Constructing a PDA for $L=${$\exists i,k\in \mathbb{N} : |w|=2k, w_i \neq w_{k+i}$}

Asked Apr 25 '23 at 11:43

Active Apr 25 '23 at 12:50

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I have the main idea, yet I'm uncertain on how to construct this PDA (in terms of states, transitions)

We can assume the alphabet $\Sigma$ is {$0,1$}, proving for $\Sigma=${$0,1$} is a sufficient proof that 'covers' all $\Sigma$'s such that $|\Sigma|\geq 2$.

I want to use the fact that the language $L=${$w\in\Sigma^* : w=u_1\circ u_2 , |u_1|=|u_2| , u_1\neq u_2$} can be constructed using a PDA, and that since $u_1\neq u_2$ there must be atleast one character which is in one sub-word, and isnt in the other.

Edit:

After some playing around, I've constructed 2 PDAs which I believe one might be correct, however I'm not sure how to 'test' them. The PDAs are as such:

And:

Edit 2:

I've managed to construct this PDA (lets call it PDA3):

I feel like PDA3 is the right answer, simply because I've tried some examples and all seem to work/fail according to $L$, yet I have no way of formally proving it.

Is there a method to proving? except for experimenting with specific examples.

edited Apr 25 '23 at 12:50

asked Apr 25 '23 at 11:43

Aishgadol

Does this answer your question? How do we check $x ≠ y$ in $PDA$ for $L = {xy | x, y \in (0 + 1)^*, |x| = |y|, x ≠ y}?$ – Hendrik Jan Apr 25 '23 at 15:45

Constructing a PDA for $L=${$\exists i,k\in \mathbb{N} : |w|=2k, w_i \neq w_{k+i}$}

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