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