Let $L = L_1 \cap L_2 $, where $L_1$ and $L_2$ are languages as defined below: $L_1= \left \{ a^m b^mca^nb^m \mid m,n \geq 0 \right \}$ $L_2=\left \{ a^i b^j c^k \mid i,j,k \geq 0 \right \}$ Then L is Not recursive Regular Context free but not regular Recursively enumerable but not context free. […]

Draw a finite state machine which will accept the regular expression: $(a^2)^* + (b^3)^*$ In particular, I am confused by the $+$ sign, what does it exactly mean? Most literature I could find about $+$ is $a^+$, which means 1 or more $a$; but here it is clearly not the same meaning.

I’ve been battling the following two questions for more than a day today. Write a regular expression (comprised of {a, b}) that contain at least two b and do not contain abb. Write a regular expression (comprised of {a, b}) that contain abba and do not contain baa. For the first part I wrote a […]

This is a NFA, I have been working to covert it to a regular expression. After I’am done, I arrive at an expression as follows $$ \left(((a\cup b)a^*b) (ba^*b)^*a\right)^* \left(((a\cup b)a^*b) (ba^*b)^*\right) $$ Now, I really doubt if the answere is correct if anyone could help. I used the state removal method.

Can anyone help me with this question: I know it before, but I have tried to solve it myself and didnt succeed. what is the regular expression for this language: L=all words that have 00 or 11 but not both. Thank you!

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