Let's solve the problem step by step: --- ## **Given Grammar** ``` S → aAb | bbB A → Baa | ba B → bB | b ``` --- ## **(a) Convert the grammar into Chomsky Normal Form (CNF)** **Rules for CNF:** - Each production is either of the form \( A \rightarrow BC \) (where B and C are non-terminals), or \( A \rightarrow a \) (where a is a terminal). **Step 1: Eliminate Terminals from Right-Hand Sides with Length > 1** Introduce new variables for terminals: Let: - \( X_a \rightarrow a \) - \( X_b \rightarrow b \) Rewrite productions: ### S productions: - \( S \rightarrow aAb \) becomes \( S \rightarrow X_a A X_b \) - \( S \rightarrow bbB \) becomes \( S \rightarrow X_b X_b B \) ### A productions: - \( A \rightarrow Baa \) becomes \( A \rightarrow B X_a X_a \) - \( A \rightarrow ba \) becomes \( A \rightarrow X_b X_a \) ### B productions: - \( B \rightarrow bB \) becomes \( B \rightarrow X_b B \) - \( B \rightarrow b \) becomes \( B \rightarrow X_b \) So the grammar is: ``` S → X_a A X_b | X_b X_b B A → B X_a X_a | X_b X_a B → X_b B | X_b X_a → a X_b → b ``` **Step 2: Ensure All Productions Have At Most Two Symbols on RHS** Break down the right sides with more than two symbols: - \( S \rightarrow X_a A X_b \) - \( S \rightarrow X_b X_b B \) - \( A \rightarrow B X_a X_a \) - \( A \rightarrow X_b X_a \) Introduce new variables: Let: - \( Y_1 \rightarrow A X_b \) - \( Y_2 \rightarrow X_b B \) - \( Y_3 \rightarrow X_a X_a \) Rewrite: - \( S \rightarrow X_a Y_1 \) - \( S \rightarrow X_b Y_2 \) - \( A \rightarrow B Y_3 \) - \( A \rightarrow X_b X_a \) So the final CNF grammar is: ``` S → X_a Y_1 | X_b Y_2 Y_1 → A X_b Y_2 → X_b B A → B Y_3 | X_b X_a Y_3 → X_a X_a B → X_b B | X_b X_a → a X_b → b ``` --- ### **Final CNF Answer:** ``` S → X_a Y_1 | X_b Y_2 Y_1 → A X_b Y_2 → X_b B A → B Y_3 | X_b X_a Y_3 → X_a X_a B → X_b B | X_b X_a → a X_b → b ``` --- ## **(b) Convert the grammar into Greibach Normal Form (GNF)** **Rules for GNF:** - Each production is of the form \( A \rightarrow a\alpha \), where \( a \) is a terminal and \( \alpha \) is a (possibly empty) string of non-terminals. Start from the original grammar: ``` S → aAb | bbB A → Baa | ba B → bB | b ``` Let’s substitute the non-terminals on the right to get GNF. ### Step 1: Replace B in A's Productions #### B → bB | b So, - \( A \rightarrow Baa \) becomes \( A \rightarrow bBaa | baa \) - \( A \rightarrow ba \) stays Now, - \( bBaa \) can be expanded as \( b(bB)aa \) and so on, so it's already in the desired form (starts with terminal). So - \( A \rightarrow bBaa | baa | ba \) But \( baa \) and \( ba \) are both in the form \( b\alpha \). ### Step 2: Make S Productions Start with a Terminal #### S → aAb | bbB - \( aAb \) starts with \( a \) — good. - \( bbB \) starts with \( b \) — good. But \( bbB \) can be rewritten as \( b bB \) or \( b b \), after expanding \( B \). ### Step 3: Express All Productions in GNF Let’s write all possible derivations, making sure each production starts with a terminal: #### S Productions: - \( S \rightarrow aAb \) - \( S \rightarrow bbB \) But \( aAb \) — after substituting \( A \): - \( aAb \rightarrow a(bBaa)b = abBaa b \) - \( aAb \rightarrow a(baa)b = abaab \) - \( aAb \rightarrow a(ba)b = abab \) But in GNF, the right of the arrow should be a terminal followed by zero or more non-terminals. But let's try to write productions for each variable: #### B Productions: - \( B \rightarrow bB \) - \( B \rightarrow b \) Both are in GNF. #### A Productions: - \( A \rightarrow bBaa \) - \( A \rightarrow baa \) - \( A \rightarrow ba \) All start with terminal 'b', which is fine. #### S Productions: - \( S \rightarrow aAb \) - \( S \rightarrow bbB \) But 'aAb' should be 'a' followed by a string of non-terminals. Let’s see if we can express 'aAb' in GNF. \( S \rightarrow aAb \) is not in GNF because of the trailing terminal 'b'. Let’s introduce a new variable \( Z \rightarrow b \): So, - \( S \rightarrow aA Z \) - \( S \rightarrow b Y \) - \( Y \rightarrow bB \) Now, for A: - \( A \rightarrow bB aa \) Let’s introduce \( W \rightarrow aa \): - \( A \rightarrow bB W \) - \( W \rightarrow a V \) - \( V \rightarrow a \) So, let's organize: #### Final GNF Productions: ``` S → a A Z | b Y Z → b Y → b B A → b B W | b a a | b a W → a V V → a B → b B | b ``` But \( A \rightarrow baa \) and \( A \rightarrow ba \) can be handled directly: - \( A \rightarrow b W' \) - \( W' \rightarrow a a \) - \( A \rightarrow b V' \) - \( V' \rightarrow a \) But for clarity, let’s keep the original and just rewrite as follows: ### **Final GNF Answer:** ``` S → a A Z | b Y Z → b Y → b B A → b B W | b a a | b a W → a a B → b B | b ``` --- ## **Summary Table** | Non-terminal | CNF Productions | GNF Productions | |--------------|--------------------------------------------------------|-----------------------------------| | S | X_a Y_1 \| X_b Y_2 | a A Z \| b Y | | Y_1 | A X_b | | | Y_2 | X_b B | | | A | B Y_3 \| X_b X_a | b B W \| b a a \| b a | | Y_3 | X_a X_a | | | B | X_b B \| X_b | b B \| b | | X_a | a | | | X_b | b | | | Z | | b | | Y | | b B | | W | | a a | --- **If you need further clarification or a specific step expanded, let me know!**

Let's carefully work through the problem step by step, providing detailed explanations and calculations to convert the given context-free grammar into both Chomsky Normal Form (CNF) and Greibach Normal Form (GNF). --- ## Given Grammar: ``` S → aAb | bbB A → Baa | ba B → bB | b ``` --- ## Part (a): Convert to **Chomsky Normal Form (CNF)** ### Step 1: Understand CNF rules - Productions must be either: - \( A \rightarrow BC \), where \( B \) and \( C \) are non-terminals, **or** - \( A \rightarrow a \), where \( a \) is a terminal. - No epsilon (\( \varepsilon \)) productions (unless the start symbol), and no unit productions unless necessary. ### Step 2: Eliminate terminals from right-hand sides with length > 1 - For productions with longer RHS, replace terminals with new non-terminals. Create new non-terminals: - \( X_a \rightarrow a \) - \( X_b \rightarrow b \) Now rewrite all productions: - \( S \rightarrow aAb \) becomes \( S \rightarrow X_a A X_b \) - \( S \rightarrow bbB \) becomes \( S \rightarrow X_b X_b B \) - \( A \rightarrow Baa \) becomes \( A \rightarrow B X_a X_a \) - \( A \rightarrow ba \) becomes \( A \rightarrow X_b X_a \) - \( B \rightarrow bB \) becomes \( B \rightarrow X_b B \) - \( B \rightarrow b \) becomes \( B \rightarrow X_b \) - Terminals: - \( X_a \rightarrow a \) - \( X_b \rightarrow b \) Updated grammar: ``` S → X_a A X_b | X_b X_b B A → B X_a X_a | X_b X_a B → X_b B | X_b X_a → a X_b → b ``` ### Step 3: Break RHS with more than 2 symbols into binary productions For productions with RHS longer than 2, introduce new variables. - \( S \rightarrow X_a A X_b \): length 3, break into two rules: Introduce \( Y_1 \rightarrow A X_b \): - \( S \rightarrow X_a Y_1 \) - \( Y_1 \rightarrow A X_b \) Similarly, for \( S \rightarrow X_b X_b B \): Introduce \( Y_2 \rightarrow X_b B \): - \( S \rightarrow X_b Y_2 \) - \( Y_2 \rightarrow X_b B \) For \( A \rightarrow B X_a X_a \): Introduce \( Y_3 \rightarrow X_a X_a \): - \( A \rightarrow B Y_3 \) - \( Y_3 \rightarrow X_a X_a \) Now, \( A \rightarrow X_b X_a \) is already binary, no change needed. ### Final CNF grammar: ``` S → X_a Y_1 | X_b Y_2 Y_1 → A X_b Y_2 → X_b B A → B Y_3 | X_b X_a Y_3 → X_a X_a B → X_b B | X_b X_a → a X_b → b ``` --- ## **Final CNF answer:** ``` S → X_a Y_1 | X_b Y_2 Y_1 → A X_b Y_2 → X_b B A → B Y_3 | X_b X_a Y_3 → X_a X_a B → X_b B | X_b X_a → a X_b → b ``` --- ## Part (b): Convert to **Greibach Normal Form (GNF)** ### Step 1: Understand GNF rules - Productions must be of the form: - \( A \rightarrow a \alpha \), where \( a \) is a terminal and \( \alpha \) is a (possibly empty) string of non-terminals. - The main goal is to ensure each production starts with a terminal. ### Step 2: Start from the original grammar ``` S → aAb | bbB A → Baa | ba B → bB | b ``` ### Step 3: Expand B's productions - \( B \rightarrow bB \) and \( B \rightarrow b \): both start with terminal \( b \). Already suitable. ### Step 4: Rewrite A's productions - \( A \rightarrow Baa \) becomes \( A \rightarrow b B a a \) - \( A \rightarrow ba \): starts with \( b \), so suitable. Now, \( b B a a \) has a non-terminal \( B \) in the middle, which violates GNF rules because the production doesn't start with a terminal. **To fix this**, we need to replace \( B \) with its productions: - \( B \rightarrow b B \) or \( B \rightarrow b \) Thus: - \( A \rightarrow b B a a \) Expanding \( B \): - \( b B a a \rightarrow b (b B) a a \), which is \( b b B a a \) - or \( b b a a \) The first form starts with terminal \( b \), which is acceptable, but the second is just terminal. ### Step 5: Define new non-terminals for sequences Create a non-terminal \( C \rightarrow B a a \): - \( A \rightarrow b C \) - \( C \rightarrow B a a \) Now, expand \( C \): - \( C \rightarrow b a a \) or \( b B a a \) But since \( B \rightarrow b B | b \): - \( C \rightarrow b a a \) or \( b B a a \) To get \( B a a \) starting with terminal \( b \), substitute: - \( B \rightarrow b B \) or \( B \rightarrow b \) Thus, \( B a a \) can be rewritten as: - \( b B a a \) which starts with \( b \), suitable for GNF. ### Step 6: Rewrite S productions - \( S \rightarrow a A b \) and \( S \rightarrow b b B \) For \( a A b \): - \( A \) can be replaced by the previous expansions. Similarly, \( S \rightarrow a A b \): - \( A \rightarrow b C \), so: \( S \rightarrow a (b C) b \) which is: - \( a b C b \) This starts with terminal \( a \), acceptable. Similarly, \( S \rightarrow b b B \): - \( B \rightarrow b B | b \) So: - \( S \rightarrow b b B \) which starts with terminal \( b \). --- ### **Summary of GNF conversions:** Define non-terminals for sequences starting with terminals: - \( A \rightarrow b C \), \( C \rightarrow B a a \), \( B \rightarrow b B | b \) - \( S \rightarrow a A' \) with \( A' \rightarrow b \), or \( S \rightarrow b B' \) Let's formalize the final GNF: ``` S → a A' | b B' A' → b B' → b B | b A → b C C → B a a B → b B | b ``` Note that: - All productions start with a terminal, followed by non-terminals or ε (if any). --- ## **Final GNF answer:** ``` S → a A' | b B' A' → b B' → b B | b A → b C C → B a a B → b B | b ``` --- ## **Summary:** - **CNF**: Productions are binary or terminal-only, achieved by introducing helper variables. - **GNF**: Productions start with a terminal, achieved by rewriting productions and introducing helper variables for sequences. --- If you'd like, I can prepare a summarized table or a diagram. Let me know!

Certainly! Here are the final answers for both parts: ### (a) **Chomsky Normal Form (CNF):** ``` S → X_a Y_1 | X_b Y_2 Y_1 → A X_b Y_2 → X_b B A → B Y_3 | X_b X_a Y_3 → X_a X_a B → X_b B | X_b X_a → a X_b → b ``` --- ### (b) **Greibach Normal Form (GNF):** ``` S → a A' | b B' A' → b B' → b B | b A → b C C → B a a B → b B | b ``` These are the final simplified forms in CNF and GNF respectively.