Project Euler Problem 059


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 cipher1.txt, 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.


In order to achieve this goal, I first divide the whole text into three lists, each of them containing the characters that are ciphered with the same character. Then, for each of these groups I search for the letters that decipher those characters into valid ones.

Once I have the possible values for each of the three characters of the key, I bruteforce all possible combinations of these and verify manually which of them is the correct key, sum and deciphered text.

I considered that a character to be valid must in one of the groups:

  • space, !, "
  • ', (, ), *, ,(comma), +, -, .(dot)
  • ;, :
  • alpha numeric character

After having that, I found that there was only one valid solution.

def valid_char(x):
    if (x >= 32 and x <= 34):
        return True
    if (x >= 39 and x <= 46):
        return True
    if (x >= 58 and x <= 59):
        return True
    return chr(x).isalnum()
if __name__ == '__main__':
    f = open("cipher1.txt")
    chars = [int(c) for c in',')]
    chars_per_key = [[],[],[]]
    for i in range(len(chars)):
        chars_per_key[i % 3].append(chars[i])
    valid_keys = [[],[],[]]
    for k in range(ord('a'), ord('z')+1):
        for i in range(3):
            decoded = filter(valid_char, map(lambda x: k ^ x, chars_per_key[i]))
            if sum(1 for _ in decoded) == len(chars_per_key[i]):
    for k1 in valid_keys[0]:
        for k2 in valid_keys[1]:
            for k3 in valid_keys[2]:
                key = (k1, k2, k3)
                suma = 0
                decoded_text = ''
                for i in range(len(chars)):
                    decoded_char = chars[i] ^ key[i % 3]
                    decoded_text += chr(decoded_char)
                    suma += decoded_char
                print("Possible solution (",
                      ''.join(map(chr,  key)),

The Python file is available for download here.

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