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    Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 324687 times)
    Jolly Jocker
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    March 12, 2022, 04:41:06 PM
    Last edit: March 13, 2022, 05:56:15 PM by Jolly Jocker
     #1781

    Test Code for the 64 bit range... ca. 310000000 keys/hr...

    Code:
    import random
    from threading import Thread
    import secp256k1 as ice
    from time import sleep
    print('\n ========= ================= ==ADDRESS HUNTER== ================= ==========\n\n')
    y=1000000
    count=-2000000
    def process1(number):
        line_count=0
        num = random.randrange(0x8000000000000000,0xffffffffffffffff)
        for i in range(number):
            num+=i
            addr = ice.privatekey_to_address(0, True, num)
            lineA = (addr, hex(num))
            line_count+=1
                        
            if '16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN' in lineA:
                file=open(u"BTH.Target.Info.txt","a")
                file.write('\n ' + str(lineA))
                file.close()
                wait = input("Press Enter to Exit.")
                exit()
                            
            if line_count == y:
                print(' current line: ', str(lineA))
                line_count=0
                        
    while True:
        count+=(2*y)
        number = y
        threads = []
        print('\n Total Scanned Lines: ' + str(count) + '\n')
        for n in range(0,2):
            t = Thread(target=process1, args=(number,))
            threads.append(t)
            t.start()
            
        for t in threads:
            t.join()

    donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
    I tested your code, but the wrong ddd code doesn't collide even with 2 hex, which is a big problem



    this is what your code should look like which works at 100% delete tqdm slowing code 10%
    import random
    from tqdm import tqdm
    import secp256k1 as ice
    for x in tqdm(range(100000000)):
     num = random.randrange(0xd2c00,0xd2cff)
     add = ice.privatekey_to_address(0,True,num)
     if "1HsMJxNiV7TLxmoF6uJNkydxPFDog4NQum" in (add):
      f=open("von.txt","a")
      f.write(str(num)+"-"+(add)+"\n")
      f.close()

    I see you didn't understand the principle of my code...

    This code (slightly modified) for the search for that Pubkey of 16jY7.... with 4 Threads make more as 380000000 Keys in 30 minutes...
    I'll keep working on it... I'm sure there's more to come^^

    Code:
    import time
    import random
    import secp256k1 as ice
    from time import sleep
    from threading import Thread
    from multiprocessing import Value
    t = time.ctime()
    print('',t)
    print('\n |========== ========== ==========! PUBKEY HUNTER !========== ========== ==========||============\n\n')
    print(' [*] create privkeys....', end='\r')

    list=[]
    a=2
    y=1048576
    counter = Value('L')
    def process1(counter,number,n):
        ran = random.randrange(0x8000000000000001,0xffffffffffffffff)
        for i in range(number):
            ran+=1
            pub = ice.privatekey_to_h160(0, True, ran).hex()
            lineA = pub, hex(ran)
            list.append(lineA)
            if len(list)==y:
                for line in list:
                    line!='\n'
                    with counter.get_lock():
                        counter.value += 1
               
                    if '3ee4133d991f52fdf6a25c9834e0745ac74248a4' in line:
                        threads.clear()
                        a=0
                        f=open("16jY7.txt","a")
                        f.write(str(line) +"\n")
                        f.close()
                        print('\n\n Target found!! \n' + str(line))
                        exit()
                   
                    if (counter.value)%y == 0:
                        print(' [ Recent-Line:', str(line),'] Total:', str(counter.value))#, end='\r')
                        list.clear()
                           
    while len(list)<=y:
        number = y
        threads = []
        for n in range(a):
            t = Thread(target=process1, args=(counter,number,n))
            threads.append(t)
            t.start()

        for t in threads:
            t.join()

    ok i take it back, I won't go into that anymore there are only 2 options 64 address has total epic fail hex or in this range is not at all happy hunting

    The range doesn't matter in principle, we're just unlucky that we came too late and the lower areas have already been solved... every higher range reduces the chance of finding the desired address. Of course, luck plays a big part...
    The "fun" factor and further learning how to use Python inspire the whole thing...

    so don't fret Smiley Happy hunting!! despite everything ...
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