https://projecteuler.net/problem=59
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 (in this directory), 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.
The following cell shows examples of how to perform XOR in Python and how to go back and forth between characters and integers:
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assert 65 ^ 42 == 107
assert 107 ^ 42 == 65
assert ord('a') == 97
assert chr(97) == 'a'
Certain functions in the itertools module may be useful for computing permutations:
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from itertools import
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cipher = open("cipher.txt", "r")
numbers = cipher.read().split(",")
a = [i for i in range(97,123)] # This is set because the keys MUST be lowercase
b = [i for i in range(97,123)]
c= [i for i in range(97,123)]
# --------
#every third number will use the same key (abcabcabc...) so I make an array of text that will use the "a", "b", and "c" keys, respectively
frst_thrd = []
snd_thrd = []
thrd_thrd = []
for i in range(len(numbers)):
if i % 3 == 0 or i == 0:
frst_thrd.append(int(numbers[i]))
elif i % 3 == 1 or i == 1:
snd_thrd.append(int(numbers[i]))
elif i % 3 == 2 or i == 2:
thrd_thrd.append(int(numbers[i]))
# --------
# Let's start with "a", our first key:
probable_keys_a = []
for possible_key in a:
for elements in frst_thrd:
if elements ^ possible_key == 32: # here I'm assuming that the original text must have at least one "space" in the (3^n - 1) number of the original text
probable_keys_a.append(possible_key)
probable_keys_a = list(set(probable_keys_a)) # these are now all my possible keys for a that include a space (ASCII 32)
for elements in frst_thrd:
for keys in probable_keys_a:
if elements ^ keys < 32 or elements ^ keys > 122: #because this is common english word, I'm removing the ASCII numbers that are not included in plain enlgish or are not printable
probable_keys_a.remove(keys)
# This works out well, and now "probable_keys_a" only has one element in it, and that is my 'a' key
#--------
# Now we do the same for the 'b' key
probable_keys_b = []
for possible_key in b:
for elements in snd_thrd:
if elements ^ possible_key == 32: # assuming there is a space somewhere
probable_keys_b.append(possible_key)
probable_keys_b = list(set(probable_keys_b))
for elements in snd_thrd:
for keys in probable_keys_b:
if elements ^ keys < 32 or elements ^ keys > 122: # eliminating unlikely ASCII numbers
probable_keys_b.remove(keys)
# once again, I have one element stored in 'probable_keys_b', this is my 'b' key
#-------
# and the same process for c
probable_keys_c = []
for possible_key in c:
for elements in thrd_thrd:
if elements ^ possible_key == 32: # assuming spaces
probable_keys_c.append(possible_key)
probable_keys_c = list(set(probable_keys_c))
for elements in thrd_thrd:
for keys in probable_keys_c:
if elements ^ keys < 32 or elements ^ keys > 122: # removing unlikely keys
probable_keys_c.remove(keys)
# and I have one key for 'c' stored in 'probable_keys_c'
# NOW WE HAVE OUR KEYS (hopefully)
key = [probable_keys_a[0], probable_keys_b[0], probable_keys_c[0]] # make a list including the single elements stored in the 'a', 'b', and 'c' lists.
import itertools
key_looped = itertools.cycle(key) # use itertools cycle to produce keys a b c a b c a b c... ad infinitem
deciphered = []
for i in range(len(numbers)):
new_element = next(key_looped) ^ int(numbers[i]) # XOR each element in my cycle to the encrypted message
deciphered.append(new_element)
print("".join(map(chr, deciphered))) # print the string equivalent of the ASCII text
print(sum(deciphered)) # print the sum of the ASCII values from the original text
# https://docs.python.org used to look up built- in python functions
# Used SJSlavin's idea to look up spaces
# Used telthien's idea to excluded unlikely characters
cipher.close()
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# This cell will be used for grading, leave it at the end of the notebook.
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