[英]AES CTR implementation
我正在尝试仅使用pycrypto的AES内置函数自己实现CTR模式(目前仅解密)。 这意味着我不应该使用mode = AES.MODE_CTR。 但是,我知道使用AES.MODE_CTR会更简单,但这是我的学习经验。
我不确定如何将AES用作PRF,以便在CTR密码算法中使用它。
我究竟做错了什么? (非平行版本)
from Crypto.Cipher import AES
ciphers = ["69dda8455c7dd4254bf353b773304eec0ec7702330098ce7f7520d1cbbb20fc3" + \
"88d1b0adb5054dbd7370849dbf0b88d393f252e764f1f5f7ad97ef79d59ce29f5f51eeca32eabedd9afa9329", \
"770b80259ec33beb2561358a9f2dc617e46218c0a53cbeca695ae45faa8952aa" + \
"0e311bde9d4e01726d3184c34451"]
key = "36f18357be4dbd77f050515c73fcf9f2"
class IVCounter(object):
def __init__(self, value):
self.value = value
def increment(self):
# Add the counter value to IV
newIV = hex(int(self.value.encode('hex'), 16) + 1)
# Cut the negligible part of the string
self.value = newIV[2:len(newIV) - 1].decode('hex') # for not L strings remove $ - 1 $
return self.value
def __repr__(self):
self.increment()
return self.value
def string(self):
return self.value
class CTR():
def __init__(self, k):
self.key = k
def __strxor(self, a, b): # xor two strings of different lengths
if len(a) > len(b):
return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a[:len(b)], b)])
else:
return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b[:len(a)])])
def __split_len(self, seq, lenght):
return [seq[i:i+lenght] for i in range(0, len(seq), lenght)]
def __AESdecryptor(self, k, cipher):
decryptor = AES.new(k, AES.MODE_ECB)
return decryptor.decrypt(cipher)
def decrypt(self, cipher):
# Split the CT in blocks of 16 bytes
blocks = self.__split_len(cipher.decode('hex'), 16)
# Takes the initiator vector
self.IV = IVCounter(blocks[0])
blocks.remove(blocks[0])
# Message block
msg = []
# Decrypt
for b in blocks:
aes = self.__AESdecryptor(self.key.decode('hex'), self.IV.string())
msg.append(self.__strxor(b, aes))
self.IV.increment()
return ''.join(msg)
def main():
decryptor = CTR(key)
for c in ciphers:
print 'msg = ' + decryptor.decrypt(c)
if __name__ == '__main__':
main()
该代码原本应该与下面的代码相同,但是并没有按照原样进行解码。
import Crypto.Util.Counter
ctr_e = Crypto.Util.Counter.new(128, initial_value=long(IV.encode('hex'), 16))
decryptor = AES.new(key.decode('hex'), AES.MODE_CTR, counter=ctr_e)
print decryptor.decrypt(''.join(blocks))
# Decrypt
for b in blocks:
aes = self.__AESdecryptor(self.IV.string(), self.key.decode('hex'))
msg.append(self.__strxor(b, aes))
self.IV.increment()
return ''.join(msg)
AES CTR模式使用AES的前向转换进行加密和解密。 也就是说,在两种情况下,都加密计数器,然后执行XOR。 当我说“正向转换”时,我的意思是您始终执行AES_Encrypt(counter)
(而从不执行AES_Decrypt(counter)
)。
无论是加密还是解密,都对纯文本和密文执行XOR。 text XOR encrypt(counter)
是加密或解密操作。 那是流密码。
self.IV.string()
不是AES密钥。 它是在密钥下加密的值。 加密后,将与{plain | cipher}文本进行XOR。
我终于使这段代码运行良好,并且错误非常简单。 我不应该使用解密AES函数,我应该使用加密AES函数(就像noloader所说的那样,而且我第一次不太了解他)。 感谢所有提供帮助的人,这里是固定代码:
from Crypto.Cipher import AES
ciphers = ["69dda8455c7dd4254bf353b773304eec0ec7702330098ce7f7520d1cbbb20fc3" + \
"88d1b0adb5054dbd7370849dbf0b88d393f252e764f1f5f7ad97ef79d59ce29f5f51eeca32eabedd9afa9329", \
"770b80259ec33beb2561358a9f2dc617e46218c0a53cbeca695ae45faa8952aa" + \
"0e311bde9d4e01726d3184c34451"]
key = "36f18357be4dbd77f050515c73fcf9f2"
class IVCounter(object):
def __init__(self, value):
self.value = value
def increment(self):
# Add the counter value to IV
newIV = hex(int(self.value.encode('hex'), 16) + 1)
# Cut the negligible part of the string
self.value = newIV[2:len(newIV) - 1].decode('hex') # for not L strings remove $ - 1 $
return self.value
def __repr__(self):
self.increment()
return self.value
def string(self):
return self.value
class CTR():
def __init__(self, k):
self.key = k.decode('hex')
def __strxor(self, a, b): # xor two strings of different lengths
if len(a) > len(b):
return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a[:len(b)], b)])
else:
return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b[:len(a)])])
def __split_len(self, seq, lenght):
return [seq[i:i+lenght] for i in range(0, len(seq), lenght)]
def __AESencryptor(self, cipher):
encryptor = AES.new(self.key, AES.MODE_ECB)
return encryptor.encrypt(cipher)
def decrypt(self, cipher):
# Split the CT into blocks of 16 bytes
blocks = self.__split_len(cipher.decode('hex'), 16)
# Takes the initiator vector
self.IV = IVCounter(blocks[0])
blocks.remove(blocks[0])
# Message block
msg = []
# Decrypt
for b in blocks:
aes = self.__AESencryptor(self.IV.string())
msg.append(self.__strxor(b, aes))
self.IV.increment()
return ''.join(msg)
def main():
decryptor = CTR(key)
for c in ciphers:
print 'msg = ' + decryptor.decrypt(c)
if __name__ == '__main__':
main()
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