这是一个相当简单的 Python 3 实现 Format-preserving encryption 使用平衡的 Feistel network，这是我大约 2 年前写的。它可以在 32 位系统上对 N 最多 264 或在 64 位 Python 构建上的 2128 执行您想要的索引排列。这是由于hash() 函数返回的整数的大小。请参阅sys.hash_info 以查找您的系统的限制。使用可以返回更大位长度的值的更高级的哈希函数并不难，但我不想让这段代码更复杂或更慢。
更新
我对之前的版本做了一些小的改进，并在 cmets 中添加了更多信息。我们不使用从哈希函数返回的低位，而是使用高位，这通常会提高随机性，尤其是对于短位长度。我还添加了另一个散列函数xxhash by Yann Collet，它比 Python 的 hash 在这个应用程序中的工作效果好多，特别是对于较短的位长度，虽然它有点慢。 xxhash 算法的avalanche effect 比内置的hash 高得多，因此得到的排列往往更容易打乱。
虽然此代码适用于 stop 的小值，但它更适合处理 stop >= 2**16。如果您需要置换更小的范围，那么在list(range(stop)) 上使用random.shuffle 可能是一个好主意。它会更快，并且不会使用那么多 RAM：list(range(2**16)) 在 32 位机器上消耗大约 1280 KB。
您会注意到我使用字符串作为随机数生成器的种子。对于这个应用程序，我们希望随机化器有足够的熵，并且使用大字符串（或bytes）是一种简单的方法，正如random module 文档提到的那样。即便如此，当stop 很大时，这个程序只能产生所有可能排列的一小部分。 stop == 35 有 35 个！ （35 阶乘）不同的排列，还有 35！ > 2132，但是我们的密钥的总比特长度只有 128，所以它们不能覆盖所有这些排列。我们可以增加 Feistel 轮数以获得更多覆盖，但显然这对于 stop 的大值是不切实际的。
''' Format preserving encryption using a Feistel network
This code is *not* suitable for cryptographic use.
See https://en.wikipedia.org/wiki/Format-preserving_encryption
https://en.wikipedia.org/wiki/Feistel_cipher
http://security.stackexchange.com/questions/211/how-to-securely-hash-passwords
A Feistel network performs an invertible transformation on its input,
so each input number produces a unique output number. The netword operates
on numbers of a fixed bit width, which must be even, i.e., the numbers
a particular network operates on are in the range(4**k), and it outputs a
permutation of that range.
To permute a range of general size we use cycle walking. We set the
network size to the next higher power of 4, and when we produce a number
higher than the desired range we simply feed it back into the network,
looping until we get a number that is in range.
The worst case is when stop is of the form 4**k + 1, where we need 4
steps on average to reach a valid n. In the typical case, where stop is
roughly halfway between 2 powers of 4, we need 2 steps on average.
Written by PM 2Ring 2016.08.22
'''
from random import Random
# xxhash by Yann Collet. Specialised for a 32 bit number
# See http://fastcompression.blogspot.com/2012/04/selecting-checksum-algorithm.html
def xxhash_num(n, seed):
n = (374761397 + seed + n * 3266489917) & 0xffffffff
n = ((n << 17 | n >> 15) * 668265263) & 0xffffffff
n ^= n >> 15
n = (n * 2246822519) & 0xffffffff
n ^= n >> 13
n = (n * 3266489917) & 0xffffffff
return n ^ (n >> 16)
class FormatPreserving:
""" Invertible permutation of integers in range(stop), 0 < stop <= 2**64
using a simple Feistel network. NOT suitable for cryptographic purposes.
"""
def __init__(self, stop, keystring):
if not 0 < stop <= 1 << 64:
raise ValueError('stop must be <=', 1 << 64)
# The highest number in the range
self.maxn = stop - 1
# Get the number of bits in each part by rounding
# the bit length up to the nearest even number
self.shiftbits = -(-self.maxn.bit_length() // 2)
self.lowmask = (1 << self.shiftbits) - 1
self.lowshift = 32 - self.shiftbits
# Make 4 32 bit round keys from the keystring.
# Create an independent random stream so we
# don't intefere with the default stream.
stream = Random()
stream.seed(keystring)
self.keys = [stream.getrandbits(32) for _ in range(4)]
self.ikeys = self.keys[::-1]
def feistel(self, n, keys):
# Split the bits of n into 2 parts & perform the Feistel
# transformation on them.
left, right = n >> self.shiftbits, n & self.lowmask
for key in keys:
left, right = right, left ^ (xxhash_num(right, key) >> self.lowshift)
#left, right = right, left ^ (hash((right, key)) >> self.lowshift)
return (right << self.shiftbits) | left
def fpe(self, n, reverse=False):
keys = self.ikeys if reverse else self.keys
while True:
# Cycle walk, if necessary, to ensure n is in range.
n = self.feistel(n, keys)
if n <= self.maxn:
return n
def test():
print('Shuffling a small number')
maxn = 10
fpe = FormatPreserving(maxn, 'secret key string')
for i in range(maxn):
a = fpe.fpe(i)
b = fpe.fpe(a, reverse=True)
print(i, a, b)
print('\nShuffling a small number, with a slightly different keystring')
fpe = FormatPreserving(maxn, 'secret key string.')
for i in range(maxn):
a = fpe.fpe(i)
b = fpe.fpe(a, reverse=True)
print(i, a, b)
print('\nHere are a few values for a large maxn')
maxn = 10000000000000000000
print('maxn =', maxn)
fpe = FormatPreserving(maxn, 'secret key string')
for i in range(10):
a = fpe.fpe(i)
b = fpe.fpe(a, reverse=True)
print('{}: {:19} {}'.format(i, a, b))
print('\nUsing a set to test that there are no collisions...')
maxn = 100000
print('maxn', maxn)
fpe = FormatPreserving(maxn, 'secret key string')
a = {fpe.fpe(i) for i in range(maxn)}
print(len(a) == maxn)
print('\nTesting that the operation is bijective...')
for i in range(maxn):
a = fpe.fpe(i)
b = fpe.fpe(a, reverse=True)
assert b == i, (i, a, b)
print('yes')
if __name__ == "__main__":
test()
输出
Shuffling a small number
0 4 0
1 2 1
2 5 2
3 9 3
4 1 4
5 3 5
6 7 6
7 0 7
8 6 8
9 8 9
Shuffling a small number, with a slightly different keystring
0 9 0
1 8 1
2 3 2
3 5 3
4 2 4
5 6 5
6 1 6
7 4 7
8 7 8
9 0 9
Here are a few values for a large maxn
maxn = 10000000000000000000
0: 7071024217413923554 0
1: 5613634032642823321 1
2: 8934202816202119857 2
3: 296042520195445535 3
4: 5965959309128333970 4
5: 8417353297972226870 5
6: 7501923606289578535 6
7: 1722818114853762596 7
8: 890028846269590060 8
9: 8787953496283620029 9
Using a set to test that there are no collisions...
maxn 100000
True
Testing that the operation is bijective...
yes
0 4
1 2
2 5
3 9
4 1
5 3
6 7
7 0
8 6
9 8
以下是如何使用它来制作生成器：
def ipermute(stop, keystring):
fpe = FormatPreserving(stop, keystring)
for i in range(stop):
yield fpe.fpe(i)
for i, v in enumerate(ipermute(10, 'secret key string')):
print(i, v)
输出
0 4
1 2
2 5
3 9
4 1
5 3
6 7
7 0
8 6
9 8
它相当快（对于 Python），但它绝对不适合密码学。 可以通过将 Feistel 轮数增加到至少 5 轮并使用合适的加密哈希函数（例如 Blake2）来实现加密级别。此外，需要使用加密方法来生成 Feistel 密钥。当然，除非您确切地知道自己在做什么，否则不应该编写加密软件，因为编写容易受到定时攻击等攻击的代码太容易了。