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# [Answers] Prove that this language is not context-free

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Problem Detail:

I'm not very comfortable with pumping lemma for context-free grammar. I understand the sufficient conditions that must hold but proving it gets me everytime. For example, I need to prove whether $L=\{0^{2^n}∣n \geq 0\}$ is not context-free.

There is no pattern of 0's that can be recreated by a pushdown automata but alas I need to prove this. I know you start off assuming it is by being able to create a substring $uvxyz$ where $v$ and $y$ are raised to the $i$th power where $i≥0$. I'm having trouble from there, any help in this and the understanding would be greatly appreciated.

#### Answered By : Yuval Filmus

A unary language (a language over a unary alphabet) is context-free if and only if it is regular. The easiest way to see this is to convert a context-free grammar to a regular grammar using the fact that $xy = yx$ over a unary alphabet.

There are many ways of showing that $L$ is not regular, and so not context-free. One way is using the Myhill–Nerode criterion. Since $0^{2^n} 0^{2^n} \in L$ while $0^{2^m} 0^{2^n} \notin L$ for $m \neq n$, we see that the words $\{ 0^{2^n} : n \geq 0 \}$ are pairwise inequivalent, and so $L$ is not regular.

A proof using the pumping lemma is also not too difficult. Given a pumping length $n$, choose $m$ so that $2^{m-1} > n$, and consider the word $0^{2^m} \in L$. According to the pumping lemma you can write $0^{2^m} = xyz$ with $|xy| \leq n < 2^{m-1}$ and $|y| \geq 1$ so that $xz \in L$. But $xz = 0^{2^m-|y|}$ cannot be in $L$ since $2^{m-1} < 2^m-|y| < 2^m$. This contradiction shows that $L$ is not regular.