Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 4H (Automata and Formal Languages)

Use state elimination for finite automata on a generalized finite automaton, whose directed edges are labelled by regular expressions. Add a fresh initial state $s$ with an $\epsilon$-edge to the old initial state and a fresh final state $f$ with $\epsilon$-edges from all old accepting states. Every missing edge has label $\varnothing$, the empty language; parallel edges are merged by union. The label $\epsilon$ denotes the language containing just the empty word.

When eliminating a state $k$, replace the label on each surviving edge $i\to j$ byHere juxtaposition means concatenation of formal languages, and the Kleene star allows zero or more traversals of the loop at $k$. Then delete $k$ and its incident edges. Eliminate every original state; the remaining $s\to f$ label is a regular expression for exactly the accepted formal language. The update accounts for paths avoiding $k$ and paths entering $k$, looping there any number of times, and leaving it.

Let $K$ be the diagonal halting set, which is undecidable. The pumping argument below actually holds for every subset $K\subseteq \mathbb{N}$; nonregularity needs the stated convention, or at least a set of exponents whose unary language is nonregular.

The pumping lemma for regular languages holds with pumping length one. For an accepted nonempty word, pump its first symbol. A word containing only $1$s remains in $1^{\ast }$ even if that symbol is deleted. If there is a zero and a nonempty prefix before its last zero, pumping the first symbol leaves the last zero and its following $n\in K$ ones unchanged. If the word is $01^n$, pumping its initial zero produces $0^i{1}^n$; this is accepted for $i\ge 1$, and for $i=0$ it belongs to $1^{\ast }$. In every case the pumped substring is nonempty and lies in the first position.

If $L$ were a regular language, closure under intersection would makeregular. A finite-state automaton for this intersection would decide membership of $n$ in the diagonal halting set by reading $01^n$, a contradiction. Thus $L$ satisfies the pumping lemma but is not regular, under the usual meaning of $\mathbb{K}$. If $K$ were arbitrary, the printed conclusion would be false, for example for $K=\mathbb{N}$.