Key frequency analysis (short keys have leading "3"'s trimmed)

So maybe they randomly scrambled the characters? ...why would they do that?

How do you know this?

if you type a key that's less than 8 digits with 3s equal to the amout of digits less than 8 before the key it will download the program. (so if you have a program which has a key of AABBCCD you can type 3AABBCCD instead and it will download it)

Ah! Thanks

np

So, what's this useful for...?

really nothing, it's just a fact.

It's useful in the case where you want to type extra threes so you can have the satisfaction of _{supporting the illuminati} adding random garbage to yet another location.

Take a key with less than 8 digits Add 3's to the front to make it 8 digits You now have an 8 digit key Any key containing 8 digits can have up to 6 of any digits added to the end of that key And it'll still accept it The public key AABBCC =3AABBCC =33AABBCC =33AABBCC_HELLO =33AABBCCDDEEFF Getting the smallest public key is all about getting multiple 3's in a row in the beginning Interesting find

So the characters 0, I (uppercase i), O, U, -, and _ never appear in keys, which means there are exactly 32 characters that do appear in the keys: 3, E, 4, K, D, N, X, V, J, 2, Q, 5, 8, A, R, S, Y, C, P, 7, W, Z, H, B, 1, F, M, L, 9, 6, G, T Considering that keys are in base-32, with 3 being unscrambled to 0, we can use the position frequency in the graph to estimate the order of unscrambling keys, to get their base-10 index

How many keys did you use to make that graph? Can you share the database with us too?

About 1000 I think. https://pastebin.com/raw/PFugbCGu There are a few duplicates

ALREADY EXPLAINED ON i's COMMENT BY ME to quote myself:"if you type a key that's less than 8 digits with 3s equal to the amout of digits less than 8 before the key it will download the program. (so if you have a program which has a key of AABBCCD you can type 3AABBCCD instead and it will download it)"

DAT:illuminati heard