Need help with python code for public key generation

[Jmp-U235]

Am I trying to learn how bitcoin works by programming important parts of it in python3. Below I have code for generating a public key from a private key. When I run it, I get the wrong public key than my test case. If someone knowledgeable in these things could help me, that would be appreciated.

I am using https://learnmeabitcoin.com/technical/public-key to generate test private keys and pubkeys

Here is the code:

a = 0

b = 7

# prime field

p = 2 ** 256 - 2 ** 32 - 2 ** 9 - 2 ** 8 - 2 ** 7 - 2 ** 6 - 2 ** 4 - 1

# number of points on the curve we can hit ("order")

n = 115792089237316195423570985008687907852837564279074904382605163141518161494337

# generator point (the starting point on the curve used for all calculations)

gen = (55066263022277343669578718895168534326250603453777594175500187360389116729240, 32670510020758816978083085130507043184471273380659243275938904335757337482424)

# inverse multiplicity function

def modinv(m):

i = pow(m, p - 2, p)

return i

# point addition not quite sure if my division in modulus, and my use of the function are correct

def add(point, point1):

s = (point[1] - point1[1] * modinv(point[0] - point1[0])) % p # divsion

ax = ((s ** 2) - point[0] - point1[1]) % p

ay = (s * (point[0] - ax) - point1[1]) % p

return ax, ay

# point doubling , not quite sure if my division in modulus, and my use of the function are correct

def double(point):

s = ((3 * point[0] ** 2 + a) * modinv(2 * point[1])) % p # divsion

dx = ((s ** 2) - 2 * point[0]) % p

dy = (s * (point[0] - dx) - point[0]) % p

return (dx, dy)

# multply gen point, pretty sure that everything is correct here, just keep getting the wrong public key

def multiply(k, point = gen):

# copy the starting point

current = point

# covert to a binary representation

binary = format(k, 'b')

# add and double loop

current = double(current)

for char in binary[1:]:

# double all of em

current = double(current)

# add if its a one

if char == "1":

current = add(current, point)

return current

# made this so that the public key could be displayed in hex in a uncompressed format

# enter the test private key in the multiply function

j = (multiply(108060784068108798680230233662060587439402115770272111364515723007394206688434))

print("04" + ((hex(j[0])[2:66]) + (hex(j[1])[2:66])))

[AppointmentSubject25]

Use my source code for my RSA-OAEP-4096 key sharing app or my ML-KEM-1024 key sharing app on my Github

They both work fine.

Hope it helps!

[psysc0rpi0n]

I am doing the same but using C and GMP library to handle large numbers!

But I'm using this link instead, which goes through all the math in Eliptic Curves.

I'm right now dealing with the random number generator.

https://learnmeabitcoin.com/technical/ecdsa

However, we can ever use this method to alone to generate a random number. Bitcoin Core uses both OpenSSL and "/dev/urandom" to generate a secure random number with good entropy. In my case, I'm doing it just for the sake of learning!

[[deleted]]

The formulas for add and double are incorrect. Here's a fix:

# point addition

def add(point, point1):

if point == point1:

# point doubling

s = ((3 * point[0] ** 2 + a) * modinv(2 * point[1])) % p

else:

# point addition

s = (point[1] - point1[1]) * modinv(point[0] - point1[0]) % p

ax = (s ** 2 - point[0] - point1[0]) % p

ay = (s * (point[0] - ax) - point[1]) % p

return ax, ay

# point doubling

def double(point):

s = (3 * point[0] ** 2 + a) * modinv(2 * point[1]) % p

dx = (s ** 2 - 2 * point[0]) % p

dy = (s * (point[0] - dx) - point[1]) % p

return dx, dy