2
$\begingroup$

I am having hard time proving that the following language,comprised from two regular languages $L_1,L_2$(over the same $\Sigma$)is indeed regular:

$$L^\frown = \{ w\in \Sigma^* | w=u\sigma_1\mu_1...\sigma_n\mu_nv\}$$

  • $u,v \in \Sigma^*$

  • $0\leq n$, for every i($1\leq i \leq n$): $\sigma_i,\mu_i \in \Sigma$

  • $\sigma_1...\sigma_n \in L_1$

  • $\mu_1...\mu_n \in L_2$

I don't understand how to prove if because of the suffix and prefix(u,v accordingly).

If it weren't for u,v what I think I would've done is building an automaton(Deterministic finite automaton): $A=\left(Σ,Q1xQ2x\left\{1,2\right\},\left(q01,q02,2\right),F1xF2x2,δ\right)$, Ending in accepting iff both automatons that accept each language end in an accepting state, and $L_2$ needs to be after $L_1$ from the language's description. However, I don't know how to deal with the u,v in the beginning and in the end. I am not sure how to configure it correctly to prove $L^\frown$ is regular.

Would very much appreciate your assistance with it.

$\endgroup$
1
  • $\begingroup$ I must be missing something. Isn't $L⌢$ just $Σ∗$? Set $u = w$ and $n =0$. $\endgroup$ Oct 23, 2019 at 7:39

1 Answer 1

2
$\begingroup$

Your language is the concatenation $\Sigma^* L \Sigma^*$, where $L$ is the so-called perfect shuffle of $L_1$ and $L_2$. There are at least two proofs that the perfect shuffle of two regular languages is regular on this site: using automata and using closure operations.

If you are hell-bent on constructing an automaton for $\Sigma^* L \Sigma^*$, start with one for $L$, then follow the automata proof of closure under concatenation (which constructs an NFA).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.