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 | ∈
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 | ε
(2) Now we will remove the null productions −
B → ε and A → ε
After removing B → ε, the production set becomes −
S0 → S, S → ASA | aB | a, A → B | S | ε, B → b
After removing A → ε, 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 |AX | YB | a | AS | SA
B → b
X → SA
Y → a
