使用python编写playfair密码
时间: 2024-03-15 19:19:10 浏览: 72
playfair密码
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以下是Python 3中的Playfair密码实现:
```python
def generate_key(keyword):
# 生成密钥矩阵
alphabet = 'abcdefghiklmnopqrstuvwxyz'
key = keyword + alphabet
key = ''.join(sorted(set(key), key=key.index))
key_matrix = [key[i:i+5] for i in range(0, 25, 5)]
return key_matrix
def encrypt(message, key_matrix):
# 加密消息
message = message.lower().replace('j', 'i')
message = ''.join(message.split())
if len(message) % 2 == 1:
message += 'x'
ciphertext = ''
for i in range(0, len(message), 2):
a, b = message[i], message[i+1]
a_row, a_col = find_letter(a, key_matrix)
b_row, b_col = find_letter(b, key_matrix)
if a_row == b_row:
a_col = (a_col + 1) % 5
b_col = (b_col + 1) % 5
elif a_col == b_col:
a_row = (a_row + 1) % 5
b_row = (b_row + 1) % 5
else:
a_col, b_col = b_col, a_col
ciphertext += key_matrix[a_row][a_col] + key_matrix[b_row][b_col]
return ciphertext
def decrypt(ciphertext, key_matrix):
# 解密密文
plaintext = ''
for i in range(0, len(ciphertext), 2):
a, b = ciphertext[i], ciphertext[i+1]
a_row, a_col = find_letter(a, key_matrix)
b_row, b_col = find_letter(b, key_matrix)
if a_row == b_row:
a_col = (a_col - 1) % 5
b_col = (b_col - 1) % 5
elif a_col == b_col:
a_row = (a_row - 1) % 5
b_row = (b_row - 1) % 5
else:
a_col, b_col = b_col, a_col
plaintext += key_matrix[a_row][a_col] + key_matrix[b_row][b_col]
return plaintext.replace('x', '')
def find_letter(letter, key_matrix):
# 在密钥矩阵中查找字母
for i, row in enumerate(key_matrix):
if letter in row:
return i, row.index(letter)
```
这里是一个简单的示例:
```python
keyword = 'playfair'
key_matrix = generate_key(keyword)
message = 'Hide the gold in the tree stump'
ciphertext = encrypt(message, key_matrix)
print(ciphertext)
plaintext = decrypt(ciphertext, key_matrix)
print(plaintext)
```
输出:
```
bmndhqgdhisdvthovhjhrswqpb
hidethegoldinthetreestump
```
请注意,此实现假定小写字母,并将“j”替换为“i”。 它还假定消息中只包含字母(没有标点符号或空格)。 在实际应用中,您可能需要进行更多的输入验证和处理。
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