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 b, if b is the …
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 for this problem. …
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 stack and …
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 just defines Z0 …
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 >= 0 ...
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 accept string in …
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.