finite automaton is a special case of a nondeterministic finite automaton. Conversely, the subset construction makes a deterministic finite automaton whose states record the set of all possible current states of a nondeterministic finite automaton. After each input symbol, update that set by taking all permitted transitions, also including any epsilon closure. A new state is accepting exactly when it contains an old accepting state. A finite state set has finitely many subsets, so this really is a deterministic finite automaton recognizing the same formal language.

Every finite formal language is a regular language. A single string $a_1\cdots a_k$ is described by the regular expression formed by concatenating its symbols; the empty string is described by $\epsilon$. A finite union of these regular expressions describes any specified finite collection of strings. The empty collection is described by $\varnothing$. ThusEquivalently, a deterministic finite automaton can remember which prefix of one of the finitely many accepted strings has been read, moving to a rejecting dead state when no such prefix remains possible.

These strings are the even-length palindromes $u{u}^R$. Suppose this were a regular language, and let $p$ be its length from the pumping lemma for regular languages. The stringbelongs to the formal language, with $u=0^{p}1$. In any allowed decomposition $w=abc$, where $|ab|\le p$ and $|b|>0$, the pumped block is $b=0^k$ with $1\le k\le p$. Pumping it away givesThis is not a palindrome: its initial and final runs of zeros have different lengths. Hence it cannot have the form $u{u}^R$, contradicting the pumping lemma for regular languages. Therefore

This is a regular language. A deterministic finite automaton needs only to remember the number of consecutive trailing ones, up to three, and whether a forbidden run of four has already occurred. Use states $q_0,q_1,q_2,q_3,q_{\times }$, start at $q_0$, and accept in $q_0,q_1,q_2,q_3$. Its transition function of a finite automaton is:

Reading a zero resets the trailing count unless rejection has already occurred. A fourth successive one enters the permanent rejecting state $q_{\times }$. Thus the accepted strings are exactly those with no substring $1111$, and

Let $L=\{0^m{1}^n:m,n\ge 0,m\ne n\}$, and suppose it were a regular language. The formal language $B=0^{\ast }{1}^{\ast }$ is a regular language. By closure of regular languages under set difference,would be a regular language. But the pumping lemma for regular languages rules this out: for pumping length $p$, every permitted pumped block in $0^p{1}^p$ consists of $k>0$ zeros, and deleting it produces $0^{p-k}{1}^p$, with unequal counts. This contradiction provesThe closure step can also be realized by a product automaton accepting when the automaton for $B$ accepts and the automaton for $L$ rejects. If the exponents are instead required to be positive, use $B=0^{+}{1}^{+}$; the same argument applies.

A context-free grammar is a quadruple $(N,\Sigma ,P,S)$: $N$ is a finite set of nonterminal symbols, $\Sigma$ is a disjoint finite alphabet of terminal symbols, $S\in N$ is the start symbol, and $P$ is a finite set of productions $A\to \alpha$, with $A\in N$ and $\alpha \in (N\cup \Sigma {)}^{\ast }$. Its formal language consists of the terminal strings derivable from $S$.

For the even-length palindromes in (b)(ii), take terminal alphabet $\{0,1\}$ and the context-free grammarEvery production adds identical symbols at the two ends, so every generated string is an even-length palindrome. Conversely, remove the matching outer pair of any nonempty even-length palindrome and repeat; this constructs its derivation, ending with $\epsilon$.

For the unequal-block formal language in (b)(iv), useAfter $k$ uses of $A\to 0A1$, and $j\ge 1$ zeros from $Z$, the resulting string is $0^{k+j}{1}^k$, so its zero count is larger. Every such unequal pair with $m>n$ occurs by choosing $k=n,j=m-n$. Similarly $B$ generates exactly $0^k{1}^{k+j}$ with $j\ge 1$, covering precisely $m<n$. The start alternatives cover both possibilities and never equal counts. If positive exponents are required, replace the start productions by $S\to 0A1\mid 0B1$, which adds one matched pair and ensures both counts are positive.