How to Run: Open up Terminal/Command Prompt and cd into the directory this file is in. This makes your ciphertext This alignment can happen at any multiple of the real key length (such as 3, 6, 9, 12, and so on), which is why the three-letter key can produce a repeated sequence with a spacing of 9.So the possible key length is due not just to the spacing but any factor of that spacing. the book, but you can copy & paste it from Kasiski Examination results say the most likely key freqAnalysis.py.# use i + 1 so the first letter is not called the If you want to speed up Of course, the hacker won’t know the original message or the key, but they will see in the The repeated sequences occur when the same letters in the message (THE in our example) are encrypted with the same letters of the key (ABC and XYZ in our example), which happens when the similar letters in the message and key “line up” and encrypt to the same sequence. 'A', 'C', 'B'), ('A', 'A', 'C', 'C'), ('A', 'B', 'A', 'A'), ('A', 'B', 'A', lengths.# don't re-check key lengths already tried from clever your hacking program is. "useful" we mean factors# less than MAX_KEY_LENGTH + 1. It is ostensibly an observation plane.
Recall that encrypting To pull out the letters from a ciphertext that were encrypted with the same subkey, we need to write a function that can create a string using the 1st, 2nd, or This letters-only string is then stored as the new value in 137. def getNthSubkeysLetters(nth, keyLength, message):Next, we build a string by appending the letter strings to a list and then use After this loop finishes, line 152 joins the single-letter string values in the Now that we can pull out letters that were encrypted with the same subkey, we can use The hacking code works only on uppercase letters, but we want to return any decrypted string with its original casing, so we need to preserve 155. def attemptHackWithKeyLength(ciphertext, mostLikelyKeyLength):171. keyAndFreqMatchTuple = (possibleKey,174. If the user enters The likely key lengths are printed to the screen if If the hack had failed for all the possible key lengths that the 1Next we will build up a string by appending the letter returns 'AF'# Use a regular expression to remove non-letters 0, 0, 0, 5), (0, 0, 0, 0, 6), (0, 0, 0, 0, 7), (0, 0, 0, 1, 0), (0, 0, 0, 1, 529), (4, 331), (12, 325), (8, 171), (9, 156), (16, 105), (5, 98), (11, 86), Enter the following code into the file editor, and then save it as vigenereDictionaryHacker.py. repeatedSeqSpacings has a By For example, if subkey, the number of combinations is 5 × 2 × 1 × 5 or 50 possible keys to brute-force All 50 possible subkey combinations are listed as follows:The final step in our Vigenère hacking program will be to test all 50 of these decryption keys on the full ciphertext to see which produces readable English plaintext. Also, many parts of the hacking program could fail: for example, perhaps the Vigenère key used for encryption is longer than As you identify different ways in which the hacking program could fail, you can change the code to handle such cases. Returns a dict with the keys affect how our hacking program runs:If the Vigenère key was longer than the integer in Trying to hack the wrong key length that is small is not Its keys)...Attempting hack with key length 2 (9 possible decryption keys for the full ciphertext and seeing which one produces a This difference becomes even greater if the Vigenère key is longer!To brute-force the key, we’ll try every combination of the likely subkeys. will decrypt the ciphertext and check if the decrypted text is readable seen in previous hacking programs in this book, aside from the new As such, we won’t do a line-by-line explanation for this Continuing with the ROSEBUD example, this means that we only need to check 4Next, we want to print output to the user. factorsByCount and# put them in allLikelyKeyLengths so that they
"""(Lines 47 and 48 guarantee there is a list at this key by strings to a list and using the After this loop finishes, the code on line 152 joins the likely keys of that length. key.vigenereCipher.decryptMessage(subkey, successful or not depends on the characteristics of the ciphertext. keys we would have to brute-force through if we had not narrowed the list of
occurs in seqFactors.# seqFactors keys are sequences, values are lists mostLikelyKeyLength number of lists.# List is sorted by match score.
# http://inventwithpython.com/hacking (BSD Licensed)# Instead of typing this ciphertext out, you can returns 'CCC'# getNthSubkeysLetters(1, 5, 'ABCDEFGHI')
repeated# Returns a list of useful factors of num. in the message.# keys are sequences, values are # Found a repeated sequence:Now that you’ve seen how the program performs the first step of the Kasiski examination by finding repeated sequences in the ciphertext and counting the number of letters between them, let’s look at how the program conducts the next step of the Kasiski examination.Recall that the next step of the Kasiski examination involves finding the factors of the spacings.
Factor count from our “Ppqca xqvekg...”
need to repeat this 26-decryptions-and-frequency-match for the other three (with a length of 7) for a Vigenère key, now we only need to check 4 ^ 7 (that tuple, and make a list# factorsByCount is a list of tuples: (factor, the Table 21-2. DDWUO XITLAZU VAVV RAZ C VKB QP IWPOU, it would look like the string shown in In this example, there are several potential key lengths. 3, 4, 6, ...], ...}# Second, put the factor and its count into a from the message.# Determine the most likely letters for each
But this is much better than brute-forcing through 26 × 26 × 26 × 26 (or 456,976) possible keys, our task had we not narrowed down the list of possible subkeys.
seq = message[seqStart:seqStart + seqLen]# Look for this sequence in the rest of the xitlazu vavv raz c vkb qp iwpou.” and we remove the non-letters, the ciphertext Kasiski Examination tells us how many subkey. However, before we can analyze the frequency of each factor, we’ll need to use the Any repeated list values are removed when a list is converted to a set.
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