Let’s go step-by-step for converting each CFG into **Chomsky Normal Form (CNF)**, starting with
the first question and continuing through the others. CNF requires that all rules must either be of the
form \( A \rightarrow BC \) (where \( B \) and \( C \) are non-terminal variables) or \( A \rightarrow
a \) (where \( a \) is a terminal), with no epsilon (ε) rules except for the start variable, and no unit
productions.
---
### Question 1: CFG
```
S → ASB
A → aASA | a | ε
B → SbS | A | bb
```
#### Step 1: Eliminate ε-productions
- \( A \rightarrow \epsilon \) is an ε-production. We need to eliminate it by substituting \( A \) in
other rules:
- \( S \rightarrow ASB \) gives \( S \rightarrow SB \) when \( A = \epsilon \).
- \( A \rightarrow aASA \) gives \( A \rightarrow aSA \) and \( A \rightarrow aA \) when one or
more \( A \)'s are replaced by ε.
- \( B \rightarrow A \) gives \( B \rightarrow \epsilon \), but we leave this as we will handle it after ε-
removal is done.
Updated grammar after ε-production removal:
```
S → ASB | SB
A → aASA | aSA | aA | a
B → SbS | ASB | SB | bb
```
#### Step 2: Eliminate unit productions
- A unit production is of the form \( A \rightarrow B \), where both \( A \) and \( B \) are non-
terminals. The grammar has the following unit productions:
- \( B \rightarrow A \). Replace \( B \) with the productions of \( A \):
- \( B \rightarrow aASA | aSA | aA | a \).
- \( S \rightarrow A \). Replace \( S \) with productions of \( A \):
- \( S \rightarrow aASA | aSA | aA | a \).
The updated grammar becomes:
```
S → ASB | SB | aASA | aSA | aA | a
A → aASA | aSA | aA | a
B → SbS | ASB | SB | bb | aASA | aSA | aA | a
```
#### Step 3: Convert to binary rules
- Convert long right-hand sides into two non-terminal symbols:
- \( S \rightarrow ASB \) becomes \( S \rightarrow AX_1 \), with \( X_1 \rightarrow SB \).
- \( S \rightarrow aASA \) becomes \( S \rightarrow aX_2 \), with \( X_2 \rightarrow SA \).
- Similarly for other rules.
Final CNF after converting to binary rules:
```
S → AX_1 | SB | aX_2 | aSA | aA | a
A → aX_2 | aSA | aA | a
B → SbS | AX_1 | SB | bb | aX_2 | aSA | aA | a
X_1 → SB
X_2 → SA
```
---
### Question 2: CFG
```
A → BAB | B | ε
B → 00 | ε
```
#### Step 1: Eliminate ε-productions
- Both \( A \rightarrow \epsilon \) and \( B \rightarrow \epsilon \) are ε-productions:
- Replace \( A \) and \( B \) in other rules:
- \( A \rightarrow BAB \) becomes \( A \rightarrow BA \), \( A \rightarrow AB \), and \( A \
rightarrow B \) after removing the ε's from \( B \).
- \( B \rightarrow 00 \) remains unchanged.
Updated grammar:
```
A → BAB | BA | AB | B
B → 00
```
#### Step 2: Eliminate unit productions
- Unit production \( A \rightarrow B \) can be replaced by \( B \)'s production:
- \( A \rightarrow 00 \).
Updated grammar:
```
A → BAB | BA | AB | 00
B → 00
```
#### Step 3: Convert to binary rules
- Split any longer productions:
- \( A \rightarrow BAB \) becomes \( A \rightarrow BX_1 \), where \( X_1 \rightarrow AB \).
Final CNF:
```
A → BX_1 | BA | AB | 00
B → 00
X_1 → AB
```
---
### Question 3: CFG
```
S → AA
A → B | BB
B → abB | b | bb
```
#### Step 1: Eliminate unit productions
- \( A \rightarrow B \) and \( B \rightarrow b \) are unit productions:
- Replace \( A \rightarrow B \) by all of \( B \)'s productions:
- \( A \rightarrow abB | b | bb \).
Updated grammar:
```
S → AA
A → abB | b | bb | BB
B → abB | b | bb
```
#### Step 2: Convert to binary rules
- Split longer right-hand sides:
- \( B \rightarrow abB \) becomes \( B \rightarrow aX_1 \), \( X_1 \rightarrow bB \).
Final CNF:
```
S → AA
A → abX_1 | b | bb | BB
B → aX_1 | b | bb
X_1 → bB
```
---
### Question 4: CFG
```
S → TaT
T → aTb | bTa | TT | ε
```
#### Step 1: Eliminate ε-productions
- \( T \rightarrow \epsilon \) is an ε-production:
- Replace \( T \) with appropriate alternatives in other rules.
Updated grammar:
```
S → TaT | aT | Ta | aa
T → aTb | bTa | TT
```
#### Step 2: Convert to binary rules
- Split longer productions:
- \( T \rightarrow aTb \) becomes \( T \rightarrow aX_1 \), \( X_1 \rightarrow Tb \).
Final CNF:
```
S → TX_2 | aT | Ta | aa
T → aX_1 | bX_2 | TT
X_1 → Tb
X_2 → aT
```
---
### Question 5: CFG
```
S → BSB | B | λ
B → bb | λ
```
#### Step 1: Eliminate ε-productions
- Both \( S \rightarrow \epsilon \) and \( B \rightarrow \epsilon \) are ε-productions:
- Replace \( S \) and \( B \) in other rules.
Updated grammar:
```
S → BSB | BS | SB | B
B → bb
```
#### Step 2: Convert to binary rules
- Convert longer productions into binary rules:
- \( S \rightarrow BSB \) becomes \( S \rightarrow BX_1 \), \( X_1 \rightarrow SB \).
Final CNF:
```
S → BX_1 | BS | SB | B
B → bb
X_1 → SB
```
---
These are the detailed steps for converting each CFG into **Chomsky Normal Form (CNF)**. Let me
know if you need further clarifications!
