Designing a PDA for a^n b^m (n ≥ 1, m greater than n) | Automata Lecture 11

How do you build a Pushdown Automaton that accepts a^n b^m only when n ≥ 1 and m is strictly greater than n? In this problem-solving session we design it step by step. The key idea: push a marker for every 'a', pop one for each matching 'b', and then make sure at least one extra 'b' remains on the stack before accepting. That leftover 'b' is what guarantees m is greater than n rather than just equal to it. We walk through the stack operations, the transition design, and a trace on sample strings so you can see exactly why the machine accepts valid inputs and rejects everything else. This is Lecture 11 in the Theory of Automata and Formal Languages series, aimed at undergraduate students. It builds on earlier lectures covering context-free languages and basic PDA construction, so by the end you should be comfortable adapting this technique to similar a^n b^m style problems. What you will learn: Translating the condition n ≥ 1 and m greater than n into stack behavior Pushing and popping symbols to compare two counts Handling the 'at least one extra b' requirement cleanly Tracing the PDA on accepting and rejecting strings SUBJECT: Theory of Automata and Formal Languages VIDEO TYPE: Problem Solving TARGET LEVEL: Undergraduate INSTRUCTOR: M. Imran Shafi Full playlist:    • Design a PDA for a^n b^(2n) | Automata Lec...   Related video:    • Design a PDA for a^n b^(2n) | Automata Lec...   If this helped, like and subscribe for the rest of the series.