# Chomsky Normal Form

Posted By on April 29, 2016

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 → aBwhere 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 → Sis 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

#### Posted by Akash Kurup

Founder and C.E.O, World4Engineers Educationist and Entrepreneur by passion. Orator and blogger by hobby

Website: http://world4engineers.com