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))
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:
