- Moore & Mealy Machines
- DFA Minimization
- NDFA to DFA Conversion
- Non-deterministic Finite Automaton
- Deterministic Finite Automaton
- Automata Theory Introduction
- Automata Theory - Home
Classification of Grammars
Regular Grammar
- DFA Complement
- Pumping Lemma for Regular Grammar
- Constructing FA from RE
- Regular Sets
- Regular Expressions
Context-Free Grammars
- Pumping Lemma for CFG
- Greibach Normal Form
- Chomsky Normal Form
- CFG Simplification
- CFL Closure Properties
- Ambiguity in Grammar
- Context-Free Grammar Introduction
Pushdown Automata
- PDA & Parsing
- PDA & Context Free Grammar
- Pushdown Automata Acceptance
- Pushdown Automata Introduction
Turing Machine
- Linear Bounded Automata
- Semi-Infinite Tape Turing Machine
- Non-Deterministic Turing Machine
- Multi-Track Turing Machine
- Multi-tape Turing Machine
- Accepted & Decided Language
- Turing Machine Introduction
Decidability
- Post Correspondence Problem
- Rice Theorem
- Turing Machine Halting Problem
- Undecidable Language
- Language Decidability
Automata Theory Useful Resources
Selected Reading
- Who is Who
- Computer Glossary
- HR Interview Questions
- Effective Resume Writing
- Questions and Answers
- UPSC IAS Exams Notes
Pumping Lemma for CFG
Lemma
If L is a context-free language, there is a pumping length p such that any string w ∈ L of length ≥ p can be written as w = uvxyz, where vy ≠ ε, |vxy| ≤ p, and for all i ≥ 0, uvixyiz ∈ L.
Apppcations of Pumping Lemma
Pumping lemma is used to check whether a grammar is context free or not. Let us take an example and show how it is checked.
Problem
Find out whether the language L = {xnynzn | n ≥ 1} is context free or not.
Solution
Let L is context free. Then, L must satisfy pumping lemma.
At first, choose a number n of the pumping lemma. Then, take z as 0n1n2n.
Break z into uvwxy, where
|vwx| ≤ n and vx ≠ ε.
Hence vwx cannot involve both 0s and 2s, since the last 0 and the first 2 are at least (n+1) positions apart. There are two cases −
Case 1 − vwx has no 2s. Then vx has only 0s and 1s. Then uwy, which would have to be in L, has n 2s, but fewer than n 0s or 1s.
Case 2 − vwx has no 0s.
Here contradiction occurs.
Hence, L is not a context-free language.
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