pycharm hill密码随机输入一段明文加密或者密文解密
时间: 2023-12-12 09:04:36 浏览: 180
为了回答你的问题,我需要先了解一下 Hill 密码。Hill 密码是一种基于矩阵运算的对称密码,它将明文分成若干个数字组合,然后通过矩阵运算进行加密。因此,我们需要使用矩阵计算的工具来进行 Hill 密码的加密和解密。
下面是使用 Python 和 NumPy 库实现 Hill 密码的加密和解密的示例代码:
```python
import numpy as np
# 加密函数
def encrypt(plain_text, key):
# 将明文转换成数字向量
plain_text = [ord(c) - ord('a') for c in plain_text]
plain_text = np.array(plain_text)
# 将密钥转换成矩阵
key = np.array(key)
# 将数字向量按照密钥的大小分成若干个块
block_size = key.shape[0]
num_blocks = int(np.ceil(len(plain_text) / block_size))
plain_text = np.pad(plain_text, (0, num_blocks * block_size - len(plain_text)), 'constant')
plain_text = plain_text.reshape(-1, block_size)
# 对每个块进行加密
cipher_text = []
for block in plain_text:
cipher_block = np.mod(key @ block, 26)
cipher_text += cipher_block.tolist()
# 将密文转换成字符串
cipher_text = ''.join([chr(c + ord('a')) for c in cipher_text])
return cipher_text
# 解密函数
def decrypt(cipher_text, key):
# 将密文转换成数字向量
cipher_text = [ord(c) - ord('a') for c in cipher_text]
cipher_text = np.array(cipher_text)
# 将密钥转换成矩阵
key = np.array(key)
# 将数字向量按照密钥的大小分成若干个块
block_size = key.shape[0]
num_blocks = int(np.ceil(len(cipher_text) / block_size))
cipher_text = np.pad(cipher_text, (0, num_blocks * block_size - len(cipher_text)), 'constant')
cipher_text = cipher_text.reshape(-1, block_size)
# 对每个块进行解密
plain_text = []
for block in cipher_text:
# 计算矩阵的逆
det = np.linalg.det(key)
inv_key = np.mod(np.round(det * np.linalg.inv(key)), 26)
plain_block = np.mod(inv_key @ block, 26)
plain_text += plain_block.tolist()
# 将明文转换成字符串
plain_text = ''.join([chr(c + ord('a')) for c in plain_text])
return plain_text
# 测试代码
key = [[3, 10], [20, 9]]
plain_text = 'hello'
cipher_text = encrypt(plain_text, key)
print(cipher_text)
plain_text = decrypt(cipher_text, key)
print(plain_text)
```
在这个示例中,我们定义了两个函数 `encrypt` 和 `decrypt` 分别实现了 Hill 密码的加密和解密。函数的参数 `plain_text` 表示明文字符串,`cipher_text` 表示密文字符串,`key` 表示 Hill 密钥矩阵。在加密和解密函数中,我们都采用了相同的方式把明文或密文分成若干个块,并对每个块进行加密或解密。
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