Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
There's at least one other address starting with "1PWo3JeB9jrGw" in the 71 range.
No matter how you search, people should reflect on this fact: there's a 50% chance you'd first find this other address, before actually finding the correct one, and way way ... way later after dealing with half of the other 395.000 unique addresses (plus or minus 2000) that match the first 1+9 base58 chars.
It's even worse if matching just the H160 bits for efficiency. The number of comparable H160s is 3 to 4 times higher than b58 matches, because a lot of those won't match the desired b58 pattern, even though they have lots of matching bits.
No matter how you search, people should reflect on this fact: there's a 50% chance you'd first find this other address, before actually finding the correct one, and way way ... way later after dealing with half of the other 395.000 unique addresses (plus or minus 2000) that match the first 1+9 base58 chars.
It's even worse if matching just the H160 bits for efficiency. The number of comparable H160s is 3 to 4 times higher than b58 matches, because a lot of those won't match the desired b58 pattern, even though they have lots of matching bits.
Code:
import base58
min = base58.b58decode('1PWo3JeB9jrGw111111111111111111111').hex()
max = base58.b58decode('1PWo3JeB9jrGwzzzzzzzzzzzzzzzzzzzzz').hex()
# Get total possibilities, and remove checksum bytes
n = (int(max, 16) - int(min, 16) + 1) >> 32
# Get AVERAGE count over ANY 40 bits set
n = n / (2**(160 - 70))
print(n)
= 2.0245494726457194min = base58.b58decode('1PWo3JeB9jrGw111111111111111111111').hex()
max = base58.b58decode('1PWo3JeB9jrGwzzzzzzzzzzzzzzzzzzzzz').hex()
# Get total possibilities, and remove checksum bytes
n = (int(max, 16) - int(min, 16) + 1) >> 32
# Get AVERAGE count over ANY 40 bits set
n = n / (2**(160 - 70))
print(n)
There are exactly only 2 1PWo3JeB9jrGw
LMAO you do realize that I'm the one who posted that code snippet, right?
Anyway, it's slightly wrong (min and max H160 need to actually be valid). The actual code that handles these issues is much longer than that snippet.
After computing a Poisson confidence interval, there are, with a 99% chance, anywhere between 2 and 6 H160 values in 2**70 hash trials, which end up with an address having the mentioned base58 prefix.
A confidence of 100% is that there are anywhere between 0 and 2**71 H160 having that property.