hxp


hxp 36C3 CTF: md15

rev (909 points, 2 solves)

Full release of our research paper on executing a “preimage attack” on the MD5 function.

The “paper” can be downloaded here.

tl;dr

Apparently our joke on a feasible preimage attack on MD5 seemed too legit, because we saw a lot of people actually reading crypto papers about MD5 at 36C3.

The short answer is that we—like everybody else on this planet—also have no feasible preimage attack on MD5. Instead, we shamelessly backdoored the binary. The backdoor consists of two main components:

  1. The __libc_csu_init function usually contains a nop that we replaced by an instruction that would shift the pointers (of non-RELRO ELF binaries) contained in the init_array by 8 bytes. This way, we could hide a jump instruction in the code directly after the default constructor (frame_dummy).
  2. The jump instruction targets code beyond the end of the segment limit. This code is not displayed by any disassembler (try objdump -D) we are currently aware of. Nevertheless, the GNU loader works at a page granularity when loading segments and therefore code beyond the end of the segment limit ends up in memory, perfectly executable. (Note that not all loaders offer this (un-)feature; the ELF loader implemented in Microsoft’s WSL honors segment boundaries causing md15 to crash immediately on program startup. We took this risk, and according to the internet it went unnoticed for a pretty long time.

The hidden code implemented a simple check on the input such that it would not reveal the backdoor functionality when sampling the MD5 implementation in the binary using GDB. Only if the check passed (and if the environment looked sane), the backdoor code would patch the MD5 implementation in such a way that md5 implementation was reduced to a 12 rounds. MD5 with 12 (instead of 64) rounds can be trivially inverted using a SMT solver. Our test solve script used z3 to calculate the flag:

import struct, z3

h = bytes.fromhex('3ed50eac373185348499454857b06fd3') # md5(flag ^ 'h')
x = bytes.fromhex('448582faa78b404a898d0532542d327b') # md5(flag ^ 'x')
p = bytes.fromhex('9973f05fde3fe6320be04a918c5b50ab') # md5(flag ^ 'p')

a0, b0, c0, d0 = 0x67452301, 0xefcdab89, 0x98badcfe, 0x10325476
ah, bh, ch, dh = struct.unpack('IIII', h)
ax, bx, cx, dx = struct.unpack('IIII', x)
ap, bp, cp, dp = struct.unpack('IIII', p)

unknown_words = z3.BitVecs('f0 f1 f2 f3', 32)
remaining_words = struct.unpack('I' * 12, b'\x80' + b'\x00' * 39 + struct.pack('Q', 128))

h_flag = [w ^ 0x68686868 for w in unknown_words] + list(remaining_words)
x_flag = [w ^ 0x78787878 for w in unknown_words] + list(remaining_words)
p_flag = [w ^ 0x70707070 for w in unknown_words] + list(remaining_words)

K = [0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee,
     0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501,
     0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be]
S = [7, 12, 17, 22] * 3
def FF(b, c, d):
    return (b & c) | ((~b) & d)
def u32(value):
    return (value + (1 << 32)) % (1 << 32) if isinstance(value, int) else value
def ror(value, shift):
    if isinstance(value, int):
        shift %= 32
        shifted = u32(value) >> shift
        excess = value & ((1 << shift) - 1)
        return shifted | (excess << (32 - shift))
    return z3.RotateRight(value, shift)
def invert_md5(a, b, c, d, values):
    a -= a0
    b -= b0
    c -= c0
    d -= d0
    for r in range(11, -1, -1):
        B, C, D = c, d, a
        a_t1 = u32(ror(b - c, S[r]) - K[r])
        a_t2 = u32(a_t1 - values[r])
        A = u32(a_t2 - FF(B, C, D))
        a, b, c, d = A, B, C, D
    print(a, b, c, d)
    return z3.And(a == a0, b == b0, c == c0, d == d0)

ss = z3.Solver()
print('Adding H flag')
ss.add(invert_md5(ah, bh, ch, dh, h_flag))
print('Adding X flag')
ss.add(invert_md5(ax, bx, cx, dx, x_flag))
print('Adding P flag')
ss.add(invert_md5(ap, bp, cp, dp, p_flag))
print('Solving')
print(ss.check())

m = ss.model()
result = struct.pack('IIII', *[int(str(m.evaluate(w))) for w in unknown_words])
print('hxp{' + result.decode() + '}')