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    Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 351103 times)
    wipall
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    April 11, 2022, 10:44:46 AM
     #1821

    little experiment...

    this code creates a sha256 key from str(random.random()), which is fed to the next "ice.get_sha256().hex()".
    This is repeated, in this case + 2 times.
    The created keys are concatenated ( keys = key1+key2+key3 ).
    In the next step, the entire key is divided into 16-digit (64-bit) pieces and read out.
    a filter keeps the keys in the desired range...

    Happy hunting! XD

    Code:
    import time
    import random
    import secp256k1 as ice
    from ctypes import c_int
    from multiprocessing import Value, Lock, Process

    y=1000000
    cores=4
    counter = Value(c_int)
    counter_lock = Lock()      
    def process1(number,counter,):
        while True:
            t = str(random.random())
            key1 = ice.get_sha256(t).hex()
            key2 = ice.get_sha256(key1).hex()
            key3 = ice.get_sha256(key2).hex()
            keys = key1+key2+key3
            for i in range(0, len(keys), 16):
                line = keys[i:i+16]
                if line.startswith("8") or line.startswith("9") or line.startswith("a") or line.startswith("b") or line.startswith("c") or line.startswith("d") or line.startswith("e") or line.startswith("f"):
                    with counter_lock:
                            counter.value += cores
                            cv = str(counter.value)
                    addr = ice.privatekey_to_address(0, True, int(line, 16))

                    if addr.startswith("16jY7q"):
                        print('\n\n Pattern:',('0x'+line),addr,'\n')
                        
                    if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                        print ('\n\n TARGET FOUND!!:',addr,'\n\n')
                        file=open(u"16jY.Info.txt","a")
                        file.write('\n ' + ('0x'+line) + ' | ' + addr)
                        file.close()
                        wait = input("Press Enter to Exit.")
                        exit()
                        
                    if (counter.value) % y == 0:
                        print(' CNT:',cv.zfill(10),' Recent:',('0x'+line),addr,end='\r')

    if __name__ == '__main__':
        t = time.ctime()
        print('',t)
        number = y
        workers = []
        print('\n K E Y   -   C H O P P E R   -   64 \n')
        print('\n ===============|        |===== KEYS =======|============ ADDR ================| \n\n')
        for r in range(0,cores):
            p = Process(target=process1, args=(number,counter,))
            workers.append(p)
            p.start()

        for worker in workers:
            worker.join()          


    donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR



    the generation of the keys on a time basis (ms) is too slow. the result was double, triple and multiple repetition of the keys in a row.
    For this reason I switched to random.random()...
    the better way ^^

     Sat Mar 26 20:40:24 2022

     K E Y   -   C H O P P E R   -   64

     =============|            |===== KEYS =======|============ ADDR ================|

     CNT: 0032000000  Recent:   0xaaf4c88e4a40a299     175NYisjDfbxUyoonY8rEbghWvBbyMGPRj

     Pattern: 0x9a5da0765a81d90f 16jY7qNwsgykvkAzm8oiwVSnC71KsZsXqM

     CNT: 0304000000  Recent:   0xb338bf39e2ba9922     1EfbHH9BPXSDuibzn3gCao9aQB6Chr7C3G

    [moderator's note: consecutive posts merged]

    hi my friend, i want to ask you a private question but i can't dm you because i am a new member. i need a code to search in the range i want. The codes here are only for puzzle 64 and I can't edit spacing.
    I want to search by typing the range I want between these two ranges, with the range 0 to 115792089237316195423570985008687907853269984665640564039457584007913129639935. I'll be glad, if you help me.

    kötü ingilizcem için özür dilerim.
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