Python Implementation of Problem 59
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Problem Solution
Project Euler Problem 59
I had a very hard time debugging this one, so sadly I cheated a little bit.
Problem:
Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.
For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both "halves", it is impossible to decrypt the message.
Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.
Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher.txt (right click and 'Save Link/Target As...'), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.
1"""
2Project Euler Problem 59
3
4I had a very hard time debugging this one, so sadly I cheated a little bit.
5
6Problem:
7
8Each character on a computer is assigned a unique code and the preferred
9standard is ASCII (American Standard Code for Information Interchange). For
10example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
11
12A modern encryption method is to take a text file, convert the bytes to ASCII,
13then XOR each byte with a given value, taken from a secret key. The advantage
14with the XOR function is that using the same encryption key on the cipher text,
15restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.
16
17For unbreakable encryption, the key is the same length as the plain text
18message, and the key is made up of random bytes. The user would keep the
19encrypted message and the encryption key in different locations, and without
20both "halves", it is impossible to decrypt the message.
21
22Unfortunately, this method is impractical for most users, so the modified
23method is to use a password as a key. If the password is shorter than the
24message, which is likely, the key is repeated cyclically throughout the
25message. The balance for this method is using a sufficiently long password key
26for security, but short enough to be memorable.
27
28Your task has been made easy, as the encryption key consists of three lower
29case characters. Using cipher.txt (right click and 'Save Link/Target As...'),
30a file containing the encrypted ASCII codes, and the knowledge that the plain
31text must contain common English words, decrypt the message and find the sum of
32the ASCII values in the original text.
33"""
34from itertools import cycle, permutations
35
36from .lib.utils import get_data_file
37
38alphabet = b'abcdefghijklmnopqrtsuvwxyz'
39
40
41def main() -> int:
42 ciphertext = bytes(int(x) for x in get_data_file('p0059_cipher.txt').split(','))
43 for key in permutations(alphabet, 3):
44 plaintext = bytes(x ^ y for x, y in zip(ciphertext, cycle(key)))
45 if b'beginning' in plaintext:
46 return sum(plaintext)
47 return -1 # pragma: no cover