PDA to accept a language of strings containing more a's than b's
Dec 13, 2015 · Your problem of "more a's than b's" can be solved by PDA. All you have to do is: When input is a and the stack is either empty or has an a on the top, push a on the stack; pop …
Pushdown Automaton (PDA) for a^(2n) b^n - Stack Overflow
Jun 1, 2025 · Deterministic push down automata for a^2nb^n n>=0 Bypass altetnate a's and push rest of a's
c++ - How to design a pushdown automata - Stack Overflow
Nov 19, 2012 · How would i design a PDA that accepts balanced parenthesis and brackets for instance ([][]), I am having a hard time getting started. I need help writing transition functions …
NPDA for L= {w ∈ {a,b}*: number of a's is twice the number of b's}
May 29, 2020 · Push down automata can be done for a^2nb^n.push down automata is nothing but the finite automata with memory (can be stack). Here, for every two a's push one 'a' into the …
can i push two symbols to the stack of a pushdown automata?
Jun 12, 2020 · What i wanted to understand is if this automaton pushes two symbols at once or just one, because in the formal definition, there isn't anything saying it pushes two symbols, it …
How to construct a pushdown automata for L= {a^nb^m where …
Nov 12, 2019 · How to construct a pushdown automata for L= {a^nb^m where n<=m<=2n}? Asked 6 years, 1 month ago Modified 5 years, 4 months ago Viewed 9k times
Pushdown Automata for Palindrones - Stack Overflow
Pushdown Automata for Palindrones Asked 10 years, 1 month ago Modified 10 years, 1 month ago Viewed 9k times
Push Down Automata for the language L = { a^i b^j c^k | i, j, k >
May 19, 2024 · Push Down Automata for the language L = { a^i b^j c^k | i, j, k >= 0 and j = i + 2k } Asked 1 year, 7 months ago Modified 1 year, 4 months ago Viewed 2k times
automata - Design a PDA of all strings of 0's and 1's so that the ...
May 16, 2015 · While practicing for my final exams I found this question in Automata Theory, Language and Computation by J. Hopcroft, R. Motwani, J. Ullman on page 222. PDA should …
Can we make pda for a^n b^n+1? - Stack Overflow
May 29, 2022 · The Context free grammar of a^n b^n+1 would be: S -> aS'bb S' -> aS'b | empty Since the CFG exists, the PDA is also possible for this language.