this post was submitted on 31 Oct 2024
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Matches strings of any character repeated a non-prime number of times

https://www.youtube.com/watch?v=5vbk0TwkokM

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[–] CanadaPlus@lemmy.sdf.org 24 points 3 weeks ago (3 children)

So this is a definite example of "regex" that's not regular, then. I really don't think there's any finite state machine that can track every possible number of string repeats separately.

[–] NateNate60@lemmy.world 20 points 3 weeks ago (1 children)

You got downvoted here but you're absolutely right. It's easy to prove that the set of strings with prime length is not a regular language using the pumping lemma for regular languages. And in typical StackExchange fashion, someone's already done it.

Here's their proof.

Claim 1: The language consisting of the character 1 repeated a prime number of times is not regular.

A further argument to justify your claim—

Claim 2: If the language described in Claim 1 is not regular, then the language consisting of the character 1 repeated a composite number of times is not regular.

Proof: Suppose the language described in Claim 2 is regular if the language described in Claim 1 is not. Then there must exist a finite-state automaton A that recognises it. If we create a new finite-state automaton B which (1) checks whether the string has length 1 and rejects it, and (2) then passes the string to automaton A and rejects when automaton A accepts and accepts when automaton A rejects, then we can see that automaton B accepts the set of all strings of non-composite length that are not of length 1, i.e. the set of all strings of prime length. But since the language consisting of all strings of prime length is non-regular, there cannot exist such an automaton. Therefore, the assumption that the language described in Claim 2 being regular is false.

[–] CanadaPlus@lemmy.sdf.org 3 points 3 weeks ago* (last edited 3 weeks ago)

By now, I have just one, so thanks for the assist. There's always that one (sometimes puzzling) downvote on anything factual.

The pumping lemma, for anyone unfamiliar. It's a consequence of the fact an FSM is finite, so you can construct a repeatable y just by exhausting the FSM's ability to "remember" how much it's seen.

[–] ProfessorScience@lemmy.world 12 points 3 weeks ago

Yeah backreferences in general are not "regular" in the mathematical sense.

[–] Rin@lemm.ee 2 points 3 weeks ago (2 children)

You can have states point to each other in a loop, no?

[–] NateNate60@lemmy.world 6 points 3 weeks ago (1 children)

If the set of all strings of composite length is a regular language, you can use that to prove the set of all strings of prime length are also a regular language.

But it's also easy to prove that the set of language of strings of prime length is not regular, and thus the language of strings of composite length also can't be regular.

A more formal proof.

[–] Rin@lemm.ee 1 points 3 weeks ago

Thank you for this. I'll review this when I can.

[–] CanadaPlus@lemmy.sdf.org 3 points 3 weeks ago* (last edited 3 weeks ago)

Yeah, but in an FSM all you have are states. To do it the obvious way, you need a loop with separate branches for every number greater than 2, or at the very least every prime number, and that's not going to be finite.