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I have to write the language of the below $NPDA$(Non-Deterministic Push Down Automata). enter image description here

I think that from $q_0$ to $q_1$ and then $q_2$, we are actually building the below all the strings of $0$'s and $1$'s of the form $a^nb^n$ with the lenght at least 2.

But there is also a transition from $q_2$ to $q_0$, which makes a cycle, and it just read a $1$ from input string. For example these strings are accepted by this machine, and the point is that in all of them the last character is $0$.

1100 1 10 1 1100

111000 1 1100 1 10

1100

But, unfortunatly I don't know how to write its language accuratly. My idea was to write something like the below language:

$L = \{ w(1^n) z (0^n) | w ∈ L \}$ and $z ∊ \{1,ɛ \}$

But it is not correct. I will be grateful for any help.

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  • $\begingroup$ Isn't a NPDA a non-deterministic pushdown (rather than finite) automaton? $\endgroup$
    – Nathaniel
    Commented Apr 30, 2022 at 16:47
  • $\begingroup$ Yes. You are right, sorry, it was a mistake.@Nathaniel $\endgroup$ Commented Apr 30, 2022 at 17:06

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Consider $L = \{1^n0^n\mid n>0\}$. It seems that the language of your PDA is $L(1L)^*$.

It is not exactly the way you wanted, but I think it is an accurate and short way to write it.

You could be a bit more verbose and write it as:

$$\{1^{n_1}0^{n_1}11^{n_2}0^{n_2}1…11^{n_k}0^{n_k}\mid k \geqslant 1 \wedge \forall i\in \{1, …, k\}, n_i > 0\}$$

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  • $\begingroup$ I think because of the cycle we should use recursion in our language. But are we allowed to use of $*$ sign in our set? $\endgroup$ Commented Apr 30, 2022 at 17:09
  • $\begingroup$ Actually, I wanted to work with an string such as $w$, because we have been told that we can not use of $*$ sign in our set🤔 $\endgroup$ Commented Apr 30, 2022 at 17:13
  • $\begingroup$ Not in the way you want, because if $u$ is a word, $u^*$ is a regular expression that is considered as a set, so something like $\{u^*\mid u \in L \}$ would be a set of sets of words. $\endgroup$
    – Nathaniel
    Commented Apr 30, 2022 at 17:14
  • $\begingroup$ @AylinNaebzadeh I edited the answer. $\endgroup$
    – Nathaniel
    Commented Apr 30, 2022 at 19:08
  • $\begingroup$ But I think each $1$ between $1^n0^n$ is not necessary. $\endgroup$ Commented May 1, 2022 at 5:28

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