# Chujowy CTF

I loved this CTF, there were some great crypto challenges, that all felt well made and interesting. For now, I’m going to write up the most interesting: Real ECC, but might come back to this and add in how I solved the three RSA challenges. Thanks to Chujowy for hosting this CTF.

Challenge Points
Real ECC 470

## Real ECC

### Disclaimer

⚠️ I didn’t realise this during the CTF, but another player DMd me after reading my write up and told me that this challenge had been copied from GoogleCTF 2017.

### Challenge

This recent change allows us to finally reduce the key sizeAuthor: @enedil

from operator import xor
from public import p, g
from secret import flag, a_s, b_s

try:
return {p1: p2, p2: p1}[-1]
except:
pass
x1, y1 = p1
x2, y2 = p2
x3 = FF(x1*x2 - x1*y2 - x2*y1 + 2*y1*y2) / FF(x1 + x2 - y1 - y2 - 1)
y3 = FF(y1*y2) / FF(x1 + x2 - y1 - y2 - 1)
return (x3, y3)

def mul(F, x, k):
acc = -1
while k:
if k & 1:
# acc = add(F, acc, acc) Typo in challenge
k //= 2
return acc

# Assume input is less than 256 bytes.
return data + bytes([length - len(data)] * (length - len(data)))

def encrypt(data, stream):
return xor(int(data), int(stream))

def encrypt_bytes(data, key):
data = pad(data, int(key).bit_length() // 8)
return encrypt(int.from_bytes(data, 'big'), key)

FF = Zmod(p)
A = mul(FF, g, a_s)
B = mul(FF, g, b_s)
print(f'g = {g}', f'p = {p}', f'A = {A}', f'b = {B}', sep='\n')
a_ms = mul(FF, B, a_s)
b_ms = mul(FF, A, b_s)
assert a_ms == b_ms
shared = a_ms[0] * a_ms[1]

print(f'len = {len(flag)}, enc = {encrypt_bytes(flag, shared)}')

# g = (375383228397780342292610905741415543021123193893202993933376546008355999579881, 125127742799260114097536968580471847673707731297734331311125515336118666526627)
# p = 1017349223066738178194531452435878724694134639196427641168991759143390320356263
# A = (603956890649406768784284509883012839855804103607835093214222589654065615494206, 749634286053611578152285189158606552324508418540678613236591040516722145253708)
# b = (890062254689797703350145707732638943570461065304155615771307683230377614308406, 467371775612631851798003093722695784734930391102311439111204608050476921601852)
# len = 42, enc = 54357159864722158692491564537102129439237984275607683326888133775459718903987000238912076679209103764


### Solution

#### Defining the problem

This is a really interesting challenge. We’re given an instance of a DH key exchange using the group formed by a point addition law for some unknown curve. The goal for the challenge is to break the discrete log in this group to recover the shared secret.

Unlike traditonal ECDLP challenges, rather than being given the curve, we’re only given the operations of point addition and we’re shown that this operation is commutative:

a_ms = mul(FF, B, a_s)
b_ms = mul(FF, A, b_s)
assert a_ms == b_ms


If we had the curve parameters, we could look at the curve order and see whether we could attack it to solve the discrete log. I spent a bit of time trying to deduce the form of the polynomial which gives rise to this point addition law, but I was unable to.

Looking at the characteristic of the curve, we see that it’s smooth

sage: p = 1017349223066738178194531452435878724694134639196427641168991759143390320356263
sage: factor(p-1)
2 * 4229971 * 4604689 * 6790219 * 9085073 * 9562139 * 9774509 * 11406217 * 20683151 * 24963011 * 26496859 * 29026219


So then the question is, can we define a map from group of points on the curve to the group $\mathbb{F}_{p}$? When an elliptic curve is singular, this is indeed the case and is an interesting problem (we have a version of this on CryptoHack). I’ve also seen this for rational curves of genus 0, such as the circle.

Generally, the method is defining an isomorpishm from the group $C (\mathbb{F}_{p})$ for some curve $C$ to the finite group $\mathbb{F}_{p^n}$ for some integer $n$. Normally I have seen this done from the curve polynomial, but we can do just as well from the point addition law itself.

What we want to show, is that for the point addition

$P = (x_0,y_0), \quad Q = (x^\prime, y^\prime) \qquad \qquad Q = k \cdot P$

we can find a function $f(x,y)$ such that

$g = f(x_0,y_0), \quad h = f(x^\prime, y^\prime), \qquad h = g^k, \qquad f(x,y) \in \mathbb{F}_{p}$

we can then find the secret integer $k$ by solving the discrete log in $\mathbb{F}_p$. This will be very quick using Pohlig-Hellman as $p$ is smooth and easy as sage implements this for us!

#### Finding the mapping

From the challenge, lets write down exactly what the point addition law is. Note that all operations are done mod $p$, which we will not include explicitly. Adding the two points $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ we get a third point

$P + Q = (x_3, y_3) = \left( \frac{x_1 x_2 + x_1 y_2 - x_2 y_1 + 2y_1 y_2}{x_1 + x_2 - y_1 - y_2 - 1}, \frac{y_1 y_2}{x_1 + x_2 - y_1 - y_2 - 1} \right)$

When $P = Q$, we can simplify the above and show that point doubling is given by

$2P = (x^\prime, y^\prime) = \left( \frac{x^2 - 2xy + 2y^2}{2(x - y) - 1}, \frac{y^2}{2(x - y) - 1} \right)$

Written in this form, we want to try and find a function $f(x,y)$ such that $f^2(x,y) = f(x^\prime,y^\prime)$. Looking at the numerator, we see the tempting simplification

$x^2 - 2xy + 2y^2 - y^2 = (x - y)^2 \quad \Rightarrow \quad x^\prime - y^\prime = \frac{(x - y)^2}{2(x - y) - 1}$

but how do we deal with the denominatior? I first decided to look at $3P$

$x^\prime - y^\prime = \frac{(x - y)^3}{1 - 3(x - y) + 3(x - y)^2} \\$

and $4P$

$x^\prime - y^\prime = \frac{(x - y)^4}{(2(x - y) - 1) (2(x - y)^2 - 2(x - y) + 1)}$

to check things were following my assumed pattern. Let us now change variables to $t = x - y$, which gives us

\begin{align} t_{2P} &= \frac{t^2}{2t - 1} \\ t_{3P} &= \frac{t^3}{3t^2 - 3t + 1} \\ t_{4P} &= \frac{t^4}{(2t - 1) (2t^2 - 2t + 1)} = \frac{t^4}{4t^3 - 6t^2 + 4t - 1} \end{align}

Written like this, we can see that the denominator is of the form

$t^k - (t-1)^k$

and we can write

$t^\prime = \frac{t^k}{t^k - (t-1)^k}$

The last step is to write this in the form such that $h = g^k$, which means we need to rearrange a little

$\frac{1}{t^\prime} = \frac{t^k - (t-1)^k}{t^k} = 1 - \left(\frac{t-1}{t}\right)^k$

which allows us to write down

$g = 1 - \frac{1}{t} \qquad h = 1 - \frac{1}{t^\prime} = \left(1 - \frac{1}{t}\right)^k, \qquad h = g^k$

We now have everything we need to solve, the generator, and Alice and Bob’s public keys can be represented as

$g = 1 - \frac{1}{x_0 - y_0}, \quad h_{A} = 1 - \frac{1}{x_A - y_A}, \quad h_{B} = 1 - \frac{1}{x_B - y_B}, \qquad g, h_a, h_B \in \mathbb{F}_p$

All that’s left is to code it up and grab the shared secret. From that we simply extract the flag from a xor with the enc from the challenge data.

### Implementation

from Crypto.Util.number import long_to_bytes

try:
return {p1: p2, p2: p1}[-1]
except:
pass
x1, y1 = p1
x2, y2 = p2
x3 = FF(x1*x2 - x1*y2 - x2*y1 + 2*y1*y2) / FF(x1 + x2 - y1 - y2 - 1)
y3 = FF(y1*y2) / FF(x1 + x2 - y1 - y2 - 1)
return (x3, y3)

def mul(F, x, k):
acc = -1
while k:
if k & 1:
k //= 2
return acc

# Begin Challenge Data
enc = 54357159864722158692491564537102129439237984275607683326888133775459718903987000238912076679209103764
p = 1017349223066738178194531452435878724694134639196427641168991759143390320356263
G = (375383228397780342292610905741415543021123193893202993933376546008355999579881, 125127742799260114097536968580471847673707731297734331311125515336118666526627)
A = (603956890649406768784284509883012839855804103607835093214222589654065615494206, 749634286053611578152285189158606552324508418540678613236591040516722145253708)
B = (890062254689797703350145707732638943570461065304155615771307683230377614308406, 467371775612631851798003093722695784734930391102311439111204608050476921601852)
# End Challenge Data

FF = GF(p)
t = G[0] - G[1]
t_a = A[0] - A[1]
t_b = B[0] - B[1]

g = 1 - FF(1 / t)
h_a = 1 - FF(1 / t_a)
h_b = 1 - FF(1 / t_b)

a_s = h_a.log(g)
b_s = h_b.log(g)

assert A == mul(FF, G, a_s)
assert B == mul(FF, G, b_s)
assert mul(FF, A, b_s) == mul(FF, B, a_s)

secret = mul(FF, A, b_s)
s = secret[0] * secret[1]

print(long_to_bytes(Integer(enc)^^Integer(s)))


### Flag

chCTF{this_wasnt_elliptic_curve_after_all}