Okay, so I come across a question: What language is represented by this regular expression:
$$(((a*b)b) \cup b)$$
An example given prior to this is: L(anguage) = $\{w | w \in \{0,1\}\}$
L(anguage) = $0*10*010*(10* \cup E)$
I am completely lost with this o.o The only thing i can think of here is... $L = \{ab, b, bb\}$ But that is a wild guess.
On
* - zero or more occurrences of followed symbol
UNION - $or$ operation
so regular expression can generate bb, abb, aabb, aaabb, b (because of $or$ operation),...
good brief explanation
Regular expression $baa \in a^*b^*a^*b^*$: is that true or false?
complete topic about regular expressions with many examples
http://www.cs.princeton.edu/courses/archive/spr05/cos126/lectures/18.pdf
On
You have the regular expression $(((a^*b)b)\cup b)$. The subexpression $(a^*b)$ generates the words $b,ab,aab,aaab$, etc., because $a^*$ generates any string of $a$’s, including none at all. These words can be described in English as the words over the alphabet $\{a,b\}$ that contain exactly one $b$ and have that $b$ at the very end. Alternatively, they are the words that start with any number of $a$’s (including none) and then have a single $b$.
Let’s build up to the next larger subexpression: $((a^*b)b)$ takes any word generated by $(a^*b)$ and tacks a $b$ on the end, so we now get $bb,abb,aabb,aaabb$, etc.; these are the words that start with any number of $a$’s, possibly none, and then have $bb$. Alternatively, they could be described as the words that end in $bb$ and have nothing but $a$’s before that.
The subexpression $b$ on the right of the union of course generates only one word, $b$.
Finally, the union in $(((a^*b)b)\cup b)$ gives you everything generated by $((a^*b)b)$ or by $b$, so you get every word that ends in $bb$ and has nothing but $a$’s before that (if anything), and in addition you get the word $b$.
You are trying to understand the regular expression $(((a^∗b)b)∪b)$ = $(a^*bb) \cup b)$. If we split this expression up we get two expressions $a^*bb$ and $b$ with the union operator. As you know the union operator is "or", so our expression can be formulated as $a^*bb$ or $b$. So L(Expression) = {w|w $\in a^*bb$ or $w \in b$} ("The language of E, with E our expression, are all strings that are in the language generated by the expression $a^*bb$ or the expression $b$.)
Now the only string that is generated by the expression $b$ is $b$ itself, so we have our first word. Now to look at the expression $a^*bb$, $a^*$ implies zero or more a's and $bb$ just implies two b's. So we can finally define L(E) = {w|w has zero or more a's followed by two b's or w is b}. Example strings are $abb, bb, aaaabb, b, ...$ but not $bbb, ba, ...$. Hope this makes it a bit more clear.