Proving that $L = \{ a^{n!} \ | \ n \geq 0 \}$ is not regular

Question

Let $L$ a language over $X = \{a\}$ defined as follow :

$$L = \{ a^{n!} \ | \ n \geq 0 \}$$

I want to prove that $L$ isn't regular, I have searched in the forum for an equivalent question, but I found nothing. If it's a duplicate, I apologize and I'll be glad if you provide me the link of the duplicated question.

I'll show here what I have done, using the pumping lemma.

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Suppose that $L$ is regular, let $n \in \mathbb{N}^*$ and $\omega = a^{n!}$. We have $\omega \in L$ and $|\omega| \geq n$. Thus, there exist $x, y, z \in X$ so as :

1. $a^{n!} = xyz$
2. $y \neq \epsilon$
3. $|xy| \leq n$
4. $xy^kz \in L, \quad \forall k \geq 0$

Because of fact 3, $xy = a^k$ and $y = a^j$ with $k \leq n$ and $1 \leq j \leq n$. Thus :

$$a^{n!} = a^{k - j}.a^{j}.a^{n! - k}$$

By fact 4, we must have $a^{k - j}.a^{n! - k} = a^{n! - j} \in L$.

In addition, because $1 \leq j \leq n$, the following is true :

$$n! - n \leq n! - j \lneq n!$$

The final result that I need is that $(n - 1)! \lneq n! - n$. Using this I can deduce :

$$(n - 1)! \lneq n! - j \lneq n!$$

Showing that $n! - j$ can't be the factorial of an integer, giving the contradiction $a^{n! - j} \notin L$. I have tried to prove the final result, but I didn't succeed. It is not true for the first values of $n$, I have tried to consider multiple cases but went for nothing.

So I ask for your advice, maybe I am going wrong and didn't applied the pumping lemma with a word that leads to contradiction.

Answer 1

The easiest way is probably to use the pumping lemma on the complement language $\overline{L} = \{ a^m : m \neq n! \}$. Suppose that $L$ were regular. Than $\overline{L}$ would be regular. According to the pumping lemma, there is a pumping length $p$ such that every word $w$ of length at least $p$ in $\overline{L}$ has a decomposition $w = xyz$ such that $|xy| \leq p$, $y \neq \epsilon$, and $xy^iz \in \overline{L}$ for all $i \geq 0$. Take $w = a^{p\cdot p!} \in \overline{L}$ (without loss of generality, $p > 1$). Suppose that the decomposition is $w=xyz$, where $y=a^t$. Since $t \leq p$, it follows that $t \mid p!$. But then $xy^{1+p!/t}z = a^{p\cdot p! + p!} = a^{(p+1)!} \notin \overline{L}$.

Answer 2

i can't comment as i haven't yet touched the benchmark of 50 reputations.

I would like to make your work a bit clearer.You have used 3 variables although even 2 are sufficient .

a^n! let n=m ,

now consider this, |w|=|xz|+i|y|

also , m! has two parts k! = |xz| and j = i.|y|

now we know j<=m , considering m is not < 3 (smaller than 3)

m!-j=k! (it is needed to make sure xz exists )

now , the same thing ,

it is impossible for it to exist,

as n!-j >= n!-n > (n-1)! (how did it happen ? see below)

n!-n = n(n-1)!-n = n((n-1)!-1) > (n-1)!

therefore it's not regular .

alternative way to check identify RL is : (credits to Seemona A for this short algo )

1. 1.If given language is finite => Regular
2. If language is infinite: Language over one symbol : Forms AP => Regular ; Doesn't form AP => Not Regular ;; Language over more than one symbol : No dependency => Regular; Dependency => Not Regular;;

another logic is L(G)+{a^n! , n>=0} has set of elements which are not periodic in nature . a grammar on single variable is CFG if it is RegG

And a grammar is RegG iff there is a periodic nature in the set of elements of its language.

similarly if we try for CFL(context free) , we can substitute (p+r) and still the logic will be same as the string to be pumped are v= a^p and y= a^r . Logic will remain the same

https://qr.ae/pvVy55 , user named "Seemona A" has nicely concluded in short (it's a "quora" site link , a general QnA networking site).

https://john.cs.olemiss.edu/~hcc/csci311/notes/ (here you can also find it in regular expression section...... Ole Miss is University of Mississippi )

Answer 3

It is a little confusing that you seem to have used $k$ in two different ways. First in the pumping lemma $xy^kz$, then to fix the length of $xy$.

Rather than trying to prove that deleting $y$ from the string $\omega$ (pumping with $k=0$) will give a string outside the language, perhaps it is easier to add a copy of $y$ (pumping with $k=2$).

$n!+j \le n!+n < n!(n+1)$