Let's solve the problem step by step as you requested: --- ### **STEP 1: Introduction to the Concept** A **Pushdown Automaton (PDA)** is a computational model similar to a finite automaton but with an added stack storage. This stack allows the PDA to recognize context-free languages by "pushing" and "popping" symbols according to the input and stack contents. The language \( L \) given here is defined by the set: \[ L = \{ b^m u^h b^k l^f e^f \mid n, h \geq 0, f \geq 1, f = m, 2n = k \} \] This means every string in \( L \) consists of: - \( m \) b's, - \( h \) u's, - \( k \) b's, - \( f \) l's, - \( f \) e's, with the constraints \( f = m \) and \( 2n = k \). --- ### **STEP 2: Construct PDA with Empty Stack** #### **a. PDA Construction** We need a PDA that accepts strings by empty stack for the described language. Let's break the constraints: - \( m \) b's, followed by \( h \) u's, followed by \( k \) b's, followed by \( f \) l's, followed by \( f \) e's, - \( f = m \), \( 2n = k \). Let's take \( n = 0 \), so \( k = 0 \). But since \( n \geq 0 \), \( k \) is always even and non-negative. **PDA Design:** - **Push** a symbol (say X) for every \( b \) in the initial segment (\( m \)); keep track of \( m \). - Ignore \( u \)'s (\( h \) times, \( h \geq 0 \)). - For every \( l \), **pop** an X (since \( f = m \)). - For every \( e \), **pop** an X. - For every \( b \) after \( u \), **push** two X's (since \( 2n = k \)). - Accept by empty stack. #### **Transitions:** 1. **Start state (q0):** On reading \( b \), push X. 2. **From q0:** On \( u \), stay in q0 (ignore, no stack change). 3. **From q0:** After \( u \), on \( b \), push two X's. 4. **From q0:** On \( l \), pop X. 5. **From q0:** On \( e \), pop X. 6. **Accepting:** If stack is empty and input is finished, accept. --- ### **STEP 3: Convert to Final-State PDA** #### **b. Conversion** To convert an empty-stack PDA to a final-state PDA: - Add a new start state with an ε-move to the old start state and push a special bottom-of-stack marker (\$). - Any transition to an empty stack (original accepting condition) is now replaced by a transition to a new final state after reading all input and popping the bottom marker. --- ### **STEP 4: Does "bubble" belong to L?** #### **c. Check "bubble"** Let's break "bubble": - "b" "u" "b" "b" "l" "e" - \( m = 1 \) ("b" at start), - \( h = 1 \) ("u"), - \( k = 2 \) ("bb" after "u"), - \( f = 1 \) ("l"), - \( f = 1 \) ("e"). Check constraints: - \( f = m \Rightarrow 1 = 1 \) ✔️ - \( 2n = k \Rightarrow 2n = 2 \implies n = 1 \geq 0 \) ✔️ Thus, "bubble" **fits the pattern**. **Sequence of PDA Moves:** 1. Read "b", push X. Stack: X 2. Read "u", do nothing. Stack: X 3. Read "b", push two X's. Stack: XXX 4. Read "b", push two X's. Stack: XXXXX 5. Read "l", pop X. Stack: XXXX 6. Read "e", pop X. Stack: XXX Stack is **not empty** after reading the input. So, "bubble" **does not belong to L**. --- ### **STEP 5: Final Short Answer** **"bubble" does not belong to L, as the PDA does not reach an empty stack/final state after processing the word.** --- ### **STEP 6: Explanation of Each Step** - **Step 1:** Introduced PDA and analyzed the constraints of the language. - **Step 2:** Designed a PDA with stack operations to recognize strings of the language. - **Step 3:** Converted the empty-stack PDA to a final-state PDA. - **Step 4:** Checked if "bubble" fits the language and simulated the PDA. - **Step 5:** Summarized the answer. - **Step 6:** Briefly explained each step performed. If you need the PDA diagram or transitions in formal notation, let me know!