How to extract the intermediate Initialization Vector from a sequence of encryption steps?

Question

I am implementing an log-structured file system and want to encrypt a series of blocks by using the .NET Cryptography namespace. I've chosen the Aes symmetric encryption, created the key and the initial, random Initialization Vector.

So far so good, using the ICryptoTransform returned by SymmetricAlgorithm.CreateEncryptor() it is possible to encrypt the individual blocks.

To enable "random access" retrieval at decryption time, the intermediate initialization vectors need to be stored alongside with the encrypted blocks.

But I can't see a way to extract the intermediate IVs? They must be stored in the instance that implements the encryption algorithm, but - as I can see - the current IV is not accessible.

Of course, this limitation can be circumvented by generating random IVs, or by misusing encrypted data to encrypt the next block in the chain. But this feels like a hack due to the fact the that block chaining is supported in most of the symmetric algorithms which - as I assume - are just reusing the resulting vector from the previous block's encryption.

Answer 1

The IV must be known to the recipient of the encrypted information to be able to decrypt it. This can be ensured in a number of ways: by transmitting the IV along with the packet, by agreeing on it beforehand during the key exchange or the handshake, by calculating it (usually incrementally), or by measuring such parameters as current time (used in hardware authentication tokens such as RSA SecurID, VASCO Digipass, etc.), IDs such as sender's and/or recipient's address or ID, file ID, the packet, sector or cluster number, etc.

Ref.

Answer 2

The CBC mode chains each block together by XORing the encryption of the previous block onto the plaintext block before encrypting. When encrypting the first block there is no previous block to use, so you provide an IV to use for the first block.

So what you call the "intermediate IV" is just the encryption of the previous block. No hack needed.