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We have two languages: $L_1,L_2$. We know that $L_1L_2$ is regular language, so my question is if $L_2L_1$ is regular too?

I try to find a way to prove it...

I can't assume of course that $L_1,L_2$ are regular...
So I look for a way to prove it.

I'd like to get any hint!

xskxzr
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stud1
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4 Answers4

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I was posting only a hint, then I saw other full answers, so this is a full (hidden) succinct solution :-)

Let $L_1 = \{ 1^p \mid p \text{ is prime}\}$, $L_2 = \{ 1^* 0 \}$; we have $L_1 L_2 = \{ 11^+ 0\}$ which is regular, but $L_2 L_1 = \{ 1^* 0 1^p \mid p \text { is prime}\}$ which is not regular.

Vor
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No, $L_2L_1$ is not necessarily regular.

Let $L_1 = \{0,1\}^*$, which is regular, and $L_2 = \{1\} \cup \{0^n1^n\mid n\geq 1\}$, which is not. Then $L_1L_2$ is the set of all strings ending with $1$, which is regular, but $L_2L_1$ is the set of all strings that either begin with $1$, begin with a nonzero number of $0$s followed by at least as many $1$s. This language is not regular, since its intersection with $\{0^m1^n\mid m,n\geq 1\}$ is $\{0^m1^n\mid 1\leq m\leq n\}$, which is non-regular.

David Richerby
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This is not a hint, but a full answer. Don't read on if you're still trying to solve.

There is no need for $L_2\cdot L_1$ to be regular.

Let $A$ be a unary (non-regular) language such that $A\cdot A$ is regular. Such languages can be found in the post here. Assume $A$ is over the alphabet $\{a\}$.

Define $L_1=\{b\}\cdot A$ and $L_2=A\cdot \{b\}$. Then, you get $L_1\cdot L_2=\{b\}\cdot A^2\cdot \{b\}$, which is regular. However, $L_2\cdot L_1=A\cdot \{bb\}\cdot A$, which can be easily proven to be non-regular, based on $A$ being non-regular.

Community
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Shaull
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The following rules define the language associated with any regular expression. Rule 1 The language associated with the regular expression that is just a single letter is that one-letter word alone and the language associated with A is just {A}, a one-word language. Rule 2 If r, is a regular expression associated with the language L, and r 2 is a regular expression associated with the language L2 then,

(i) The regular expression (rl) (r2) is associated with the language L, times L 2. language (r, r2) = L1L 2 (ii) The regular expression r, + r2 is associated with the language formed by the union of the sets L1 and L2. language (rl + r2) = L, + L2 (iii) The language associated with the regular expression (rl)* is LI*, the Kleene closure of the set LI as a set of words. language (rl*) = L1*

Shashi Kumar
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