this post was submitted on 17 Sep 2025
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[–] Ghoelian@lemmy.dbzer0.com 2 points 1 day ago* (last edited 1 day ago) (1 children)

I think that's just 6^32, no? (Amount of options^string length). Which is 7958661109E24.

[–] some_kind_of_guy@lemmy.world 2 points 1 day ago* (last edited 1 day ago) (1 children)

Yeah, calculation of the amount of possible strings containing only a-f is trivial. But the idea is for addresses to be memorable. So I'm wondering how many strings which are valid IPv6 addresses are possible if you are limited to actual English (or, pick a language) 4-letter words containing only a-f. As someone mentioned, this could be expanded with 1337-speak.

[–] Ghoelian@lemmy.dbzer0.com 1 points 1 day ago

Ahh right, that would be a bit more difficult to calculate.

I guess you could make a script which just bruteforces all combinations of a-f against an English dictionary. I might try to do that tonight.