The crypto challenges in DownUnderCTF are very intriguing. Unfortunately I have very limited time during the contest and could only solve apbq rsa ii (26 solves), hhhhh (14 solves) and handshake (4 solves). I will be covering the latter two challenges in this writeup.

## hhhhh⌗

### Challenge Summary⌗

The code given in this challenge is pretty short:

#!/usr/bin/env python3
from os import getenv as hhhhhhh
from hashlib import md5 as hhhhh

def hhhhhh(hhh):
h = hhhhh()
hh = bytes([0] * 16)
for hhhh in hhh:
h.update(bytes([hhhh]))
hh = bytes([hhhhhhh ^ hhhhhhhh for hhhhhhh, hhhhhhhh in zip(hh, h.digest())])
return hh

print('hhh hhh hhhh hhh hhhhh hhhh hhhh hhhhh hhhh hh hhhhhh hhhh?')

h = bytes.fromhex(input('h: '))

if hhhhhh(h) == b'hhhhhhhhhhhhhhhh':
print('hhhh hhhh, hhhh hh hhhh hhhh:', hhhhhhh('FLAG'))
else:
print('hhhhh, hhh hhhhh!')


We are given a custom hash algorithm, namely $\mathcal{H}$ (or hhhhhh), which is based on MD5. Let each of $x_1, x_2, …, x_n$ represent a byte. $\mathcal{H}$ is defined by

$$\mathcal{H}(x_1x_2…x_n) = \text{MD5}(x_1) \oplus \text{MD5}(x_1x_2) \oplus … \oplus \text{MD5}(x_1x_2…x_n).$$

The objective is to find a preimage $x_1x_2…x_n$ such that

$$\mathcal{H}(x_1x_2…x_n) = \texttt{hhhhhhhhhhhhhhhh}.$$

### Solution⌗

#### Part I: Generating MD5 Multicollisions⌗

It is well-known that MD5 is vulnerable to collision attacks. We can easily generate collisions with a particular prefix using FastColl. Let $B_1, B_2, …, B_k$ be $k$ 128-byte blocks, we can generate two 128-byte blocks, $X_{k+1,1}$ and $X_{k+1,2}$, such that

$$\text{MD5}(B_1B_2…B_kX_{k+1}) = \text{MD5}(B_1B_2…B_kY_{k+1}).$$

To attack, we can generate a bunch of 128-byte blocks $X_{11}, X_{12}, X_{21}, X_{22}, …$ such that

• $\text{MD5}(X_{11}) = \text{MD5}(X_{12})$ (generated using the prefix $\varepsilon$, i.e., the empty string)
• $\text{MD5}(X_{11}X_{21}) = \text{MD5}(X_{11}X_{22})$ (generated using the prefix $X_{11}$)
• $\text{MD5}(X_{11}X_{21}X_{31}) = \text{MD5}(X_{11}X_{21}X_{32})$ (generated using the prefix $X_{11}X_{21}$)
• $\text{MD5}(X_{11}X_{21}X_{31}…X_{m-1,1}X_{m1}) = \text{MD5}(X_{11}X_{21}…X_{m-1,1}X_{m2})$.

Since MD5 is an implementation of the Merkle-Damgard construction,

$$\text{MD5}(X_{1i_1}X_{2i_2}…X_{ki_k}) = \text{MD5}(X_{1j_1}X_{2j_2}…X_{kj_k})$$

holds for any $k = 1, 2, …, m$ and $i_1, j_1, i_2, j_2, …, i_m, j_m \in \{0, 1\}$. In short, we can freely swap those blocks and they have the same MD5 digests. We found a $2^m$-way collision.

#### Part II: From multicollisions of MD5 to preimage attack of $\mathcal{H}$⌗

For us to compute a preimage of $\mathcal{H}$, we would like to show that

$$\mathcal{H}(X_{1,i_1}…X_{k,i_k}) \oplus \mathcal{H}(X_{1,1}…X_{k,1}) = \bigoplus_{j=1}^ k \left[\mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,i_j}) \oplus \mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,1})\right].$$

Proof.

\begin{aligned} & \mathcal{H}(X_{1,i_1}X_{2,i_2}…X_{k,i_k}) \\ &= \mathcal{H}(x_{1,i_1}…x_{128,i_1}x_{129,i_2}…x_{256,i_2}…x_{128k-127,i_k}…x_{128k,i_k}) \\ &= [\text{MD5}(x_{1,i_1}) \oplus … \oplus \text{MD5}(X_{1,i_1})] \\ & \qquad \oplus [\text{MD5}(X_{1,i_1}x_{129,i_2}) \oplus … \oplus \text{MD5}(X_{1,i_1}X_{2,i_2})] \oplus … \\ & \qquad \oplus [\text{MD5}(X_{1,i_1}…X_{k-1,i_{k-1}}x_{128k-127,i_k}) \oplus … \oplus \text{MD5}(X_{1,i_1}…X_{k-1,i_{k-1}}X_{k,i_k})] \\ &= [\text{MD5}(x_{1,i_1}) \oplus … \oplus \text{MD5}(X_{1,i_1})] \\ & \qquad \oplus [\text{MD5}(X_{1,1}x_{129,i_2}) \oplus … \oplus \text{MD5}(X_{1,1}X_{2,i_2})] \oplus … \\ & \qquad \oplus [\text{MD5}(X_{1,1}…X_{k-1,1}x_{128k-127,i_k}) \oplus … \oplus \text{MD5}(X_{1,1}…X_{k-1,1}X_{k,i_k})] \\ &= \mathcal{H}(X_{1,i_1}) \oplus [\mathcal{H}(X_{1,1}) \oplus \mathcal{H}(X_{1,1}X_{2,i_2})] \oplus … \oplus [\mathcal{H}(X_{1,1}…X_{k-1,1}) \oplus \mathcal{H}(X_{1,1}…X_{k-1,1}X_{k,i_k})] \\ &= [\mathcal{H}(X_{1,i_1}) \oplus \mathcal{H}(X_{1,1})] \oplus [\mathcal{H}(X_{1,1}X_{2,i_2}) \oplus \mathcal{H}(X_{1,1}X_{2,1})] \oplus … \\ & \qquad \oplus [\mathcal{H}(X_{1,1}…X_{k-2,1}X_{k-1,i_{k-1}}) \oplus \mathcal{H}(X_{1,1}…X_{k-1,1})] \oplus \mathcal{H}(X_{1,1}…X_{k-1,1}X_{k,i_k}) \\ &= [\mathcal{H}(X_{1,i_1}) \oplus \mathcal{H}(X_{1,1})] \oplus [\mathcal{H}(X_{1,1}X_{2,i_2}) \oplus \mathcal{H}(X_{1,1}X_{2,1})] \oplus … \\ & \qquad \oplus [\mathcal{H}(X_{1,1}…X_{k-2,1}X_{k-1,i_{k-1}}) \oplus \mathcal{H}(X_{1,1}…X_{k-1,1})] \\ & \qquad \oplus [\mathcal{H}(X_{1,1}…X_{k-1,1}X_{k,i_k}) \oplus \mathcal{H}(X_{1,1}…X_{k,1})] \oplus \mathcal{H}(X_{1,1}…X_{k,1}) \\ & = \mathcal{H}(X_{1,1}…X_{k,1}) \oplus \bigoplus_{j=1}^ k \left[\mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,i_j}) \oplus \mathcal{H}(X_{11}…X_{j-1,1}X_{j,1})\right] \qquad \blacksquare \end{aligned}

We can rewrite it the equation with $k$ variables $x_1, x_2, …, x_k$:

$$\mathcal{H}(X_{1,i_1}…X_{k,i_k}) \oplus \mathcal{H}(X_{1,1}…X_{k,1}) = \bigoplus_{j=1}^k x_j \left[\mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,1}) \oplus \mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,2})\right],$$

here $x_r = 0$ if $i_r = 1$ and $x_r = 1$ otherwise. Since we can compute $\mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,1})$ and $\mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,2})$ for $j = 1, 2, …, k$. Also, since $\mathcal{H}(X_{1,i_1}X_{2,i_2}…X_{k,i_k})$ is the target MD5 digest we want to obtain, thus it is also known.

For $j = 1, 2, …, k$, denote

\begin{aligned} & a_j := \mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,1}) \oplus \mathcal{H}(X_{1,1}…X_{j-1,1}X_{j,2}) \\ & b := \mathcal{H}(X_{1,i_1}…X_{k,i_k}) \oplus \mathcal{H}(X_{1,1}…X_{k,1}). \end{aligned}

The equation is now further converted to $a_1x_1 \oplus … \oplus a_kx_k = b$, which is essentially a system of 128 linear equations and $k$ unknowns. If $k$ is large enough (say $\geq 128$), we are confident to find a set of $(x_1, x_2, …, x_k)$.

In my experiment, I used $k = 256$ and the below 32768-byte message $m$ (hex-encoded), would yield $\mathcal{H}(m) = \texttt{hhhhhhhhhhhhhhhh}$.

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9aaf66db585eff78da724b05efd0e4c567647f4fad121bde22b3d1c67423d81ab7e142a225ca7b8ecbfc081dae11abf27a321ae0eb6fd2bbbb8be7fb17e59a1ded87d4bde0f55cd53dcdd4c01359ba7982b670972dbc805ebd473a6824a5c14918f9605ecddb15c15d597daeb74956be6e049341a357d1cf80457bc6da028df10c2aee21d2227f792cb1e55beba8afdd38d9653a092f5ba84617660ee28ce0cd2e29180c490e7a2ee92cb7196ac3c6066a4af936e7e35e3486e53b393856105602127e653b31f07cb62ef05de4af1037039cc8dab7ce8bb4edfd7b8e1850b2a8dc33c9eb22691f47cb79cded459f335e980dd5e86270c9718c70551c0b996e9386cf9340f31fece535a17ce9


Sending the above message to the server, we have the flag:

hhh hhh hhhh hhh hhhhh hhhh hhhh hhhhh hhhh hh hhhhhh hhhh?
h: d39710...a17ce9
hhhh hhhh, hhhh hh hhhh hhhh: DUCTF{hhh.hhh_hh_hhhhhh_hhh,hh_hh?h_hh_hhhh_hhh-hh,hhh-hhhhh_hhhhh_hh.}


## handshake⌗

### Challenge Summary⌗

We are able to connect to a remote service which performs the below protocol:

Server->Client: Server's public key Client->Server: Client's certificate Note left of Server: authenticates the\nclient's certificate Note over Server, Client: compute the shared secret Client->Server: Client's nonce Note left of Server: generates a nonce and\ncomputes a shared nonce Server->Client: Server's nonce, Shared nonce Note over Server, Client: compute a derived key Note left of Server: encrypts a message with\nthe derived key Server->Client: Encrypted message

Denote $a, b$ be the private keys of the server and the client and $A, B$ be their public counterpart. The shared secret $s$ is given by the x-coordinate of $a \cdot B = b \cdot A$, and the shared nonce $\eta_\text{Shared}$ is computed by

$$\eta_\text{Shared} := \text{SHA256}(\eta_\text{Client} \ || \ s \ || \ \eta_\text{Server}).$$

The derived key $k$ is computed by

$$k := \text{HKDF}({\color{gray}\text{ikm}=}\ s \ || \ \eta_\text{Client} \ || \ \eta_\text{Server} \ || \ \eta_\text{Shared}, {\color{gray}\text{salt}=}\ \texttt{DUCTF-2023}).$$

On the other hand, we are given

1. an unprivileged client’s certificate (with the private key),
2. a privileged client’s certificate.
3. the certificate authority’s public key, and
4. the server’s public key.

Additionally, the flag will be included in the message if we are using a privileged client’s certificate. The objective is to recover the flag.

### Solution⌗

There are a lot of red herrings in this challenge. I thought of

1. Forge a privileged client’s certificate. The private key that signs the client certificates is disposed after generation, so it is impossible.
2. Recover the shared secret. Remarkably, if we are using the same client certificate, the shared secret will remain unchanged.

There are two issues that made the challenge vulnerable:

1. The shared secret is computed solely with the server’s and client’s static keys, which will be a constant for the same server-client pair.
2. The derived key $k$ is maliciously constructed.

Given that $s$ is unchanged, we can derive $k$ by crafting multiple $\eta_\text{Client}$s and observing the corresponding $\eta_\text{Shared}$s. To begin, we will look at what is under the hood of HKDF.

We can see that $\eta_\text{Shared}$ and $k$ are computed in a similar fashion (what?). Let’s unwrap the HKDF function and express $k$ in terms of HMAC in the first stage. We should end up presenting $k$ in terms of a bunch of SHA256 calls:

\begin{aligned} k &= \text{HKDF}({\color{gray}\text{ikm}=}\ s \ || \ \eta_\text{Client} \ || \ \eta_\text{Server} \ || \ \eta_\text{Shared}, {\color{gray}\text{salt}=}\ \texttt{DUCTF-2023}) \\ &= \text{HMAC}({\color{gray}\text{msg}=}\ \texttt{[01]}, {\color{gray}\text{key}=}\ \text{HMAC}({\color{gray}\text{msg}=}\ s \ || \ \eta_\text{Client} \ || \ \eta_\text{Server} \ || \ \eta_\text{Shared}, {\color{gray}\text{key}=}\ \texttt{DUCTF-2023})) \end{aligned}

Furthermore, denote the key of the HMAC outside by $\text{prk}$. We then have

\begin{aligned} \text{prk} &= \text{HMAC}({\color{gray}\text{msg}=}\ s \ || \ \eta_\text{Client} \ || \ \eta_\text{Server} \ || \ \eta_\text{Shared}, {\color{gray}\text{key}=}\ \texttt{DUCTF-2023}) \\ &= \text{SHA256}(t_1 \ || \ \text{SHA256}(t_2 \ || \ s \ || \ \eta_\text{Client} \ || \ \eta_\text{Server} \ || \ \eta_\text{Shared})). \end{aligned}

Here $t_1$ and $t_2$ are two strings based on the HMAC key. Since the HMAC key is known, we have those $t_i$’s. Look into the SHA256 function inside:

• $\eta_\text{Client}$ is controllable,
• $t_2$ is known,
• $s$ is unknown but it is constant for all connections,
• $\eta_\text{Server}$ and $\eta_\text{Shared}$ are known after we send $\eta_\text{Client}$ to the server.

Let’s revisit on how $\eta_\text{Shared}$ is generated:

$$\eta_\text{Shared} = \text{SHA256}(\eta_\text{Client} \ || \ s \ || \ \eta_\text{Server}).$$

If the internal digest for $\text{prk}$ is known, we can readily recover $\text{prk}$ and thus $k$. One idea I had during this moment is to set $\eta^{(1)}_\text{Client}$ to be $t_2$. By sending this client nonce to the server, we will obtain $\eta^{(1)}_\text{Shared} = \text{SHA256}(t_2 \ || \ s \ || \ \eta^{(1)}_\text{Server})$ as the shared nonce.

Since we do not have $s$, we cannot compute the internal digest of $\text{prk}$. Fortunately, since $s$ is unchanged for every connection, we might be able to compute $\text{prk}$ because SHA256 is constructed by the Merkle-Damgard scheme. For the second connection, we can send $\eta'_\text{Client}$ defined below to the server.

$$\eta^{(2)}_\text{Client} := \eta^{(1)}_\text{Server} \ || \ \texttt{80 00 00 00 00 00 00 00 00 00 00 00 00 00 03 80}$$

Notably, $\texttt{80 00 00 … 00 03 80}$ is the padding (assuming $s$ is 32-byte long) before computing the SHA256 digest for $\eta^{(1)}_\text{Client} \ || \ s \ || \ \eta^{(1)}_\text{Server}$. We can recover

\begin{aligned} & \text{SHA256}(t_2 \ || \ s \ || \ \eta^{(2)}_\text{Client} \ || \ \eta^{(2)}_\text{Server} \ || \ \eta^{(2)}_\text{Shared}) \\ &= \text{SHA256}(t_2 \ || \ s \ || \ \eta^{(1)}_\text{Server} \ || \ \texttt{80 00 00 … 00 03 80} \ || \ \eta^{(2)}_\text{Server} \ || \ \eta^{(2)}_\text{Shared}) \end{aligned}

by length extension attack using $\eta^{(1)}_\text{Shared}$, $\eta^{(2)}_\text{Server}$ and $\eta^{(2)}_\text{Shared}$. With that, we can recover $k^{(2)}$, effectively decrypting the ciphertext.

DUCTF{k3y_der1v3d_fr0m_ov3rsh4r3d_mat3r1al}