Prove that regular languages and context-free languages aren't closed under $Perm(L)$

Question

Let the operation $$Perm(L) = \{ w | \exists u \in L \text{ such that } u \text{ is a permutation of } w \}$$

Prove that both regular languages and CFLs aren't closed under $Perm(L)$.

I've tried to use several well-known languages (like $\{0^n1^n\}$) and applying $Perm(L)$ and afterward manipulate them or using the pumping lemma in order to get a contradiction, but nothing worked out.

You are aware that you should do your homework on your own? — Raphael, May 19 '15 at 14:04
See also: Are permutations of context-free languages context-free?. — Hendrik Jan, May 24 '15 at 02:14

score 6 · Accepted Answer · answered May 19 '15 at 13:15

6

Hints:

For regular languages, consider $Perm((01)^*) \cap 0^* 1^*$.
For context-free languages, consider $Perm(0^n 1^n 2^m 3^m) \cap 0^* 2^* 1^* 3^*$.

answered May 19 '15 at 13:15

Yuval Filmus

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And also $Perm((012)^) \cap 0^1^2^$, which jumps from regular ouside context-free. – Hendrik Jan May 19 '15 at 13:59

Prove that regular languages and context-free languages aren't closed under $Perm(L)$

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