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* added doctest to playfair_cipher.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Added newline to EOF andremoved trailing whitespace * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Apply suggestions from code review --------- Co-authored-by: Keyboard-1 <142900182+Keyboard-1@users.noreply.github.com> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Tianyi Zheng <tianyizheng02@gmail.com>
160 lines
4.4 KiB
Python
160 lines
4.4 KiB
Python
"""
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https://en.wikipedia.org/wiki/Playfair_cipher#Description
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The Playfair cipher was developed by Charles Wheatstone in 1854
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It's use was heavily promotedby Lord Playfair, hence its name
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Some features of the Playfair cipher are:
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1) It was the first literal diagram substitution cipher
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2) It is a manual symmetric encryption technique
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3) It is a multiple letter encryption cipher
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The implementation in the code below encodes alphabets only.
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It removes spaces, special characters and numbers from the
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code.
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Playfair is no longer used by military forces because of known
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insecurities and of the advent of automated encryption devices.
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This cipher is regarded as insecure since before World War I.
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"""
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import itertools
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import string
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from collections.abc import Generator, Iterable
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def chunker(seq: Iterable[str], size: int) -> Generator[tuple[str, ...], None, None]:
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it = iter(seq)
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while True:
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chunk = tuple(itertools.islice(it, size))
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if not chunk:
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return
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yield chunk
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def prepare_input(dirty: str) -> str:
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"""
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Prepare the plaintext by up-casing it
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and separating repeated letters with X's
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"""
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dirty = "".join([c.upper() for c in dirty if c in string.ascii_letters])
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clean = ""
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if len(dirty) < 2:
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return dirty
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for i in range(len(dirty) - 1):
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clean += dirty[i]
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if dirty[i] == dirty[i + 1]:
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clean += "X"
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clean += dirty[-1]
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if len(clean) & 1:
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clean += "X"
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return clean
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def generate_table(key: str) -> list[str]:
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# I and J are used interchangeably to allow
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# us to use a 5x5 table (25 letters)
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alphabet = "ABCDEFGHIKLMNOPQRSTUVWXYZ"
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# we're using a list instead of a '2d' array because it makes the math
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# for setting up the table and doing the actual encoding/decoding simpler
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table = []
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# copy key chars into the table if they are in `alphabet` ignoring duplicates
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for char in key.upper():
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if char not in table and char in alphabet:
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table.append(char)
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# fill the rest of the table in with the remaining alphabet chars
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for char in alphabet:
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if char not in table:
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table.append(char)
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return table
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def encode(plaintext: str, key: str) -> str:
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"""
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Encode the given plaintext using the Playfair cipher.
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Takes the plaintext and the key as input and returns the encoded string.
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>>> encode("Hello", "MONARCHY")
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'CFSUPM'
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>>> encode("attack on the left flank", "EMERGENCY")
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'DQZSBYFSDZFMFNLOHFDRSG'
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>>> encode("Sorry!", "SPECIAL")
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'AVXETX'
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>>> encode("Number 1", "NUMBER")
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'UMBENF'
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>>> encode("Photosynthesis!", "THE SUN")
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'OEMHQHVCHESUKE'
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"""
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table = generate_table(key)
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plaintext = prepare_input(plaintext)
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ciphertext = ""
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for char1, char2 in chunker(plaintext, 2):
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row1, col1 = divmod(table.index(char1), 5)
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row2, col2 = divmod(table.index(char2), 5)
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if row1 == row2:
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ciphertext += table[row1 * 5 + (col1 + 1) % 5]
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ciphertext += table[row2 * 5 + (col2 + 1) % 5]
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elif col1 == col2:
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ciphertext += table[((row1 + 1) % 5) * 5 + col1]
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ciphertext += table[((row2 + 1) % 5) * 5 + col2]
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else: # rectangle
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ciphertext += table[row1 * 5 + col2]
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ciphertext += table[row2 * 5 + col1]
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return ciphertext
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def decode(ciphertext: str, key: str) -> str:
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"""
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Decode the input string using the provided key.
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>>> decode("BMZFAZRZDH", "HAZARD")
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'FIREHAZARD'
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>>> decode("HNBWBPQT", "AUTOMOBILE")
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'DRIVINGX'
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>>> decode("SLYSSAQS", "CASTLE")
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'ATXTACKX'
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"""
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table = generate_table(key)
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plaintext = ""
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for char1, char2 in chunker(ciphertext, 2):
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row1, col1 = divmod(table.index(char1), 5)
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row2, col2 = divmod(table.index(char2), 5)
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if row1 == row2:
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plaintext += table[row1 * 5 + (col1 - 1) % 5]
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plaintext += table[row2 * 5 + (col2 - 1) % 5]
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elif col1 == col2:
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plaintext += table[((row1 - 1) % 5) * 5 + col1]
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plaintext += table[((row2 - 1) % 5) * 5 + col2]
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else: # rectangle
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plaintext += table[row1 * 5 + col2]
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plaintext += table[row2 * 5 + col1]
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return plaintext
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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print("Encoded:", encode("BYE AND THANKS", "GREETING"))
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print("Decoded:", decode("CXRBANRLBALQ", "GREETING"))
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