Let's break down and solve each part step by step: ### **Given:** \( L = \{ b^m u^h b^k l^f e^f \mid n, h \geq 0, f \geq 1, f = m, \text{ and } 2n = k \} \) #### **a. Construct a PDA with empty stack that recognizes the strings of L. (20 points)** #### **Step 1: Understanding the Language** - The string is of the form \( b^m u^h b^k l^f e^f \) - Constraints: - \( n, h \geq 0 \) - \( f \geq 1 \) - \( f = m \) ⇒ number of initial 'b's = number of 'l's and 'e's - \( 2n = k \) ⇒ number of 'b's after 'u's is twice the number of 'u's #### **Step 2: PDA Construction (Empty Stack Acceptance)** **States:** - \( q_0 \): Start state - \( q_1 \): Counting 'b's (for \( m \)) - \( q_2 \): Counting 'u's (for \( n \)) - \( q_3 \): Counting 'b's (for \( 2n \)) - \( q_4 \): Counting 'l's (for \( f \)) - \( q_5 \): Counting 'e's (for \( f \)) - \( q_{accept} \): Accept state **Stack Alphabet:** - \( Z_0 \): Initial stack symbol - \( B \): Used to count \( m \) 'b's and then match with 'l's and 'e's - \( U \): Used to count 'u's, then expect double 'b's after 'u' **Transitions:** 1. **From \( q_0 \) to \( q_1 \):** Start reading 'b's and push \( B \) for each. - \((q_0, b, Z_0) \to (q_1, BZ_0)\) - \((q_1, b, B) \to (q_1, BB)\) 2. **From \( q_1 \) to \( q_2 \):** Start reading 'u's and push \( U \) for each. - \((q_1, u, B) \to (q_2, UB)\) - \((q_2, u, U) \to (q_2, UU)\) 3. **From \( q_2 \) to \( q_3 \):** For each pair of 'b's after 'u's, pop a \( U \) for every two 'b's. - \((q_2, b, U) \to (q_3, U)\) [first 'b', stay in \( q_3 \)] - \((q_3, b, U) \to (q_2, \epsilon)\) [second 'b', pop \( U \)] - [Repeat for all \( n \) pairs, ensuring \( k = 2n \)] 4. **From \( q_2 \) or \( q_3 \) to \( q_4 \):** After all 'b's are consumed, start reading 'l's and pop \( B \) for each. - \((q_2, l, B) \to (q_4, \epsilon)\) - \((q_4, l, B) \to (q_4, \epsilon)\) 5. **From \( q_4 \) to \( q_5 \):** Read 'e's for each 'l' (at least one each). - \((q_4, e, \epsilon) \to (q_5, \epsilon)\) - \((q_5, e, \epsilon) \to (q_5, \epsilon)\) 6. **Accept:** If stack is empty after reading all input, accept. **Formal Definition:** \( M = (Q, \Sigma, \Gamma, \delta, q_0, Z_0) \) - \( Q = \{ q_0, q_1, q_2, q_3, q_4, q_5 \} \) - \( \Sigma = \{ b, u, l, e \} \) - \( \Gamma = \{ Z_0, B, U \} \) - \( q_0 \) is the start - \( Z_0 \) is the initial stack symbol - \( \delta \) as described above --- #### **b. Convert the generated empty-stack PDA into a final-state PDA using the approach studied in lectures. (15 points)** **Conversion (Sketch):** - Add a new accept state \( q_f \). - For every transition in the original PDA where the stack would be empty (just before acceptance), add a transition to \( q_f \) upon reading \( \epsilon \) and popping \( Z_0 \). - \( q_f \) is the only accept state. **Formal Description:** - \( Q' = Q \cup \{ q_f \} \) - Add transition: For all states \( q \) where \( \delta(q, \epsilon, Z_0) \) is defined, add \( (q, \epsilon, Z_0) \to (q_f, \epsilon) \). - Accept by final state \( q_f \). --- #### **c. Does the word "bubble" belong to L? Justify your answer. (15 points)** Let’s analyze "bubble": - "bubble" = b u b b l e - Let's parse: \( b^m u^h b^k l^f e^f \) - \( m = 1 \) (first 'b') - \( h = 1 \) ('u') - \( k = 2 \) (next two 'b's) - \( f = 1 \) ('l') - \( f = 1 \) ('e') Now, check constraints: - \( f = m \): **1 = 1** ✔️ - \( 2n = k \): \( n = h = 1 \), \( k = 2 \): **2×1 = 2** ✔️ - \( f \geq 1 \): **1 ≥ 1** ✔️ **So yes, "bubble" belongs to L.** **Sequence of PDA Moves:** 1. Push \( B \) for first 'b' (\( q_0, b, Z_0 \to q_1, BZ_0 \)) 2. Push \( U \) for 'u' (\( q_1, u, B \to q_2, UB \)) 3. For the next two 'b's: - First 'b': (\( q_2, b, U \to q_3, U \)) - Second 'b': (\( q_3, b, U \to q_2, \epsilon \)) [pops U] 4. 'l' (\( q_2, l, B \to q_4, \epsilon \)) [pops B] 5. 'e' (\( q_4, e, \epsilon \to q_5, \epsilon \)) 6. Stack is empty (except \( Z_0 \)), transition to accept state. **Indication:** "bubble" is accepted by both the empty-stack and final-state PDA, so **it belongs to L**. --- ### **Summary Table** | Part | Answer/Approach | |------|-----------------| | a | PDA construction and transitions as above | | b | Convert to final-state PDA by adding \( q_f \) and new transitions | | c | "bubble" belongs to L. Sequence of moves shows acceptance. | --- #### **If you need the full formal description (states, transition functions) written out in detail, let me know!**