Finding simpler equivalent regular expressions

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?

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

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).