0
$\begingroup$

Let $\Sigma = \{\alpha, \beta \}$ and $L = \{\alpha (\beta \alpha)^*\}$

I don't understand the notation for the language $L$. Does it mean that every string must start with $\alpha$, and then have an arbitrary number of ($\beta\alpha$) after it?

$\endgroup$
1
  • $\begingroup$ The notation is incorrect. It should be $L = \alpha (\beta \alpha)^*$. $\endgroup$
    – J.-E. Pin
    Jun 22, 2017 at 7:40

1 Answer 1

1
$\begingroup$

That's exactly right. And the 'arbitrary number' is 0 or more. So, just having $\alpha$ would be a word in this language.

$\endgroup$
1
  • $\begingroup$ Is there a different notation to show that any order can be used? Such as $\beta$ beginning before $\alpha$ $\endgroup$
    – Vermillion
    Jun 21, 2017 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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