9/16/2023 0 Comments Adobe illustrator crack sypher![]() ![]() Filling in the letters for wednesday and updating the keyword yielded QUARTZ. At the start I see something that might be usual, and further in I see int-cept and near the end w-nesd- at mi-ight. O-usua-ropo-loca-onha-eenc-prom-edth-eama-eint-cept-gour-mmun-atio-wewi-beus-gthe-cond-yloc-ionf-mnow-thed-poff-mere-insw-nesd-atmi-ight I think updated the plaintext to see the results. I replaced the corresponding hyphen with u and clicked update keyword to find what letter would have produced that combination. At the end, there's ght - this made me think the prior letter is likely a vowel. You start recognizing bits of the plaintext. Here's where I applied a bit of brain power. ![]() O-sua-opo-oca-nha-enc-rom-dth-ama-int-ept-our-mun-tio-ewi-eus-the-ond-loc-onf-now-hed-off-ere-nsw-esd-tmi-ght Update the key, adding only what you know Q-TZ and click 'update plaintext'. What's nice about the approach here is that unknowns in either the plaintext or key are left as hyphens. Provide the ciphertext, a key length of 6 and hit initialize. I started by using this page (now not functional) and the information you supplied. I don't have a programmatic solution for cracking the original ciphertext, but I was able to solve it with a little mind power and some helpful JavaScript. Ideally, you would use a random key that is longer than the plaintext, so it never repeats, which guarantees that the ciphertext cannot be decoded without the key.Ĭopying my answer over from StackOverflow: If you have a short text (like here), or a longer key, then it gets harder. In general, the Vigenere cipher can be cracked easily for longer texts with shorter keys. The only way I can think of would be to check for dictionary words in both the key and the resulting output, ignoring any keys which do not make sense. I am not sure how you managed to decode this automatically. In addition, because the text is so short, the letter frequencies at each of the positions of the key are even more skewed. Using the cleartext from your question, the most frequent letter is, in fact, 'O', not 'E'. However, this only works for longer texts, since the letter distribution is easy enough to mess up in shorter ones. Knowing that the letter 'A' will give you a offset of 0, and the letter 'Z' will give you a offset of 25, it is trivial to guess the original key. This means that you can find each letter of the key using the most frequent letter at a position and its offset from the letter 'E'. The most frequent letter in each position is not the letter of the keyword. With a long enough text, you can assume that this is the letter 'E'. This means that the most frequent letter in each block will (or should) correspond to the most frequent letter in the cleartext. I just hoped one of you might be able to offer some clue as to where I am going wrong. I guess I must be approaching this the wrong way. There are online applets that can crack this no problem, so it mustn't be too short a text to yield decent frequency counts from the blocks. Which yields QCNVHM, or QUNVHM (being generous), neither of which are that close to QUARTZ. ![]() My idea was to calculate the highest-frequency letter in each block, hoping that the most frequent letter would give me some clue as to how to find 'U,' 'A' and 'R.' However, the most frequent letters in these blocks are: KeyPos=0: Q,4 T,3 E,3, J,3 So the first slice is 0, k, 2k, 3k, 4k the second is 1, k + 1, 2k + 1, 3k + 1 et cetera. So far I have split the ciphertext into slices using this awesome applet. However, in this question we are only told the first and last two letters of the Key - 'Q' and 'TZ.' The index of coincidence gave a shift of six (6): I know this is right (I used an online Java applet to decrypt the whole thing using the key 'QUARTZ'). MINKA DTLOG HEVNI DXQUG AZGRM YDEXR TUYRM LYXNZ SNXVY FLSXT ICNXM GUMET HMTUR PENSU TZHMV LODGN WSXYD VTLQS BVBMJ YRTXO JCNXH THWOD FTDCC RMHEH So this is a Vigenere cipher-text EORLL TQFDI HOEZF CHBQN IFGGQ MBVXM SIMGK NCCSV ![]()
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