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Seven Codes You'll Never Ever Break
- jzig 14y agoThe Dorabella cipher has been solved: http://unsolvedproblems.org/S12x.pdf http://unsolvedproblems.org/S12x.pdf
- jgrahamc 14y agoNo, it hasn't. Did you read his 'explanation'? First the decrypted code is close to gibberish and second he starts by creating the key by removing duplicated letters and yet the key has two Ns, three Is, two Ys, two Ws, etc.
- jgrahamc 14y agoStupid title. Just because Kryptos Part IV hasn't been broken so far doesn't mean that it won't be. It was designed to be broken. Oh, wait, the actual title is "7 Codes You’ll Never Ever Break" instead of "Codes Hackers Will Never Ever Break". The actual title is more likely to be true, especially given the complete crackpots that appear in groups like Kryptos with outlandish 'breaks' to the codes that are complete crap. Breaking codes is hard, takes a long time and requires method. Most people won't break these codes.
- stephengillie 14y agoAt some point, someone will just brute-force these.
- achille 14y agoYou can't brute force if you don't know the cipher. Heres an example: Imagine a code that smartly encodes and hides spoken voice. No amount of brute force would be able to crack that.
- stephengillie 14y agoWe can brute force all possible ciphers. Then brute force all of the outcomes that are 100% dictionary words. If those all fail, then we go back to the ones with a few dictionary words missed, and brute-force those. For computers and humans, this is just pattern matching, and brute-force means we have unlimited time and resources to map the entire patternspace find the correct pattern. You're going to have to quantify "smartly" into something more objective before you can use it in this argument.
- iso-8859-1 14y agoBrute-force all possible ciphers? We don't know the input, only the output. But brute-forcing in this context would mean trying out every cipher with every input, and that's not possible since there is an infinite amount of inputs.
- stephengillie 14y agoI'm not sure where you're getting an infinite number of inputs. The way I understand one-time pads, based on a skim of the wiki article you linked, is that you're taking two strings of letters, one is your message, the other is a random, one-time string, adding them, and taking the modulus of 26. Where else do we have 2 mixed streams of information? Music. Optics. So we take the input and reverse the mod26, one step at a time, basically creating an array of "demodulated" input values, then I feel as though we should be able to do a Fourier analysis to separate the 2 strings of numbers.
- iso-8859-1 14y ago"a cipher (or cypher) is an algorithm for performing encryption or decryption" So when you say "cipher", that means you want every possible algorithm. Since algorithms can produce output shorter than the input, you get the infinite amount of inputs. With one-time pads, fourier analysis would only work if the key is not truly random. How do you reverse a "mod26"? You can't. I give you the number 17. You know this number is produced using the equation: "SECRET mod 26 = 17". How do you know if 43 or 69 was the input?
- iso-8859-1 14y agoHow do you know they're not one-time pads?
- stephengillie 14y agoWhat if we try every possible combination of one-time pads? This is like sculpting with Michelangelo - the correct pattern exists, we just have to remove all of the incorrect patterns from the patternspace.
- iso-8859-1 14y agoWith one time pads, you can decrypt it to anything, it all depends on the key. See http://en.wikipedia.org/wiki/One-time_pad#Attempt_at_cryptanalysis http://en.wikipedia.org/wiki/One-time_pad#Attempt_at_cryptan... What am I missing here?
- stephengillie 14y agoSo use all keys. Then sort the result and continue to brute-force the correct output. How many billions of NTLM keys can a Geforce crack in a second? How is this cryptography problem so greatly different than that one?
- perokreco 14y agoYou are missing that when bruteforcing a one time pad you will get all possible arrangements of all letters. How do you tell all the plausible looking strings apart? What kind of a sort function could you apply?
- marshray 14y ago> How many billions of NTLM keys can a Geforce crack in a second? According to https://hashcat.net/oclhashcat-plus/ https://hashcat.net/oclhashcat-plus/ , it's approximately 2.5 B. This corresponds to a one time pad message of just under 4 bytes long. The difference, of course, with NTLM is you know when you've found the right value. With OTPs, all decryptions are possible and equally valid.
- podperson 14y agoOne of the codes may be a one-time-pad reference to a book which may no longer exist OR the ciphertext may simply be a meaningless fraud. Another is probably written in an invented, private language. Two others are short enough that brute force approaches will almost certainly create false positives. In general, brute force can only work if you have some idea of what to expect the clear text to look like.
- stephengillie 14y agoSo 2 of these probably aren't even codes, just gibberish. So what if false positives are created? We can just brute-force the analysis of those to find the right one. Are you forgetting the sheer magnitude of quantum, digital, & human-analog computing power we have available today?
- podperson 14y agoI think you're forgetting to check your hubris at the door. Given the folks who've attacked these problems, the ones that have yielded to brute force thus far: 0.
- marshray 14y agoI know, would anyone like an 8th? 641648018FD15E9157AB76299C47FFCC9974EB401494D13398A9F2635F66AD83 Spoiler alert: It's just /dev/random
- Dylan16807 14y agoSpoiler alert: That makes it not a code.
- marshray 14y agoWell, I could at any point in the future produce a one-time-pad keystream that decrypts it to a plaintext message. Also, my comment about it being /dev/urandom may have been pure misdirection.
- Tichy 14y agoGrr, of course the link to "drawings of naked women" is broken. Anybody has the correct URL?
- ColinWright 14y agoFor your delight and delectation, here's a link: http://brbl-zoom.library.yale.edu/viewer/1006200 http://brbl-zoom.library.yale.edu/viewer/1006200 For reference, a simple Google search found it in about 30 seconds.
- Tichy 14y agoDelightful, thanks!
- dutchbrit 14y agoYou can find a PDF version here: http://archive.org/details/TheVoynichManuscript http://archive.org/details/TheVoynichManuscript
- ColinWright 14y agoUnpaginated: http://www.wired.com/dangerroom/2012/12/codes/?pid=1708&viewall=true http://www.wired.com/dangerroom/2012/12/codes/?pid=1708&... Added in edit: Interesting that someone downvoted this comment. Did someone fat-finger it, or do some people really think it's wrong to give a link to the unpaginated version? I'd like to know. If you feel unable to reply here, then feel free to email - address is in my profile. Thanks.
- Surio 14y agoI, for one, appreciate unpaginated links, so thank you.
- deleted 14y ago[deleted]
- dutchbrit 14y agoOn decoding the Voynich manuscript, http://www.edithsherwood.com/voynich_decoded/ http://www.edithsherwood.com/voynich_decoded/
- nnq 14y agosemi-offtopic: how could one crack a cyphered text if it was written in a 100% invented/artificial language, with no relation to any natural language, not even good word-to-word mapping (imagine a pictografic language like the assian ones, but unrelated)? where would one start without word-frequency analysis or something similar to begin with? Maybe the Voynich manuscript has such a completely imagined language, and that's why nothing worked (if it's not a "hoax:, of course).
- MichaelGG 14y agoA 100% artificial language can be equivalent to a one-time pad, so there's no guarantee it could be cracked.
- stephengillie 14y agoWhy can't we brute-force one-time pads? Did someone figure out how to create infinite combinations of 26 letters? As far as I understand, this space is finite, and thus can be brute-forced.
- mtinkerhess 14y agoThe thing about a one-time pad is that the size in bits of the pad is the same as the size in bits of the message (before and after encryption). So given an encrypted message, the number of possible one-time pads is equal to the number of possible corresponding cleartext messages. There's no way to brute force that -- if you tried every single possible one-time pad to decrypt the ciphertext, you'd end up with every single possible cleartext message, with no way to distinguish which was the original cleartext.
- stephengillie 14y agoSo for every iteration, we get a cleartext message? So every cleartext message is 100% dictionary words and absolutely no gibberish? So every cleartext message has complete, logical sentences? So the complete, logical sentences in each message make sense in the context of the message? I do not believe this to be the case. Only one cleartext message would actually make sense.
- piqufoh 14y agoBruce Schneier broke these codes before breakfast, and now he's gonna break you! http://www.schneierfacts.com/ http://www.schneierfacts.com/