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Crypto - Steps

Alice and Bob have taken steps to communicate securely. - pot

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Full Implementationarrow-up-right

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Challenge

from Crypto.Util.number import getPrime
from random import randint
from hashlib import sha512
from secret import FLAG

p = getPrime(1024)
Pair = tuple[int, int]

def apply(x: Pair, y: Pair) -> Pair:
    z0 = x[0] * y[1] + x[1] * y[0] - x[0] * y[0]
    z1 = x[0] * y[0] + x[1] * y[1]
    return z0 % p, z1 % p

def calculate(n: int) -> Pair:
    out = 0, 1
    base = 1, 1
    while n > 0:
        if n & 1 == 1: out = apply(out, base)
        n >>= 1
        base = apply(base, base)
    return out

def step(x: Pair, n: int):
    '''Performs n steps to x.'''
    return apply(x, calculate(n))
def main() -> None:
    g = tuple(randint(0, p - 1) for _ in range(2))
    a = randint(0, p)
    b = randint(0, p)

    A = step(g, a)
    B = step(g, b)

    print(p)
    print(g)
    print(A)
    print(B)
    
    shared = step(A, b)
    assert shared == step(B, a)
    pad = sha512(str(shared).encode()).digest()
    print(xor(FLAG, pad))

We are given p,g,A,B and we need to find the shared secret by step(A,b) or step(B,a)

step(A,b) is the same as apply(A,calculate(b)). If we could somehow reverse the output of apply to get the input then we could use it on step(g,a) and find the value of calculate(a)

In apply, given the output values z0,z1 and the first input values x0,x1, we need to find the values of the second input values y0,y1 .

For this, we employ the use of Groebner basis

x02∗y1+x0∗x1∗y1−x12∗y1−x0∗z0−x0∗z1+x1∗z1=0y1∗(x02+x0∗x1−x12)=x0∗z0+x0∗z1−x1∗z1y1=(x02+x0∗x1−x12)−1∗(x0∗z0+x0∗z1−x1∗z1)(modp)x_0^2 * y_1 + x_0*x_1*y_1 - x_1^2 *y_1 - x_0*z_0 - x_0*z_1 + x_1 * z_1 = 0 \\ y_1*(x_0^2 + x_0*x_1 - x_1^2) = x_0*z_0 + x_0*z_1-x_1*z_1 \\ y_1 =(x_0^2 + x_0*x_1 - x_1^2) ^{-1} * ( x_0*z_0 + x_0*z_1-x_1*z_1 ) \pmod p
x0∗y0+x1∗y1−z1=0y0=(z1−x1∗y1)∗x0−1(modp)x_0*y_0 + x_1*y_1 - z_1 = 0 \\ y_0 = (z_1-x_1*y_1)* x_0^{-1} \pmod p

Now we have everything that we need, we can now decrypt.

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