Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 4F (Automata and Formal Languages)

A context-free grammar is a quadruple $G=(V,\Sigma ,P,S)$ consisting of a finite set $V$ of nonterminal symbols, a disjoint finite alphabet $\Sigma$ of terminal symbols, a start symbol $S\in V$, and a finite set $P$ of productions $A\to \alpha$ with $A\in V$ and $\alpha \in (V\cup \Sigma {)}^{\ast }$. A sentence is a terminal word over an alphabet $w\in {\Sigma }^{\ast }$ with $S{\Rightarrow }^{\ast }w$, and the generated formal language is $\mathcal{L}(G)=\{w\in {\Sigma }^{\ast }:S{\Rightarrow }^{\ast }w\}$.

The first language is context-free. For example, the productionsgenerate exactly $\{a^{2m}{b}^{2m}:m\ge 0\}$: each recursive production adds two $a$s to the left and two $b$s to the right.

The language $L=\{a^{m^2}{b}^{m^2}:m\ge 0\}$ is not context-free. Suppose it had pumping length $p$ in the pumping lemma for context-free languages, and apply the lemma to $w=a^{p^2}{b}^{p^2}$. Write $w=uvxyz$ with $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z\in L$ for every $i\ge 0$.

Let $\alpha$ and $\beta$ be the respective numbers of $a$s and $b$s in $vy$. If $\alpha \ne \beta$, pumping with $i=0$ or $i=2$ destroys equality of the two block lengths. Hence a valid decomposition would require $\alpha =\beta =d>0$. Pumping with $i=2$ would then produce $a^{p^2+d}{b}^{p^2+d}$, where $1\le d\le p$. Butso $p^2+d$ is not a square number. This contradiction proves the square-count language is not context-free.