Quote from: digaran on Today at 06:50:17 PM

Quote from: mcdouglasx on Today at 04:58:29 PM

it is more likely that the pub starts with "03". I would omit the ones that start with "02".

Nice approach, but do you think there is any possibility to determine how often permutations occur and if there is a solution to find out any pattern of y parity, like if we could figure out how many 02, and 03 keys exist in a certain range and if that could give us a pattern being repeated in all ranges.

That way we could use strides to ignore at least %40 of the whole bit range.

I suppose that to verify this you would have to store a large amount of data to look for a repetition pattern.
We change "02" to 1 and "03" to 0 (as binary).

Code:

import secp256k1 as ice

start= 1
end= 10000
for i in range (start, end+1):

A1 = ice.scalar_multiplication(i)
B0 = ice.to_cpub(A1.hex())

data = open("test03.txt","a")
data.write(str(B0)[ :2])
data.close()

with open('test03.txt', 'r') as file:
filedata = file.read()

# Replace the target string
filedata = filedata.replace('02', '1')

with open('fileP.txt', 'w') as file:
file.write(filedata)

filedata = filedata.replace('03', '0')

with open('binary.txt', 'w') as file:
file.write(filedata)

Then you look for some pattern in the binary sequence.

And if there were, you would only sequentially subtract or add "x" amount to your target, compare it with the pattern and you would know at what point in your pattern the target is, which would mean the end of ecc, if said pattern existed.