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Let $\Sigma=\{0,1\}$ and $L$ be a regular language. Prove that $$Z(L) = \{x \in \Sigma^* : \exists w \in \Sigma^* \ xww \in L \}$$ is a regular language.

I tried to build a NFA based on the DFA that accepts $L$ but failed to do so. I don't know how to ensure the $\, ww \,$ part. Please advise.

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  • $\begingroup$ Related: the language $\{ w \mid ww\in L\}$ is sometimes called the root of $L$. The root of a regular language is again regular. See If L is a regular language then so is √L={w:ww∈L} $\endgroup$
    – Hendrik Jan
    May 21, 2021 at 12:35
  • $\begingroup$ @HendrikJan I don't see how this helps (though it looks close). Any hint? $\endgroup$
    – Dennis
    May 21, 2021 at 12:41
  • $\begingroup$ My aim was to indicate that the $ww$ suffixes can be recognized using a finite state construction. Nathaniel made that explicit in his answer. In the $\sqrt L$ construction you keep $w$, in Nathaniels construction it is omitted from the string. $\endgroup$
    – Hendrik Jan
    May 21, 2021 at 18:41

2 Answers 2

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Let $A = (Q, \delta, q_0, F)$ a DFA that accepts $L$. For $q, q' \in Q$, define:

  • $L_{q,q'} = \{u\in \Sigma^*, \delta^*(q, u) = q'\}$;
  • $L_{q',F} = \{u\in \Sigma^*, \delta^*(q', u) \in F\}$.

It is quite easy (can you prove it?) to see that those languages are regular.

Now the language $\{x\in \Sigma^*\mid \exists w\in\Sigma^*, xww\in L\}$ can be written as: $$\bigcup\limits_{q\in Q}L_{q_0,q}\cdot M(q)$$ Where $M(q) = \left\{\begin{array}{rl}\emptyset&\text{if }\bigcup\limits_{q'\in Q}L_{q,q'}\cap L_{q',F}=\emptyset\\\{\varepsilon\}&\text{otherwise}\end{array}\right.$

$M(q)$ is regular since it is either $\emptyset$ or $\{\varepsilon\}$, so that means that $Z(L)$ is regular.

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  • $\begingroup$ Thank you @Nathaniel, can you please explain the intuition behind $M(q)$? I'm trying to understand what it means. I can't see why the original language can be expressed in that way. $\endgroup$
    – Dennis
    May 21, 2021 at 12:45
  • $\begingroup$ We want to find the states $q$ such that there is a "double" path to a final state of $A$, that is states $q$ such that $\exists w\in \Sigma^*, \delta^*(q, ww)\in F$. Those $w$ are expressed by $\bigcup\limits_{q'\in Q}L_{q,q'}\cap L_{q',F}$, so we just need to now if this set is empty or not, and that is what is expressed by $M(q)$. $\endgroup$
    – Nathaniel
    May 21, 2021 at 12:50
  • $\begingroup$ Thank you. If this set is empty, shouldn't we accept only $x \in L$? In this case the union would accept any path from $q_0$ to $q$, wouldn't it? $\endgroup$
    – Dennis
    May 21, 2021 at 12:56
  • $\begingroup$ If $M(q) = \emptyset$, that means that $q$ is not a good state to end with when reading $x$. In this case, $L_{q_0,q}\cdot M(q) = \emptyset$ (because the concatenation with the empty set is the empty set), so those paths are not considered. $\endgroup$
    – Nathaniel
    May 21, 2021 at 12:58
  • $\begingroup$ Oh, I didn't know that the concatenation with the empty set if the empty set. Now it's clear, thank you! $\endgroup$
    – Dennis
    May 21, 2021 at 13:00
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The operator $Z(L) = \{ x \mid xww\in L \text{ for some } w\}$ takes strings from the original language $L$ but keeps only prefixes $x$ that are obtained by chopping a suffix of the form $ww$.

This is a special application of the right quotient operation: $L/K = \{ x \mid xy\in L \text{ for some } y\in K\}$.

It is known that the family of regular languages is closed under right quotient of arbitrary languages, see Show L1 /L2 is regular. For your problem take $K = \{ww\mid w\in \Sigma^*\}$.

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  • $\begingroup$ Thank you, that's a neat solution :) $\endgroup$
    – Dennis
    May 22, 2021 at 5:40

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