Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 12F (Automata and Formal Languages)

A parse tree has the start variable at its root; the ordered children of a variable spell the right side of the production used at that occurrence, and its leaves, read left to right, spell the derived word.

For a grammar in Chomsky normal form with $r$ variables, take pumping length $p=2^r$. A parse tree for a word of length at least $p$ has a root-to-leaf path containing one variable twice. The yield can therefore be writtenwhere $|vxy|\le p$, $|vy|>0$, and replacing the subtree between the repeated variables zero, one, or several times provesThis is the pumping lemma for context-free languages. Without Chomsky normal form but with no epsilon or unit productions, use the maximum production length to choose a larger exponential $p$; bounded branching and the same repeated-variable argument apply.

The language $\{a^i{b}^i{a}^i:i\ge 0\}$ is not context free. A pumped window of bounded length cannot alter all three long blocks equally, so pumping breaks one of the required equalities.

The language $\{a^i{b}^j:i\ge j\ge 0\}$ is context free. For example the grammargenerates $a^{j+k}{b}^j$ with $j,k\ge 0$, and every required word has this form.

The language $\{wabw:w\in \{a,b{\}}^{\ast }\}$ is not context free. Intersecting with a suitable regular language that forces both copies into long homogeneous blocks reduces to a marked copy language; the context-free pumping lemma then places the two pumped pieces within only finitely many local blocks and cannot change the two copies identically. Equivalently, the standard copy-language proof applies with the fixed marker $ab$.