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A safe is obviously a safe. It obviously contains something. Encrypted data isn't so clear cut. It's trivial to make a datastore that has several encryption ke

A safe is obviously a safe. It obviously contains something. Encrypted data isn't so clear cut. It's trivial to make a datastore that has several encryption keys, so that you could give out one key, and it'd "decrypt" to some boring stuff, whilst keeping the real data, and the alternate key, secret. It'd also be trivial to devise a decryption algorithm, and key, which "decrypts" anyones hard drive to reveal illegal images even when none are really there... So I don't think it's a good analogy. It's quite obvious when you have successfully got into a safe, but how do you know when you have successfully decrypted something, to the real stuff that is important and being hidden in it?

chc 15y ago
Is it actually trivial to encrypt arbitrary text in such a way that it could be decrypted to the source text or a different but still meaningful alternate text? That sounds really hard to do. Is this indeed a solved problem and I just don't know about it?
- lisper 15y ago
  It is trivial. Just encrypt the true plaintext and the alternate plaintext with separate keys and place the results in a container. To decrypt, try both ciphertexts and return the one that validly decrypts with the provided key.
  - windsurfer 15y ago
    Or better yes, 50'000 containers of varying sizes, 49'998 of which are filled with random data.
  - dedward 15y ago
    It's trivial cryptographically, but from a security point of view, it's not necessarily trivial at all. If we assume the courts can order you to decrypt the drive (and without debating that point) - one has to consider that the court may be fully aware that the system has multiple hidden volumes, either by eyewitness testimony, 3rd party evidence (check out truecrypt's warnings on their site about full system encryption and what to watch out for. Things like finding the same windows installation doing every update twice. There are all kinds of information leaks that COULD pop up. I'm not saying it's impossible - just as strong cryptography, which is easy and is all over, doesn't mean all our data is secure, neither would a more complex system like this protect someone from the legal system.
- bhickey 15y ago
  The short answer on this is no. Check out unicity distance. http://en.wikipedia.org/wiki/Unicity_distance http://en.wikipedia.org/wiki/Unicity_distance
  - dryicerx 15y ago
    That's assuming the same algorithm is used throughout. You can use one algorithm to encrypt/decrypt the original content. But you can use a different algorithm (with a different key) that would output a different output. The secondary algorithm would be one that given a some text (ciphertext from the original encryption) along with the desired output, would return a suitable key. The most basic example to prove the point would be XOR.
    - loup-vaillant 15y ago
      That probably won't work: you need the combination of the algorithm + the key to get your data out. Of the 4 possible combinations, only 2 will yield valid data. If you give a password, the cops will know what algorithm they must use (2 trials at most). Even if they don't know before hand which algorithm points to the real data, they can notice that it doesn't use all data. With your method, you can at best cast doubt: is the data not extracted real data encrypted differently (algorithm or key), or random data that the software insert by default to give everyone plausible deniability?
- andrewcooke 15y ago
  that's now what is being suggested (i think). typically how it's done is to have separate pieces of information, related to different keys. you can then make a system that (1) produces the data associated with a valid key and (2) does not reveal how many valid keys there are.
chopsueyar 15y ago
A safe is obviously a safe. It obviously contains something. http://en.wikipedia.org/wiki/The_Mystery_of_Al_Capone%27s_Vaults http://en.wikipedia.org/wiki/The_Mystery_of_Al_Capone%27s_Va...
pavel_lishin 15y ago
> It'd also be trivial to devise a decryption algorithm, and key, which "decrypts" anyones hard drive to reveal illegal images even when none are really there... How trivial? Remember that the whole hard drive must be consistent, including a file system. Fifty gigabytes of garbage followed by a 2-megabyte photograph followed by fifty more gigabytes of noise is not plausible.
marshray 15y ago
It'd also be trivial to devise a decryption algorithm, and key, which "decrypts" anyones hard drive to reveal illegal images even when none are really there... Well the revealed images could not be significantly larger in size than then the "key material" you supplied plus the most concise description of the algorithm. http://en.wikipedia.org/wiki/Kolmogorov_complexity http://en.wikipedia.org/wiki/Kolmogorov_complexity

chc 15y ago

Is it actually trivial to encrypt arbitrary text in such a way that it could be decrypted to the source text or a different but still meaningful alternate text? That sounds really hard to do. Is this indeed a solved problem and I just don't know about it?

lisper 15y ago
It is trivial. Just encrypt the true plaintext and the alternate plaintext with separate keys and place the results in a container. To decrypt, try both ciphertexts and return the one that validly decrypts with the provided key.
- windsurfer 15y ago
  Or better yes, 50'000 containers of varying sizes, 49'998 of which are filled with random data.
- dedward 15y ago
  It's trivial cryptographically, but from a security point of view, it's not necessarily trivial at all. If we assume the courts can order you to decrypt the drive (and without debating that point) - one has to consider that the court may be fully aware that the system has multiple hidden volumes, either by eyewitness testimony, 3rd party evidence (check out truecrypt's warnings on their site about full system encryption and what to watch out for. Things like finding the same windows installation doing every update twice. There are all kinds of information leaks that COULD pop up. I'm not saying it's impossible - just as strong cryptography, which is easy and is all over, doesn't mean all our data is secure, neither would a more complex system like this protect someone from the legal system.
bhickey 15y ago
The short answer on this is no. Check out unicity distance. http://en.wikipedia.org/wiki/Unicity_distance http://en.wikipedia.org/wiki/Unicity_distance
- dryicerx 15y ago
  That's assuming the same algorithm is used throughout. You can use one algorithm to encrypt/decrypt the original content. But you can use a different algorithm (with a different key) that would output a different output. The secondary algorithm would be one that given a some text (ciphertext from the original encryption) along with the desired output, would return a suitable key. The most basic example to prove the point would be XOR.
  - loup-vaillant 15y ago
    That probably won't work: you need the combination of the algorithm + the key to get your data out. Of the 4 possible combinations, only 2 will yield valid data. If you give a password, the cops will know what algorithm they must use (2 trials at most). Even if they don't know before hand which algorithm points to the real data, they can notice that it doesn't use all data. With your method, you can at best cast doubt: is the data not extracted real data encrypted differently (algorithm or key), or random data that the software insert by default to give everyone plausible deniability?
andrewcooke 15y ago
that's now what is being suggested (i think). typically how it's done is to have separate pieces of information, related to different keys. you can then make a system that (1) produces the data associated with a valid key and (2) does not reveal how many valid keys there are.

chopsueyar 15y ago

A safe is obviously a safe. It obviously contains something. http://en.wikipedia.org/wiki/The_Mystery_of_Al_Capone%27s_Vaults http://en.wikipedia.org/wiki/The_Mystery_of_Al_Capone%27s_Va...

pavel_lishin 15y ago

> It'd also be trivial to devise a decryption algorithm, and key, which "decrypts" anyones hard drive to reveal illegal images even when none are really there... How trivial? Remember that the whole hard drive must be consistent, including a file system. Fifty gigabytes of garbage followed by a 2-megabyte photograph followed by fifty more gigabytes of noise is not plausible.

marshray 15y ago

It'd also be trivial to devise a decryption algorithm, and key, which "decrypts" anyones hard drive to reveal illegal images even when none are really there... Well the revealed images could not be significantly larger in size than then the "key material" you supplied plus the most concise description of the algorithm. http://en.wikipedia.org/wiki/Kolmogorov_complexity http://en.wikipedia.org/wiki/Kolmogorov_complexity