# Tag Info

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### Why is { w | |w| mod 3 = #_a(w) mod 3 } a Regular Language?

$\newcommand{\m}{\operatorname{\%}}$ Let $d(w)=(|w|-\#_a(w))\m3$, where $n\m 3$ is the remainder of dividing $n$ by $3$ as defined in almost every programming language. Note $L=\{ w\mid d(w)=0\}$. ...

### Why is { w | |w| mod 3 = #_a(w) mod 3 } a Regular Language?

The following language is not regular $L = \{a^n b^m c^n \mid m = n \bmod 2\}$. To see that $L$ is not regular, suppose towards a contradiction that $L$ is regular and let $p$ be its pumping length. ...

### Why is { w | |w| mod 3 = #_a(w) mod 3 } a Regular Language?

So far every language that I saw containing modulo was a regular language. As John L. notes, that's a very good observation. Indeed, any language where the only constraint on words is that some ...

### Why is { w | |w| mod 3 = #_a(w) mod 3 } a Regular Language?

Lucky enough, your case is quite easy. The language is defined by the rule "total number of letters, modulo 3, equals total number of a's, modulo 3". This is equivalent to "number of ...
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### DFA and NFA with 2 Substrings

Find an automaton for $L_1= \{uopv\mid u,v\in \Sigma^*\}$ and an automaton for $L_2 = \{upqv\mid u,v\in\Sigma^*\}$ (this should be easy enough). Then, you can compute the product automaton of the two ...
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### Why 2- way DFA is equivalent to NFA (and thus DFA)?

The language $L=\{ (u\#,v\#) \mid |u|=|2v|\}$ from your question is actually a two-dimensional language, that is a relation between two strings, each written on their own input tape. In that way the ...
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### Prove that if C is a regular language, then the language $\{x x^R : x\in C\}$ is context-free

Recall that every finite state automaton can be changed into a rightlinear grammar which has productions like $X\to aY$ and $X\to \varepsilon$. Your language can be generated using the same technique,...