class TestDecryptString(unittest.TestCase):
def setUp(self):
self.msg = VigenereCipher(KEY, CIPHERTEXT)
def test_isinstance(self):
self.assertIsInstance(self.msg, VigenereCipher)
def test_isequals(self):
check = self.msg.decrypt()
expect = PLAINTEXT
self.assertEquals(check, expect)
columnFrequencies = letFreq.getFrequencies(''.join(column))[1]
mostFrequentLetTuple = columnFrequencies[0]
mostFreqLet = mostFrequentLetTuple[0]
# for tuple in columnFrequencies[:4]:
# print vigCi.vigenereSquareDecrypt(tuple[0], 'e')
# print "options above"
print "Column", idx+1, [vigCi.vigenereSquareDecrypt(tuple[0], 'e') for tuple in columnFrequencies[:6]]
# assuming this letter corresponds to 'e', use the vigenere square
# to discover the original keyword letter
keywordLet = vigCi.vigenereSquareDecrypt(mostFreqLet, 'e')
keyword += keywordLet
# I discovered this to be the keyword through analyzing the possible combinations across the columns
# and put it in manually to make the output more interesting
#keyword = 'funny'
print "\nPotential keyword: %s" % keyword
print "Message deciphered using keyword '%s':\n%s\n" % (keyword, vigCi.decrypt(msg,keyword))
else:
print "\nKasiski test failed (no repeated substrings with length >= 3)"
kasiskiFailed = True
mostLikelyKeyIC = polyCi.getKeywordLength(len(msg), msgIC)
print "\nMost likely key length using IC test: %0.4f" % (mostLikelyKeyIC)
if kasiskiFailed:
print "\nAssuming that message was enciphered using a hill cipher."
HillSystem().super_decrypt(msg, None, zeroSystem=True)
# 2. For the ones which you guess to be monoalphabetic, do the following:
else:
# (a) Run frequency analysis to determine which letters most likely correspond to "e" and "t".
# (b) Use these correspondences to find the key.
# (c) Decrypt the message.
plaintext = monCi.decrypt(msg, verbose=True)
