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Let $L$ be a regular language over alphabet $\Sigma$. $L$ is the result of merging $2$ languages letter by letter that is for $a_1a_2...a_n\in L_1, b_1b_2...b_n\in L_2, L=a_1b_1a_2b_2...a_nb_n$. $\epsilon \in L \iff\epsilon \in L_1,L_2$. Does this mean that $L_1$ is necessarily regular?

At first I tried to prove that this is true using closure properties of regular languages before finding out that the claim is not true (e.g. $L_1=\{a^ib^i|i\ge 0\}, L_2=\{ab\}$).

I'd like to understand though what's wrong with my proof:

1) Define homomorphism $h:\Sigma\to \Sigma^*\cup \Sigma'^*$ as follows: $h(\sigma)=h(\sigma')=\sigma$ for $\sigma,\sigma'\in \Sigma.$

2) Let $X=h^{-1}(L)\cap(\Sigma\Sigma^*)$

3) Define another homomorphism: $f_1:\Sigma^*\cup \Sigma'^*\to \Sigma$ like this: $$ f_1(\sigma)=\sigma\\f_1(\sigma')=\epsilon $$ 4) Define another homomorphism: $f_2:\Sigma^*\cup \Sigma'^*\to \Sigma$ like this: $$ f_2(\sigma')=\sigma'\\f_2(\sigma)=\epsilon $$ Then $L_1=f_1(X), L_2=f_2(X)$. Now if either $L_1$ or $L_2$ is not regular than it's contradiction because by closure properties they should've been regular. Where is my mistake?


EDIT: following one of the comments, the problem with the attempt to use closure properties is that apply the homomorphisms $f_1,f_2$ on $X$ we may get a different language from the original one.

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    $\begingroup$ You have a counterexample in your question. What do you learn when you apply this counterexample in the construction of your suggested proof? $\endgroup$ – Hendrik Jan Jan 29 at 14:42
  • $\begingroup$ Now I see, if I apply my construction I get a different language. $\endgroup$ – Yos Jan 29 at 14:56
  • $\begingroup$ @Yos, write an answer? $\endgroup$ – Apass.Jack Jan 29 at 18:51
  • $\begingroup$ @Apass.Jack I added an edit section in the OP, the answer seems to be so minimal that I don't think it warrants an answer. Let me know if the rules encourage adding an official answer in such situations. $\endgroup$ – Yos Jan 29 at 18:56
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    $\begingroup$ For this kind of exercises it is often useful to check for trivial cases, first. E.g. one might assume the conjecture, choose (say) $L_1$ to be the empty (or full) language, and simplify the statement accordingly. Sometimes, a counterexample can now be seen. $\endgroup$ – chi Jan 30 at 11:28
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Your construction outputs the language $$ \{ w \in L_1 : \exists z \in L_2 \text{ s.t. } |z|=|w| \}. $$ In words, it outputs those words in $L_1$ whose length matches the length of some word in $L_2$. In general, this could be much smaller than $L_1$.

In particular, if $L_2$ is a finite language, then the result of merging $L_1$ and $L_2$ is finite, and so regular, for every language $L_1$.

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