"""
https://en.wikipedia.org/wiki/Playfair_cipher#Description
The Playfair cipher was developed by Charles Wheatstone in 1854
It's use was heavily promotedby Lord Playfair, hence its name
Some features of the Playfair cipher are:
1) It was the first literal diagram substitution cipher
2) It is a manual symmetric encryption technique
3) It is a multiple letter encryption cipher
The implementation in the code below encodes alphabets only.
It removes spaces, special characters and numbers from the
code.
Playfair is no longer used by military forces because of known
insecurities and of the advent of automated encryption devices.
This cipher is regarded as insecure since before World War I.
"""
import itertools
import string
from collections.abc import Generator, Iterable
def chunker(seq: Iterable[str], size: int) -> Generator[tuple[str, ...], None, None]:
it = iter(seq)
while True:
chunk = tuple(itertools.islice(it, size))
if not chunk:
return
yield chunk
def prepare_input(dirty: str) -> str:
"""
Prepare the plaintext by up-casing it
and separating repeated letters with X's
"""
dirty = "".join([c.upper() for c in dirty if c in string.ascii_letters])
clean = ""
if len(dirty) < 2:
return dirty
for i in range(len(dirty) - 1):
clean += dirty[i]
if dirty[i] == dirty[i + 1]:
clean += "X"
clean += dirty[-1]
if len(clean) & 1:
clean += "X"
return clean
def generate_table(key: str) -> list[str]:
# I and J are used interchangeably to allow
# us to use a 5x5 table (25 letters)
alphabet = "ABCDEFGHIKLMNOPQRSTUVWXYZ"
# we're using a list instead of a '2d' array because it makes the math
# for setting up the table and doing the actual encoding/decoding simpler
table = []
# copy key chars into the table if they are in `alphabet` ignoring duplicates
for char in key.upper():
if char not in table and char in alphabet:
table.append(char)
# fill the rest of the table in with the remaining alphabet chars
for char in alphabet:
if char not in table:
table.append(char)
return table
def encode(plaintext: str, key: str) -> str:
"""
Encode the given plaintext using the Playfair cipher.
Takes the plaintext and the key as input and returns the encoded string.
>>> encode("Hello", "MONARCHY")
'CFSUPM'
>>> encode("attack on the left flank", "EMERGENCY")
'DQZSBYFSDZFMFNLOHFDRSG'
>>> encode("Sorry!", "SPECIAL")
'AVXETX'
>>> encode("Number 1", "NUMBER")
'UMBENF'
>>> encode("Photosynthesis!", "THE SUN")
'OEMHQHVCHESUKE'
"""
table = generate_table(key)
plaintext = prepare_input(plaintext)
ciphertext = ""
for char1, char2 in chunker(plaintext, 2):
row1, col1 = divmod(table.index(char1), 5)
row2, col2 = divmod(table.index(char2), 5)
if row1 == row2:
ciphertext += table[row1 * 5 + (col1 + 1) % 5]
ciphertext += table[row2 * 5 + (col2 + 1) % 5]
elif col1 == col2:
ciphertext += table[((row1 + 1) % 5) * 5 + col1]
ciphertext += table[((row2 + 1) % 5) * 5 + col2]
else: # rectangle
ciphertext += table[row1 * 5 + col2]
ciphertext += table[row2 * 5 + col1]
return ciphertext
def decode(ciphertext: str, key: str) -> str:
"""
Decode the input string using the provided key.
>>> decode("BMZFAZRZDH", "HAZARD")
'FIREHAZARD'
>>> decode("HNBWBPQT", "AUTOMOBILE")
'DRIVINGX'
>>> decode("SLYSSAQS", "CASTLE")
'ATXTACKX'
"""
table = generate_table(key)
plaintext = ""
for char1, char2 in chunker(ciphertext, 2):
row1, col1 = divmod(table.index(char1), 5)
row2, col2 = divmod(table.index(char2), 5)
if row1 == row2:
plaintext += table[row1 * 5 + (col1 - 1) % 5]
plaintext += table[row2 * 5 + (col2 - 1) % 5]
elif col1 == col2:
plaintext += table[((row1 - 1) % 5) * 5 + col1]
plaintext += table[((row2 - 1) % 5) * 5 + col2]
else: # rectangle
plaintext += table[row1 * 5 + col2]
plaintext += table[row2 * 5 + col1]
return plaintext
if __name__ == "__main__":
import doctest
doctest.testmod()
print("Encoded:", encode("BYE AND THANKS", "GREETING"))
print("Decoded:", decode("CXRBANRLBALQ", "GREETING"))
Playfair 密码是由 查尔斯·惠斯通 于 1854 年发明的,但以推广使用该密码的普莱费尔勋爵命名。
Playfair 密码是第一个实用的二字母替换密码。与传统的密码不同,在 Playfair 密码中,我们加密的是一对字母(二字母组),而不是单个字母。一个 5 × 5
的字母网格被用作密钥方格。25 个字母中的每一个都是唯一的,其中一个字母(通常是 J)被省略。如果明文中包含 J,则将其替换为 I 或反之。密钥方格中的初始字母是密钥中唯一字母的顺序,然后是剩余字母的顺序。
假设我们以以下为例
明文(PT):instruments,密钥:monarchy
Diagraph: "me"
Encrypted Text: cl
Encryption:
m -> c
e -> l
Diagraph: "st"
Encrypted Text: tl
Encryption:
s -> t
t -> l
Diagraph: "nt"
Encrypted Text: rq
Encryption:
n -> r
t -> q
上述规则用于加密。可以反过来应用于解密。
5 × 5
矩阵,如下所示 [m o n a r]
[c h y b d]
[e f g i k]
[l p q s t]
[u v w x z]
'in' 'st' 'ru' 'me' 'nt' 'sx'
Plain Text: instrumentsx
key: monarchy
Encryption:
i -> g
n -> a
s -> t
t -> l
r -> m
u -> z
m -> c
e -> l
n -> r
t -> q
s -> x
x -> a
因此,我们将获得加密文本 gatlmzclrqxa。
5 × 5
矩阵,如下所示 [m o n a r]
[c h y b d]
[e f g i k]
[l p q s t]
[u v w x z]
'ga' 'tl' 'mz' 'cl' 'rq' 'xa'
Plain Text: gatlmzclrqtx
key: monarchy
Decryption:
ga -> in
tl -> st
mz -> ru
cl -> me
rq -> nt
xa -> sx
因此,我们将获得解密文本 instrumentsx。