Practice Question - 1
Duration: 5 min
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This educational video features a lecture by Sanchit Jain from Knowledge Gate, focusing on automata theory. The specific problem involves analyzing a given Pushdown Automaton (PDA) defined by a set of transition mappings to identify the correct language it accepts. The instructor systematically breaks down the transition function, interpreting stack operations to deduce the relationship between input symbol counts. The problem presents four multiple-choice options involving languages of the form $a^m b^n$, requiring the student to map the PDA's behavior to one of these formal language definitions.
Chapters
0:00 – 2:00 00:00-02:00
The instructor begins by presenting the problem: "consider the following mapping and find the correct language?" He lists six transition rules, starting with $\delta(q_0, 1, z_0) = (q_0, xz_0)$ and $\delta(q_0, 1, x) = (q_0, xx)$. He explains that these transitions indicate that for every input symbol '1' read while in state $q_0$, an 'x' is pushed onto the stack. He draws a visual representation of the stack growing with 'x's. He then points to the transition $\delta(q_0, 0, x) = (q_1, \epsilon)$, explaining that reading a '0' causes the machine to pop an 'x' and transition to state $q_1$. This establishes a counting mechanism where '1's are pushed and '0's are popped. He notes that the input alphabet appears to be {0, 1}, while the options use {a, b}, implying a mapping between them.
2:00 – 4:41 02:00-04:41
The instructor continues by analyzing the transitions in state $q_1$. He highlights $\delta(q_1, 0, x) = (q_1, \epsilon)$, confirming that '0's continue to pop 'x's. Crucially, he circles $\delta(q_1, \epsilon, x) = (q_1, \epsilon)$, explaining that this epsilon transition allows the machine to pop remaining 'x's from the stack without consuming any input. This means if the number of '1's (pushed) is greater than the number of '0's (popped), the extra 'x's can be cleared. He also notes the final transition $\delta(q_1, \epsilon, z_0) = (q_f, z_0)$ which moves the machine to the final state $q_f$ when the stack is empty (except for the bottom marker $z_0$). He explains that if there are more '0's than '1's, the machine would crash because there is no transition to pop $z_0$ with input '0'. He concludes that the language accepts strings where the count of the first symbol is greater than or equal to the count of the second symbol. He writes $L = \{1^m 0^n \mid m \ge n\}$ on the board and selects option (c) $L = \{a^m b^n \mid m \ge n\}$ as the correct answer, eliminating options (a), (b), and (d) which represent equality, inequality, and less-than conditions respectively.
The lecture demonstrates a standard method for solving PDA language problems: tracing stack operations. By identifying push phases, pop phases, and epsilon transitions, the instructor deduces that the PDA accepts strings where the number of 'a's is greater than or equal to the number of 'b's, corresponding to option (c). The instructor emphasizes that the epsilon transition is the key differentiator that allows for the 'greater than' case, distinguishing it from languages requiring exact equality.