print "\nMost likely key lengths using Kasiski test:\n%s\n" % (mostLikelyKeysKasiski)
# initialize the letter frequency class
letFreq = LetterFrequency()
# loop through the top 4 most likely key lengths according to the kasiski test
for keyLen in mostLikelyKeysKasiski[:2]:
# Split the message into keyLen columns
msgColumns = polyCi.msgSplit(msg, keyLen)
# holds the potential keyword
keyword = ""
# perform frequency analysis on each column
print "Using the Vigenere Square on the top 4 most frequent letters in each column assuming that each corresponds to 'e' results in the following options:"
for idx, column in enumerate(msgColumns):
# get the most frequent letter in the column
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))
def decrypt(self, msg, mfc=None, mfp=None, verbose=False, showDigraphFrequencies=False, auto=False, bruteAmt=None):
# remove spaces from the message
msg = ''.join(msg.split())
if verbose:
print "Attempting to decipher:\n", msg, "\n"
# get the message and default letter frequencies
letFreq = LetterFrequency()
(freqDict, freqTuples) = letFreq.getFrequencies(msg)
(stdDict, stdTuples) = letFreq.getStandardFrequencies()
# if the mfc or mfp are None, use the top two most frequent letters
# in the ciphertext and english language respectively as defaults
mfc = tuple(mfc) if mfc else (freqTuples[0][0], freqTuples[1][0])
mfp = tuple(mfp) if mfp else (stdTuples[0][0], stdTuples[1][0])
if verbose:
print "Ciphertext Letter Frequencies:\n%s\n" % freqTuples
print "English Language Letter Frequencies:\n%s\n" % stdTuples
if showDigraphFrequencies:
diFreq = DigraphFrequency()
diFreqTuples = diFreq.getFrequencies(msg)[1]
diStdTuples = diFreq.getStandardFrequencies()[1]
print "Ciphertext Digraph Frequencies:\n%s\n" % diFreqTuples
print "English Language Digraph Frequencies:\n%s\n" % diStdTuples
print "Using letter in ciphertext: %s" % mfc[0]
print "\twith frequency: %.2f%%" % freqDict[mfc[0]]
print "Using letter in ciphertext: %s" % mfc[1]
print "\twith frequency: %.2f%%\n" % freqDict[mfc[1]]
print "Assuming that", mfc[0], "is equivalent to", mfp[0]
print "Assuming that", mfc[1], "is equivalent to", mfp[1], "\n"
# get the affine key using the mfc letters and mfp letters
key = self.getAffineKey(*tuple(map(self.letToInt, mfp+mfc)), verbose=verbose)
if auto:
iEnd = jEnd = 10
if bruteAmt:
iEnd = int(bruteAmt[0])
jEnd = int(bruteAmt[1])
maxIdx = len(freqTuples)-1
for i in range(0, iEnd):
for j in range(0, jEnd):
if i != j and i <= maxIdx and j <= maxIdx:
self.decrypt(msg, mfc=[freqTuples[i][0], freqTuples[j][0]], mfp=mfp, verbose=False, showDigraphFrequencies=showDigraphFrequencies)
if key is None:
if verbose:
print "Unable to decipher using '%s'->'%s' and '%s'->'%s'" % (mfc[0], mfp[0], mfc[1], mfp[1])
return None
# Use the key to decrypt
plaintext = ""
for ch in msg.lower():
cipherInt = ((ord(ch) - 96)*self.multInv[key[0]] - key[1]) % 26
# cover the 'z' case
if cipherInt == 0:
cipherInt = 26
plaintext += chr(cipherInt + 96)
if plaintext and not verbose:
print "%s %s %s %s " % (str(key).ljust(8), mfc[0], mfc[1], plaintext)
return plaintext
