如果一种正则语言只包含 Kleene star，那么它是否可能来自两种非常规语言的串联？

Question

I want to know that given a regular language L that only contains Kleene star operator (eg (ab)*), is it possible that L can be generated by the concatenation of two non-regular languages? I try to prove that L can be only generated by the concatenation of two regular languages.

Thanks.

Answer 1

This statement is false. Consider these two languages over Σ = {a}:

L ₁ = { a ⁿ | n is a power of two } ∪ { ε }

L ₂ = { a ⁿ | n is not a power of two } ∪ { ε }

Neither of these languages are regular (the first one can be proven to be nonregular by using the Myhill-Nerode theorem, and the second is closely related to the complement of L ₁ and can also be proven to be nonregular.

However, I'm going to claim that L ₁ L ₂ = a*. First, note that any string in the concatenation L ₁ L ₂ has the form a ⁿ and therefore is an element of a*. Next, take any string in a*; let it be a ⁿ . If n is a power of two, then it can be formed as the concatenation of a ⁿ from L ₁ and ε from L ₂ . Otherwise, n isn't a power of two, and it can be formed as the concatenation of ε from L ₁ and a ⁿ from L ₂ . Therefore, L ₁ L ₂ = a*, so the theorem you're trying to prove is false.

Hope this helps!