● 12-char prefix
20d45a6a7625355df5e5fc41b358e6f8581991e2
13zb1hQbWVsc1111111111111111111111 => 20d45a6a76253570014f1a30288bd88dc3ff6ffb
13zb1hQbWVsczzzzzzzzzzzzzzzzzzzzzz => 20d45a6a76253571d7012502fabee39e6bbaf2b4
1d5b20ad2d2330b10a7bb82b9
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF / 1D5B20AD2D2330B10A7BB82B9 = 0x8B8758F7C8AD0286
(every 0x8B8758F7C8AD0286 will find 1 address 12-char prefix)
Did you realize that 0x8B8758F7C8AD0286 is near 1/3 of the 66 bit space?
>>> 2**65 // 0x8B8758F7C8AD0286
3
>>> 2**65 / 0x8B8758F7C8AD0286
3.6694959191703664
It just interesting
It seems to me that these calculations are very abstract, we are missing several steps for ripemd160. If this were really the case, then any addresses could be taken away without problems. I know you Alberto, we corresponded several times in tg. I am grateful to you in many ways, most likely thanks to you I started this activity. I have no goal of emptying other people’s wallets, but if one is found ownerless or from a puzzle, I will definitely send you a significant reward, rly. But, over time, I realized that this is more of a study, most likely this will never happen, maybe. Once again I apologize, I'm using a translator.
If you are talking addresses or better yet, address prefixes (which both are a reflection of their H160), you don't really need to know all of the steps, really any of them, the H160 or the SHA256, etc. You just have to know what comprises an address, in this case base58. And from there you can get a rough idea of how many specific prefixes will be in a specific, certain sized range.
First, the leading 1, isn't part of the prefix, it is a gimme because all of these type addresses start with 1.
Someone correct me if I am wrong, but couldn't we just take range size / 58^number of leading prefixes (minus the starting "1")
If everything is uniformly spread out, over the range/curve (as the experts like to say, but it's not proven, but is a good "guesstimator" of sorts), an 11-char prefix (1 + 11 characters = total of 12 characters), in a 2^65 bit range, there should be around 1.4 of them. 2^65 / 58^11
2^65 / 58^11 = 1.4
2^65 / 58^10 = 85
2^65 / 58^9 = 4,967
2^65 / 58^8 = 288,088
and I guess you could apply a similar logic to H160s as well, but use 16^number of leading H160 prefixes, instead of 58^number of leading prefixes (minus the starting "1")
Anyone?
