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.

# Tag: $ab$

## 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.

## 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.

## 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?

## 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*}

## 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.

## 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?