# Fowler–Noll–Vo hash function¶

## 2018 网鼎杯ashcoll¶

In fact, this question was copied from NSU Crypto, https://nsucrypto.nsu.ru/archive/2017/problems_solution, the specific wp before hellman also wrote, https://gist.github.com/hellman/9bf8376cd04e7a8dd2ec7be1947261e9 .

Simply look at the topic

h0 = 45740974929179720441799381904411404011270459520712533273451053262137196814399

# 2**168 + 355

g = 374144419156711147060143317175368453031918731002211L

def shitty_hash(msg):

h = h0

msg = map(ord, msg)

for i in msg:

h = (h + i) * g

# This line is just to screw you up :))

h = h &amp; 0xffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

return h - 0xe6168647f636


The topic hopes that we give two messages with the same hash value. If we expand the function, then

$hash(m)=h_0g^n+x_1g^n+x_2g_{n-1}+...+x_ng \bmod 2^{256}$

Suppose the hash values of the two messages are the same.

$h_0g ^ n + x_1g ^ n + x_2g_ {n-1} + ... + x_ng equiv h_0g ^ n + y_1g ^ n + y_2g_ {n-1} + ... + y_ng bmod 2 ^ {256}$

and then

$(x_1-y_1) g ^ {n-1} + (x_2-y_2) g ^ {n-2} + ... + (x_n-y_n) g ^ 0 equiv 0 bmod 2 ^ {256}$

That is, we only need to find an n-dimensional vector $z_i=x_i-y_i$, which satisfies the above equation, we can further convert it into

$z_1g ^ {n-1} + z_2g ^ {n-2} + ... + z_ng ^ 0-k * 2 ^ {256} = 0$

That is, a set of vectors is found to satisfy the above formula. This can be thought of as a simple case of the second example in LLL Paper (see the Lattice Question section).

Then we can quickly construct the matrix as follows

A = \left[ \begin{matrix} 1 & 0 & 0 & \cdots & 0 & Kg^{n-1} \\ 0 & 1 & 0 & \cdots & 0 & Kg^{n-2} \\ 0 & 0 & 1 & \cdots & 0 & Kg^{n-3} \\\vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 &0 & \cdots & 1 & K*mod \\ \end{matrix} \right]

Then we can get two identical hashes using the LLL algorithm.

from sage.all import *

mod = 2 ** 256
h0 = 45740974929179720441799381904411404011270459520712533273451053262137196814399

g = 2**168 + 355

def shitty_hash(msg):

h = h0

msg = map(ord, msg)

for i in msg:

h = (h + i) * g

# This line is just to screw you up :))

h = h &amp; 0xffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

return h - 0xe6168647f636

K = 2 ** 200
N = 50

base_str = 'a' * N

base = map (word, base_str)
m = Matrix(ZZ, N + 1, N + 2)

for i in xrange(N + 1):

ge = ZZ(pow(g, N - i, mod))

m[i, i] = 1

m[i, N + 1] = ZZ(ge * K)

m[i, N + 1] = ZZ(K * mod)

ml = m.LLL()

ttt = ml.rows()[0]

print "result:", ttt

if ttt[-1] != 0:

print "Zero not reached, increase K"

exit()

else:

msg = []

for i in xrange(N):

msg.append(base[i] + ttt[i])

if not (0 <= msg[i] <= 255):

print "Need more bytes!"

quit()

print msg

other = ''.join(map(chr, msg))

print shitty_hash(base_str)

print shitty_hash(other)


Note that you can't just use pow(g, N - i, mod) directly, otherwise the generated number will be in the domain corresponding to mod, which is really a big pit.

as follows

➜  hashcoll sage exp.sage

result: (15, -14, 17, 14, 6, 0, 12, 21, 8, 29, 6, -4, -9, 10, -2, -12, -6, 0, -12, 13, -28, -28, -24, -3, 6, -5, -16, 15, 17, -14, 3, -2, -16, -25, 3, -21, -27, -9, 16, 5, -1, 0, -3, -4, -4, -19, 6, 8, 0, 0, 0, 0)

[112, 83, 114, 111, 103, 97, 109, 118, 105, 126, 103, 93, 88, 107, 95, 85, 91, 97, 85, 110, 69, 69, 73, 94, 103, 92, 81, 112, 114, 83, 100, 95, 81, 72, 100, 76, 70, 88, 113, 102, 96, 97, 94, 93, 93, 78, 103, 105, 97, 97]

106025341237231370726407656306665079105509255639964756437758376184556498283725

106025341237231370726407656306665079105509255639964756437758376184556498283725


That is success.