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4 votes
Accepted

$L'=\left \{ z^Rx : xyz\in L \right \}$ is regular

Let $A = (\Sigma, Q, q_0, \delta, F)$ be a DFA that recognizes $L$. For states $p, q\in Q$, let $A_{p, q}$ denote the DFA obtained from $A$ by letting $p$ be the initial state and $q$ be the only ...
Bader Abu Radi's user avatar
0 votes

Myhill-Nerode sentence and the relation $R_L$

Look at a string x. If x contains some letter twice, then xy is in the language for every string y. Or x consists of k different letters in any order. Then xy is in the language if either y contains ...
gnasher729's user avatar
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1 vote
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Confusion regarding the conversion of epsilon-NFAs to non-epsilon NFAs

It is not clear where you got stuck, but generally speaking, every $\epsilon$-NFA $A = (\Sigma, Q, Q_0, \delta, F)$ can be converted into an equivalent NFA $B$ in polynomial time. Here is one way to ...
Bader Abu Radi's user avatar
3 votes
Accepted

Myhill-Nerode sentence and the relation $R_L$

Note that $u, v$ and $w$ can contain any letter. Therefore, $L$ is the language of all words that contain some letter at least twice. In particular, a word that contains some letter 3 times is in $L$. ...
Bader Abu Radi's user avatar
0 votes

Convert finite automaton to regular expression

3rd one will be correct, you can also solve this by minimal sting which is abd and 3rd option is the one that generates abd
user173344's user avatar
1 vote
Accepted

What is the connection between a regular language's pumping number, and the number of states of an equivalent deterministic automaton?

Usually, we care about a pumping constant as the size of every automaton (including nondeterministic ones) for the language can be used as such. If your goal is to find the minimal natural $p$ that ...
Bader Abu Radi's user avatar
2 votes

Complexity of deciding if a DFA is counter-free

I've found it in the literature [1], it's PSPACE-complete even for DFAs. Sang Cho, Dung T. Huynh Finite-automaton aperiodicity is PSPACE-complete, Theoretical Computer Science, Volume 88, Issue 1, ...
Nicola Gigante's user avatar

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