Simple regular expression problem

Question

new to automata theory and this is quite difficult for me so I'd appreciate a hint in the right direction if possible.

We want to find a regular expression for the language $L$ over $\Sigma = {a,b,c}$ where $L$ is the language of all strings of even length, contain $a$ at least once, and after the final $a$ letter there are no $b$s.

for example - $bcbabaca$ is a valid word.

Best thing I could come up with is somewhat akin to $R = ((a+b+c)(a+b+c))^*a((a+b+c)(a+b+c))^*$ and it has to be something like that since we need to control the length of the word to make it even, but still its no good and im having difficulties.

It's sometimes easier to start with an automaton and then translate it to a regular expression. — adrianN, Jun 16 '16 at 20:20
@adrianN It's certainly an option, but it tends to give rather complicated regular expressions that are impossible to check for correctness (and which are likely to be wrong because there are a hundred and one mistakes you could make on the way). — David Richerby, Jun 16 '16 at 20:30

score 2 · Accepted Answer · answered Jun 16 '16 at 20:25

You're almost there.

$((a+b+c)(a+b+c))^∗$ gives you strings of even length.

$((a+b+c)(a+b+c))^∗a$ gives you strings of even length plus an $a$. That is, odd-length strings that contain at least one $a$ and end with an $a$.

Now, we're told that there are no $b$s after the last $a$. And there are certainly no $a$s after the last $a$, because that's what "last" means! So there can only be $c$s after the last $a$.

So, the answer is going to have to look something like $((a+b+c)(a+b+c))^∗ac^*$. That's not quite right, because it doesn't force the whole string to be even-length and it does force the last $a$ to be in an odd-numbered position in the string (whereas, hint, it should be able to be even or odd).

I'm going to stop there because you showed enough in the question that I think you can finish this yourself, now.

Simple regular expression problem

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