维吉尼亚密码（又译维热纳尔密码）是使用一系列凯撒密码组成密码字母表的加密算法，属于多表密码的一种简单形式。
在一个凯撒密码中，字母表中的每一字母都会作一定的偏移，例如偏移量为3时，A就转换为了D、B转换为了E……而维吉尼亚密码则是由一些偏移量不同的恺撒密码组成。
为了生成密码，需要使用表格法。这一表格（如图1所示）包括了26行字母表，每一行都由前一行向左偏移一位得到。具体使用哪一行字母表进行编译是基于密钥进行的，在过程中会不断地变换。
例如，假设明文为：ATTACKATDAWN
选择某一关键词并重复而得到密钥，如关键词为LEMON时，密钥为：LEMONLEMONLE
对于明文的第一个字母A，对应密钥的第一个字母L，于是使用表格中L行字母表进行加密，得到密文第一个字母L。类似地，明文第二个字母为T，在表格中使用对应的E行进行加密，得到密文第二个字母X。以此类推，可以得到：
明文：ATTACKATDAWN密钥：LEMONLEMONLE密文：LXFOPVEFRNHR
解密的过程则与加密相反。例如：根据密钥第一个字母L所对应的L行字母表，发现密文第一个字母L位于A列，因而明文第一个字母为A。密钥第二个字母E对应E行字母表，而密文第二个字母X位于此行T列，因而明文第二个字母为T。以此类推便可得到明文。
用数字0-25代替字母A-Z，维吉尼亚密码的加密文法可以写成同余的形式：
解密方法则能写成：
简而言之，使用一个单词作为密钥，将密钥不断重复并与明文逐一对齐，将明文的字符序加上密钥的字符序得到密文的字符序。每个明文字符用于加密的密钥不同。
1 # Vigenere Cipher (Polyalphabetic Substitution Cipher) 2 # http://inventwithpython.com/hacking (BSD Licensed) 3 4 import pyperclip 5 6 LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' 7 8 def main(): 9 # This text can be copy/pasted from http://invpy.com/vigenereCipher.py10 myMessage = """Alan Mathison Turing was a British mathematician, logician, cryptanalyst, and computer scientist. He was highly influential in the development of computer science, providing a formalisation of the concepts of "algorithm" and "computation" with the Turing machine. Turing is widely considered to be the father of computer science and artificial intelligence. During World War II, Turing worked for the Government Code and Cypher School (GCCS) at Bletchley Park, Britain's codebreaking centre. For a time he was head of Hut 8, the section responsible for German naval cryptanalysis. He devised a number of techniques for breaking German ciphers, including the method of the bombe, an electromechanical machine that could find settings for the Enigma machine. After the war he worked at the National Physical Laboratory, where he created one of the first designs for a stored-program computer, the ACE. In 1948 Turing joined Max Newman's Computing Laboratory at Manchester University, where he assisted in the development of the Manchester computers and became interested in mathematical biology. He wrote a paper on the chemical basis of morphogenesis, and predicted oscillating chemical reactions such as the Belousov-Zhabotinsky reaction, which were first observed in the 1960s. Turing's homosexuality resulted in a criminal prosecution in 1952, when homosexual acts were still illegal in the United Kingdom. He accepted treatment with female hormones (chemical castration) as an alternative to prison. Turing died in 1954, just over two weeks before his 42nd birthday, from cyanide poisoning. An inquest determined that his death was suicide; his mother and some others believed his death was accidental. On 10 September 2009, following an Internet campaign, British Prime Minister Gordon Brown made an official public apology on behalf of the British government for "the appalling way he was treated." As of May 2012 a private member's bill was before the House of Lords which would grant Turing a statutory pardon if enacted."""11 myKey = 'ASIMOV'12 myMode = 'encrypt' # set to 'encrypt' or 'decrypt'13 14 if myMode == 'encrypt':15 translated = encryptMessage(myKey, myMessage)16 elif myMode == 'decrypt':17 translated = decryptMessage(myKey, myMessage)18 19 print('%sed message:' % (myMode.title()))20 print(translated)21 pyperclip.copy(translated)22 print()23 print('The message has been copied to the clipboard.')24 25 26 def encryptMessage(key, message):27 return translateMessage(key, message, 'encrypt')28 29 30 def decryptMessage(key, message):31 return translateMessage(key, message, 'decrypt')32 33 34 def translateMessage(key, message, mode):35 translated = [] # stores the encrypted/decrypted message string36 37 keyIndex = 038 key = key.upper()39 40 for symbol in message: # loop through each character in message41 num = LETTERS.find(symbol.upper())42 if num != -1: # -1 means symbol.upper() was not found in LETTERS43 if mode == 'encrypt':44 num += LETTERS.find(key[keyIndex]) # add if encrypting45 elif mode == 'decrypt':46 num -= LETTERS.find(key[keyIndex]) # subtract if decrypting47 48 num %= len(LETTERS) # handle the potential wrap-around49 50 # add the encrypted/decrypted symbol to the end of translated.51 if symbol.isupper():52 translated.append(LETTERS[num])53 elif symbol.islower():54 translated.append(LETTERS[num].lower())55 56 keyIndex += 1 # move to the next letter in the key57 if keyIndex == len(key):58 keyIndex = 059 else:60 # The symbol was not in LETTERS, so add it to translated as is.61 translated.append(symbol)62 63 return ''.join(translated)64 65 66 # If vigenereCipher.py is run (instead of imported as a module) call67 # the main() function.68 if __name__ == '__main__':69 main()
破译方法：
频率分析
通过频率分析，英语中每个字母出现的频率不同，使用替换的方法后字母的出现频率改变了
代码模块
getLetterCount() 这个函数接受一个字符串参数，并返回一个字典，里面包含了这个字符串的每个字符出现频率
getFrequencyOrder() 接受一个字符串，返回这个字符串里的26个字母出现频率从高到低排序之后的字符串
englishFreqMathScore() 接受一个字符串，返回这个字符串的字母频率配分值，是一个从0到12的整数
1 # Frequency Finder 2 # http://inventwithpython.com/hacking (BSD Licensed) 3 4 5 6 # frequency taken from http://en.wikipedia.org/wiki/Letter_frequency 7 englishLetterFreq = { 'E': 12.70, 'T': 9.06, 'A': 8.17, 'O': 7.51, 'I': 6.97, 'N': 6.75, 'S': 6.33, 'H': 6.09, 'R': 5.99, 'D': 4.25, 'L': 4.03, 'C': 2.78, 'U': 2.76, 'M': 2.41, 'W': 2.36, 'F': 2.23, 'G': 2.02, 'Y': 1.97, 'P': 1.93, 'B': 1.29, 'V': 0.98, 'K': 0.77, 'J': 0.15, 'X': 0.15, 'Q': 0.10, 'Z': 0.07} 8 ETAOIN = 'ETAOINSHRDLCUMWFGYPBVKJXQZ' 9 LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'10 11 12 13 def getLetterCount(message):14 # Returns a dictionary with keys of single letters and values of the15 # count of how many times they appear in the message parameter.16 letterCount = { 'A': 0, 'B': 0, 'C': 0, 'D': 0, 'E': 0, 'F': 0, 'G': 0, 'H': 0, 'I': 0, 'J': 0, 'K': 0, 'L': 0, 'M': 0, 'N': 0, 'O': 0, 'P': 0, 'Q': 0, 'R': 0, 'S': 0, 'T': 0, 'U': 0, 'V': 0, 'W': 0, 'X': 0, 'Y': 0, 'Z': 0}17 18 for letter in message.upper():19 if letter in LETTERS:20 letterCount[letter] += 121 22 return letterCount23 24 25 def getItemAtIndexZero(x):26 return x[0]27 28 29 def getFrequencyOrder(message):30 # Returns a string of the alphabet letters arranged in order of most31 # frequently occurring in the message parameter.32 33 # first, get a dictionary of each letter and its frequency count34 letterToFreq = getLetterCount(message)35 36 # second, make a dictionary of each frequency count to each letter(s)37 # with that frequency38 freqToLetter = {}39 for letter in LETTERS:40 if letterToFreq[letter] not in freqToLetter:41 freqToLetter[letterToFreq[letter]] = [letter]42 else:43 freqToLetter[letterToFreq[letter]].append(letter)44 45 # third, put each list of letters in reverse "ETAOIN" order, and then46 # convert it to a string47 for freq in freqToLetter:48 freqToLetter[freq].sort(key=ETAOIN.find, reverse=True)49 freqToLetter[freq] = ''.join(freqToLetter[freq])50 51 # fourth, convert the freqToLetter dictionary to a list of tuple52 # pairs (key, value), then sort them53 freqPairs = list(freqToLetter.items())54 freqPairs.sort(key=getItemAtIndexZero, reverse=True)55 56 # fifth, now that the letters are ordered by frequency, extract all57 # the letters for the final string58 freqOrder = []59 for freqPair in freqPairs:60 freqOrder.append(freqPair[1])61 62 return ''.join(freqOrder)63 64 65 def englishFreqMatchScore(message):66 # Return the number of matches that the string in the message67 # parameter has when its letter frequency is compared to English68 # letter frequency. A "match" is how many of its six most frequent69 # and six least frequent letters is among the six most frequent and70 # six least frequent letters for English.71 freqOrder = getFrequencyOrder(message)72 73 matchScore = 074 # Find how many matches for the six most common letters there are.75 for commonLetter in ETAOIN[:6]:76 if commonLetter in freqOrder[:6]:77 matchScore += 178 # Find how many matches for the six least common letters there are.79 for uncommonLetter in ETAOIN[-6:]:80 if uncommonLetter in freqOrder[-6:]:81 matchScore += 182 83 return matchScore
注：加密密钥不可用英语单词，会受到字典攻击
1 # Vigenere Cipher Dictionary Hacker 2 # http://inventwithpython.com/hacking (BSD Licensed) 3 4 import detectEnglish, vigenereCipher, pyperclip 5 6 def main(): 7 ciphertext = """Tzx isnz eccjxkg nfq lol mys bbqq I lxcz.""" 8 hackedMessage = hackVigenere(ciphertext) 9 10 if hackedMessage != None:11 print('Copying hacked message to clipboard:')12 print(hackedMessage)13 pyperclip.copy(hackedMessage)14 else:15 print('Failed to hack encryption.')16 17 18 def hackVigenere(ciphertext):19 fo = open('dictionary.txt')20 words = fo.readlines()21 fo.close()22 23 for word in words:24 word = word.strip() # remove the newline at the end25 decryptedText = vigenereCipher.decryptMessage(word, ciphertext)26 if detectEnglish.isEnglish(decryptedText, wordPercentage=40):27 # Check with user to see if the decrypted key has been found.28 print()29 print('Possible encryption break:')30 print('Key ' + str(word) + ': ' + decryptedText[:100])31 print()32 print('Enter D for done, or just press Enter to continue breaking:')33 response = input('> ')34 35 if response.upper().startswith('D'):36 return decryptedText37 38 if __name__ == '__main__':39 main()
暴力破解：
1 # Vigenere Cipher Hacker 2 # http://inventwithpython.com/hacking (BSD Licensed) 3 4 import itertools, re 5 import vigenereCipher, pyperclip, freqAnalysis, detectEnglish 6 7 LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' 8 SILENT_MODE = False # if set to True, program doesn't print attempts 9 NUM_MOST_FREQ_LETTERS = 4 # attempts this many letters per subkey 10 MAX_KEY_LENGTH = 16 # will not attempt keys longer than this 11 NONLETTERS_PATTERN = re.compile('[^A-Z]') 12 13 14 def main(): 15 # Instead of typing this ciphertext out, you can copy & paste it 16 # from http://invpy.com/vigenereHacker.py 17 ciphertext = """Adiz Avtzqeci Tmzubb wsa m Pmilqev halpqavtakuoi, lgouqdaf, kdmktsvmztsl, izr xoexghzr kkusitaaf. Vz wsa twbhdg ubalmmzhdad qz hce vmhsgohuqbo ox kaakulmd gxiwvos, krgdurdny i rcmmstugvtawz ca tzm ocicwxfg jf "stscmilpy" oid "uwydptsbuci" wabt hce Lcdwig eiovdnw. Bgfdny qe kddwtk qjnkqpsmev ba pz tzm roohwz at xoexghzr kkusicw izr vrlqrwxist uboedtuuznum. Pimifo Icmlv Emf DI, Lcdwig owdyzd xwd hce Ywhsmnemzh Xovm mby Cqxtsm Supacg (GUKE) oo Bdmfqclwg Bomk, Tzuhvif'a ocyetzqofifo ositjm. Rcm a lqys ce oie vzav wr Vpt 8, lpq gzclqab mekxabnittq tjr Ymdavn fihog cjgbhvnstkgds. Zm psqikmp o iuejqf jf lmoviiicqg aoj jdsvkavs Uzreiz qdpzmdg, dnutgrdny bts helpar jf lpq pjmtm, mb zlwkffjmwktoiiuix avczqzs ohsb ocplv nuby swbfwigk naf ohw Mzwbms umqcifm. Mtoej bts raj pq kjrcmp oo tzm Zooigvmz Khqauqvl Dincmalwdm, rhwzq vz cjmmhzd gvq ca tzm rwmsl lqgdgfa rcm a kbafzd-hzaumae kaakulmd, hce SKQ. Wi 1948 Tmzubb jgqzsy Msf Zsrmsv'e Qjmhcfwig Dincmalwdm vt Eizqcekbqf Pnadqfnilg, ivzrw pq onsaafsy if bts yenmxckmwvf ca tzm Yoiczmehzr uwydptwze oid tmoohe avfsmekbqr dn eifvzmsbuqvl tqazjgq. Pq kmolm m dvpwz ab ohw ktshiuix pvsaa at hojxtcbefmewn, afl bfzdakfsy okkuzgalqzu xhwuuqvl jmmqoigve gpcz ie hce Tmxcpsgd-Lvvbgbubnkq zqoxtawz, kciup isme xqdgo otaqfqev qz hce 1960k. Bgfdny'a tchokmjivlabk fzsmtfsy if i ofdmavmz krgaqqptawz wi 1952, wzmz vjmgaqlpad iohn wwzq goidt uzgeyix wi tzm Gbdtwl Wwigvwy. Vz aukqdoev bdsvtemzh rilp rshadm tcmmgvqg (xhwuuqvl uiehmalqab) vs sv mzoejvmhdvw ba dmikwz. Hpravs rdev qz 1954, xpsl whsm tow iszkk jqtjrw pug 42id tqdhcdsg, rfjm ugmbddw xawnofqzu. Vn avcizsl lqhzreqzsy tzif vds vmmhc wsa eidcalq; vds ewfvzr svp gjmw wfvzrk jqzdenmp vds vmmhc wsa mqxivmzhvl. Gv 10 Esktwunsm 2009, fgtxcrifo mb Dnlmdbzt uiydviyv, Nfdtaat Dmiem Ywiikbqf Bojlab Wrgez avdw iz cafakuog pmjxwx ahwxcby gv nscadn at ohw Jdwoikp scqejvysit xwd "hce sxboglavs kvy zm ion tjmmhzd." Sa at Haq 2012 i bfdvsbq azmtmd'g widt ion bwnafz tzm Tcpsw wr Zjrva ivdcz eaigd yzmbo Tmzubb a kbmhptgzk dvrvwz wa efiohzd.""" 18 hackedMessage = hackVigenere(ciphertext) 19 20 if hackedMessage != None: 21 print('Copying hacked message to clipboard:') 22 print(hackedMessage) 23 pyperclip.copy(hackedMessage) 24 else: 25 print('Failed to hack encryption.') 26 27 28 def findRepeatSequencesSpacings(message): 29 # Goes through the message and finds any 3 to 5 letter sequences 30 # that are repeated. Returns a dict with the keys of the sequence and 31 # values of a list of spacings (num of letters between the repeats). 32 33 # Use a regular expression to remove non-letters from the message. 34 message = NONLETTERS_PATTERN.sub('', message.upper()) 35 36 # Compile a list of seqLen-letter sequences found in the message. 37 seqSpacings = {} # keys are sequences, values are list of int spacings 38 for seqLen in range(3, 6): 39 for seqStart in range(len(message) - seqLen): 40 # Determine what the sequence is, and store it in seq 41 seq = message[seqStart:seqStart + seqLen] 42 43 # Look for this sequence in the rest of the message 44 for i in range(seqStart + seqLen, len(message) - seqLen): 45 if message[i:i + seqLen] == seq: 46 # Found a repeated sequence. 47 if seq not in seqSpacings: 48 seqSpacings[seq] = [] # initialize blank list 49 50 # Append the spacing distance between the repeated 51 # sequence and the original sequence. 52 seqSpacings[seq].append(i - seqStart) 53 return seqSpacings 54 55 56 def getUsefulFactors(num): 57 # Returns a list of useful factors of num. By "useful" we mean factors 58 # less than MAX_KEY_LENGTH + 1. For example, getUsefulFactors(144) 59 # returns [2, 72, 3, 48, 4, 36, 6, 24, 8, 18, 9, 16, 12] 60 61 if num < 2: 62 return [] # numbers less than 2 have no useful factors 63 64 factors = [] # the list of factors found 65 66 # When finding factors, you only need to check the integers up to 67 # MAX_KEY_LENGTH. 68 for i in range(2, MAX_KEY_LENGTH + 1): # don't test 1 69 if num % i == 0: 70 factors.append(i) 71 factors.append(int(num / i)) 72 if 1 in factors: 73 factors.remove(1) 74 return list(set(factors)) 75 76 77 def getItemAtIndexOne(x): 78 return x[1] 79 80 81 def getMostCommonFactors(seqFactors): 82 # First, get a count of how many times a factor occurs in seqFactors. 83 factorCounts = {} # key is a factor, value is how often if occurs 84 85 # seqFactors keys are sequences, values are lists of factors of the 86 # spacings. seqFactors has a value like: {'GFD': [2, 3, 4, 6, 9, 12, 87 # 18, 23, 36, 46, 69, 92, 138, 207], 'ALW': [2, 3, 4, 6, ...], ...} 88 for seq in seqFactors: 89 factorList = seqFactors[seq] 90 for factor in factorList: 91 if factor not in factorCounts: 92 factorCounts[factor] = 0 93 factorCounts[factor] += 1 94 95 # Second, put the factor and its count into a tuple, and make a list 96 # of these tuples so we can sort them. 97 factorsByCount = [] 98 for factor in factorCounts: 99 # exclude factors larger than MAX_KEY_LENGTH100 if factor <= MAX_KEY_LENGTH:101 # factorsByCount is a list of tuples: (factor, factorCount)102 # factorsByCount has a value like: [(3, 497), (2, 487), ...]103 factorsByCount.append( (factor, factorCounts[factor]) )104 105 # Sort the list by the factor count.106 factorsByCount.sort(key=getItemAtIndexOne, reverse=True)107 108 return factorsByCount109 110 111 def kasiskiExamination(ciphertext):112 # Find out the sequences of 3 to 5 letters that occur multiple times113 # in the ciphertext. repeatedSeqSpacings has a value like:114 # {'EXG': [192], 'NAF': [339, 972, 633], ... }115 repeatedSeqSpacings = findRepeatSequencesSpacings(ciphertext)116 117 # See getMostCommonFactors() for a description of seqFactors.118 seqFactors = {}119 for seq in repeatedSeqSpacings:120 seqFactors[seq] = []121 for spacing in repeatedSeqSpacings[seq]:122 seqFactors[seq].extend(getUsefulFactors(spacing))123 124 # See getMostCommonFactors() for a description of factorsByCount.125 factorsByCount = getMostCommonFactors(seqFactors)126 127 # Now we extract the factor counts from factorsByCount and128 # put them in allLikelyKeyLengths so that they are easier to129 # use later.130 allLikelyKeyLengths = []131 for twoIntTuple in factorsByCount:132 allLikelyKeyLengths.append(twoIntTuple[0])133 134 return allLikelyKeyLengths135 136 137 def getNthSubkeysLetters(n, keyLength, message):138 # Returns every Nth letter for each keyLength set of letters in text.139 # E.g. getNthSubkeysLetters(1, 3, 'ABCABCABC') returns 'AAA'140 # getNthSubkeysLetters(2, 3, 'ABCABCABC') returns 'BBB'141 # getNthSubkeysLetters(3, 3, 'ABCABCABC') returns 'CCC'142 # getNthSubkeysLetters(1, 5, 'ABCDEFGHI') returns 'AF'143 144 # Use a regular expression to remove non-letters from the message.145 message = NONLETTERS_PATTERN.sub('', message)146 147 i = n - 1148 letters = []149 while i < len(message):150 letters.append(message[i])151 i += keyLength152 return ''.join(letters)153 154 155 def attemptHackWithKeyLength(ciphertext, mostLikelyKeyLength):156 # Determine the most likely letters for each letter in the key.157 ciphertextUp = ciphertext.upper()158 # allFreqScores is a list of mostLikelyKeyLength number of lists.159 # These inner lists are the freqScores lists.160 allFreqScores = []161 for nth in range(1, mostLikelyKeyLength + 1):162 nthLetters = getNthSubkeysLetters(nth, mostLikelyKeyLength, ciphertextUp)163 164 # freqScores is a list of tuples like:165 # [( , ), ... ]166 # List is sorted by match score. Higher score means better match.167 # See the englishFreqMatchScore() comments in freqAnalysis.py.168 freqScores = []169 for possibleKey in LETTERS:170 decryptedText = vigenereCipher.decryptMessage(possibleKey, nthLetters)171 keyAndFreqMatchTuple = (possibleKey, freqAnalysis.englishFreqMatchScore(decryptedText))172 freqScores.append(keyAndFreqMatchTuple)173 # Sort by match score174 freqScores.sort(key=getItemAtIndexOne, reverse=True)175 176 allFreqScores.append(freqScores[:NUM_MOST_FREQ_LETTERS])177 178 if not SILENT_MODE:179 for i in range(len(allFreqScores)):180 # use i + 1 so the first letter is not called the "0th" letter181 print('Possible letters for letter %s of the key: ' % (i + 1), end='')182 for freqScore in allFreqScores[i]:183 print('%s ' % freqScore[0], end='')184 print() # print a newline185 186 # Try every combination of the most likely letters for each position187 # in the key.188 for indexes in itertools.product(range(NUM_MOST_FREQ_LETTERS), repeat=mostLikelyKeyLength):189 # Create a possible key from the letters in allFreqScores190 possibleKey = ''191 for i in range(mostLikelyKeyLength):192 possibleKey += allFreqScores[i][indexes[i]][0]193 194 if not SILENT_MODE:195 print('Attempting with key: %s' % (possibleKey))196 197 decryptedText = vigenereCipher.decryptMessage(possibleKey, ciphertextUp)198 199 if detectEnglish.isEnglish(decryptedText):200 # Set the hacked ciphertext to the original casing.201 origCase = []202 for i in range(len(ciphertext)):203 if ciphertext[i].isupper():204 origCase.append(decryptedText[i].upper())205 else:206 origCase.append(decryptedText[i].lower())207 decryptedText = ''.join(origCase)208 209 # Check with user to see if the key has been found.210 print('Possible encryption hack with key %s:' % (possibleKey))211 print(decryptedText[:200]) # only show first 200 characters212 print()213 print('Enter D for done, or just press Enter to continue hacking:')214 response = input('> ')215 216 if response.strip().upper().startswith('D'):217 return decryptedText218 219 # No English-looking decryption found, so return None.220 return None221 222 223 def hackVigenere(ciphertext):224 # First, we need to do Kasiski Examination to figure out what the225 # length of the ciphertext's encryption key is.226 allLikelyKeyLengths = kasiskiExamination(ciphertext)227 if not SILENT_MODE:228 keyLengthStr = ''229 for keyLength in allLikelyKeyLengths:230 keyLengthStr += '%s ' % (keyLength)231 print('Kasiski Examination results say the most likely key lengths are: ' + keyLengthStr + '\n')232 233 for keyLength in allLikelyKeyLengths:234 if not SILENT_MODE:235 print('Attempting hack with key length %s (%s possible keys)...' % (keyLength, NUM_MOST_FREQ_LETTERS ** keyLength))236 hackedMessage = attemptHackWithKeyLength(ciphertext, keyLength)237 if hackedMessage != None:238 break239 240 # If none of the key lengths we found using Kasiski Examination241 # worked, start brute-forcing through key lengths.242 if hackedMessage == None:243 if not SILENT_MODE:244 print('Unable to hack message with likely key length(s). Brute forcing key length...')245 for keyLength in range(1, MAX_KEY_LENGTH + 1):246 # don't re-check key lengths already tried from Kasiski247 if keyLength not in allLikelyKeyLengths:248 if not SILENT_MODE:249 print('Attempting hack with key length %s (%s possible keys)...' % (keyLength, NUM_MOST_FREQ_LETTERS ** keyLength))250 hackedMessage = attemptHackWithKeyLength(ciphertext, keyLength)251 if hackedMessage != None:252 break253 return hackedMessage254 255 256 # If vigenereHacker.py is run (instead of imported as a module) call257 # the main() function.258 if __name__ == '__main__':259 main()