Take the 2-minute tour ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

Considering a symmetric cipher (i.e : AES in counter mode), is it possible for any given key to be randomly decomposed into other keys without knowing the message or the ciphertext, such as :

100 = 45+55 (or) 30+70 (or) 25*4 (or)...

Key1 = Key2|Key3 (or) Key4|Key5 (or)...

For any message, applying Key1 or Key2|Key3 would have the same encryption/decryption result.

One-time pad would be an option but it is not very practical because it would take the same length as the message. I am open to all suggestion. Thanks !

share|improve this question
1  
What do you actually want to do? Split an existing key (for existing ciphertext) to two parties so both are needed for decryption (one after another)? –  Paŭlo Ebermann Aug 7 '12 at 17:11
    
Does the description for the secret-sharing tag sound like what you are trying to do? –  David Cary Aug 14 '12 at 0:15
add comment

1 Answer

It might help if you explain what problem you're trying to solve.

For example, what exactly do you mean "Key1 = Key2|Key3"? If you mean (say) taking a 256 bit AES key, and splitting it up into two halves (where you need to present both halves to encrypt/decrypt), well, that's easy; however, I doubt that's what you're asking.

Alternatively, when you say "Key2|Key3", you might be asking to split the key into two halves; where Bob has one half (and performs the operation on it), and Carol has the other hand (and she takes Bob's result and her own key, performs the operation on it, and her result is exactly what Alice got with the original key. If so, we can consider (for example) 2AES; that is, you encrypt the message with AES key 1, and then with AES key 2. So, the full key can be considered AES key 1 concatenated with AES key 2; obviously, we can split it into two halves. However, I suspect that this won't address your problem (whatever it is) either.

Assuming that the second guess was on the right track, and you insist that the operation on key 1 are exactly the same as the operations on key 2 and key 3, well, we run into a problem. Generally speaking, symmetric ciphers aren't closed operations; there usually aren't any key 2's and key 3's which are jointly perform the same operation as key 1. However, one example does come to mind: Pohlig-Hellman; with that, given a key, you really can split it randomly into two keys. However, I again can't be quite certain if that's what you're looking for either.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.