Quote from: zahid888 on May 08, 2021, 04:30:20 PM

Quote from: NotATether on May 07, 2021, 08:31:38 AM

Quote from: zahid888 on May 06, 2021, 10:51:49 PM

Is there any way to guess any letters of key 64, As on the basis of similarities between these two keys ?
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN -
16jY7qLJnxLQQRYPX5BLuCtcBs6tvXz8BE - 3AD3536AEA8CEB2B0
16jY7qLJnx2EZZumnYFke3GutCrRnHKs1M - 7013536A91A6441B0

Theoretically we can replace the black characters with 1111 and zzzz so that we have two strings, then we convert them into numbers. Basic sequential counting is cheap compared to the cryptographic hashing done on each PK so with a fast Base58-to-hex converter, we can successively increment those black characters from 1 to z.

Quote from: blm98 on May 06, 2021, 11:01:36 PM

hello,

you mean to say something like that ??

Code:

from bit import *
from bitcoin import *
import random

list = ["16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN","13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9","1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ"]
while 1==1:
   a = random.randrange(144,255)
   while a <= 4095:
   a+=1
   b = ("3536A")
   c = random.randrange(16777216,268435455)
   d = ("b0")
   myhexa = "%050x" % a
   myhexc = "%0x" % c
   j = (myhexa+b+myhexc+d)
   priv = j
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)#compressed
   s1 = j
   s2 = addr
   if addr in list:
   print ("...............FOUND.............")
   f=open(u"C:/found.txt","a")
   f.write(s2+" ")
   #f.write(s3+" ")
   f.write(s1+"\n")
   f.close()
   break
   if addr.startswith("16j")or addr.startswith("13z")or addr.startswith("1B8")or addr.startswith("1MV"):
   print(j,addr)

Here is few results..

000000000000000000000000000000000000000000000002813536A3855b0ab0 1MVBvZWijEX6jLHqvc62ycQmAwQF2MGtn1
000000000000000000000000000000000000000000000003323536A37ecc11b0 13z3UU2fSmoHFsnkc3PNVEtz5fGunJyxWk
000000000000000000000000000000000000000000000004163536Af36acf8b0 16jzjW8DghfsvE9s5oYVdQjdCL68KfQWCD
000000000000000000000000000000000000000000000004763536A4d3f7d8b0 16j4ZG3tQJdCNNB7FxAXEX2pddUbcVQurc
0000000000000000000000000000000000000000000000061f3536A3b9564bb0 16jKBfsPZpqTr3Wf68UpG9GXFzGy5tvMLD
0000000000000000000000000000000000000000000000062f3536Af43f269b0 1B8fpioKSuEHWh6GNXzKYT2FhnEcisi7TS
00000000000000000000000000000000000000000000000a883536A5c18381b0 1MVAj6xiQtR9aZMvoAvCXaKKapXx5mZAoj
00000000000000000000000000000000000000000000000df93536Ab136b68b0 1B886nbuaUaXYCtYxXVU4ykrhXDM5up71v
00000000000000000000000000000000000000000000000e4a3536A58eb38bb0 16jjxYEA2ccW8pTiSzozPCX7u4m6AwCWMc
00000000000000000000000000000000000000000000000e813536A42123d2b0 13zV6pBA4aZZzKGj8dJALiW7KAtiRq5SP7
00000000000000000000000000000000000000000000000f413536Aabe707bb0 13zQuhCczRWABxgXGw1WtDAK1e1z9PXJtk
000000000000000000000000000000000000000000000001563536A67fee37b0 1MV7GUZu4efY1hRbTJm4qEUWYtN42tKgdh
000000000000000000000000000000000000000000000004073536Ac381924b0 1B8ZsYYvyqjyndX8mg3QNb3NXnPX1FeK8B
000000000000000000000000000000000000000000000004853536Ade61185b0 1B8Q2s1hKF6B8GdymtfSVUrQGan1TEBd9a
000000000000000000000000000000000000000000000008db3536A6c40e08b0 1MVTW3PdbmkF57pruNJckLNFiz6cJyoyHn
00000000000000000000000000000000000000000000000e133536A4f412b7b0 13zs9rHG9A41ih8UNDZzrA4RcTAiqfSz15

hi there zahid,
wanted to ask, how is that code also searching for 64th puzzle address, or i am seeing something wrong, one 0 gonna shift bits up, so no more 64 bit,
and how to say all those addresses, in this range you set, thanks for explaining,