This document discusses the process of converting a context-free grammar (CFG) into Chomsky normal form (CNF) in multiple steps. It first recalls the definition of CNF and the theorem that every context-free language minus the empty string has a CFG in CNF. It then outlines the steps to convert a CFG to CNF: 1) remove epsilon productions, 2) remove unit rules, 3) break productions with more than two variables into chains of productions with two variables, and 4) ensure all productions are in the forms A->BC or A->a. The document provides two examples showing the full conversion process.

1. Chomsky Normal Form
Continuation
By: Jasmine Peniel Combo
BSCS 2-3

2. Recall that:
• A CFG is said to be in Chomsky Normal Form
if every production is of one of these two
forms:
1. A -> BC (body is two variables).
2. A -> a (body is a single terminal).
• Theorem: If L is a CFL, then L – {ε} has a CFG
in CNF.

3. Recall the first steps:
• Step 1
Make sure the start symbol (S) doesn't
appear on right hand side. If so, add new
Start symbol.
Step 2
Remove ε-productions.
Step 3
Remove unit rules

4. Step 4: Break right sides longer than two into
a chain of productions with right sides of two
variables
So with the continuation of the previous example,
• S0 → ASA | aB | a | SA | AS
• S → ASA | aB | a | SA | AS
• A → b | ASA | aB | a | SA | AS
• B → b
Introduce new variables.
So for S0 → ASA & S → ASA , we replace SA with a new
variable, A1 therefore:
• S0 → ASA | aB | a | SA | AS
• S → ASA | aB | a | SA | AS
• A → b | ASA | aB | a | SA | AS
• B → b
S0 → A A1 | aB | a | SA | AS
S → A A1 | aB | a | SA | AS
A → b | A A1 | aB | a | SA | AS
A1 → SA
B → b
Recall:
1. A -> BC (body is two
variables).

5. Step 5: Make sure that the productions are in
either these two forms:
1. A -> BC (body is two variables).
2. A -> a (body is a single terminal).
• S0 → A A1 | aB | a | SA | AS
• S → A A1 | aB | a | SA | AS
• A → b | A A1 | aB | a | SA | AS
• A1 → SA
• B → b
So we replace right hand sides with the wrong form by introducing a
new variable just like the previous step. In this example, A2
• S0 → A A1 | aB | a | SA | AS
• S → A A1 | aB | a | SA | AS
• A → b | A A1 | aB | a | SA | AS
• A1 → SA
• B → b
S0 → A A1 | A2 B| a | SA | AS
S → A A1 | A2 B | a | SA | AS
A → b | A A1 | A2 B | a | SA | AS
A1 → SA
A2 → a
B → b

6. Example #2:
• S → aXbX
• X → aY | bY | ε
• Y → X | c
Remove ε-productions. Then since the right hand of S
contains X,
X → ε
S → aXbX
Consider all cases,
Case 1: S → a ε bX
S → abX
Case 2: S → aXb ε
S → aXb
Case 3: S → a ε b ε
S → ab
The variable X is nullable and
therefore, also Y so,
X → ε
Y → X
Y → ε
X → aY | bY
X → a | b add this new
production to X

7. Example #2
• From previous step we obtain,
S → aXbX | abX | aXb | ab
X → aY | bY | a | b
Y → X | c
Now we eliminate unit productions such as
Y → X so,
S → aXbX | abX | aXb | ab
X → aY | bY | a | b
Y → aY | bY | a | b | c
Recall that:
If A =>* B by a series of unit
productions, and B ->  is a
non-unit-production, then
add production A -> .

8. Example #2
S → aXbX | abX | aXb | ab
X → aY | bY | a | b
Y → aY | bY | a | b | c
Now we replace longer productions with shorter ones
by adding new variables.
D → aX
E → bX
New grammar:
S → DE | aE |Db| ab
X → aY | bY | a | b
Y → aY | bY | a | b | c
D → aX
E → bX

9. Example #2
S → DE | aE |Db| ab
X → aY | bY | a | b
Y → aY | bY | a | b | c
D → aX
E → bX
Then we correct the productions with wrong forms and
introduce new variables F & G:
F → a
G → b

10. Example #2
Final CNG in Chomsky Normal Form:
S → DE | FE |DF| FG
X → FY | GY | a | b
Y → FY | GY | a | b | c
D → FX
E → GX
F → a
G → b