Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Jolly Jocker
on 16/03/2022, 05:54:37 UTC

Code:
import sys
import time
import random
#from tqdm import tqdm
from time import sleep
import secp256k1 as ice

print('\n      B I N A R Y   S E E D   S C A N   Q U A D R O   \n')
print(' 11_________________0____________________________________________\n')
x=16000
y=131071
list=[]
x0=0b100000000000000000000000000000000000000000000
for n in range(y):
    x1 = (x0+n)
    a1 = bin(random.randrange(131072,262143))#[2:].zfill(18)
    a1 = (a1)[2:]
    for i in range(x):
        a0 = str("1")           
        a2 = str("0")
        a3 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a3 = (a3)[3:]#.zfill(44)
        a4 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a4 = (a4)[3:]#.zfill(44)
        a5 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a5 = (a5)[3:]#.zfill(44)
        a6 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a6 = (a6)[3:]#.zfill(44)
        binb = "".join(a0+a1+a2+a3)
        binc = "".join(a0+a1+a2+a4)
        bind = "".join(a0+a1+a2+a5)
        bine = "".join(a0+a1+a2+a6)
        list.append(binb)
        list.append(binc)
        list.append(bind)
        list.append(bine)
        if len(list)==x:
            line_count=0
            for bina in (list):
                bina!='\n'
                addr = ice.privatekey_to_address(0, True, int(bina,2))
                line_count+=1
                                 
                if addr.startswith('16jY7'): # 16jY7
                    ran = int(bina,2)
                    print('',bina, '\n')
                    sys.stdout.write('\r ' + hex(ran))
                    sys.stdout.write(' ' + '| ' + addr + ' ' + str(n))
                    sleep(1)
                    sys.stdout.write('\n\n continue...\n\n')
                    sleep(1)

                if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                    ran = int(bina,2)
                    print('\n\n Target found!!' + hex(ran) + ' \n ' + addr,'\n')
                    file=open(u"16j.Target.Info.txt","a")
                    file.write('\n ' + hex(ran) + ' | ' + addr)
                    file.close()
                    sleep(2)
                    wait = input("Press Enter to Exit.")
                    sleep(1)
                    exit()
                   
                if line_count==x:
                    list.clear()
                   
        if len(list)==0:
            print(' scan... ', str(n), end='\r')                       

Ca 42000 keys/s from c000000000000000 to ffffffffffffffff... The sector between  11_________________0 comprise a range of 131071 and will be processed.
The rest of this line are random numbers from 4 different generators..
The counter reflects the number of loops...

      B I N A R Y   S E E D   S C A N   Q U A D R O

 11_________________0____________________________________________

 1110010110110011011010011110001111111110100110111011100101000111

 0xe5b369e3fe9bb947 | 16jY7mM6wrpuAy53vKkh7QXm1JNdWK9Myu 19

 continue...

 1110010011000111001010001111000110100111001000001001110001000100

 0xe4c728f1a7209c44 | 16jY7Ezk5et98CVBWYEigGRwWovzhEtaQ3 121

 continue...

 scan...  192

Hi. I think your idea interesting. In this article another method how to recover privkey with add pubkeys what you think about this ?
https://toadstyle.org/cryptopals/66.txt

K is prrivkey, b is a picies of privkey:

For example, with n = 6, trace(58) gives me:

  # k = 111010
  # i = 2, b = 1
  add(1Q, 1Q)
  add(2Q, 1Q)
  # i = 3, b = 1
  add(3Q, 3Q)
  add(6Q, 1Q)
  # i = 4, b = 0
  add(7Q, 7Q)
  # i = 5, b = 1
  add(14Q, 14Q)
  add(28Q, 1Q)
  # i = 6, b = 0
  add(29Q, 29Q)

And here's trace(62):

  # k = 111110
  # i = 2, b = 1
  add(1Q, 1Q)
  add(2Q, 1Q)
  # i = 3, b = 1
  add(3Q, 3Q)
  add(6Q, 1Q)
  # i = 4, b = 1
  add(7Q, 7Q)
  add(14Q, 1Q)
  # i = 5, b = 1
  add(15Q, 15Q)
  add(30Q, 1Q)
  # i = 6, b = 0
  add(31Q, 31Q)





asic miner
github.com/achow1o1

update: Kiss
1J8wBriLsHvE52Jah75bKU1VSxj9btzLeq
1EhRDeGsoQVkAsiCtbDUbNnRW8a977bduS

Update2
2021.03.02
1CAsB6DdK6ezYzefzRe3SuQ1bZztsgaiQS
12JQGbJUHohtfoPbbZKbF8Z8hW5wYde1KN
1DKsKYzpqaTAg3Q4KrCBJ7AS4R1YyQrisE
1GymA7nkPZVBz3crj2NuFfGZsngTTX51Sb

How you crack them bro ? Thx



Enother interesting code - https://github.com/jairopaiva/BitcoinExprCracker

Search metadata for finding privkey... Needs modification for work

[moderator's note: consecutive posts merged]


Thanks a lot Bro!

But, I have to fight my way through this first... and it's not so easy, especially since English is not my mother tongue... so it needs a while ^^

you know, g00gle translater .... lol XD