Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/12g/a/solution
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 3 12G a Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
The pumping lemma for regular languages states that for every regular language there is an integer such that every with can be writtenwith
To prove it, let a deterministic finite automaton with states recognize . During the reading of the first symbols of an accepted word , the automaton visits states, including the initial state. By the pigeonhole principle, two of them agree, say with . Let be the first symbols, the next symbols, and the remaining suffix. Reading is a nonempty closed path from to itself. Traversing this path any number of times and then reading therefore reaches the same accept state. This proves all three assertions.
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