Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 4H / a

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.