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I have a PDA A = ({q0, q1}, Σ = {a, b}, Γ = {a}, δ, F = {q1}), with these transition functions δ:

((q0,a,ε),(q0,a));

((q0,b,ε),(q0,a));

((q0,a,ε),(q1,ε));

((q1,a,a),(q1,ε));

((q1,b,a),(q1,ε)).

The exercise asks me to explain the structure of the words accepted. I have designed the PDA but i'm having problems understanding what kind of words the PDA accepts. Any kind of help is really appreciated.

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  • $\begingroup$ I think it accepts all words with length bigger than $1$. This is probably not what you wanted this PDA to do... $\endgroup$
    – nir shahar
    Commented Oct 31, 2021 at 14:40
  • $\begingroup$ This PDA is given to me by the exercise, i have to design it and explain what kind of words it accepts $\endgroup$
    – rsky
    Commented Oct 31, 2021 at 15:17

1 Answer 1

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I suppose that your acceptance mode is by final state and empty stack.

The loops around $q_0$ add one stack symbol for each input letter that is read. After this, the third transition reads an $a$ without touching the stack. Finally, all transitions around $q_1$ remove one stack symbol for each input letter that is read.

To reach an empty stack in $q_1$, the number of loops around it has to be the same as around $q_0$. Exactly in the middle there is one $a$ that takes you from $q_0$ to $q_1$. Thus the language would be $$\{w_1\cdot a \cdot w_2: |w_1| = |w_2|\}.$$

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