Determining whether $ L = \{ 0^n1^{n^2} | n \ge 0 \} $ is a CFL

Question

Assuming $L$ is defined as follows:

$$ L = \{ 0^n1^{n^2} | n \ge 0 \} $$

I'm trying to either prove/disprove whether $L$ is CFL or not.

My intuition tells me its not CFL since I cannot express the power $n^2$ by itself using CFL, but I'm having trouble proving so using the pumping lemma. Also, I failed to create any such PDA (which my intuition says its impossible).

Possible duplicate of How to prove that a language is not context-free? — David Richerby, Feb 20 '17 at 08:32

score 1 · Accepted Answer · answered Feb 20 '17 at 01:23

Let $h\colon \{0,1\} \to \{0\}$ be the homomorphism given by $h(0) = \epsilon$ and $h(1) = 0$. If $L$ were context-free, so would $h(L) = \{ 0^{n^2} : n \geq 0 \}$ be. Since $h(L)$ is a unary language, it would be regular. However, a unary language is regular iff it is eventually periodic, which it isn't. Therefore $L$ cannot have been context-free.

More generally, if $L$ is context-free then:

The set of lengths of words in $L$ is eventually periodic.
The set of number of occurrences of a symbol in $L$ is eventually periodic.

The most general statement of this kind is Parikh's theorem.

Determining whether $ L = \{ 0^n1^{n^2} | n \ge 0 \} $ is a CFL

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