Quote from: digaran on September 03, 2023, 11:33:05 PM

Quote from: mcdouglasx on September 03, 2023, 11:30:10 PM

Quote from: digaran on September 03, 2023, 11:16:54 PM

Quote from: mcdouglasx on September 03, 2023, 10:34:23 PM

Quote from: james5000 on September 03, 2023, 10:20:38 PM

I'm not sure if it helps, but for 130 bits, we only need to divide the point by 2, 101 times. The issue is figuring out where to subtract 1 to avoid floating-point errors.

the problem of dividing by 2 is that you need 2**101 pubkeys (according to your approach).
update:
dividing by 3 you need 3**14 pubkeys, to reduce puzzle130 down to the equivalent of puzzle 105.

So reducing 25 bits, if dividing by 2 we need 2^25 public keys with 1 one of them to be the correct result, but dividing by 3 we need 3^14 keys, one of them would be correct, how did you calculate it?

I hope your scrip saves the results to a file, because I only see print, are we supposed to print thousands of keys on screen? 😅

He says he wants to divide 101 times, and since he doesn't know the pk he needs to do pk/2 and ( pk-1)/2 which results in 2x2x2x2.... 101 times (2**101)
by 3 is 3x3x3x3.... 14 times (3**14) equivalent to puzzle 105

Ok, can you tell us where to put our target public key in your script, where to put the number of times etc, please?

This script is just theory, not done to create massive continuous divisions, when I have time I'll do it.