The document describes the step-by-step process of converting a context-free grammar to Chomsky Normal Form (CNF) which has the following properties: 1) Every production has exactly two variables on the right hand side or one terminal symbol. 2) The start symbol S does not appear on the right hand side of any production. The process involves removing epsilon productions, unit productions, and introducing new variables to break up productions that are longer than two symbols until the grammar is in CNF. An example demonstrates applying these steps to fully convert a sample grammar.

1. • 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 productions.
Step 4: Break right sides longer than two into a chain of
productions with right sides of two variables.
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).

2. 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 SAwith 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
Recal
l:
1. A -> BC (body is
two
variables).

3. 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 → AA1 | 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

4. 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

5. 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 -> .

6. 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

7. 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

8. 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