Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 4H (Automata and Formal Languages)

Use $L=\{a^n{b}^n:n\ge 0\}$. The context-free grammar $S\to aSb\mid \epsilon$ generates exactly $L$, so it is a context-free language. It is not a regular language: if $p$ is a pumping length, any allowed decomposition of $a^p{b}^p$ has a nonempty pumped segment consisting only of $a$'s. Pumping down leaves fewer $a$'s than $b$'s, contradicting the pumping lemma for regular languages. Thus $L$ is a context-free language but not a regular language.

Use $L=\{a^n{b}^n{c}^n:n\ge 0\}$. A halting algorithm checks that a word consists of three consecutive blocks in the correct order and compares their lengths, so $L$ is a decidable language. If it were a context-free language, apply the pumping lemma for context-free languages to $a^p{b}^p{c}^p$. The segment containing both pumped pieces has length at most $p$, so it cannot meet both the $a$ and $c$ blocks. Pumping down therefore changes at least one block count but leaves another equal to $p$, or destroys the required block order. Either way the resulting word is not in $L$, a contradiction. Thus $L$ is a decidable language but not a context-free language.

Fix an effective numbering of Turing machines and use the diagonal halting setIt is computably enumerable: dovetail all the computations and output each index when its computation halts. It is not a recursive set. Otherwise a machine could, on input $e$, loop if the alleged decider says $e\in H$ and halt otherwise. Applying this machine to its own index contradicts either answer. Hence $H$ is computably enumerable but not a recursive set.

With the diagonal halting set $H$, defineIf $A$ were computably enumerable, its odd-index inverse image would make $\mathbb{N}\setminus H$ computably enumerable. If $\mathbb{N}\setminus A$ were computably enumerable, its even-index inverse image would do the same. But a set and its complement being computably enumerable would make the set a recursive set: run the two enumerations in parallel until the input appears in one. This contradicts the undecidability of $H$. Thus neither $A$ nor its complement is computably enumerable.