How to prove {a^(n^2) | n>0} is not context-free?

Question

So I have a language: $$ L = \{a^{n^2} \mid n > 0\} $$ I need to prove that this language isn't context-free using the pumping lemma. I have a vague thought process as to how to do the proof but I'm sort of doubting its validity.

So I take a pumping length $p$ such that a word $a^{p^2}$ can be split into 5 parts $uvxyz$. I need to assume that $L$ is context-free and run through some cases and check if the word violates the pumping lemma or if it doesn't violate the pumping lemma but isn't in the language definition.

I know the concept of proving languages aren't context free and the pumping lemma but I'm very stuck applying it to this particular language.

What do I do?

Answer 1

[Since (a) this is an instance of a standard problem and (b) I wasn't able to find it in the archives, I'll expand my comment into a hinted solution.]

As usual in problems like this, we assume that the language $L$ is context free, so the Pumping Lemma applies, meaning that there is an integer $p>0$ such that we can write the string $a^{p^2}$ as the concatenation of strings $uvxyz$ with $|vy|>0$ and $|vxy|=t\le p$. Hence, we'll have $$ 0 < |vy|=t\le p $$ This means that when we pump $uvxyz$ to $uv^ixy^iz$ we'll have $$ p^2<|uv^ixy^iz|=p^2+(i-1)t\le p^2+(i-1)p $$ Choosing $i=2$ gives us $$ p^2<|uv^2xy^2z|\le p^2+p $$ Now, in length order, the next string in $L$ after $a^{p^2}$ will be $a^{(p+1)^2}$. Here's the hint: use this fact to show that $uv^ixy^iz$ can't possibly be in $L$, contradicting the Pumping Lemma consequence that all the pumped strings are in $L$, so consequently $L$ cannot be a CFL.

---

This idiom, BTW, can be used to show that other languages over a one-symbol alphabet aren't context-free, like the $a^{n^5}$ language or the $a^p$ language, where $p$ ranges over the primes. There are more general results, as well, like the one that says that any CFL over a one-symbol language is regular. That, along with another theorem that says, roughly, that a regular language over a one-symbol language can't be too "sparse", gives a higher-level proof of your question.

Answer 2

If still have your question, I thought this might help why using pump up i=2 we can show the language is not Context Free.

Let $s=a^{p^2}$

next string is for (n+1) : $s=a^{(p+1)^2}$

$s=uvxyz$$; |vxy|\leq P$$;|vy|\geq1$

at most length $|vy|=p$ pump up i=2 so the pumped string $p^2+p$ which is not in the language because: $ (p+1)^2 $>$ (p^2+p)$ so the language given is not context-free.

Answer 3

A unary language (a language over a unary alphabet) is context-free iff it is regular. Moreover, such a language is context-free (or regular) iff the set of lengths of words in the language is eventually periodic.

The set $\{n^2 : n \in \mathbb{N}\}$ is not eventually periodic: the gap between adjacent elements increases without bound. So the language $\{a^{n^2} : n \in \mathbb{N}\}$ isn't regular or context-free.

Answer 4

Assume that the language $L = \{a^{(n^2)} | n > 0\}$ is context-free and let N be the pumping length given by the Pumping Lemma.

Choose the string $u = a^{(N^2)}$, which is in the language L. According to the Pumping Lemma, u can be decomposed into vwxyz such that |wxy| ≤ N, |wy| > 0, and $vw^mxy^mz$ is in L for every $m ≥ 0$.

When you choose m = 2, you pump the substring wxy, which increases the length of the string.

The length of the pumped string $vw^2xy^2z$ becomes $(N + n)^2$, where $n \in \mathbb{N}$. Now we now that |wy| = j + k, |w| = j, and |y| = k.

Now we know that this must be true $(N + n)^2 = N^2 + j+ k$.

From this equation, you deduce that $j + k = 2Nn + n^2$, which is greater than N since n > 0.

This contradicts the Pumping Lemma's requirement that |wxy| ≤ N.

Therefore, the assumption that L is context-free must be incorrect.Thus, you conclude that the language $L = \{a^{(n^2)} | n > 0\}$ is not context-free.