// you’re reading...
1 Star2 Stars3 Stars4 Stars5 Stars (9 votes, average: 5.00 out of 5)

Project Euler Solutions

Project Euler 59 Solution

Project Euler 59 Solution

Project Euler 59: Using a brute force attack, decrypt the cipher using XOR encryption

Problem Description

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 (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.


Following the instructions, we read the file into a list of integers. The decryption process is simplified by limiting the key to three characters from the set {a..z} for only 17,576 possibilities.

Taking the hint that the message contains common English words we look for the string ‘ the ‘ after each attempt at decryption. Surrounding our word with spaces guarantees it’s not part of another word, such as ‘they’. White space could also be something to look for.

We have to repeat each key cyclically until it’s the same length as the message. For example, if the message is “gduii9iu” and the password is ‘abc’ we must decrypt with ‘abcabcab’.

Project Euler 59 Solution

Runs < 1.1 seconds in Python 2.7.
import itertools, urllib2

file_url = 'https://projecteuler.net/project/resources/p059_cipher.txt'
cipher_text = map(int, urllib2.urlopen(file_url).read().split(','))

def decode(cipher_text, key_length, key_set, morsel):
    for key in itertools.product(key_set, repeat=key_length):
        msg =  [x^y for x, y in zip(cipher_text, itertools.cycle(key))]
        if morsel in ''.join(map(chr, msg)):
            return sum(msg)
    return "No solution"

print "Project Euler 59 Solution =", decode(cipher_text, 3, range(97, 123),' the ')
download arrowUse this link to get the Project Euler 59 Solution Python 2.7 source.


Slowly swipe from either end beginning with the white vertical bar to get an idea of the starting or ending digits. For less drama, just double click the answer area. The distance between the two bars will give you an idea of the magnitude. Touch devices can tap and hold the center of the box between the two bars and choose define to reveal the answer.


  • ‘the’ was selected because it is the most common English word.
Project Euler 59 Solution last updated


One Response to “Project Euler 59 Solution”

  1. the problem also falls rather easily to a frequency analysis attack. split the list in python with [0::3], [1::3], [2::3].. and find largest count occurrences of each num. listing largest counts from each split list show nearly twice as much of the most frequent number as any other. assume that is space and XOR with the actual number that occurs in each to get the password letters

    Posted by speiberg0 | April 15, 2010, 7:03 AM

Post a comment