- Automata Theory Tutorial
- Automata Theory - Home
- Automata Theory Introduction
- Deterministic Finite Automaton
- Non-deterministic Finite Automaton
- NDFA to DFA Conversion
- DFA Minimization
- Moore & Mealy Machines
- Classification of Grammars
- Introduction to Grammars
- Language Generated by Grammars
- Chomsky Grammar Classification
- Regular Grammar
- Regular Expressions
- Regular Sets
- Arden's Theorem
- Constructing FA from RE
- Pumping Lemma for Regular Grammar
- DFA Complement
- Context-Free Grammars
- Context-Free Grammar Introduction
- Ambiguity in Grammar
- CFL Closure Properties
- CFG Simplification
- Chomsky Normal Form
- Greibach Normal Form
- Pumping Lemma for CFG
- Pushdown Automata
- Pushdown Automata Introduction
- Pushdown Automata Acceptance
- PDA & Context Free Grammar
- PDA & Parsing
- Turing Machine
- Turing Machine Introduction
- Accepted & Decided Language
- Multi-tape Turing Machine
- Multi-Track Turing Machine
- Non-Deterministic Turing Machine
- Semi-Infinite Tape Turing Machine
- Linear Bounded Automata
- Decidability
- Language Decidability
- Undecidable Language
- Turing Machine Halting Problem
- Rice Theorem
- Post Correspondence Problem
- Automata Theory Useful Resources
- Automata Theory - Quick Guide
- Automata Theory - Useful Resources
- Automata Theory - Discussion
Chomsky Normal Form
A CFG is in Chomsky Normal Form if the Productions are in the following forms −
- A → a
- A → BC
- S → ε
where A, B, and C are non-terminals and a is terminal.
Algorithm to Convert into Chomsky Normal Form −
Step 1 − If the start symbol S occurs on some right side, create a new start symbol S’ and a new production S’→ S.
Step 2 − Remove Null productions. (Using the Null production removal algorithm discussed earlier)
Step 3 − Remove unit productions. (Using the Unit production removal algorithm discussed earlier)
Step 4 − Replace each production A → B1…Bn where n > 2 with A → B1C where C → B2 …Bn. Repeat this step for all productions having two or more symbols in the right side.
Step 5 − If the right side of any production is in the form A → aB where a is a terminal and A, B are non-terminal, then the production is replaced by A → XB and X → a. Repeat this step for every production which is in the form A → aB.
Problem
Convert the following CFG into CNF
S → ASA | aB, A → B | S, B → b | ε
Solution
(1) Since S appears in R.H.S, we add a new state S0 and S0→S is added to the production set and it becomes −
S0→S, S→ ASA | aB, A → B | S, B → b | ∈
(2) Now we will remove the null productions −
B → ∈ and A → ∈
After removing B → ε, the production set becomes −
S0→S, S→ ASA | aB | a, A → B | S | ∈, B → b
After removing A → ∈, the production set becomes −
S0→S, S→ ASA | aB | a | AS | SA | S, A → B | S, B → b
(3) Now we will remove the unit productions.
After removing S → S, the production set becomes −
S0→S, S→ ASA | aB | a | AS | SA, A → B | S, B → b
After removing S0→ S, the production set becomes −
S0→ ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA
A → B | S, B → b
After removing A→ B, the production set becomes −
S0 → ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA
A → S | b
B → b
After removing A→ S, the production set becomes −
S0 → ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA
A → b |ASA | aB | a | AS | SA, B → b
(4) Now we will find out more than two variables in the R.H.S
Here, S0→ ASA, S → ASA, A→ ASA violates two Non-terminals in R.H.S.
Hence we will apply step 4 and step 5 to get the following final production set which is in CNF −
S0→ AX | aB | a | AS | SA
S→ AX | aB | a | AS | SA
A → b |AX | aB | a | AS | SA
B → b
X → SA
(5) We have to change the productions S0→ aB, S→ aB, A→ aB
And the final production set becomes −
S0→ AX | YB | a | AS | SA
S→ AX | YB | a | AS | SA
A → b A → b |AX | YB | a | AS | SA
B → b
X → SA
Y → a