## With Σ = {a,b}, give a dfa for L= w1aw2 : |w1 |≥ 3, |w2 |≤ 5}

I racked my brain,I saw other people’s solutions and it don’t make sense. I think my biggest problem is I don’t know when one string ends,like for example w1 is >=3 and it can have how many ever b’s or a’s as long as it’s 3 and above,so when do we get to have the First a that’s after the W1?how do I make sure it’s not a part of the W1? that’s whats so confusing the W1 string can have as many a’s as it likes which means it should have an a loop,but you can’t have an a loop cause YOU NEED 1 A EXACTLY after the W1 string. this is a catch 22,damned if you don’t damned if you do..help please I spent an hour on this to no avail. I tried every possible thing and saw people’s solutions and I feel like when they solved it they are either dismissing what I mentioned above,or something is going COMPLETELY over my head.

## Computing automaton for $L(A) / L(B)$ gives ones for $A,B$

I’m trying to figure out whether infinite language change the answer.

Show that the following language is decidable: $$L=\{\langle A,B \rangle : \text{ A,B are DFAs, L(B) is finite, and L(A)/ L(B)=L(0^*1^*) }\}.$$

(I am talking about right division.)

We know how to check whether the language of a DFA is finite, and given two DFAs, we know how to check whether their languages are equal. The algorithms I know to the above problems uses the DFA’s, so it is necessary having the DFA’s in order to decide those problems.

I’m trying to figure out whether $$|L(B)|=\infty$$ changes the answer. To the best of my understanding, because $$|L(B)|<\infty$$, we can explicitly construct a DFA that accepts $$L(A)/ L(B)$$, whereas if $$L (B)=\infty$$ all we know is about the existence of $$DFA$$ that accepts $$L(A)/ L(B)$$.

However, even if $$L(B)$$ is an infinite language, since there is a finite number of DFAs, one of which accepts $$L(A) / L(B)$$, I can certainly know that there is a Turing machine that decides the language $$L$$. Right?

## If I can efficiently uniformly sample both $A$ and $B\subset A$, can I efficiently uniformly sample $A-B$?

As posed in the question; the statement naively seems like it should be self-evident but there are no algorithms that come immediately to mind. Suppose I have some domain $$A$$ (in my case a subset of $$\mathbb{Z}^n$$ for some $$n$$, but I would expect any answer to be independent of that structure), and a domain $$B\subset A$$. If I have an efficient algorithm for uniformly sampling from $$A$$, and an efficient algorithm for uniformly sampling from $$B$$, is there any way of ‘combining’ these to get an efficient algorithm for uniformly sampling $$A-B$$? I can certainly rejection-sample, but if $$|A-B|\ll|A|$$ then there’s no guarantee that that will be efficient.

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## Is this language L = {w $\in$ {a,b}$^*$ : ($\exists n \in \mathbb{N}$)[$w|_b = 5^n$]} regular?

Let’s say we have the language L = {w $$\in$$ {a,b}$$^*$$ : ($$\exists n \in \mathbb{N}$$)[$$w|_b = 5^n$$]}. I want to know if this is a regular language or not. How do I go about doing this? I’m familiar with the Myhill-Nerode theorem but I don’t know how to apply it.

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## Design a CFG that generates the language { x in {a,b}* | the length of x is odd and its middle symbol is a b }

I am trying to design a context-free grammar that generates the language { x in {a,b}* | the length of x is odd and its middle symbol is a b }.

This is really confusing me, I’m having trouble with making sure that b is always in the middle. Any help?

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## Convert PDA to CFG – Language of Palindromes over $\{a,b\}^+$

I tried to convert the following PDA $$M$$ to a CFG and got the following results. However I’m not quite sure if I did the algorithm the right way. I feel like I maybe misunderstood some the algorithm when it comes to the $$A_{p,q} \to xA_{r,s}y$$ part. Could you help me to clear things up?

$$G = (V,\Sigma,P,S)$$ with \begin{align*} V = \{&A_{0,0},A_{0,1},A_{0,2},A_{0,3},A_{0,4},A_{1,0},A_{1,1},A_{1,2},A_{1,3},A_{1,4},A_{2,0},A_{2,1},A_{2,2},A_{2,3},A_{2,4},\ &A_{3,0},A_{3,1},A_{3,2},A_{3,3},A_{3,4},A_{4,0},A_{4,1},A_{4,2},A_{4,3},A_{4,4},S\}, \end{align*} $$\Sigma = \{a,b\}$$, $$S = A_{0,4}$$ and the following productions: \begin{align*} S &\to A_{0,4}\ A_{0,0} &\to \epsilon\ A_{1,1} &\to \epsilon\ A_{2,2} &\to \epsilon\ A_{3,3} &\to \epsilon\ A_{4,4} &\to \epsilon\ A_{0,0} &\to A_{0,0}A_{0,0} \vert A_{0,1} A_{1,0} \vert A_{0,2}A_{2,0} \vert A_{0,3}A_{3,0} \vert A_{0,4}A_{4,0}\ A_{0,1} &\to A_{0,0}A_{0,1} \vert A_{0,1} A_{1,1} \vert A_{0,2}A_{2,1} \vert A_{0,3}A_{3,1} \vert A_{0,4}A_{4,1}\ A_{0,2} &\to A_{0,0}A_{0,2} \vert A_{0,1} A_{1,2} \vert A_{0,2}A_{2,2} \vert A_{0,3}A_{3,2} \vert A_{0,4}A_{4,2}\ &\vdots\ A_{4,4} &\to A_{4,0}A_{0,4} \vert A_{4,1} A_{1,4} \vert A_{4,2}A_{2,4} \vert A_{4,3}A_{3,4} \vert A_{4,4}A_{4,4}\ A_{0,4} &\to A_{1,3}\ A_{1,3} &\to aA_{1,3}a\ A_{1,3} &\to bA_{1,3}b\ A_{1,3} &\to aA_{1,1}\ A_{1,3} &\to bA_{1,1} \end{align*}

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## What does {a,b}* for DFA’s mean?

For instance when the question contains {a,b}* does this mean that the DFA must have at least 1 a and 1 b on top of whatever conditions it has? For example a DFA that accepts {w E {a,b}* : w contains bbb} should it reject the actual string bbb because it does not contain an a?

## Turing Machine where the number of a’s and b’s are either both even or both odd over {a,b}

Turing Machine where the number of a’s and b’s are either both even or both odd.

## Use the pumping lemma for context free languages to prove L = {w#w | w \in {a,b}*} is not context free

I know the basics of using the pumping lemma for CFG to prove a language L is not context-free, however, the # symbol seems to be throwing me off or my understanding is not complete.

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## Prove that there isn’t a DFA characterized by (a*)+(b*) for which |Q| = 3

Let $$R=(a*)+(b*)$$ be a regular expression. Prove that there cannot exist any DFA $$M=(Q,\Sigma,\delta , q0, A)$$ such that $$|Q| = 3$$ and $$L(M)=L(R)$$.

The problem is, I think IT IS possible to construct such a dfa that has 3 states in total. Am I missing something?