Bitcoin puzzle transaction ~32 BTC prize to who solves it

[7isce]

[TARGET: 1] [SPEED: 1632701.66 MKey/s] [TOTAL: 48,937,041,920] [00:05:18]
search 48,937,041,920 keys in 318 seconds

48,937,041,920 / 318
So the real speed is 153,890,068 ~= 153 MKey/s

[1714631787]

1714631787

[1714631787]

1714631787

[1714631787]

1714631787

[hari1987]

...

it would be awesome if there's a CUDA version for it.

I can give you this same script that works in C++ but you have to do the GPU part yourself.CUDA programming can be complex, and proper error handling and synchronization are crucial. Also, not all parts of your program may benefit from GPU acceleration, so it's essential to profile and optimize as needed.

Code:

import bit
import hashlib, random
import platform
from time import time
import os
import sys
import ctypes

nbits = 130
low = 2**(nbits-1)
high = -1+2**nbits
diff = high - low

filename ='tes.bin'
with open(filename,'rb') as f:
    add = f.read()#.split()
#add = set(add)

if platform.system().lower().startswith('win'):
    dllfile = 'ice_secp256k1.dll'
    if os.path.isfile(dllfile) == True:
        pathdll = os.path.realpath(dllfile)
        ice = ctypes.CDLL(pathdll)
    else:
        print('File {} not found'.format(dllfile))

elif platform.system().lower().startswith('lin'):
    dllfile = 'ice_secp256k1.so'
    if os.path.isfile(dllfile) == True:
        pathdll = os.path.realpath(dllfile)
        ice = ctypes.CDLL(pathdll)
    else:
        print('File {} not found'.format(dllfile))
else:
    print('[-] Unsupported Platform currently for ctypes dll method. Only [Windows and Linux] is working')
    sys.exit()

ice.scalar_multiplication.argtypes = [ctypes.c_char_p, ctypes.c_char_p]            # pvk,ret
ice.point_subtraction.argtypes = [ctypes.c_char_p, ctypes.c_char_p, ctypes.c_char_p, ctypes.c_char_p, ctypes.c_char_p]  # x1,y1,x2,y2,ret
ice.init_secp256_lib()

def scalar_multiplication(kk):
    res = (b'\x00') * 65
    pass_int_value = hex(kk)[2:].encode('utf8')
    ice.scalar_multiplication(pass_int_value, res)
    return res

def point_subtraction(pubkey1_bytes, pubkey2_bytes):
    x1 = pubkey1_bytes[1:33]
    y1 = pubkey1_bytes[33:]
    x2 = pubkey2_bytes[1:33]
    y2 = pubkey2_bytes[33:]
    res = (b'\x00') * 65
    ice.point_subtraction(x1, y1, x2, y2, res)
    return res

def new_pos(full_bytes):
    pos = hashlib.sha256(full_bytes).digest()
    return pos

def fixrange(full_bytes):
    t = low + int(full_bytes.hex(), 16) % diff
    return t

def pub2upub(pub_hex):
    x = int(pub_hex[2:66], 16)
    if len(pub_hex) < 70:
        y = bit.format.x_to_y(x, int(pub_hex[:2], 16) % 2)
    else:
        y = int(pub_hex[66:], 16)
    return bytes.fromhex('04' + hex(x)[2:].zfill(64) + hex(y)[2:].zfill(64))

def upub2cpub(upub_bytes):
    x1 = upub_bytes[1:33]
    prefix = str(2 + int(upub_bytes[33:].hex(), 16) % 2).zfill(2)
    return bytes.fromhex(prefix) + x1

st = time()
key_seed = b''
m = 1
while True:
    pubkey = "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852"
    P = pub2upub(pubkey)
    key_seed = new_pos(key_seed)
    qfix = fixrange(key_seed)
    #qfix = m * 1000000  # Use an interval of 1000000 for qfix (stride)
    tpub = bytes(bytearray(scalar_multiplication(qfix)))
    subP = bytes(bytearray(point_subtraction(P, tpub)))
    cpub = bytes(upub2cpub(subP))
    m += 1

    msg = 'Test Cpub : {total}, {num}, {password} '.format(total=m, num=qfix, password=bytes(cpub).hex())
    sys.stdout.write('\r' + msg)
    sys.stdout.flush()

    if cpub in add:
        print("Winner Found!:{num}, {password} ".format(num=qfix, password=bytes(cpub).hex()))
        f = open (u"Winner.txt","a")
        f.write("num:" + str(qfix) +'\n' +
                "cpub:" + str(bytes(cpub).hex())+ '\n\n')
        f.close()
        break
print('[-] Completed in {0:.2f} sec'.format(time() - st))

can you give me the same script that works in cpp? Thanks

[mcdouglasx]

Code:

import secp256k1 as ice
import random

print("scaning pub 03...")

target = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"

start= 50000000000000000000
end=   70000000000000000000
while True:
    
    A0 = random.randint(start, end)
    A1 = ice.scalar_multiplication(A0)
    B0 = ice.to_cpub(A1.hex())
    
    if B0.startswith("03"):
        A2 = ice.pubkey_to_address(0,1, A1)
        print(A2)
        if target in A2:
            print("venga rata")
            data = open("Win.txt","a")
            data.write(str(A0)+" = "+A2+"\n")
            data.close()

Same script, but with concurrent.futures.

Code:

import concurrent.futures
import sys
import os
import time
import secp256k1 as ice
import random
import multiprocessing


os.system("clear");t = time.ctime();sys.stdout.write(f"\033[?25l")
sys.stdout.write(f"\033[01;33m[+] {t}\n")

def generate_key():
    start = 36893488147419103231
    end = 73786976294838206463
    
    A0 = random.randint(start, end)
    A1 = ice.scalar_multiplication(A0)
    B0 = ice.to_cpub(A1.hex())
    
    if B0.startswith("03"):
        A2 = ice.pubkey_to_address(0, 1, A1)
        message = "[+] {}".format(A2);messages = [];messages.append(message);output = ''.join(messages) + "\r";sys.stdout.write(output);sys.stdout.flush()
        return A0, A2
    else:
        return None, None

def main():
    target = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"
    
    num_cores = multiprocessing.cpu_count()
    
    with concurrent.futures.ProcessPoolExecutor(max_workers=num_cores) as executor:
        while True:
            futures = [executor.submit(generate_key) for _ in range(num_cores)]
            
            for future in concurrent.futures.as_completed(futures):
                A0, A2 = future.result()
                
                if A2 and target in A2:
                    print("venga rata")
                    with open("Win.txt", "a") as data:
                        data.write(f"{A0} = {A2}\n")
                    break

if __name__ == "__main__":
    main()

You can imagine how this works on a 128 Core machine

I use Python for testing or when speed is not required, C is better when it comes to speed, I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds, really this part is not the curious one, the curious thing My script has a speed of 2048keys/s, how does it manage to find a key in that range in such a short time with that speed? There is a mathematical trick involved in it, I'm starting to write it in C, to focus on the puzzle 130, once I finish it, if this does not pose a security risk for bitcoin I will make it free.

[AlanJohnson]

I will make it free.

And if it does, what would you do then, sell it? Well you can't trade diamond with glass. If you found it, we will make a double team to take over the world. Lol I will bring coffee, make food, clean your office etc.
Over my dead body if I let anyone pose a threat to Bitcoin.😉

I swear this thread gets weirder day by day...

[mcdouglasx]

I will make it free.

And if it does, what would you do then, sell it? Well you can't trade diamond with glass. If you found it, we will make a double team to take over the world. Lol I will bring coffee, make food, clean your office etc.
Over my dead body if I let anyone pose a threat to Bitcoin.😉

I meant that if the script does not represent a threat, I will share it.
because if with a python script at a speed of 2048 keys/s it finds publick key in the range of puzzle #30 in a few seconds, with C I have no idea where this can go at the moment, I have to do tests in C.

[mcdouglasx]

Well this was supposed to be a community effort, now that you got something promising, it requires more tests? Just gimme it man, I promise not to share it with anyone else.

We go like this > you take 130, after 1/5 year I take 135, after 2 years after that you take 140, and when we empty 160, we come here and ask for more puzzles if by then we were still alive and kicking.😂

That's how it works, you discover a method, you claim everything you can, when you can't claim more, you share it.
Why does it happen? Because the majority of those who benefit from this do not look back to thank them with a simple tip.

The tycoons roam around here and don't even comment, they don't contribute anything, they just wait for something new to apply their great computing power, and they don't even bother to reveal the keys.

It is not worth revealing something here and having none of those present take advantage of it because we lack computing power.
Basically at this point it is giving more money to the rich.

[nomachine]

I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds

Just 10 seconds?

Here's a script that does it in 2 seconds. In Python.

Code:

import time, random, sys, os, secp256k1 as ice
puzzle = 30
target = "1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
if os.name=='nt':os.system('cls')
else:os.system('clear')
t = time.ctime();sys.stdout.write(f"\033[?25l");sys.stdout.write(f"\033[01;33m[+] STARTED: {t}\n")
sys.stdout.write(f"[+] Puzzle: {puzzle}\n")
sys.stdout.write(f"[+] Lower range limit: {lower_range_limit}\n")
sys.stdout.write(f"[+] Upper range limit: {upper_range_limit}\n")
while True:
    constant_prefix = b'yx\xcb\x08\xb70'
    prefix_length = len(constant_prefix);length = 8
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    caddr = ice.privatekey_to_address(0, True, dec)
    message = "\r[+] {}".format(dec);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if caddr == target:
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX);t = time.ctime()
        print(f'[+] SOLVED:  {t}');print(f'[+] Bitcoin address Compressed: {caddr}')
        print(f'[+] Private key (wif) Compressed: {wifc}');print(f'[+] Random Seed: {random_bytes}')
        break

• STARTED: Sat Nov 25 05:07:51 2023
• Puzzle: 30
• Lower range limit: 536870912
• Upper range limit: 1073741823
• SOLVED:  Sat Nov 25 05:07:52 2023
• Bitcoin address Compressed: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
• Private key (wif) Compressed: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M8diLSC5MyERoW
• Random Seed: b'yx\xcb\x08\xb70l\xf1'

 

[mcdouglasx]

I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds

Just 10 seconds?

Here's a script that does it in 2 seconds. In Python.

Code:

import time, random, sys, os, secp256k1 as ice
puzzle = 30
target = "1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
if os.name=='nt':os.system('cls')
else:os.system('clear')
t = time.ctime();sys.stdout.write(f"\033[?25l");sys.stdout.write(f"\033[01;33m[+] STARTED: {t}\n")
sys.stdout.write(f"[+] Puzzle: {puzzle}\n")
sys.stdout.write(f"[+] Lower range limit: {lower_range_limit}\n")
sys.stdout.write(f"[+] Upper range limit: {upper_range_limit}\n")
while True:
    constant_prefix = b'yx\xcb\x08\xb70'
    prefix_length = len(constant_prefix);length = 8
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    caddr = ice.privatekey_to_address(0, True, dec)
    message = "\r[+] {}".format(dec);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if caddr == target:
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX);t = time.ctime()
        print(f'[+] SOLVED:  {t}');print(f'[+] Bitcoin address Compressed: {caddr}')
        print(f'[+] Private key (wif) Compressed: {wifc}');print(f'[+] Random Seed: {random_bytes}')
        break

• STARTED: Sat Nov 25 05:07:51 2023
• Puzzle: 30
• Lower range limit: 536870912
• Upper range limit: 1073741823
• SOLVED:  Sat Nov 25 05:07:52 2023
• Bitcoin address Compressed: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
• Private key (wif) Compressed: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M8diLSC5MyERoW
• Random Seed: b'yx\xcb\x08\xb70l\xf1'

 

I think you're missing the point, obviously there are tools that analyze millions of keys per second and will almost instantly find the key...

but my script is not based on speed, 2048 keys/s is nothing, but nevertheless it successfully finds the keys in less than 10 seconds.

That's why I say there is mathematics behind this, I think you don't understand the magnitude of this.

If with 2048keys/s I find bit 30, what do you think will happen when I analyze 10mkeys/s, 100mkeys/s.

I hope this time I explained it well.

[nomachine]

I think you're missing the point, obviously there are tools that analyze millions of keys per second and will almost instantly find the key...

but my script is not based on speed, 2048 keys/s is nothing, but nevertheless it successfully finds the keys in less than 10 seconds.

That's why I say there is mathematics behind this, I think you don't understand the magnitude of this.

If with 2048keys/s I find bit 30, what do you think will happen when I analyze 10mkeys/s, 100mkeys/s.

I hope this time I explained it well.

You explained well. The point is that I still think that Random Seed is the solution to solve any Puzzle in a few seconds.

[mcdouglasx]

I think you're missing the point, obviously there are tools that analyze millions of keys per second and will almost instantly find the key...

but my script is not based on speed, 2048 keys/s is nothing, but nevertheless it successfully finds the keys in less than 10 seconds.

That's why I say there is mathematics behind this, I think you don't understand the magnitude of this.

If with 2048keys/s I find bit 30, what do you think will happen when I analyze 10mkeys/s, 100mkeys/s.

I hope this time I explained it well.

You explained well. The point is that I still think that Random Seed is the solution to solve any Puzzle in a few seconds.

ok, delete the "while true:" and write "for i in range(1,20480):"
Then tell me if your search is more effective.

[citb0in]

Code:

import secp256k1 as ice
import random

print("scaning pub 03...")

target = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"

start= 50000000000000000000
end=   70000000000000000000
while True:
    
    A0 = random.randint(start, end)
    A1 = ice.scalar_multiplication(A0)
    B0 = ice.to_cpub(A1.hex())
    
    if B0.startswith("03"):
        A2 = ice.pubkey_to_address(0,1, A1)
        print(A2)
        if target in A2:
            print("venga rata")
            data = open("Win.txt","a")
            data.write(str(A0)+" = "+A2+"\n")
            data.close()

Same script, but with concurrent.futures.

Code:

import concurrent.futures
import sys
import os
import time
import secp256k1 as ice
import random
import multiprocessing


os.system("clear");t = time.ctime();sys.stdout.write(f"\033[?25l")
sys.stdout.write(f"\033[01;33m[+] {t}\n")

def generate_key():
    start = 36893488147419103231
    end = 73786976294838206463
    
    A0 = random.randint(start, end)
    A1 = ice.scalar_multiplication(A0)
    B0 = ice.to_cpub(A1.hex())
    
    if B0.startswith("03"):
        A2 = ice.pubkey_to_address(0, 1, A1)
        message = "[+] {}".format(A2);messages = [];messages.append(message);output = ''.join(messages) + "\r";sys.stdout.write(output);sys.stdout.flush()
        return A0, A2
    else:
        return None, None

def main():
    target = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"
    
    num_cores = multiprocessing.cpu_count()
    
    with concurrent.futures.ProcessPoolExecutor(max_workers=num_cores) as executor:
        while True:
            futures = [executor.submit(generate_key) for _ in range(num_cores)]
            
            for future in concurrent.futures.as_completed(futures):
                A0, A2 = future.result()
                
                if A2 and target in A2:
                    print("venga rata")
                    with open("Win.txt", "a") as data:
                        data.write(f"{A0} = {A2}\n")
                    break

if __name__ == "__main__":
    main()

You can imagine how this works on a 128 Core machine

I use Python for testing or when speed is not required, C is better when it comes to speed, I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds, really this part is not the curious one, the curious thing My script has a speed of 2048keys/s, how does it manage to find a key in that range in such a short time with that speed? There is a mathematical trick involved in it, I'm starting to write it in C, to focus on the puzzle 130, once I finish it, if this does not pose a security risk for bitcoin I will make it free.

those scripts (both) don't work as expected on my side. I tested with puzzle 30, even with 15 which should be cracked in fractions of seconds.

Puzzle 15
address = 1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW
start = 16384
end = 32768

Puzzle 30
address = 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
start = 536870912
end = 1073741824

no hit at all

[nomachine]

ok, delete the "while true:" and write "for i in range(1,20480):"
Then tell me if your search is more effective.

This is not the way how random seed can be hacked. .

Code:

import secp256k1 as ice
import random

print("scaning pub 03...")

target = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"

start= 50000000000000000000
end=   70000000000000000000
while True:
    
    A0 = random.randint(start, end)
    A1 = ice.scalar_multiplication(A0)
    B0 = ice.to_cpub(A1.hex())
    
    if B0.startswith("03"):
        A2 = ice.pubkey_to_address(0,1, A1)
        print(A2)
        if target in A2:
            print("venga rata")
            data = open("Win.txt","a")
            data.write(str(A0)+" = "+A2+"\n")
            data.close()

Same script, but with concurrent.futures.

Code:

import concurrent.futures
import sys
import os
import time
import secp256k1 as ice
import random
import multiprocessing


os.system("clear");t = time.ctime();sys.stdout.write(f"\033[?25l")
sys.stdout.write(f"\033[01;33m[+] {t}\n")

def generate_key():
    start = 36893488147419103231
    end = 73786976294838206463
    
    A0 = random.randint(start, end)
    A1 = ice.scalar_multiplication(A0)
    B0 = ice.to_cpub(A1.hex())
    
    if B0.startswith("03"):
        A2 = ice.pubkey_to_address(0, 1, A1)
        message = "[+] {}".format(A2);messages = [];messages.append(message);output = ''.join(messages) + "\r";sys.stdout.write(output);sys.stdout.flush()
        return A0, A2
    else:
        return None, None

def main():
    target = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"
    
    num_cores = multiprocessing.cpu_count()
    
    with concurrent.futures.ProcessPoolExecutor(max_workers=num_cores) as executor:
        while True:
            futures = [executor.submit(generate_key) for _ in range(num_cores)]
            
            for future in concurrent.futures.as_completed(futures):
                A0, A2 = future.result()
                
                if A2 and target in A2:
                    print("venga rata")
                    with open("Win.txt", "a") as data:
                        data.write(f"{A0} = {A2}\n")
                    break

if __name__ == "__main__":
    main()

You can imagine how this works on a 128 Core machine

I use Python for testing or when speed is not required, C is better when it comes to speed, I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds, really this part is not the curious one, the curious thing My script has a speed of 2048keys/s, how does it manage to find a key in that range in such a short time with that speed? There is a mathematical trick involved in it, I'm starting to write it in C, to focus on the puzzle 130, once I finish it, if this does not pose a security risk for bitcoin I will make it free.

those scripts (both) don't work as expected on my side. I tested with puzzle 30, even with 15 which should be cracked in fractions of seconds.

Puzzle 15
address = 1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW
start = 16384
end = 32768

Puzzle 30
address = 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
start = 536870912
end = 1073741824

no hit at all

You have to change   if B0.startswith("03"):   to    if B0.startswith("02"):  for Puzzle 15  

It might get better

[citb0in]

good catch

[Feron]

I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds

Just 10 seconds?

Here's a script that does it in 2 seconds. In Python.

Code:

import time, random, sys, os, secp256k1 as ice
puzzle = 30
target = "1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
if os.name=='nt':os.system('cls')
else:os.system('clear')
t = time.ctime();sys.stdout.write(f"\033[?25l");sys.stdout.write(f"\033[01;33m[+] STARTED: {t}\n")
sys.stdout.write(f"[+] Puzzle: {puzzle}\n")
sys.stdout.write(f"[+] Lower range limit: {lower_range_limit}\n")
sys.stdout.write(f"[+] Upper range limit: {upper_range_limit}\n")
while True:
    constant_prefix = b'yx\xcb\x08\xb70'
    prefix_length = len(constant_prefix);length = 8
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    caddr = ice.privatekey_to_address(0, True, dec)
    message = "\r[+] {}".format(dec);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if caddr == target:
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX);t = time.ctime()
        print(f'[+] SOLVED:  {t}');print(f'[+] Bitcoin address Compressed: {caddr}')
        print(f'[+] Private key (wif) Compressed: {wifc}');print(f'[+] Random Seed: {random_bytes}')
        break

• STARTED: Sat Nov 25 05:07:51 2023
• Puzzle: 30
• Lower range limit: 536870912
• Upper range limit: 1073741823
• SOLVED:  Sat Nov 25 05:07:52 2023
• Bitcoin address Compressed: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
• Private key (wif) Compressed: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M8diLSC5MyERoW
• Random Seed: b'yx\xcb\x08\xb70l\xf1'

 

nice code, this version goes 3 times faster

Code:

import random, os, secp256k1 as ice
puzzle = 30
#target = "1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps"
target = "d39c4704664e1deb76c9331e637564c257d68a08"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
for x in range(1000000):
 constant_prefix = b'yx\xcb\x08\xb70'
 prefix_length = len(constant_prefix);length = 8
 ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
 random_bytes = constant_prefix + ending_bytes
 random.seed(random_bytes)
 dec = random.randint(lower_range_limit, upper_range_limit)
 caddr = ice.privatekey_to_h160(0, True, dec).hex()
 if caddr == target:
  print(caddr,dec)

[nomachine]

I am currently working on a method that allows you to find a key in bit30 in less than 10 seconds

Just 10 seconds?

Here's a script that does it in 2 seconds. In Python.

Code:

import time, random, sys, os, secp256k1 as ice
puzzle = 30
target = "1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
if os.name=='nt':os.system('cls')
else:os.system('clear')
t = time.ctime();sys.stdout.write(f"\033[?25l");sys.stdout.write(f"\033[01;33m[+] STARTED: {t}\n")
sys.stdout.write(f"[+] Puzzle: {puzzle}\n")
sys.stdout.write(f"[+] Lower range limit: {lower_range_limit}\n")
sys.stdout.write(f"[+] Upper range limit: {upper_range_limit}\n")
while True:
    constant_prefix = b'yx\xcb\x08\xb70'
    prefix_length = len(constant_prefix);length = 8
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    caddr = ice.privatekey_to_address(0, True, dec)
    message = "\r[+] {}".format(dec);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if caddr == target:
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX);t = time.ctime()
        print(f'[+] SOLVED:  {t}');print(f'[+] Bitcoin address Compressed: {caddr}')
        print(f'[+] Private key (wif) Compressed: {wifc}');print(f'[+] Random Seed: {random_bytes}')
        break

• STARTED: Sat Nov 25 05:07:51 2023
• Puzzle: 30
• Lower range limit: 536870912
• Upper range limit: 1073741823
• SOLVED:  Sat Nov 25 05:07:52 2023
• Bitcoin address Compressed: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
• Private key (wif) Compressed: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M8diLSC5MyERoW
• Random Seed: b'yx\xcb\x08\xb70l\xf1'

 

nice code, this version goes 3 times faster

Code:

import random, os, secp256k1 as ice
puzzle = 30
#target = "1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps"
target = "d39c4704664e1deb76c9331e637564c257d68a08"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
for x in range(1000000):
 constant_prefix = b'yx\xcb\x08\xb70'
 prefix_length = len(constant_prefix);length = 8
 ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
 random_bytes = constant_prefix + ending_bytes
 random.seed(random_bytes)
 dec = random.randint(lower_range_limit, upper_range_limit)
 caddr = ice.privatekey_to_h160(0, True, dec).hex()
 if caddr == target:
  print(caddr,dec)
  break

The problem is that there has to be a method to find out what a random seed is.
For each Puzzle i start from 0 - constant_prefix = b''
for x in range won't work above puzzle 40 - it needs at least 200 million seeds to go through the script.

For example here I am hunting random seed for Puzzle 65:

Code:

import time, random, sys, os, secp256k1 as ice
puzzle = 65
target = 30568377312064202855
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1

while True:
    constant_prefix = b'\xc9\xd9\x1d\xbc\x16\x9d'
    prefix_length = len(constant_prefix);length = 8
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    if str(dec).startswith("30568377"):
        message = "\r[+] {} , {}".format(dec, random_bytes);messages = []
        messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
        sys.stdout.write(output);sys.stdout.flush()
        if dec == target:
            caddr = ice.privatekey_to_address(0, True, dec)
            HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX);t = time.ctime()
            print(f'[+] SOLVED:  {t}');print(f'[+] Bitcoin address Compressed: {caddr}')
            print(f'[+] Private key (wif) Compressed: {wifc}');print(f'[+] Random Seed: {random_bytes}')
            break

First you search for the seed  with - constant_prefix = b'' - for the first 8 numbers -  if str(dec).startswith("30568377") - then you write the result in the script itself - constant_prefix = b'\xc9\xd9\x1d\xbc\x16\x9d'
Then you go to 10 (removing manually the last two bytes at the end) and so on until you hit  full WIF.

The closest I've come up  to 2 ** 65

30568377238562584866, b'\xc9\xd9\x1d\xbc\x16\x9d\xdb\x86'

But what  if is not  length = 8 ? can be 9, 10, 11 ....and so on  

What if hypothetically there is a common seed for all puzzles, let's say on the seed  length = 28  ?
I'll never know because I've never even attempted that length

[nomachine]

The only promising code I have seen in a long time is the one mc posted,  searching for either 02, or 03 public keys, if it was 02, ignore and discard, if it was 03, generate rmd160 to compare. That's the half of key space.

You can combine that method together with seed search. Seed can be found even faster. There is NO limitation in terms of what can be added as an automation of this process.

Code:

import time, random, sys, os, secp256k1 as ice
puzzle = 65
target_h160 = "52e763a7ddc1aa4fa811578c491c1bc7fd570137"
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
while True:
    constant_prefix = b''
    prefix_length = len(constant_prefix);length = 9 #let's play with 9 
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    A1 = ice.scalar_multiplication(dec)
    B0 = ice.to_cpub(A1.hex())    
    if B0.startswith("02"):  #"02" for Puzzle 65
        if str(dec).startswith("30568377"):
           message = "\r[+] {} , {}".format(dec, random_bytes);messages = []
           messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
           sys.stdout.write(output);sys.stdout.flush()
           h160 = ice.privatekey_to_h160(0, True, dec).hex()
           if h160 == target_h160:
               HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX);t = time.ctime()
               print(f'[+] SOLVED:  {t}')
               print(f'[+] Private key (wif) Compressed: {wifc}');print(f'[+] Random Seed: {random_bytes}')
               break

[mcdouglasx]

ok, delete the "while true:" and write "for i in range(1,20480):"
Then tell me if your search is more effective.

This is not the way how random seed can be hacked. .

is what I'm talking about, imagine finding publikeys in bit30 with less than 20480 keys scanned, conventional methods are no longer convenient due to the difficulty of the puzzle.
It's time for new things.

[nomachine]

It's time for new things.

I'm here for it!
Is this where I get my cape and superhero mask for the new adventure?

[mcdouglasx]

The only promising code I have seen in a long time is the one mc posted,  searching for either 02, or 03 public keys, if it was 02, ignore and discard, if it was 03, generate rmd160 to compare. That's the half of key space.

If we want to try our luck, we can add to this a function that only searches for even or odd numbers in the Pk + search for "02" or "03" and we would be dividing the area into 4.
but it's lucky that you find the right combination

[nomachine]

The only promising code I have seen in a long time is the one mc posted,  searching for either 02, or 03 public keys, if it was 02, ignore and discard, if it was 03, generate rmd160 to compare. That's the half of key space.

search for "02" or "03" and we would be dividing the area into 4.
but it's lucky that you find the right combination

I have that option in this script.  

You can change as you like:
public_key_hex = ('02' and '03' if compressed else '04') + format(public_key[0], '064x')  '02' and '03' together
public_key_hex = ('02' or '03' if compressed else '04') + format(public_key[0], '064x')  '02' or '03' /  '03' or '02'
public_key_hex = ('02' if compressed else '04') + format(public_key[0], '064x')  only '02'
public_key_hex = ('03' if compressed else '04') + format(public_key[0], '064x')  only '03'

No secp256k1 as ice, no ecdsa here.... Even encode base58 is custom with mpz.

And it's not slower than iceland - it's the same...

Code:

import hashlib, sys, os, time, random, gmpy2
from functools import lru_cache

# Constants as mpz
Gx = gmpy2.mpz('0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798', 16)
Gy = gmpy2.mpz('0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8', 16)
p = gmpy2.mpz('0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F', 16)
n = gmpy2.mpz('0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141', 16)

@lru_cache(maxsize=None)
def private_key_to_public_key(private_key):
    Q = point_multiply(Gx, Gy, private_key, p)
    return Q

def point_multiply(x, y, k, p):
    result = (gmpy2.mpz(0), gmpy2.mpz(0))
    addend = (x, y)
   
    while k > 0:
        if k & 1:
            result = point_add(result, addend, p)
        addend = point_double(addend, p)
        k >>= 1

    return result

def point_double(point, p):
    x, y = point
    lmbda = (3 * x * x * gmpy2.powmod(2 * y, -1, p)) % p
    x3 = (lmbda * lmbda - 2 * x) % p
    y3 = (lmbda * (x - x3) - y) % p
    return x3, y3

def point_add(point1, point2, p):
    x1, y1 = point1
    x2, y2 = point2

    if point1 == (gmpy2.mpz(0), gmpy2.mpz(0)):
        return point2
    if point2 == (gmpy2.mpz(0), gmpy2.mpz(0)):
        return point1

    if point1 != point2:
        lmbda = ((y2 - y1) * gmpy2.powmod(x2 - x1, -1, p)) % p
    else:
        lmbda = ((3 * x1 * x1) * gmpy2.powmod(2 * y1, -1, p)) % p

    x3 = (lmbda * lmbda - x1 - x2) % p
    y3 = (lmbda * (x1 - x3) - y1) % p
    return x3, y3

def encode_base58(byte_str):
    __b58chars = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'
    __b58base = len(__b58chars)
    long_value = gmpy2.mpz(int.from_bytes(byte_str, byteorder='big'))
    result = ''
    while long_value >= __b58base:
        div, mod = gmpy2.f_divmod(long_value, __b58base)
        result = __b58chars[int(mod)] + result
        long_value = div
    result = __b58chars[int(long_value)] + result

    # Add leading '1's for zero bytes
    nPad = 0
    for byte in byte_str:
        if byte == 0:
            nPad += 1
        else:
            break

    return __b58chars[0] * nPad + result

def public_key_to_hex(public_key, compressed=True):
    x_hex = format(public_key[0], '064x')[2:]  # Remove '0x' prefix
    if compressed:
        return ('02' if public_key[1] % 2 == 0 else '03') + x_hex

def public_key_to_address(public_key, compressed=True):
    public_key_hex = ('02' and '03' if compressed else '04') + format(public_key[0], '064x')
    sha256_hash = hashlib.sha256(bytes.fromhex(public_key_hex)).digest()
    ripemd160_hash = hashlib.new('ripemd160', sha256_hash).digest()
    versioned_hash = (b'\x00' if compressed else b'\x04') + ripemd160_hash
    checksum = hashlib.sha256(hashlib.sha256(versioned_hash).digest()).digest()[:4]
    address_bytes = versioned_hash + checksum
    return encode_base58(address_bytes)

# Configuration for the puzzle
puzzle = 30
add = '1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps'
lower_range_limit = 2 ** (puzzle - 1);upper_range_limit = (2 ** puzzle) - 1
if os.name=='nt':os.system('cls')
else:os.system('clear')
t = time.ctime();sys.stdout.write(f"\033[?25l");sys.stdout.write(f"\033[01;33m[+] STARTED: {t}\n")
sys.stdout.write(f"[+] Puzzle: {puzzle}\n")
sys.stdout.write(f"[+] Lower range limit: {lower_range_limit}\n")
sys.stdout.write(f"[+] Upper range limit: {upper_range_limit}\n")

while True:
    constant_prefix = b'yx\xcb\x08\xb70'
    prefix_length = len(constant_prefix);length = 8
    ending_length = length - prefix_length;ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    key = random.randint(lower_range_limit, upper_range_limit)
    public_key = private_key_to_public_key(key)
    bitcoin_address = public_key_to_address(public_key, compressed=True)
    public_key = public_key_to_hex(public_key)
    message = "[+] {}".format(key);messages = [];messages.append(message);output = ''.join(messages) + "\r";sys.stdout.write(output);sys.stdout.flush()
    if bitcoin_address == add:
        t = time.ctime()
        sys.stdout.write("\n")
        sys.stdout.write(f"\033[01;32m[+] SOLVED:  {t}\n[+] Private Key (dec): {key}\n[+] Random Seed: {random_bytes}\033[0m\n")
        with open('KEYFOUNDKEYFOUND.txt', 'a') as file:
            file.write(f"Private Key (dec): {key}\nBitcoin Address (Compressed): {bitcoin_address}\nPublic Key: {public_key}\nRandom Seed: {random_bytes}\n\n")
        break

• STARTED: Sat Nov 25 16:05:19 2023
• Puzzle: 30
• Lower range limit: 536870912
• Upper range limit: 1073741823
• 1033162084
• SOLVED:  Sat Nov 25 16:05:21 2023
• Private Key (dec): 1033162084