是语言{0 ^ n 1 ^ n 0 ^ k | k！= n}上下文免费？

Question

I believe this language isn't context free, because there is no chance that a PDA could compare 2 blocks of 0's and 1's of the same length and also remember it's length for later use.

Unfortunately, I have no idea how to prove it.

I tried using the pumping lemma to no avail...

I've also tried to assume by contradiction that the language is context free and use the fact that the intersection of a context free language with a regular language is also context free (by finding some mysterious regular language L), and surprisingly (or not) - all my efforts were in vain...

Any help would be appreciated

Answer 1

Is the language { 0 ⁿ 1 ⁿ 0 ^k | k != n} context free?

No , the language L = { 0 ⁿ 1 ⁿ 0 ^k | k!=n } is not a context free language . Also, Class of Regular Languages is subset of class Context free languages .

You Idea using PDA is correct and obvious way to show that language is not context free. And we can't draw PDA for language 0 ⁿ 1 ⁿ 0 ^k because after matching prefix 0 ⁿ to 1 ⁿ stack become empty, then we don't have stored information to check weather suffix 0 ^K are equal to n or not.

HINT: For formal proofs

• First

L = {0 ⁿ 1 ⁿ 0 ^k | k!=n } now complement of L is L ^' .

L ^' = {{0 ⁿ 1 ⁿ 0 ⁿ } that is well known context sensitive language (can be proof).

And the complement of a context-sensitive language is itself context-sensitive.

• Second

For Pumping Lemma:

L = {0 ⁿ 1 ⁿ 0 ^k | k!=n } is Union of L ₁ and L ₂ , Where
L ₁ = {0 ⁿ 1 ⁿ 0 ^k | k > n } and L ₂ = {0 ⁿ 1 ⁿ 0 ^k | k < n },

L = L ₁ UL ₂

L ₁ and L ₂ both are non-context-free Language. and union of two non-context free languages are non-context-free. ( that can easily proof by grammar)

---

Also, The union, concatenation of two context-sensitive languages is context-sensitive.