Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 148. (Read 215539 times)

class EllipticCurve:
    def __init__(self, p, a, b, g_x, g_y, n):
        self.p = p
        self.a = a
        self.b = b
        self.g_x = g_x
        self.g_y = g_y
        self.n = n 

    def add_points(self, p1, p2):
        # Point addition logic here
        pass 

    def subtract_points(self, p1, p2):
        # Point subtraction logic here
        pass 

    def scalar_multiply(self, point, scalar):
        # Scalar multiplication logic here
        pass 

# Replace these values with your desired parameters
N = ...
P = ...
G = (..., ...)  # (x-coordinate, y-coordinate) 

# Create an instance of the EllipticCurve class with your parameters
curve = EllipticCurve(P, 0, 7, G[0], G[1], N) 

# Replace with your target public key and desired number of subtractions
target_public_key = "..."
num = 100 

# Example of point subtraction loop
subtract_point = G  # Initialize subtract_point with base point G 

for t in range(num + 1):
    # Perform point subtraction
    result = curve.subtract_points(subtract_point, G) 

    # Convert the result to hexadecimal representation
    h = (result[0], result[1]) 

    # Print or store the result as needed
    print("Subtraction result for iteration {}: {}".format(t, h)) 

    # Update subtract_point for the next iteration
    subtract_point = result 

# Note: You need to implement the add_points and subtract_points methods
# with the actual point addition and subtraction logic for your curve.

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


for x in range(1,65536):
         f= (ters(pub,x))
         data= open(“halfpub.txt”,”a”)
         data.write(f+”\n”)
         data.close()

key: 111
found: 00000000000000000000000000000000000000000000000000000004aed21170
Total time: 0 h, 0 m, 2 s

key: 111101
found: 00000000000000000000000000000000000000000000000000000022382facd0
Total time: 0 h, 0 m, 13 s