DiceCTF 2022

I played DiceCTF with my team Organisers. We came first place. I wrote up two challenges for our website, I’m including them here just for completeness. To read more writeups for DiceCTF, have a look at the Organisers’ write-ups.


It’s a free flag, all you have to do is wait! Verifiably.

nc mc.ax 31337



from hashlib import shake_128

# from Crypto.Util.number import getPrime
# p = getPrime(1024)
# q = getPrime(1024)
# n = p*q
n = 20074101780713298951367849314432888633773623313581383958340657712957528608477224442447399304097982275265964617977606201420081032385652568115725040380313222774171370125703969133604447919703501504195888334206768326954381888791131225892711285554500110819805341162853758749175453772245517325336595415720377917329666450107985559621304660076416581922028713790707525012913070125689846995284918584915707916379799155552809425539923382805068274756229445925422423454529793137902298882217687068140134176878260114155151600296131482555007946797335161587991634886136340126626884686247248183040026945030563390945544619566286476584591
T = 2**64

def is_valid(x):
    return type(x) == int and 0 < x < n

def encode(x):
    return x.to_bytes(256, 'big')

def H(g, h):
    return int.from_bytes(shake_128(encode(g) + encode(h)).digest(16), 'big')

def prove(g):
    h = g
    for _ in range(T):
        h = pow(h, 2, n)
    m = H(g, h)
    r = 1
    pi = 1
    for _ in range(T):
        b, r = divmod(2*r, m)
        pi = pow(pi, 2, n) * pow(g, b, n) % n
    return h, pi

def verify(g, h, pi):
    assert is_valid(g)
    assert is_valid(h)
    assert is_valid(pi)
    assert g != 1 and g != n - 1
    m = H(g, h)
    r = pow(2, T, m)
    assert h == pow(pi, m, n) * pow(g, r, n) % n

if __name__ == '__main__':
    g = int(input('g: '))
    h = int(input('h: '))
    pi = int(input('pi: '))
    verify(g, h, pi)
    with open('flag.txt') as f:


The challenge presents us with a verifiable delay function (VDF), which (if correctly implemented) requires us to compute

\[h \equiv g^{2^T} \pmod n.\]

This requires us to perform $T = 2^{64}$ squares of $g \pmod n$, which is totally infeasible for a weekend CTF! If we could factor $n$, we could first compute $a \equiv 2^T \pmod{\phi(n)}$, but as the challenge is set up, it’s obvious we can’t factor the 2048 bit modulus.

Another option would be to pick a generator $g$ of low order, for the RSA group $\mathcal{G} = (\mathbb{Z}/n\mathbb{Z})^*$, two easy options are $g=1$ or $g=-1$. However, looking at verify(g,h,pi), we see that these elements are explicitly excluded from being considered

def is_valid(x):
    return type(x) == int and 0 < x < n

def verify(g, h, pi):
    assert is_valid(g)
    assert is_valid(h)
    assert is_valid(pi)
    assert g != 1 and g != n - 1
    m = H(g, h)
    r = pow(2, T, m)
    assert h == pow(pi, m, n) * pow(g, r, n) % n

First is_valid(x) ensures that $g,h,\pi \in \mathcal{G}$ and then the additional check assert g != 1 and g != n - 1 ensures that $g$ has unknown order.

So if we can’t run prove(g) in a reasonable amount of time, and we can’t cheat the VDF by factoring, or selecting an element of known order, then there must be something within verify we can cheat.

First, let’s look at what appears in verify(g,h,pi) and what we have control over.

We choose as input any $g,h,\pi \in \mathcal{G}$ and from $g,h$ shake128 is used as a pseudorandom function to generate $m$. Finally, from $m$ we find $r \equiv 2^T \pmod m$.

To pass the test in verify, naively we need to send integers from the output of h, pi = prove(g) such that the following congruence holds:

\[h \equiv g^r \cdot \pi^m \pmod n.\]

Although this congruence assumes the input $(g,h,\pi)$ have the relationship established by prove(g), what if we instead view this as a general congruence? Let’s try by assuming all variables can be expressed as a power of a generator $b$ and attempt to forget about prove(g) altogether! For our implementation, we make the choice $b = 2$, but this is arbitary.

\[g \equiv b^M \pmod n, \quad h \equiv b^A \pmod n, \quad \pi \equiv b^B \pmod n.\]

From this point of view, we need to try and find integers $(M,A,B)$ such that

\[b^A \equiv b^{rM} \cdot b^{mB} \pmod n \Leftarrow A = rM + mB\]

The integers $(m,r)$ are generated from

def H(g, h):
    return int.from_bytes(shake_128(encode(g) + encode(h)).digest(16), 'big')

# We can pick these
M, A, B = ?, ?, ?
g = pow(2,M,n)
h = pow(2,A,n)
pi = pow(2,B,n)

# Effectively random
m = H(g, h)
r = pow(2, T, m)

and we can effectively treat these integers as totally random. More importantly, the values are unknown until we make a choice for both $g,h$ (and therefore $M,A$).

Our first simplification will be $A = 0 \Rightarrow h = 1$, which simplifies our equation and is a valid input for $h$. Now we need to pick $(M,B)$ such that

\[0 = rM + mB,\]

where we remember that the values of $(r,m)$ are only known after selecting $M$, but $B$ can be set afterwards. It then makes sense to rearrange the above equation into the form:

\[B = -\frac{rM}{m}\]

To find an integer solution $B$, we then need to find some $rM$ which is divisible by a random integer $m$.

The VDF function which appears in the challenge is based off work by Wesolowski, reviewed in a paper by Boneh, Bünz and Fisch. There is a key difference though between the paper and the challenge. In Wesolowski’s work, $m$ is prime, and finding a $M$ divisible by some large, random prime is computationally hard. The challenge becomes solvable because $m$ is totally random and so can be composite.

To find an integer $M \equiv 0 \pmod m$, the best chance we have is to use some very smooth integer, such as $M = n!$, or $M = \prod_i^n p_i$ as the product of the first $n$ primes. In the challenge author’s write-up, they pick

\[M = 256! \prod_i^n p_i,\]

where they consider all primes $p_i < 10^{20}$. Including $256!$ allows for repeated small factors in $m$. In our solution, we find it is enough to simply take the product of all primes below $10^6$.

To then solve the congruence, we first generate a very smooth integer $M$ and set $g \equiv b^M \pmod n$. From this, we compute $m = H(g,1)$. If $M \equiv 0 \pmod m$ we break the loop, compute $r$ from $m$, then $B(M,r,m)$. Finally setting $\pi \equiv b^B \pmod n$ for our solution $(g,h,\pi)$. If the congruence doesn’t hold, we square $g \equiv g^2 \pmod n$ and double $M = 2M$ for bookkeeping, and try again.

Sending our specially crafted $(g,h,\pi) = (g,1,\pi)$ to the server, we get the flag.


Note: We use gmpy2 to speed up all the modular maths we need to do, but you can do this using python’s int type and solve in a reasonable amount of time.

from gmpy2 import mpz, is_prime
from hashlib import shake_128

# Challenge Data #

n = mpz(20074101780713298951367849314432888633773623313581383958340657712957528608477224442447399304097982275265964617977606201420081032385652568115725040380313222774171370125703969133604447919703501504195888334206768326954381888791131225892711285554500110819805341162853758749175453772245517325336595415720377917329666450107985559621304660076416581922028713790707525012913070125689846995284918584915707916379799155552809425539923382805068274756229445925422423454529793137902298882217687068140134176878260114155151600296131482555007946797335161587991634886136340126626884686247248183040026945030563390945544619566286476584591)
T = mpz(2**64)

def is_valid(x):
    return type(x) == int and 0 < x < n

def encode(x):
    if type(x) == int:
        return x.to_bytes(256, 'big')
        return int(x).to_bytes(256, 'big')

def H(g, h):
    return int.from_bytes(shake_128(encode(g) + encode(h)).digest(16), 'big')

def verify(g, h, pi):
    assert is_valid(g)
    assert is_valid(h)
    assert is_valid(pi)
    assert g != 1 and g != n - 1
    m = H(g, h)
    r = pow(2, T, m)
    # change assert to return bool for testing
    return h == pow(pi, m, n) * pow(g, r, n) % n

#    Solution    #

def gen_smooth(upper_bound):
    M = mpz(1)
    for i in range(1, upper_bound):
        if is_prime(i):
            M *= i
    return M

def gen_solution(M):
    # We pick a generator b = 2
    g = pow(2, M, n)
    h = 1
    while True:
        m = mpz(H(g, h))
        if M % m == 0:
            r = pow(2, T, m)
            B = -r*M // m
            pi = pow(2, B, n)
            return int(g), int(h), int(pi) 
        M = M << 1
        g = pow(g,2,n)

print(f"Generating smooth value M")
M = gen_smooth(10**6)

print(f"Searching for valid m")
g, h, pi = gen_solution(M)

assert verify(g, h, pi)
print(f"g  = {hex(g)}")
print(f"h  = {hex(h)}")
print(f"pi = {hex(pi)}")



Commitment Issues

I created a new commitment scheme, but commitment is scary so I threw away the key.


from random import randrange
from Crypto.Util.number import getPrime, inverse, bytes_to_long, GCD

flag = b'dice{?????????????????????????}'
n = 5

def get_prime(n, b):
    p = getPrime(b)
    while GCD(p - 1, n) != 1:
        p = getPrime(b)
    return p

p = get_prime(n, 1024)
q = get_prime(n, 1024)
N = p*q
phi = (p - 1)*(q - 1)

e = 0xd4088c345ced64cbbf8444321ef2af8b
d = inverse(e, phi)

def sign(message):
    m = bytes_to_long(message)
    return pow(m, d, N)

def commit(s, key, n):
    return (s + key) % N, pow(key, n, N)

def reveal(c1, c2, key, n):
    assert pow(key, n, N) == c2
    return (c1 - key) % N

r = randrange(1, N)
s = sign(flag)
c1, c2 = commit(s, r, n)

print(f'N = {hex(N)}')
print(f'c1 = {hex(c1)}')
print(f'c2 = {hex(c2)}')

Reading the Challenge

This challenge is based on a custom commitment scheme for RSA signatures. Before diving into the solution, let’s break down what we’re given and try and identify the insecure part of the scheme.

The RSA modulus $N=pq$ has 2048 bits, and is the product of two 1024 bit primes, which are generated such that $n = 5$ is not a factor of $(p-1)$ or $(q-1)$. From this alone, we will not be able to factor $N$.

The public exponent is unusual: e = 0xd4088c345ced64cbbf8444321ef2af8b, but it’s prime and not so large as to cause much suspicion. So far, so good (or bad for finding a solution, i suppose…).

We are given the length of the flag , which is 31 bytes or 248 bits long. The signature of the flag is $s = m^d \pmod N$, where $m$ is not padded before signing. This means that $m$ is relatively small compared to the modulus (Coppersmith should start being a thought we have now). However, we don’t have the value of the signature, only the commitment.

The commitment gives us two values, $c_1$ and $c_2$. Let’s look at how the commitment is made.

def commit(s, key, n):
    return (s + key) % N, pow(key, n, N)

r = randrange(1, N)
s = sign(flag)
c1, c2 = commit(s, r, n)

First a random number $r$ is generated from r = randrange(1, N) as the key. The flag is signed and so we are left with two integers $(r,s)$ both approximately of size $N$. The commitment is made by adding together these integers modulo $N$:

\[c_1 = (s + r) \pmod N.\]

We can understand $r$ here as effectively being a OTP, obscuring the signature $s$. We cannot recover $s$ from $c_1$ without knowing $r$ and we cannot recover $r$ without knowing $s$.

The second part of the commitment depends only on the random number $r$ and is given by

\[c_2 = r^5 \pmod N.\]

Obtaining $r$ from $c_2$ is as hard as breaking RSA with the public key $(e=5,N)$. If $r$ was small, we could try taking the fifth root, but as it of the size of $N$, we cannot break $c_2$ to recover $r$.

So… either the challenge is impossible, or there’s a way to use our knowledge of $(c_1,c_2)$ together to recover the flag.

Combining Commitments

Let’s write down what we know algebraically:

\[\begin{aligned} s &= m^d &&\pmod N, \\ c_1 &= s + r &&\pmod N, \\ c_2 &= r^5 &&\pmod N. \end{aligned}\]

Additionally, we know that $m$ is small with respect to $N$, so if we could write down a polynomial $g(m) = 0 \pmod N$, we could use Coppersmith’s small roots to recover $m$ and hence the flag!

Note: The following solution was thought up by my teammate, Esrever, so all credit to him.

Consider the polynomial in the ring $R = (\mathbb{Z}/N\mathbb{Z})[X]$:

\[f(X) = (c_1 - X)^e \pmod N,\]

we have the great property that $f(r) = m$. However, written like this, the polynomial will be enormous, as $e$ is a (moderately) large prime [Maybe this is the reason $e$ was chosen to be in the form we see in the challenge].

Esrever’s great idea was to work in the quotient ring $K = R[X] / (X^5 - c_2)$, using the additional information we get from $c_2$. This allows us to take the $e$ degree polynomial $f(X)$ and recover a (at most) degree four polynomial by repeatedly substituting in $X^5 = c_2$.

Taking powers of the polynomial, we have that

\[m^k = f^k(r) = (c_1 - r)^{e\cdot k} \pmod N\]

The hope was that by taking a set of these polynomials, we could write down a linear combination of $m^k$ such that all $r$ cancel, leaving a univariate polynomial in $m$. This is exactly what we need to find if we hope to solve using small roots.

We were able to accomplish this with a bit of linear algebra. Let’s go through step by step.

Linear Algebra to the Rescue

First let us write the $k^{\text{th}}$ power of $f(X)$ as $f^k(X)$ with coefficients $b_{ki}$:

\[f^k(X) = \sum_{i=0}^{4} b_{ki} \cdot X^i\]

Taking $k \in \{1,\ldots 5 \}$ we can write down five degree four polynomials using a $5\times5$ matrix and column vector:

\[\mathbf{M} = \begin{pmatrix} b_{10} & b_{11} & b_{12} & b_{13} & b_{14} \\ b_{20} & b_{21} & b_{22} & b_{32} & b_{24} \\ b_{30} & b_{31} & b_{32} & b_{33} & b_{34} \\ b_{40} & b_{41} & b_{42} & b_{43} & b_{44} \\ b_{50} & b_{51} & b_{52} & b_{53} & b_{54} \\ \end{pmatrix} \quad \mathbf{x} = \begin{pmatrix} X^0 \\ X^1 \\ X^2 \\ X^3 \\ X^4 \\ \end{pmatrix}.\]

With these, our polynomials can be recovered from matrix multiplication:

\[\mathbf{F} = \mathbf{M}(\mathbf{x}) = \begin{pmatrix} f^1(X) \\ f^2(X) \\ f^3(X) \\ f^4(X) \\ f^5(X) \\ \end{pmatrix}\]

To solve the challenge, our goal is to find a vector $\mathbf{a} = (\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5)^\top$ such that

\[\mathbf{M}^\top(\mathbf{a}) = (1,0,0,0,0)^\top.\]

This is equivalent to finding simultaneous solutions to

\[\sum_{k=1}^5 \alpha_k \cdot b_{k0} = 1, \quad \sum_{k=1}^5 \alpha_k \cdot b_{kj} = 0, \quad j \in \{1,\ldots 4\}\]

Practically, finding this vector $\mathbf{a}$, allows us to derive the linear combination

\[g(m) = \sum_{i=1}^5 \alpha_i f^i(X) = \sum_{i=1}^5 \alpha_i \cdot m^i.\]

with no dependency on the variable $X$, allowing us to understand $g(m)$ as a univariate polynomial in $m$, precisely what we need for small roots!!

Recovering $\mathbf{a}$ is possible as long as $\mathbf{M}$ has an inverse, as we can write

\[\mathbf{a} = (\mathbf{M}^\top)^{-1} (1,0,0,0,0)^\top\]

Using SageMath, this is as easy as

M = ... # Matrix of coefficients
v = vector(Zmod(N), [1,0,0,0,0])
a = M.transpose().solve_right(v)

With the polynomial $g(m)$ recovered, we can apply SageMath’s .small_roots() method on our univariate polynomial and recover the flag!


# Challenge Data #

N  = 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
e  = 0xd4088c345ced64cbbf8444321ef2af8b
c1 = 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
c2 = 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

#    Solution    #

R.<X> = PolynomialRing(Zmod(N))
R.<X> = R.quo(X^5 - c2)

f1 = (c1 - X)^e
f2 = f1^2
f3 = f1^3
f4 = f1^4
f5 = f1^5

M = Matrix(Zmod(N), 

v = vector(Zmod(N), [1,0,0,0,0])

sol = list(M.solve_right(v))

K.<m> = PolynomialRing(Zmod(N), implementation='NTL')
g = -1
for i,v in enumerate(sol):
    g += v*m^(i+1)

flag = g.monic().small_roots(X=2**(31*8), beta=1, epsilon=0.05)[0]
print(int(flag).to_bytes(31, 'big'))