Describe a language accepted by a pushdown automata.

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SUMMARY

The language accepted by the described pushdown automata is defined as $\{ab\}$. The automaton begins in the initial state with an empty stack, where the input 'A' pushes 'A' onto the stack and remains in the initial state. Upon receiving 'B', the automaton pops 'A' from the stack and transitions to the final state. Once in the final state, no further transitions are possible, confirming that the only accepted string is 'ab'.

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View attachment 3640

That's the question. I've had a go at drawing a diagram to help me explain it.

View attachment 3641

My understand is that from the start state, an A pushes an A to the stack and stays in the initial state s. Getting a B whilst in state s pops an A from the stack and moves to final state f. Getting a B whilst in state s also pops an A from the stack and remains in state f. And getting an A in the final state whilst A is on the stack remains pops an A from the stack.

How do I describe this language?

I initially thought it might be:

${a^n b^n ∈ {a, b}∗| n ≥ 0}$

but I'm pretty sure that's wrong now.
 

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JaAnTr said:
View attachment 3640

That's the question. I've had a go at drawing a diagram to help me explain it.

View attachment 3641

My understand is that from the start state, an A pushes an A to the stack and stays in the initial state s. Getting a B whilst in state s pops an A from the stack and moves to final state f. Getting a B whilst in state s also pops an A from the stack and remains in state f. And getting an A in the final state whilst A is on the stack remains pops an A from the stack.

How do I describe this language?

I initially thought it might be:

${a^n b^n ∈ {a, b}∗| n ≥ 0}$

but I'm pretty sure that's wrong now.

Hi JaAnTr,

Your drawing looks good to me!

Initially the stack is empty.
That means only $a$ will be accepted, which will also push $a$ on the stack.
After that $a$ will not be accepted any more, since the top of the stack holds $a$.

So we can only continue with $b$, which will pop $a$ from the stack, leaving the stack empty.

From there there is no valid transition any more, since the stack is empty.
Luckily we're in a final state when that happens.
 
How would I go about describing the language?

I think it needs to be in the same form as this $ {a^m {b}^{2m} ∈ {a, b}∗| m ≥ 0}$ but obviously it will be a different language.

Thanks!
 
Last edited:
JaAnTr said:
How would I go about describing the language?

I think it needs to be in the same form as this $ {a^m {b}^{2m} ∈ {a, b}∗| m ≥ 0}$ but obviously it will be a different language.

Thanks!

At the start we can only accept $a$, and after that only $b$, and then nothing anymore.
So I believe that the language is $\{ab\}$.
 
Oh ok thanks, that makes sense. :D
 

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