Pollard Rho Kangaroo Algorithm for 256 bits

[krashfire]

Hi, im hoping to get some advice on how i can fine tune, optimized or enhance my code further to calculate the 256 bit private key.

your advice is very much appreciated.

Code:

import random
from math import gcd

# Function to extract public key coordinates from hex string
def extract_public_key_coordinates(public_key):
    try:
        x, y = public_key.split(',')
        print("Public key coordinates extracted successfully.")
        return int(x, 16), y
    except ValueError:
        print("Error: Invalid public key format.")
        return None, None

# Function for Pollard's Rho factorization
def pollard_rho_factorization(n, limit=1000000):
    print("Performing Pollard's Rho factorization...")
    def f(x):
        return (x ** 2 + 1) % n

    x = random.randint(1, n - 1)
    y = x
    d = 1

    while d == 1:
        x = f(x)
        y = f(f(y))
        d = gcd(abs(x - y), n)

    return d if d != n else None

# Function to perform the Kangaroo algorithm with factors
def kangaroo_algorithm(g, h, n, factors, B):
    print("Performing Kangaroo algorithm...")
    P = 0
    Q = 0
    wild_kangaroos = [(random.randint(1, n - 1), random.randint(1, n - 1)) for _ in range(10)]  # Multiple parallel wild kangaroos
    wild_jump_size = random.randint(1, B // 10)  # Randomized wild kangaroo step size

    while True:
        for wild_x, wild_y in wild_kangaroos:
            u = wild_x
            v = wild_y
            for _ in range(wild_jump_size):
                u = (u * g) % n
                v = (v * h) % n
            d = gcd(u - v, n)
            if 1 < d < n:
                return d
            if d == 1:
                alpha = kangaroo_attack(P, Q, g, n, h, B)  # Using kangaroo_attack
                if alpha is not None:
                    return alpha
                P += B
                Q += B
            else:
                return d

# Simple Kangaroo Attack (Floyd's Cycle Detection Algorithm)
def kangaroo_attack(P, Q, g, n, h, B):
    print("Performing Kangaroo attack...")
    u = P
    v = Q
    u_iterations = 0
    v_iterations = 0

    while True:
        # Tame Kangaroo
        u = (u * g) % n
        u_iterations += 1

        # Wild Kangaroo
        v = (v * h) % n
        v_iterations += 1

        if u == v:
            # Collision detected, solve for alpha
            alpha = (P - Q) * pow(g, -1, n) % n  # Inverse of g modulo n
            return alpha

        if max(u_iterations, v_iterations) > B:
            # Maximum iterations reached without collision
            return None

if __name__ == "__main__":
    # Example usage
    public_key = "d6xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx,3xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
    x, y_hex = extract_public_key_coordinates(public_key)

    if x is not None and y_hex is not None:
        # Convert y-coordinate from hex to integer
        try:
            y = int(y_hex, 16)
            print("Y-coordinate converted to integer successfully.")
        except ValueError:
            print("Error: Unable to convert y-coordinate to integer.")

        # Perform factorization using Pollard's Rho
        try:
            factors = [pollard_rho_factorization(number) for number in [x, y, x + y]]
            print("Factorization complete:", factors)
        except Exception as e:
            print("Error during factorization:", e)

        # Parameters for Kangaroo algorithm
        p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
        n = 115792089237316195423570985008687907852837564279074904382605163141518161494337
        g = 55066263022277343669578718895168534326250603453777594175500187360389116729240
        h = y  # Use the integer value of y
        B = 2**42  # Adjust based on your system's memory constraints

        # Find private key using the Kangaroo algorithm with factors
        try:
            private_key = kangaroo_algorithm(g, h, n, factors, B)
            print("Private Key:", private_key)
        except Exception as e:
            print("Error during Kangaroo algorithm:", e)
    else:
        print("Error: Public key extraction failed.")

[COBRAS]

Imposible to dolve 256 bit

and in python imposible to solve 100 bit

[satashi_nokamato]

How do you set the range to search for the key? Also what is h and which y should we use in this line  "h = y  # Use the integer value of y" ?

[krashfire]

How do you set the range to search for the key? Also what is h and which y should we use in this line  "h = y  # Use the integer value of y" ?

Here's a simplified explanation of how the Kangaroo algorithm works in this context:

Tame Kangaroos: These kangaroos make small jumps and keep track of their positions as they move along the curve. They're responsible for exploring the region near the expected position of
𝑘
k and
𝑄
Q.

Wild Kangaroos: These kangaroos make large jumps and are used to cover a wider range of possible positions of
𝑘
k and
𝑄
Q.

Collision Detection: The algorithm looks for collisions, which occur when a tame kangaroo and a wild kangaroo land on the same point on the curve. This indicates a potential match for
𝑘
k and
𝑄
Q.

Private Key Extraction: Once a collision is detected, you can extract the private key
𝑘
k from the positions of the kangaroos at the collision point.

The key concept here is that you're not searching through a numerical range of
𝑘
k values linearly.

Instead, you're using kangaroos to explore the curve in a smart way, leveraging collisions to pinpoint the private key without exhaustively trying every possible
𝑘
k value.

h = y means the public key xy coordinates. in this part we are just using y.

[nomachine]

I have 205288 hops per second in python on single core my friend......

It will work from 1 - 256 bit . But it is impossible to solve any Puzzle above 50 with this for a lifetime.

Code:

import time
import os
import sys
import random
import gmpy2

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[+] Kangaroo: {t}\n")
sys.stdout.flush()

modulo = gmpy2.mpz(0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F)
order = gmpy2.mpz(0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141)
Gx = gmpy2.mpz(0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798)
Gy = gmpy2.mpz(0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)

# Define Point class
class Point:
    def __init__(self, x=0, y=0):
        self.x = gmpy2.mpz(x)
        self.y = gmpy2.mpz(y)

PG = Point(Gx, Gy)
Z = Point(0, 0)  # zero-point, infinite in real x, y - plane

def add(P, Q, p=modulo):
    if P == Z:
        return Q
    elif Q == Z:
        return P
    elif P.x == Q.x and (P.y != Q.y or P.y == 0):
        return Z
    elif P.x == Q.x:
        m = (3 * P.x * P.x) * gmpy2.invert(2 * P.y, p) % p
    else:
        m = (Q.y - P.y) * gmpy2.invert(Q.x - P.x, p) % p
  
    x = (m * m - P.x - Q.x) % p
    y = (m * (P.x - x) - P.y) % p
    return Point(x, y)

def mul2(P, p=modulo):
    if P == Z:
        return Z
    m = gmpy2.f_mod(3 * P.x * P.x * gmpy2.invert(2 * P.y, p), p)
    x = gmpy2.f_mod(m * m - 2 * P.x, p)
    y = gmpy2.f_mod(m * (P.x - x) - P.y, p)
    return Point(x, y)

def mulk(k, P=PG, p=modulo):
    if k == 0:
        return Z
    elif k == 1:
        return P
    elif k % 2 == 0:
        return mulk(k // 2, mul2(P, p), p)
    else:
        return add(P, mulk((k - 1) // 2, mul2(P, p), p), p)

def X2Y(X, y_parity, p=modulo):
    X_cubed = gmpy2.powmod(X, 3, p)
    X_squared = gmpy2.powmod(X, 2, p)
    tmp = gmpy2.f_mod(X_cubed + 7, p)
    Y = gmpy2.powmod(tmp, gmpy2.f_div(gmpy2.add(p, 1), 4), p)
    if y_parity == 1:
        Y = gmpy2.f_mod(-Y, p)
    return Y


def comparator(A, Ak, B, Bk):
    result = set(A).intersection(set(B))
    if result:
        sol_kt = A.index(next(iter(result)))
        sol_kw = B.index(next(iter(result)))
        HEX = "%064x" % abs(Ak[sol_kt] - Bk[sol_kw])
        dec = int(HEX, 16)
        total_time = time.time() - starttime
        print('\n[+] total time: %.2f sec' % (total_time))
        t = time.ctime()
        print(f"\033[32m[+] PUZZLE SOLVED: {t} \033[0m")
        print(f"\033[32m[+] Private key (dec) : {dec} \033[0m")
        with open("KEYFOUNDKEYFOUND.txt", "a") as file:
            file.write("\n\nSOLVED " + t)
            file.write(f"\nTotal Time: {total_time:.2f} sec")
            file.write(f"\nRandom seed: {seed}")
            file.write("\nPrivate Key (decimal): " + str(dec))
            file.write("\nPrivate Key (hex): " + HEX)
            file.write(
                "\n-------------------------------------------------------------------------------------------------------------------------------------------\n"
            )
        file.close()
        return True
    else:
        return False

def check(P, Pindex, DP_rarity, A, Ak, B, Bk):
    modulo_val = P.x % DP_rarity
    if modulo_val == 0:
        A.append(gmpy2.mpz(P.x))
        Ak.append(gmpy2.mpz(Pindex))
        return comparator(A, Ak, B, Bk) 
    else:
        return False

# Generate a list of powers of two for faster access

def generate_powers_of_two(hop_modulo):
    return [gmpy2.mpz(1 << pw) for pw in range(hop_modulo)]

def search(P, W0, DP_rarity, Nw, Nt, hop_modulo, upper_range_limit, lower_range_limit, powers_of_two):
    solved = False
    t = [gmpy2.mpz(lower_range_limit + gmpy2.mpz(random.randint(0, upper_range_limit - lower_range_limit))) for _ in range(Nt)]
    T = [mulk(ti) for ti in t]
    dt = [gmpy2.mpz(0) for _ in range(Nt)]
    w = [gmpy2.mpz(random.randint(0, upper_range_limit - lower_range_limit)) for _ in range(Nw)]
    W = [add(W0, mulk(wk)) for wk in w]
    dw = [gmpy2.mpz(0) for _ in range(Nw)]
    print('[+] tame and wild herds are prepared')
    Hops, Hops_old = 0, 0
    t0 = time.time()  

    # Memoization dictionary
    memo = {}

    while not solved:
        for k in range(Nt):
            Hops += 1
            pw = T[k].x % hop_modulo
            if pw not in memo:
                memo[pw] = powers_of_two[pw]
            dt[k] = memo[pw]
            solved = check(T[k], t[k], DP_rarity, T, t, W, w)
            if solved: break
            t[k] += dt[k]
            T[k] = add(P[int(pw)], T[k])
        if solved: break
        for k in range(Nw):
            Hops += 1
            pw = W[k].x % hop_modulo
            if pw not in memo:
                memo[pw] = powers_of_two[pw]
            dw[k] = memo[pw]
            solved = check(W[k], w[k], DP_rarity, W, w, T, t)
            if solved: break
            w[k] += dw[k]
            W[k] = add(P[int(pw)], W[k])
        if solved: break
        t1 = time.time()
        if (t1 - t0) > 5:
            print('\r[+] Hops: %.0f h/s' % ((Hops - Hops_old) / (t1 - t0)), end='', flush=True)
            t0 = t1
            Hops_old = Hops
    print('[+] Hops:', Hops)
    return 'sol. time: %.2f sec' % (time.time() - starttime)

# Configuration for the puzzle
puzzle = 40
compressed_public_key = "03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4"
kangaroo_power = 4  
lower_range_limit = 2 ** (puzzle - 1)
upper_range_limit = (2**puzzle) - 1

DP_rarity = 1 << int(((puzzle -  2*kangaroo_power)/2 - 2))
hop_modulo = ((puzzle - 1) // 2) + kangaroo_power
Nt = Nw = 2**kangaroo_power

# Precompute powers of two for faster access
powers_of_two = generate_powers_of_two(hop_modulo)

T, t, dt = [], [], []
W, w, dw = [], [], []

if len(compressed_public_key) == 66:
    X = gmpy2.mpz(compressed_public_key[2:66], 16)
    Y = X2Y(X, gmpy2.mpz(compressed_public_key[:2]) - 2)
else:
    print("[error] pubkey len(66/130) invalid!")

W0 = Point(X,Y)
starttime = oldtime = time.time()

print(f"[+] [Puzzle]: {puzzle}")
print(f"[+] [Lower range limit]: {lower_range_limit}")
print(f"[+] [Upper range limit]: {upper_range_limit}")

#Random seed Config
seed = os.urandom(9).hex()
print(f"[+] [Random seed]: {seed}")
random.seed(seed)

Hops = 0

P = [PG]
for k in range(255): P.append(mul2(P[k]))  
print('[+] P-table prepared')

solved = False
search(P, W0, DP_rarity, Nw, Nt, hop_modulo, upper_range_limit, lower_range_limit, powers_of_two)

print('[+] Average time to solve: %.2f sec' % ((time.time()-starttime)))

• Kangaroo: Thu May 16 13:33:48 2024
• [Puzzle]: 40
• [Lower range limit]: 549755813888
• [Upper range limit]: 1099511627775
• [Random seed]: 9b76a9623bd76fb8cc
• P-table prepared
• tame and wild herds are prepared
• Hops: 205288 h/s
• total time: 6.92 sec
• PUZZLE SOLVED: Thu May 16 13:33:55 2024
• Private key (dec) : 1003651412950
• Hops: 1416585
• Average time to solve: 6.92 sec

It needs to be converted into C++ or Rust so that it can solve problems up to puzzle 120.

[krashfire]

i have created twist attack on ecdsa secp256k1.

can read about Twist Attack on ECDSA SECP256k1 here. https://github.com/demining/Twist-Attack

It gets the partial private key correctly.

my issue now is how to write the code to get to the full private key. PM Me Bruhhh.... im lost.

Code:

from fastecdsa.curve import secp256k1
from fastecdsa.point import Point
from fastecdsa.keys import gen_private_key, get_public_key
from sympy import mod_inverse

# Generate a private key
private_key = gen_private_key(secp256k1)

# Get the corresponding public key
public_key = get_public_key(private_key, secp256k1)

# Print the private key and its bit length
print("Original Private Key:", hex(private_key))
print("Private Key Bit Length:", private_key.bit_length())

# Print the public key coordinates (x, y)
print("Original Public Key (x, y):", public_key.x, public_key.y)

# Twist Attack
n = secp256k1.q  # Order of the curve
Gx, Gy = public_key.x, public_key.y

# Calculate the twist point
twist_x = Gx
twist_y = (-Gy) % secp256k1.p
print("twist_y:", hex(twist_y))

# Perform twist attack to find the partial private key
partial_private_key = (twist_y * mod_inverse(Gy, n)) % n

# Print the partial private key and its bit length
print("Partial Private Key:", hex(partial_private_key))
print("Partial Private Key Bit Length:", partial_private_key.bit_length())

[greenAlien]

I have 205288 hops per second in python on single core my friend......

It will work from 1 - 256 bit . But it is impossible to solve any Puzzle above 50 with this for a lifetime.

Code:

import time
import os
import sys
import random
import gmpy2

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[+] Kangaroo: {t}\n")
sys.stdout.flush()

modulo = gmpy2.mpz(0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F)
order = gmpy2.mpz(0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141)
Gx = gmpy2.mpz(0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798)
Gy = gmpy2.mpz(0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)

# Define Point class
class Point:
    def __init__(self, x=0, y=0):
        self.x = gmpy2.mpz(x)
        self.y = gmpy2.mpz(y)

PG = Point(Gx, Gy)
Z = Point(0, 0)  # zero-point, infinite in real x, y - plane

def add(P, Q, p=modulo):
    if P == Z:
        return Q
    elif Q == Z:
        return P
    elif P.x == Q.x and (P.y != Q.y or P.y == 0):
        return Z
    elif P.x == Q.x:
        m = (3 * P.x * P.x) * gmpy2.invert(2 * P.y, p) % p
    else:
        m = (Q.y - P.y) * gmpy2.invert(Q.x - P.x, p) % p
  
    x = (m * m - P.x - Q.x) % p
    y = (m * (P.x - x) - P.y) % p
    return Point(x, y)

def mul2(P, p=modulo):
    if P == Z:
        return Z
    m = gmpy2.f_mod(3 * P.x * P.x * gmpy2.invert(2 * P.y, p), p)
    x = gmpy2.f_mod(m * m - 2 * P.x, p)
    y = gmpy2.f_mod(m * (P.x - x) - P.y, p)
    return Point(x, y)

def mulk(k, P=PG, p=modulo):
    if k == 0:
        return Z
    elif k == 1:
        return P
    elif k % 2 == 0:
        return mulk(k // 2, mul2(P, p), p)
    else:
        return add(P, mulk((k - 1) // 2, mul2(P, p), p), p)

def X2Y(X, y_parity, p=modulo):
    X_cubed = gmpy2.powmod(X, 3, p)
    X_squared = gmpy2.powmod(X, 2, p)
    tmp = gmpy2.f_mod(X_cubed + 7, p)
    Y = gmpy2.powmod(tmp, gmpy2.f_div(gmpy2.add(p, 1), 4), p)
    if y_parity == 1:
        Y = gmpy2.f_mod(-Y, p)
    return Y


def comparator(A, Ak, B, Bk):
    result = set(A).intersection(set(B))
    if result:
        sol_kt = A.index(next(iter(result)))
        sol_kw = B.index(next(iter(result)))
        HEX = "%064x" % abs(Ak[sol_kt] - Bk[sol_kw])
        dec = int(HEX, 16)
        total_time = time.time() - starttime
        print('\n[+] total time: %.2f sec' % (total_time))
        t = time.ctime()
        print(f"\033[32m[+] PUZZLE SOLVED: {t} \033[0m")
        print(f"\033[32m[+] Private key (dec) : {dec} \033[0m")
        with open("KEYFOUNDKEYFOUND.txt", "a") as file:
            file.write("\n\nSOLVED " + t)
            file.write(f"\nTotal Time: {total_time:.2f} sec")
            file.write(f"\nRandom seed: {seed}")
            file.write("\nPrivate Key (decimal): " + str(dec))
            file.write("\nPrivate Key (hex): " + HEX)
            file.write(
                "\n-------------------------------------------------------------------------------------------------------------------------------------------\n"
            )
        file.close()
        return True
    else:
        return False

def check(P, Pindex, DP_rarity, A, Ak, B, Bk):
    modulo_val = P.x % DP_rarity
    if modulo_val == 0:
        A.append(gmpy2.mpz(P.x))
        Ak.append(gmpy2.mpz(Pindex))
        return comparator(A, Ak, B, Bk) 
    else:
        return False

# Generate a list of powers of two for faster access

def generate_powers_of_two(hop_modulo):
    return [gmpy2.mpz(1 << pw) for pw in range(hop_modulo)]

def search(P, W0, DP_rarity, Nw, Nt, hop_modulo, upper_range_limit, lower_range_limit, powers_of_two):
    solved = False
    t = [gmpy2.mpz(lower_range_limit + gmpy2.mpz(random.randint(0, upper_range_limit - lower_range_limit))) for _ in range(Nt)]
    T = [mulk(ti) for ti in t]
    dt = [gmpy2.mpz(0) for _ in range(Nt)]
    w = [gmpy2.mpz(random.randint(0, upper_range_limit - lower_range_limit)) for _ in range(Nw)]
    W = [add(W0, mulk(wk)) for wk in w]
    dw = [gmpy2.mpz(0) for _ in range(Nw)]
    print('[+] tame and wild herds are prepared')
    Hops, Hops_old = 0, 0
    t0 = time.time()  

    # Memoization dictionary
    memo = {}

    while not solved:
        for k in range(Nt):
            Hops += 1
            pw = T[k].x % hop_modulo
            if pw not in memo:
                memo[pw] = powers_of_two[pw]
            dt[k] = memo[pw]
            solved = check(T[k], t[k], DP_rarity, T, t, W, w)
            if solved: break
            t[k] += dt[k]
            T[k] = add(P[int(pw)], T[k])
        if solved: break
        for k in range(Nw):
            Hops += 1
            pw = W[k].x % hop_modulo
            if pw not in memo:
                memo[pw] = powers_of_two[pw]
            dw[k] = memo[pw]
            solved = check(W[k], w[k], DP_rarity, W, w, T, t)
            if solved: break
            w[k] += dw[k]
            W[k] = add(P[int(pw)], W[k])
        if solved: break
        t1 = time.time()
        if (t1 - t0) > 5:
            print('\r[+] Hops: %.0f h/s' % ((Hops - Hops_old) / (t1 - t0)), end='', flush=True)
            t0 = t1
            Hops_old = Hops
    print('[+] Hops:', Hops)
    return 'sol. time: %.2f sec' % (time.time() - starttime)

# Configuration for the puzzle
puzzle = 40
compressed_public_key = "03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4"
kangaroo_power = 4  
lower_range_limit = 2 ** (puzzle - 1)
upper_range_limit = (2**puzzle) - 1

DP_rarity = 1 << int(((puzzle -  2*kangaroo_power)/2 - 2))
hop_modulo = ((puzzle - 1) // 2) + kangaroo_power
Nt = Nw = 2**kangaroo_power

# Precompute powers of two for faster access
powers_of_two = generate_powers_of_two(hop_modulo)

T, t, dt = [], [], []
W, w, dw = [], [], []

if len(compressed_public_key) == 66:
    X = gmpy2.mpz(compressed_public_key[2:66], 16)
    Y = X2Y(X, gmpy2.mpz(compressed_public_key[:2]) - 2)
else:
    print("[error] pubkey len(66/130) invalid!")

W0 = Point(X,Y)
starttime = oldtime = time.time()

print(f"[+] [Puzzle]: {puzzle}")
print(f"[+] [Lower range limit]: {lower_range_limit}")
print(f"[+] [Upper range limit]: {upper_range_limit}")

#Random seed Config
seed = os.urandom(9).hex()
print(f"[+] [Random seed]: {seed}")
random.seed(seed)

Hops = 0

P = [PG]
for k in range(255): P.append(mul2(P[k]))  
print('[+] P-table prepared')

solved = False
search(P, W0, DP_rarity, Nw, Nt, hop_modulo, upper_range_limit, lower_range_limit, powers_of_two)

print('[+] Average time to solve: %.2f sec' % ((time.time()-starttime)))

• Kangaroo: Thu May 16 13:33:48 2024
• [Puzzle]: 40
• [Lower range limit]: 549755813888
• [Upper range limit]: 1099511627775
• [Random seed]: 9b76a9623bd76fb8cc
• P-table prepared
• tame and wild herds are prepared
• Hops: 205288 h/s
• total time: 6.92 sec
• PUZZLE SOLVED: Thu May 16 13:33:55 2024
• Private key (dec) : 1003651412950
• Hops: 1416585
• Average time to solve: 6.92 sec

It needs to be converted into C++ or Rust so that it can solve problems up to puzzle 120.

not a bad working code! Are you thinking on implementing multi threading? I know this is not comparable with GPU but its not bad at all Congrats!

[COBRAS]

i have created twist attack on ecdsa secp256k1.

can read about Twist Attack on ECDSA SECP256k1 here. https://github.com/demining/Twist-Attack

It gets the partial private key correctly.

my issue now is how to write the code to get to the full private key. PM Me Bruhhh.... im lost.

Code:

from fastecdsa.curve import secp256k1
from fastecdsa.point import Point
from fastecdsa.keys import gen_private_key, get_public_key
from sympy import mod_inverse

# Generate a private key
private_key = gen_private_key(secp256k1)

# Get the corresponding public key
public_key = get_public_key(private_key, secp256k1)

# Print the private key and its bit length
print("Original Private Key:", hex(private_key))
print("Private Key Bit Length:", private_key.bit_length())

# Print the public key coordinates (x, y)
print("Original Public Key (x, y):", public_key.x, public_key.y)

# Twist Attack
n = secp256k1.q  # Order of the curve
Gx, Gy = public_key.x, public_key.y

# Calculate the twist point
twist_x = Gx
twist_y = (-Gy) % secp256k1.p
print("twist_y:", hex(twist_y))

# Perform twist attack to find the partial private key
partial_private_key = (twist_y * mod_inverse(Gy, n)) % n

# Print the partial private key and its bit length
print("Partial Private Key:", hex(partial_private_key))
print("Partial Private Key Bit Length:", partial_private_key.bit_length())

Explain were you see partial privkey correctly ?

I not see any parts of privkey in partial privkey

ps as I know twist attack and defining twist,-attack he is bla bla bla about special base point or public keys I not remember this exact, but in your dcrypt you have only one possible twist point but attack and defining use many this points

[nomachine]

not a bad working code! Are you thinking on implementing multi threading? I know this is not comparable with GPU but its not bad at all Congrats!

I already have a (several different) multicore versions somewhere on my profile. Solves up to Puzzle 50.
Here for example:
https://bitcointalk.org/index.php?topic=1306983.msg62884789#msg62884789

But I won't give you false hope that this script can solve anything above 50. It's simply beyond the reach of python's capabilities.

You have to program in C++ yourself if the current scripts do not meet your needs.

Even in C++ you have to be realistic - you need thousands of GPUs to solve puzzle 130.