"""
Hill Cipher:
The 'HillCipher' class below implements the Hill Cipher algorithm which uses
modern linear algebra techniques to encode and decode text using an encryption
key matrix.
Algorithm:
Let the order of the encryption key be N (as it is a square matrix).
Your text is divided into batches of length N and converted to numerical vectors
by a simple mapping starting with A=0 and so on.
The key is then multiplied with the newly created batch vector to obtain the
encoded vector. After each multiplication modular 36 calculations are performed
on the vectors so as to bring the numbers between 0 and 36 and then mapped with
their corresponding alphanumerics.
While decrypting, the decrypting key is found which is the inverse of the
encrypting key modular 36. The same process is repeated for decrypting to get
the original message back.
Constraints:
The determinant of the encryption key matrix must be relatively prime w.r.t 36.
Note:
This implementation only considers alphanumerics in the text. If the length of
the text to be encrypted is not a multiple of the break key(the length of one
batch of letters), the last character of the text is added to the text until the
length of the text reaches a multiple of the break_key. So the text after
decrypting might be a little different than the original text.
References:
https://apprendre-en-ligne.net/crypto/hill/Hillciph.pdf
https://www.youtube.com/watch?v=kfmNeskzs2o
https://www.youtube.com/watch?v=4RhLNDqcjpA
"""
import string
import numpy as np
from maths.greatest_common_divisor import greatest_common_divisor
class HillCipher:
key_string = string.ascii_uppercase + string.digits
# This cipher takes alphanumerics into account
# i.e. a total of 36 characters
# take x and return x % len(key_string)
modulus = np.vectorize(lambda x: x % 36)
to_int = np.vectorize(round)
def __init__(self, encrypt_key: np.ndarray) -> None:
"""
encrypt_key is an NxN numpy array
"""
self.encrypt_key = self.modulus(encrypt_key) # mod36 calc's on the encrypt key
self.check_determinant() # validate the determinant of the encryption key
self.break_key = encrypt_key.shape[0]
def replace_letters(self, letter: str) -> int:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.replace_letters('T')
19
>>> hill_cipher.replace_letters('0')
26
"""
return self.key_string.index(letter)
def replace_digits(self, num: int) -> str:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.replace_digits(19)
'T'
>>> hill_cipher.replace_digits(26)
'0'
"""
return self.key_string[round(num)]
def check_determinant(self) -> None:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.check_determinant()
"""
det = round(np.linalg.det(self.encrypt_key))
if det < 0:
det = det % len(self.key_string)
req_l = len(self.key_string)
if greatest_common_divisor(det, len(self.key_string)) != 1:
msg = (
f"determinant modular {req_l} of encryption key({det}) "
f"is not co prime w.r.t {req_l}.\nTry another key."
)
raise ValueError(msg)
def process_text(self, text: str) -> str:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.process_text('Testing Hill Cipher')
'TESTINGHILLCIPHERR'
>>> hill_cipher.process_text('hello')
'HELLOO'
"""
chars = [char for char in text.upper() if char in self.key_string]
last = chars[-1]
while len(chars) % self.break_key != 0:
chars.append(last)
return "".join(chars)
def encrypt(self, text: str) -> str:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.encrypt('testing hill cipher')
'WHXYJOLM9C6XT085LL'
>>> hill_cipher.encrypt('hello')
'85FF00'
"""
text = self.process_text(text.upper())
encrypted = ""
for i in range(0, len(text) - self.break_key + 1, self.break_key):
batch = text[i : i + self.break_key]
vec = [self.replace_letters(char) for char in batch]
batch_vec = np.array([vec]).T
batch_encrypted = self.modulus(self.encrypt_key.dot(batch_vec)).T.tolist()[
0
]
encrypted_batch = "".join(
self.replace_digits(num) for num in batch_encrypted
)
encrypted += encrypted_batch
return encrypted
def make_decrypt_key(self) -> np.ndarray:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.make_decrypt_key()
array([[ 6, 25],
[ 5, 26]])
"""
det = round(np.linalg.det(self.encrypt_key))
if det < 0:
det = det % len(self.key_string)
det_inv = None
for i in range(len(self.key_string)):
if (det * i) % len(self.key_string) == 1:
det_inv = i
break
inv_key = (
det_inv * np.linalg.det(self.encrypt_key) * np.linalg.inv(self.encrypt_key)
)
return self.to_int(self.modulus(inv_key))
def decrypt(self, text: str) -> str:
"""
>>> hill_cipher = HillCipher(np.array([[2, 5], [1, 6]]))
>>> hill_cipher.decrypt('WHXYJOLM9C6XT085LL')
'TESTINGHILLCIPHERR'
>>> hill_cipher.decrypt('85FF00')
'HELLOO'
"""
decrypt_key = self.make_decrypt_key()
text = self.process_text(text.upper())
decrypted = ""
for i in range(0, len(text) - self.break_key + 1, self.break_key):
batch = text[i : i + self.break_key]
vec = [self.replace_letters(char) for char in batch]
batch_vec = np.array([vec]).T
batch_decrypted = self.modulus(decrypt_key.dot(batch_vec)).T.tolist()[0]
decrypted_batch = "".join(
self.replace_digits(num) for num in batch_decrypted
)
decrypted += decrypted_batch
return decrypted
def main() -> None:
n = int(input("Enter the order of the encryption key: "))
hill_matrix = []
print("Enter each row of the encryption key with space separated integers")
for _ in range(n):
row = [int(x) for x in input().split()]
hill_matrix.append(row)
hc = HillCipher(np.array(hill_matrix))
print("Would you like to encrypt or decrypt some text? (1 or 2)")
option = input("\n1. Encrypt\n2. Decrypt\n")
if option == "1":
text_e = input("What text would you like to encrypt?: ")
print("Your encrypted text is:")
print(hc.encrypt(text_e))
elif option == "2":
text_d = input("What text would you like to decrypt?: ")
print("Your decrypted text is:")
print(hc.decrypt(text_d))
if __name__ == "__main__":
import doctest
doctest.testmod()
main()
希尔密码是由Lester S. Hill 发明的。
希尔密码是一种基于线性代数的多字母替代密码。每个字母用一个模 26 的数字表示。通常使用简单的方案 A = 0, B = 1, …, Z = 25
,但这并不是密码的必要特征。为了加密一条消息,每个包含 n
个字母的块(被视为一个 n 维向量)乘以一个可逆的 n × n
矩阵,模 26。为了解密消息,每个块乘以用于加密的矩阵的逆矩阵。
假设我们以以下为例:明文 (PT):ACT 密钥:GYBNQKURP
n × n
矩阵,如下所示: [6 24 1]
[13 16 10]
[20 17 15]
[0]
[2]
[19]
[6 24 1] [0] [67] [15]
[13 16 10] * [2] = [222] ≈ [4] (mod 26)
[20 17 15] [19] [319] [7]
因此,我们将得到加密后的文本为 POH。
-1
[6 24 1] [8 5 10]
[13 16 10] ≈ [21 8 21] (mod 26)
[20 17 15] [21 12 8]
[8 5 10] [15] [260] [0]
[21 8 21] * [14] ≈ [574] ≈ [2] (mod 26) ≈ ACT
[21 12 8] [7] [539] [19]