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So I've been scratching my head over this problem for a couple of days now. Given some language $A$ and $B$ that is regular, show that the language $L$ which consists of all strings in $A$ whose length is equal to some string in $B$ is a regular language.

In equation form:

$$L = \{x \in A \mid \exists y \in B \text{ s.t. } |x| = |y| \}$$

My initial thought was to try and come up with some DFA for both languages $A$ and $B$ and map the two states to each other and hopefully get a 1:1 ratio that way I can generate a new DFA which proves that $L$ is regular. But then I realized that $A$ and $B$ don't have to be over the same set of symbols.

I think the correct way to solve this is to use the closure properties of regular language, but I'm not sure of how to begin/use the properties for "lengths" of strings instead of strings themselves.

Could someone point me in the right direction?

Raphael
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Jubilous
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3 Answers3

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Remember (or come up with) the proof for

$\qquad \displaystyle L_1, L_2 \in \mathsf{REG} \implies L_1 \cap L_2 \in \mathsf{REG}$.

Do you see how to modify the proof for your setting?

Abstracting the equality of lengths thing, come up with a construction for an automaton for

$\qquad \displaystyle L_l = \{ w \in \Sigma^* \mid \exists x \in L.\, |x|=|w|\}$

for given, arbitrary $L \in \mathsf{REG}$ over $\Sigma$.

Do you see the connection?

Now note that $L = A \cap B_l$.

Raphael
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Hints: Let's assume you know all the different lengths of words in $B$, $len(B) = \{ \ell_1, \ell_2, \ell_3, ... \}$. For the time being, let's assume it is finite.

Can you use this knowledge to construct a DFA for $A$?
(hint: intersection, or "cross-product" construction)

Does the alphabet of $B$ even matters?

Next, it might be the the set of lengths $len(B)$ is infinite. Then look at this question which should resolve that matter as well.

Community
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Ran G.
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The closure property way, so no (explicit) automata. Regular languages are closed under morphisms, inverse morphisms and intersection (with regular languages).

Let $\Sigma_A$ and $\Sigma_B$ be the alphabets of $A$ and $B$. Let $h_X:\Sigma_X^* \to \{1\}^*$ be the morphism that maps every letter $a\in\Sigma_X$ to $1$. Then $h_B(B) \subseteq \{1\}^*$ codes the length set of $B$, and $h_A^{-1}(h_B(B))$ consists of (all) strings that are of the same length as strings in $B$, but over the alphabet of $A$. Finally, we observe that $L= A\cap h_A^{-1}(h_B(B))$.

Now for the bonus. It also works for context-free $A$ and $B$. We need however an additional property: if $B$ is context-free then $h_B(B)$ is regular (!) as all 'unary' (=single letter alphabet) context-free languages are regular, a consequence of Parikh's theorem. Thus also $h_A^{-1}(h_B(B))$ is regular, and $L$ is context-free.

Hendrik Jan
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