Half of any bitcoin (crypto) public key - (public key half)

[BTC_Backdoor]

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???

[Lolo54]

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-Calculator
On 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

[unpluggedcoin]

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-Calculator
On 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

[brainless]

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-Calculator
On 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

[whanau]

chatGPT suggest this..

It won't work because the math to do it (correctly) is not known

Code:

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()))

[digaran]

chatGPT suggest this..

It won't work because the math to do it (correctly) is not known

Code:

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.😅

[lordfrs]

chatGPT suggest this..

It won't work because the math to do it (correctly) is not known

Code:

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.