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1  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: May 03, 2022, 11:58:19 PM
Target found!!0xcb8f4f9ff46b0325
 16jY77KNQXnocoT8HkgTJ9SfasE9biH1Fh
               --------
a bit of an exciting address Sad  Cool

2  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: April 12, 2022, 03:22:32 PM
Quote
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.


The range for 64 bit is 9223372036854775808 to 18446744073709551615


this is easy to show.. an example for it:

use this web-calculator
https://web2.0rechner.de/

and just enter 2^64 for 64bit
2^66 for 66bit.. etc..

greetings! Smiley




Thank u brother.

Have you checked which address type the codes you have written give you? compressed? not compressed?

I'm new to this puzzle. but
For 4 years I started to think that's why puzzle 64 was not solved.

3  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: April 11, 2022, 10:44:46 AM
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.
4  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 01, 2022, 04:21:06 PM
I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:
from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass

Thank you very much for your attention, my friend.
always a pleasure dude!! Smiley


i managed to run the code sequentially.
I scan an average of 500,000,000 addresses a day.
and although the computer is very old

do you think it's too slow?

How many addresses do you scan on average per day?
5  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 25, 2022, 08:35:08 PM
I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:
from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass

Thank you very much for your attention, my friend.
6  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 25, 2022, 03:55:49 PM

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



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

 11_________________0____________________________________________

Traceback (most recent call last):
  File "11.py", line 41, in <module>
    addr = ice.privatekey_to_address(0, True, int(bina,2))
AttributeError: module 'secp256k1' has no attribute 'privatekey_to_address'
7  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 22, 2022, 10:36:38 AM
Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?


Why would you want the binary value over hex?

I think the scanning process will be faster.

No it won't, python interpreter represents these numbers in the exact same way.

arrow. So do you have a python code that I can scan in order? a sequential code where I can specify the scan interval.
8  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 22, 2022, 10:09:51 AM
Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?


Why would you want the binary value over hex?

I think the scanning process will be faster.
9  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 22, 2022, 07:12:52 AM
Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?
10  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 21, 2022, 09:45:28 PM
Hi,
Where can I find python software to create private key with publickey?
It may be in other software.

thank you.



I wonder if there is bruteforce software available for python by giving short distances (hex or binnary).
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