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So if I want to encrypt all the files in a folder with AES (same password) I take each file and generate the key schedule using PBKDF2. As the PBKDF2 algorithm takes in a salt then this salt should change (and thus the keyschedule) for each new file being encrypted. Now if I want to decrypt then I load the plaintext salt and generate the different key schedule for each ciphertext file.

Is this correct? If so the problem I'm seeing is that the recommended number of iterations to make the encryption more secure (0.5 seconds e.g.) will considerably slow down the encryption process as each file requires a lengthy key derivation procedure. Is this just a necessary part of a secure encryption process?


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Why not just encrypt the entire drive with a different hash for each 4k sector while your at it? ;) – user3997 Dec 7 '12 at 17:49

You are correct that using a different salt for each file will slow down encryption and decryption, in proportion to the number of files. But it is not useful to do so. An adversary is not helped if she knows that the salt is common to several files hashed with the same password (under the assumption that she can recognize a correct password with fair probability from one ciphertext, and cost dominated by PBKDF2). Both with common and different salt per file, the hard work for the attacker is finding the password; the cost of recomputing PBKDF2 for other files when salt is different will be comparably negligible.

Thus an option if to have the enciphering program generate the salt at random common to every files, and appropriate IV for each file (e.g. 48-bit file number, and 80-bit zeros; this is enough for $2^{48}$ files of $2^{84}$ bytes each, encrypted using AES-CTR); compute only one key using PBKDF2; and store salt and IV as header of each enciphered file. The deciphering program can cache the key in RAM for reuse when deciphering files with identical salt (it should not assume all files have identical salt, in case files separately enciphered are merged into the same directory).

When using other block-cipher operating modes such as CBC, CFB or OFB for enciphering multiple huge files, you may approach the limit of reasonable usage of one key. For this reason, or just for added peace of mind, you might want to have one key per file. This can be done while still amortizing the derivation of key from password over multiple files, by generating a common "slow" master key from password and master salt (e.g. random) for each encryption session, and a "fast" derived key from master key and secondary salt (e.g. file index) for each file, with both salts in the encrypted file's header. The "fast" derivation could be with PBKDF2 and a minimal iteration count c=1, or just one HMAC.

Also: you might want to consider scrypt instead of PBKDF2 (at least for the master key), for it provides better security for a given time overhead, as summarized in this table from the scrypt paper:

Estimated cost of hardware to crack a password in one year

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Thank you for your help! – winterfell Dec 3 '12 at 20:52
+1 i love that scrypt paper. – Rook Dec 4 '12 at 1:44
You mispelled PBKDF2 a couple of times in your answer - no big deal but thought you'd like to know (it is awkward to type quickly). +1 for answer though. – Thomas Dec 4 '12 at 2:45
@Thomas: hopefully fixed, thanks for the note. – fgrieu Dec 4 '12 at 3:36

There really isn't any measurable security in that. It's also inconsistent. If you had thirty text 1KB files, you'd be generating thirty new schedules, but if it were one 20MB video, you'd be using one schedule for all of the data.

Really it depends on your mode of encryption. For example, if you're using CFB the iv would change depending on the previous block anyway, so there's no reason to keep switching schedules.

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I think the files are encrypted independently, so the mode of operation wouldn't matter here. The question seems to be about whether it's necessary to use a slow PBKDF2 to derive an encryption key for each file in the folder or if a "slow" master key derivation from the user password along with a fast derivation (e.g. HMAC) to get the encryption keys is enough. – Thomas Dec 4 '12 at 3:42
Ahh. I was under the impression he was generating some sort of archive to pack them all together. – Dakota West Dec 4 '12 at 19:36

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