can someone please explain this . why option B is the correct answer why not option A .

Comments

• KG AI MentorKG AI Mentor5 Mar 2026

  Quick AnswerThe stack empties because the machine pushes two 1s for every a but pops only one 1 for every b. To clear the $2n$ symbols pushed by $n$ as, you must read exactly $2n$ bs. Key StepsPush Phase (Reading a): For every a read, the transition pushes two symbols (11). If input is $a^n$, the stack grows to height $2n$. $$ \text{Stack Height} = 2 \times (\text{number of } a\text{'s}) = 2n $$ Pop Phase (Reading b): For every b read, the transition pops exactly one symbol (1). To reduce the stack height from $2n$ to $0$, you need to pop $2n$ times. $$ \text{Required } b\text{'s} = \text{Stack Height} = 2n $$ Why Option B Works: Option B represents the language $L = \{a^n b^{2n}\}$. - Input $a^n$ creates $2n$ items. - Input $b^{2n}$ removes those $2n$ items one by one. - The stack becomes empty exactly when the input ends. Why Option A Fails: Option A ($a^{2n}b^n$) would require the stack to hold $n$ items (if pushing 1 per 2 as) or require popping 2 items per b. Since our machine pushes 2 per a and pops 1 per b, the counts would not match (you would have $2n$ items left over or run out of stack too early). — KG AI Mentor 🤖

• Vivek Penuli5 Mar 2026

  @kg_ai_mentor But how do we empty the stack in option B

• KG AI MentorKG AI Mentor5 Mar 2026

  Quick AnswerOption B is correct because the NPDA pushes two symbols (11) for every a read, and pops one symbol for every b. This creates a stack height of 2n for n inputs of a, which matches the language $L = \{a^n b^{2n} : n \ge 0\}$. Key StepsLet's break down the logic of the NPDA transitions to see why it accepts $a^n b^{2n}$ and not $a^{2n} b^n$. 1. Analyze the "Push" Phase (Reading 'a's) Look at the transition for state $q0$ reading an a: $$ \delta(q0, a, z) = \{(q1, 11z)\} $$ Action: When the machine reads an a, it moves to state $q1$ and pushes two 1s onto the stack (replacing the top symbol $z$ with $11z$). Result: If you read $n$ as, you push $2n$ symbols onto the stack. 2. Analyze the "Pop" Phase (Reading 'b's) Look at the transition for state $q1$ reading a b: $$ \delta(q1, b, 1) = \{(q1, \lambda)\} $$ Action: When the machine reads a b, it stays in state $q1$ and pops one 1 from the stack ($\lambda$ means empty string, i.e., remove the top symbol). Result: To empty the stack of $2n$ symbols, you need to read exactly $2n$ bs. 3. Compare the Options Option A ($a^{2n}b^n$): This would require pushing 1 symbol for every 2 as, or popping 2 symbols for every b. Our machine does the opposite. Option B ($a^n b^{2n}$): This matches our logic perfectly. Input: $a^n$ $\rightarrow$ Stack grows by $2n$. Input: $b^{2n}$ $\rightarrow$ Stack shrinks by $2n$ (emptying it). Conclusion Since the machine pushes 2 items per a and pops 1 item per b, the number of bs must be double the number of as. Therefore, the language is $a^n b^{2n}$. — KG AI Mentor 🤖