3
$\begingroup$

I'm doing an exercise from my book that says:

Let $r$ and $s$ be arbitrary regular expressions over the alphabet $\Sigma$. Find a simpler equivalent regular expression:

a. $r(r^*r + r^*) + r^*$

b. $(r + \Lambda)^*$

c. $(r + s)^*rs(r + s)^* + s^*r^*$

The book doesn't cover how to simplify regular expressions, so I searched online and I presumed you would use the algebraic laws for regular expressions. I was able to use these laws to come up with something for part a. only:

a. $r(r^*r + r^*) + r^*$

$r(r^+ + r^*)+r^*$

$r(r^+ + r^+ + \Lambda) + r^*$

$r(r^++\Lambda)+r^*$

$rr^+ + r\Lambda + r^*$

$rr^+ + r + r^*$

I don't know how to approach b. or c., because b. has $(r + \Lambda)^*$ and c. has $(r+s)^*$, and I couldn't find how to deal with these. Any hints?

$\endgroup$
  • 1
    $\begingroup$ $(r + \Lambda)^* = r^*$, as you can check (or think what they mean!). And $(r + s)^* = (r^* s^*)^*$ is sometimes handy. $\endgroup$ – vonbrand Feb 25 '13 at 4:33
  • $\begingroup$ Apparently, this is a hard problem, algorithmically. $\endgroup$ – Raphael Feb 25 '13 at 7:30
6
$\begingroup$

A better approach would be to understand what words do these regular expressions represent. For example, what words are in $(r+\Lambda)^*$? They look something like $r_1r_2 \Lambda r_3 \Lambda = r_1r_2r_3$, where $r_1,r_2,r_3$ are generated by $r$. It seems that $\Lambda$ isn't making much of a difference. This should help you simplify $(r+\Lambda)^*$. The same approach works for the other expressions (including the first one, which you haven't simplified completely).

$\endgroup$

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.