# pushdown automata

140 views
what will be the pushdown automata for the language, L=a^n b^m where n=2m+1.
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I think it will be the pda 0

Sir, this also can be pda of this??? @Prateek Raghuvanshi 0
Connecting state q1 with final state will take care when m=0..
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@MiNiPanda if m>0 then it is correct right??

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@arya_stark it is not correct because in language a's are double than b's ,so we have to take care of that ,see my dpda.

## Related questions

1
71 views
Let P be a non-deterministic pushdown automaton (NPDA) with exactly one state, q, and exactly one symbol, Z, in its stack alphabet. State q is both the starting as well as the accepting state of the PDA. The stack is initialized with one Z before the start of the operation of the PDA. ... but L(P) is not necessarily ∑* Both L(P) and N(P) is necessarily ∑* Neither L(P) nor N(P) are necessarily ∑*
2
80 views
given a pushdown automata that receives L by getting to an accepting state, how can a pushdown automata be built, so that it accepts L*? (might use a “double bottom” if needed)?? i don’t know how to solve it and would appreciate any kind of help! studying for exam and must learn how to solve it Options: $\left \{ \left ( b^{n}ab^{n}a\right )^{m} | n,m \geq 0 \right \}$ $\left \{ \left ( b^{n}ab^{n}a\right )^{m} | n,m \geq 0 \right \} \cup \left \{ b^{n} | n\geq 0 \right \}$ $\left \{ \left ( b^{n}ab^{n}\right )^{m}a | n,m \geq 0 \right \}$ $NONE$