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Topic: 🖤 (Read 211 times)

copper member
Activity: 1330
Merit: 899
🖤😏
September 23, 2023, 03:35:16 AM
#4
This script can generate any point based on your desired curve parameters, you can set p, n, gx, gy, and private key then run to see the results. Note that you can't select random p and n, you could play with parameters and see the result.
Changing curve parameters
Code:
#!/usr/bin/env python3

import collections
import hashlib

# ...

def point_add(point1, point2):
    if point1 is None:
        return point2
    if point2 is None:
        return point1

    x1, y1 = point1
    x2, y2 = point2

    if x1 == x2 and y1 != y2:
        return None

    if point1 == point2:
        m = (3 * x1 * x1 + curve.a) * pow(2 * y1, -1, curve.p)
    else:
        m = (y2 - y1) * pow(x2 - x1, -1, curve.p)

    x3 = (m * m - x1 - x2) % curve.p
    y3 = (m * (x1 - x3) - y1) % curve.p

    return (x3, y3)

# ... (Rest of your code remains the same)


EllipticCurve = collections.namedtuple('EllipticCurve', 'name p a b g n h')

curve = EllipticCurve(
    'secp256k1',
    # Field characteristic.
    p=0x7027,
    # Curve coefficients.
    a=0,
    b=7,
    # Base point.
    g=(0x1,
       0x2),
    # Subgroup order.
    n=0x6997,
    # Subgroup cofactor.
    h=1,
)

# Functions that work on curve points #

def is_on_curve(point):
    """Returns True if the given point lies on the elliptic curve."""
    if point is None:
        # None represents the point at infinity.
        return True

    x, y = point

    return (y * y - x * x * x - curve.a * x - curve.b) % curve.p == 0

def scalar_mult(k, point):
    """Returns k * point computed using the double and point_add algorithm."""

    if k % curve.n == 0 or point is None:
        return None

    if k < 0:
        # k * point = -k * (-point)
        return scalar_mult(-k, point_neg(point))

    result = None
    addend = point

    while k:
        if k & 1:
            # Add.
            result = point_add(result, addend)

        # Double.
        addend = point_add(addend, addend)

        k >>= 1


    return result

# Private key
private_key = 3

# Generate public key from the private key
public_key = scalar_mult(private_key, curve.g)

print('Curve: ', curve.name)
print('Generator point:\n(0x{:x},\n 0x{:x})'.format(*curve.g))
print('\n\n')
print('Private key:\n', private_key)
print('\n')
print('Public key:\n(0x{:x},\n 0x{:x})'.format(*public_key))

It will print g, private key and new x, y public key coordinates for your input key.

Credit: I just know I changed a few things, but I didn't write equations.



Edit:
Finding prime (P) for 0, 7, a, b curve
This one is much faster in finding prime candidates for a given curve.

Code:
import random
import math

def is_prime(n, k=10000):
    """Test if a number is prime using Miller-Rabin test."""
    if n < 2:
        return False
    # Check for small primes
    for p in (2, 3, 5, 7, 11, 13, 17, 19, 23, 29):
        if n % p == 0:
            return n == p
    # Perform Miller-Rabin test
    s, d = 0, n-1
    while d % 2 == 0:
        s, d = s+1, d//2
    for _ in range(k):
        a = random.randint(2, n-2)
        x = pow(a, d, n)
        if x == 1 or x == n-1:
            continue
        for _ in range(s-1):
            x = pow(x, 2, n)
            if x == n-1:
                break
        else:
            return False
    return True

def generate_prime(n):
    """Generate a random n-bit prime using Miller-Rabin test."""
    while True:
        p = random.getrandbits(n)
        # Set the MSB and LSB to 1 to ensure n bits and odd number
        p |= (1 << n-1) | 1
        if is_prime(p):
            return p

# Generate any bit prime
p = generate_prime(256)
print("Generated prime:", p)



Ps, I'm working on a theory to speed up the brute force operations by thousands of %s. By using a small p and converting secp256k1 points in to a much more smaller points representing in a new curve in order to compute +, -, *, / faster. God willing when I find a working script, I will share it here.😉
member
Activity: 189
Merit: 16
September 19, 2023, 09:58:32 AM
#2
You might want to look into SageMath as it's really powerful for experimenting with cryptographic attacks.
copper member
Activity: 1330
Merit: 899
🖤😏
September 18, 2023, 02:13:39 AM
#1
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