Yeah .. He, most likely, generated 256 keys and manually updated the first character on the left whenever necessary.
One other thing could be placing a key outside a bit range, I'm curious did Satoshi ever confirmed that the keys are truly in the assumed bit ranges or we just hope the amounts of transactions are enough of a proof that the keys are exactly there?
But you know what I'd like to see? A large amount in a key with no exactly known bit range, somewhere between 160 and 180 bit not lower and not higher, then solving that key would be a global challenge, though not any amount which someone could spend half of it to use special tools and grab it, something which could only be solved by math and new methods.
For example, I haven't seen any tool/ algorithm capable of adding or subtracting 1 to a key and then divide it by 2, kangaroo engages square root calculations, BSGS looks for a match after adding/subtracting calculations, but no tool does + or - 1 then divide!
Now I look at it from a different angle. In binary, the program is more practical, for example, I know the last 8 bits of the private key anyway, because 8 bits equals a maximum of 256 digits. There was a friend who said that if the curve math is 0, multiply, if 1, add, knowing the last bit means solving ecdsa, the last 8 bits are 1 in 256, you can find the rest. If you say the last 16 bits, there is 1 possibility in 65536, these numbers can be tried, you will reduce 125 bits -16 bits =109bits.
This is the code
Code:
from sympy import mod_inverse
import secp256k1 as ice
k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar):
ScalarBin = str(bin(Scalar))[2:]
le=len(ScalarBin)
for i in range (0,le):
if ScalarBin[le-i] == "0":
Qx=ice.point_multiplication(k,Qx)
else:
Qx=ice.point_addition(Qx,neg1)
Qx=ice.point_multiplication(k,Qx)
return ice.point_to_cpub(Qx)
for x in range(65536):
print(ters(pub,x))
If the last bit is 1, it moves the point forward when divided by (2).
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1 +
if the last bit is 0 then normal divides
You trying to do mod_inverse using N. But N what is?
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
The code doesn't work.
NameError: name 'pub' is not defined
How to fix?
pub=ice.pub2upub('0433709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e2a1c304a39a77 775d3579d077b6ee5e4d26fd3ec36f52ad674a9b47fdd999c48')
