BTC_Backdoor
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May 05, 2023, 04:51:57 AM |
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Guys, I hope you are all doing great! Can anyone please refer some python script that can apply this halving thing on public keys and calculates its half? ??
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Lolo54
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May 05, 2023, 09:18:25 AM |
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Guys, I hope you are all doing great! Can anyone please refer some python script that can apply this halving thing on public keys and calculates its half? ?? this from MrMaxwell might suit you very simple to use https://github.com/MrMaxweII/Secp256k1-CalculatorOn the other hand if you want to divide a pubkey which corresponds to an odd pk you will have its half etc etc... deleting the float is another much more complicated story. In short, dividing a pubkey in half will not help you much unless you know the starting pk
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unpluggedcoin
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May 05, 2023, 02:12:47 PM |
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Guys, I hope you are all doing great! Can anyone please refer some python script that can apply this halving thing on public keys and calculates its half? ?? this from MrMaxwell might suit you very simple to use https://github.com/MrMaxweII/Secp256k1-CalculatorOn the other hand if you want to divide a pubkey which corresponds to an odd pk you will have its half etc etc... deleting the float is another much more complicated story. In short, dividing a pubkey in half will not help you much unless you know the starting pk BRO, would you please give a little help how to use this calculator that you've referred above? I mean how to calculate half point, where should I put my public key whose half I am interested in.... Much appreciate your above help though
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brainless
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Activity: 329
Merit: 34
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May 05, 2023, 06:32:20 PM |
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Guys, I hope you are all doing great! Can anyone please refer some python script that can apply this halving thing on public keys and calculates its half? ?? this from MrMaxwell might suit you very simple to use https://github.com/MrMaxweII/Secp256k1-CalculatorOn the other hand if you want to divide a pubkey which corresponds to an odd pk you will have its half etc etc... deleting the float is another much more complicated story. In short, dividing a pubkey in half will not help you much unless you know the starting pk BRO, would you please give a little help how to use this calculator that you've referred above? I mean how to calculate half point, where should I put my public key whose half I am interested in.... Much appreciate your above help though https://rawcdn.githack.com/nlitsme/bitcoinexplainer/aa50e86e8c72c04a7986f5f7c43bc2f98df94107/ecdsacrack.html
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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whanau
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May 06, 2023, 11:57:06 PM |
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chatGPT suggest this.. It won't work because the math to do it (correctly) is not known import os import sys from hashlib import sha256 from binascii import hexlify, unhexlify from ecdsa import SECP256k1, SigningKey, VerifyingKey from ecdsa.ecdsa import generator_secp256k1
def find_generator_point(P): count = 0 G = generator_secp256k1 while P != G: P = P.__rmul__(2) count += 1 with open("divisions.txt", "w") as f: f.write(str(count)) return P
if __name__ == '__main__': # Generate a random private key sk = SigningKey.generate(curve=SECP256k1) # Derive the public key from the private key vk = sk.get_verifying_key() # Convert the public key to an EC point P = vk.pubkey.point # Find the generator point by dividing P repeatedly G = find_generator_point(P) print("Generator Point: ({}, {})".format(hexlify(G.x().to_bytes(32, 'big')).decode(), hexlify(G.y().to_bytes(32, 'big')).decode()))
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digaran
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🖤😏
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May 08, 2023, 12:54:50 AM |
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chatGPT suggest this.. It won't work because the math to do it (correctly) is not known import os import sys from hashlib import sha256 from binascii import hexlify, unhexlify from ecdsa import SECP256k1, SigningKey, VerifyingKey from ecdsa.ecdsa import generator_secp256k1
def find_generator_point(P): count = 0 G = generator_secp256k1 while P != G: P = P.__rmul__(2) count += 1 with open("divisions.txt", "w") as f: f.write(str(count)) return P
if __name__ == '__main__': # Generate a random private key sk = SigningKey.generate(curve=SECP256k1) # Derive the public key from the private key vk = sk.get_verifying_key() # Convert the public key to an EC point P = vk.pubkey.point # Find the generator point by dividing P repeatedly G = find_generator_point(P) print("Generator Point: ({}, {})".format(hexlify(G.x().to_bytes(32, 'big')).decode(), hexlify(G.y().to_bytes(32, 'big')).decode()))
It won't work because that AI is an idiot, lol there is no SHA256 involved in finding k from p, and there is no "dividing" repeatedly until finding G, what AI can suggest is the worst way possible to brute force because it's an imbecile computer.😅
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🖤😏
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lordfrs
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May 11, 2023, 06:08:26 PM |
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chatGPT suggest this.. It won't work because the math to do it (correctly) is not known import os import sys from hashlib import sha256 from binascii import hexlify, unhexlify from ecdsa import SECP256k1, SigningKey, VerifyingKey from ecdsa.ecdsa import generator_secp256k1
def find_generator_point(P): count = 0 G = generator_secp256k1 while P != G: P = P.__rmul__(2) count += 1 with open("divisions.txt", "w") as f: f.write(str(count)) return P
if __name__ == '__main__': # Generate a random private key sk = SigningKey.generate(curve=SECP256k1) # Derive the public key from the private key vk = sk.get_verifying_key() # Convert the public key to an EC point P = vk.pubkey.point # Find the generator point by dividing P repeatedly G = find_generator_point(P) print("Generator Point: ({}, {})".format(hexlify(G.x().to_bytes(32, 'big')).decode(), hexlify(G.y().to_bytes(32, 'big')).decode()))
It won't work because that AI is an idiot, lol there is no SHA256 involved in finding k from p, and there is no "dividing" repeatedly until finding G, what AI can suggest is the worst way possible to brute force because it's an imbecile computer.😅 point division. but you won't be splitting the private key when applied.
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If you want to buy me a coffee
Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7
Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
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