Made Easy Test Series: TOC

Question

The pushdown automata $M=\left \{ \left ( q_{0},q_{1},q_{2} \right ),\left ( a,b \right ) ,\left ( 0,1 \right ),\partial ,q_{0},0,\left \{ q_{0} \right \}\right \}$

$\partial \left ( q_{0},a,0 \right )=\left ( q_{1},10 \right )$

$\partial \left ( q_{1},a,1 \right )=\left ( q_{1},11 \right )$

$\partial \left ( q_{1},b,1 \right )=\left ( q_{2},\lambda \right )$

$\partial \left ( q_{2},b,1 \right )=\left ( q_{2},\lambda \right )$

$\partial \left ( q_{2},\lambda ,0 \right )=\left ( q_{0},\lambda \right )$

Accepts language 

$l=\left \{ a^{n}b^{n}\mid n\geq 0\right \}$

$l=\left \{ a^{n}b^{n}\mid n> 0\right \}$

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Please tell me with these two options which one is correct? I think here $q_{2}$ is accepting state , not $q_{0}$ 

As last transition going from $q_{2}$ to $q_{0}$ and not $q_{0}$ to $q_{2}$

Am I right?

Comments

This is PDA right, and "Set of Final states" has q0 which is also initial state here, so if only these two options are there we don't need to even check anything as it will accept empty string i.e, a^0b^0. So the answer would definitely be the first one

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 "Set of Final states" has q0 

how?

last transition is q2 to q0, and not q0 to q2 

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$\left ( Q,\sum,\tau ,\partial ,q_{0},Z_{0},F\right )$

So, here $F=q_{0}$

right?

I thought $q_{2}$ final state

but that is not the case, right?

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here goes q2 to q0, so accept the empty state

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Yes exactly what I was going to say too😃

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thanks a lot

Made Easy diagram showing q2 final :(

So, getting in this trouble