# Big block cipher as memory-hard function

I'm wondering, if something like block cipher with big block size is a good memory-hard function?

All memory-hard key derivation functions I've seen look more complex than that, which made me question my understanding of memory-hardness.

The following function looks memory-hard (memory tradeoff looks expensive) to me:

size = 16MB / (F() word size in bytes)
rounds = 1024
state = constants ⊕ (key || salt)

for (i = 0; i < rounds * size; i++)
state[i mod size] = state[i mod size] ⊕ F(state[i + 1 mod size])

result = state ⊕ key


How memory-hard is this type-I generalized Feistel network key derivation function?

I've been toying around with your function, and I've come to the conclusion it's not memory hard. The amount of required memory can be reduced to at maximum digestsize * 3 * rounds.

The first problem is that the entropy does not avalanche throughout the state, but stays localized. For example, after 1 round the state of the 2nd block only depends on the state of the 3rd block. After 2 rounds it (indirectly) only depends on the initial state of the 3rd and the 4th block. This means that an attacker could precompute 1024 rounds on a large portion of the constant and not need per-password memory for that.

A possible improvement would be to turn around the mixing to state[i mod size] = state[i mod size] ⊕ F(state[i - 1 mod size]) so that each block depends on the previous instead of the next, but the fact that you're generating & hashing in the same order will always make tradeoffs possible. Changing that order, preferably in a data-indepent way (e.g. by reversing the order of the blocks after each round), would make that way more difficult.

To illustrate my point, here are 2 codes: 1 with, and 1 without a memory tradeoff on your function:

#!/usr/bin/env python3
from hashlib import sha512

blocks = 1000
rounds = 100
# pseudorandom constants created by hashing the literals "0", "1", ... "1000"
constants = [sha512(str(i).encode()).digest() for i in range(0, blocks)]

def mix(a, b):
global operations
operations += 1  # statistics
b = sha512(b).digest()
return bytearray(a ^ b for (a, b) in zip(a, b))

def more_memory(constant):
state = list(constant)

for i in range(0, rounds * blocks):
state[i % blocks] = mix(state[i % blocks], state[(i + 1) % blocks])

# Make a checksum for the entire state
state_hash = sha512()
for s in state:
state_hash.update(s)

return state_hash.hexdigest()[:32]

def less_memory(constant):
def perform_round(sequence):
first = None
cur = next(sequence)

try:
while True:
after = next(sequence)
result = mix(cur, after)
yield result
if not first:
first = result
cur = after
except StopIteration:
pass

yield mix(cur, first)

current = (s for s in constant)
for _ in range(0, rounds):
current = perform_round(current)

# Make a checksum for the entire state
state_hash = sha512()
for s in current:
state_hash.update(s)

return state_hash.hexdigest()[:32]

operations = 0
print('Using a lot of memory: {} ({} mixing-operations)'.format(more_memory(constants), operations))
operations = 0
print('Using little memory: {} ({} mixing-operations)'.format(less_memory(constants), operations))

# Prints:
# Using a lot of memory: e58f5665a2c388fc1420d4d1667f83df (100000 mixing-operations)
# Using little memory: e58f5665a2c388fc1420d4d1667f83df (100000 mixing-operations)

• i + 1 was actually a typo. – LightBit Apr 6 '16 at 17:34