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.
Just make your IP addresses pronouncable words like
feed:deaf:babe:beef:cafe::problem solved ez (working 2023!)But you’re limited to a-f. I wonder if anyone’s figured out how many addresses are actually possible with that system.
throw some 1337 speek in there and you’re all set!
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.