If an adversary holds thousands of encrypted files that were encrypted with the
ID: 649475 • Letter: I
Question
If an adversary holds thousands of encrypted files that were encrypted with the same cipher, can he determine which cipher method had been used on the files? Assume that the adversary has no prior knowledge about the encryption environment.
Details: Let's say I have a encryption program that allows me to select 1 out of 5 well established and well implemented encryption algorithms. I use that program to encrypt thousands of file, using always the same block cipher, while following proper cryptographic procedures (e.g. use different passwords every n files encrypted). The files are stored off-site, and an adversary gets a hold of the files. I wonder if by selecting a less common algorithm, e.g. Serpent, I would increase the workload of the adversary by factor of 5, or encryption algorithms do leak some information, some recurring properties, that will help the adversary to identify the encryption method I had used?
Explanation / Answer
Note that a block cipher is not an encryption algorithm. It's a building block. To get encryption you need a mode-of-operation that uses that block cipher. For example CBC-AES128 (and even here, it's not clear how the IV is being generated).
Many encryption algorithms overtly leak their identity. Some have packaging, they put the IV in a weird place, they use padding. It is not a requirement for security that the algorithm used be obscured whatsoever.
Now we can re-interpret your question differently and say, "Suppose you take a list of 128-bit block ciphers and you generate a list of 1 billion random 128-bit plaintexts, enciphering each under a random key with every block cipher in the list. Can you tell which cipher is which simply by examining the ciphertexts?" In other words, does some block cipher have a "signature format" to its output? No, it isn't possible. Because block ciphers are permutations (given a fixed key) so worrying about "block cipher XYZ tends to output a lot more 1's than 0's" cannot happen.
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