Creation
March 3, 2009
A coded message appears on the next page. The cipher-text was created by XORing the characters of a password, repeated as necessary, with the characters of the plain-text. The numbers are the ascii ordinals of the cipher-text characters; the cipher-text is given in that form in order to bypass the limitations of some browsers. The password consists of less than twenty alphanumeric characters (upper-case letters, lower-case letters, and digits). The plain-text consists of ascii text (printable characters, plus spaces and newlines) in English.
Determine the password and read the coded message.
Programming Praxis 
Can’t believe nobody else commented on this… That one was a lot of fun!
I took a different approach from the proposed Scheme solution, went for a dictionary attack:
#!/usr/bin/python def read_cipher(textFile): '''Takes the name of the file containing the encrypted text as argument, returns the list of ascii ordinals''' f = open(textFile, 'r') returnList = [int(a) for a in f.read().split()] f.close() return returnList def decrypt(ordList, password): '''This function XORs a text with the code (both as list of ordinals!!) and returns the encoded/decoded list of ordinals''' from itertools import cycle decryptedList = [a^ord(b) for a,b in zip(ordList, cycle(password))] return decryptedList def make_attack_dictionary(dictFile): f = open(dictFile,'r') returnDict = attack_dictionary(f.read().split()) f.close() return returnDict def attack_dictionary(rawDict): '''Takes a list of words as argument, and produces an attack dictionary from it: with added case variations (also capitalizes the words that only exist as small case in the initial list), and sorted by length''' from string import capitalize # Generate a dict of already capitalized words in the raw dictionary capWords = {} for word in rawDict: if word == capitalize(word): capWords[word] = 1 # Now capitalize the rest of them; this works in O(n) because capWords # is a dict complementDict = [capitalize(word) for word in rawDict if not word in capWords] # Finally, merge and sort the attack dictionary by word length return sorted(rawDict + complementDict, key=len) def words_found(text, dictionary): '''Returns the number of known words (from dictionary) found in text''' from re import findall wordCount = 0 for word in findall(r'\w\w+', text): if word in dictionary: wordCount +=1 return wordCount def attack(cipherText): '''Performs a dictionary attack on an encrypted text, passed as a list of integers''' attackDict = make_attack_dictionary('/usr/share/dict/words') # We'll also build a dict purely for performance reasons compareDict = {} for word in attackDict: compareDict[word] = 1 # Tweak these parameters for the dictionary attack attackDepth = 50 attackLengthMax = 8 attackThreshold = 6 # Now try all passwords of length <= attackLengthMax, # scanning the first attackDepth characters in cipherText # and print something when finding more than attackThreshold known words for password in attackDict: if len(password) > attackLengthMax: break decryptedText = "".join([chr(a) for a in decrypt(cipherText[:attackDepth], password)]) wordsFound = words_found(decryptedText, compareDict) if wordsFound >= attackThreshold: print '%d matches found with %s' % (wordsFound, password) if __name__ == '__main__': cipherText = read_cipher('text') attack(cipherText) print 'Password to try?' password = raw_input() print "".join([chr(a) for a in decrypt(cipherText, password)])Basically, I start by building an attack dictionary, sorted by length. Then I brute-force my way through this attack dictionary, trying to decrypt the beginning of the encrypted text using each one of the possible passwords, and I count the number of real words I can find in the decrypted version.
Of course it could probably be automated completely by simply returning the password yielding the highest number of real words, but it works for me like that. Running it as it is shown here (scanning the first 50 letters of the encrypted text, and trying passwords up to 8 letters long), it shows me the list of following possible keys (only showing the keys producing 6 or more known words):
6 matches found with imber
6 matches found with infer
6 matches found with inker
6 matches found with ceresin
6 matches found with chooser
6 matches found with geronto
6 matches found with oenolin
6 matches found with Allower
6 matches found with Ceresin
6 matches found with Chooser
6 matches found with Foresin
8 matches found with Genesis
6 matches found with Geronto
6 matches found with Getaway
6 matches found with Subjoin
6 matches found with Thrower
6 matches found with Vetiver
I followed the ideas of the solution here, but wished to automate the process
a bit more. I had my code try different passwords (given by assuming that space
is the most common character), checking those words against membership in my
system’s dict file. It outputs all possibilities, so it still requires a human
eye to sort through the gobbledygook.
#!/usr/bin/env python from collections import Counter from pprint import pprint def split_n(seq, n): """Split sequence into groupings by every nth item""" return (seq[i::n] for i in xrange(n)) def most_freq(seq): """Most frequent item in sequence""" return Counter(seq).most_common(1)[0][0] def pass_n(c_text, n): """Assume that most frequent item is a space (with ord 32); return a password of length n from those most frequent items in """ most_freqs = (most_freq(seq) for seq in split_n(c_text, n)) return ''.join(chr(x ^ 32) for x in most_freqs) def decipher(c_text, password): """Given a password, use it to give clear text""" return ''.join(chr(x ^ ord(y)) for (x, y) in zip(c_text, password * (len(c_text) / len(password)))) def decrypt(c_text, dictionary): """Given a dictionary of words, try successive ns in pass_n until the password is in dictionary. Then, return deciphered text via decipher.""" possibles = [] for n in xrange(len(max(dictionary, key=lambda word: len(word)))): password = pass_n(c_text, n) if password.lower() in dictionary: possibles.append((n, password, decipher(c_text, password))) return possibles if __name__ == "__main__": with open("creation_ctext.txt") as C_FILE: # Numbers in single line file C_TEXT = [int(x) for x in C_FILE.read().split()] with open("/usr/share/dict/web2") as DICT_FILE: DICTIONARY = set(line.strip() for line in DICT_FILE.readlines()) pprint(decrypt(C_TEXT, DICTIONARY))If you are really interested in automating this program fully, Google for ‘Kasiski examination’ and go from there. Like a lot of these things, it’s named for the wrong man; Charles Babbage figured it out twenty years before Kasiski.
Interesting stuff! It’s reminiscent of some of Simon Singh’s “The Code Book,” a decent source and fun read.
Like others I want to automate the process as much as possible.
So my slightly different implementation returns a password and a “confidence number” based on the frequencies of “” and “e”.
Then we can search for all passwords given a maximum length and a confidence threshold.
Clojure code:
(defn read-cipher [filepath] (map #(Integer/parseInt %) (seq (.split (slurp filepath) "\\s+")))) (defn stripe [text n] (for [i (range 0 (count text) n)] (nth text i))) (defn rsort-freq [text] (reverse (sort-by second (frequencies text)))) (defn char-xor [ord1 ord2] (char (bit-xor ord1 ord2))) (defn guess-pass [text nchar] (loop [text text pwd [] same 0] (if (= nchar (count pwd)) (vector (apply str pwd) (float (/ same nchar))) (let [strp (stripe text nchar) [[ord1 _] [ord2 _]] (take 2 (rsort-freq strp)) letter1 (char-xor (int \space) ord1) letter2 (char-xor (int \e) ord2) same (if (= letter1 letter2) (inc same) same)] (recur (next text) (conj pwd letter1) same))))) (defn search-pass [filepath max-nchar confidence] (let [text (read-cipher filepath)] (filter #(< confidence (second %)) (map #(guess-pass text %) (range 1 (inc max-nchar))))))For example, we can search for all passwords less than 20 chars with at least 80% confidence:
This one is funny:
[…] In their book Software Tools, Brian Kernighan and P. J. Plauger describe a simple command for encrypting files. It works by xor-ing each byte of a file with a byte of the key, extending the key cyclically until it is the same length as the text. The xor operation is symmetric, so only one program is needed to perform both encryption and decryption. This isn’t a particularly secure encryption algorithm; we showed how to break it in one of our earliest exercises. […]
I had the similar ideas with Graham, and I improve my script when I saw Graham’s since I am not familiar with Python.
from collections import Counter
import string
def crack(text, n):
pwd = []
for seq in [text[i::n] for i in xrange(n)]:
p = chr(ord(‘ ‘) ^ int(Counter(seq).most_common(1)[0][0]))
if is_pwd_char(p):
pwd.append(p)
else:
return None
return pwd
def is_pwd_char(ch):
return ch in string.digits or ch in string.ascii_letters
def decrypt(text, pwd):
return [chr(ord(pwd[i % len(pwd)]) ^ int(x)) for (i, x) in enumerate(text)]
if __name__ == ‘__main__’:
with open(‘/tmp/cipher-text’) as f:
text = f.read().split()
for i in range(1, 21):
pwd = crack(text, i)
if pwd != None:
print ‘decrypt with pwd: %s’ % pwd
print ”.join(decrypt(text, pwd))
https://github.com/ftt/programming-praxis/blob/master/20090303-creation/creation.py