0
$\begingroup$

What will happen if the Initialiazation Vector (IV), i.e, the initial randomness fed in to the first block arrives mangled at receiver end? How will this affect the decryption? Will the receiver be abl to decrypt all blocks correctly? Or none? Or some?

$\endgroup$
  • $\begingroup$ Also Bit Flipping Attack on CBC Mode $\endgroup$ – kelalaka Sep 17 at 12:34
  • $\begingroup$ So this question is a duplicate of another question which is also duplicate? That's deep! $\endgroup$ – hola Oct 14 at 10:57
0
$\begingroup$

For the first block, assuming the noise in the IV is not much, it may be possible to recover part of the plaintext. Say, only one bit of the IV is flipped. In this case, all but one bit (of the first block of) plaintext can be recovered, as the with usual notations $C=E_K(P\oplus IV \oplus e)$, $e$ is a vector of small Hamming weight. By decrypting $C$, Bob gets $P\oplus IV \oplus e$. By the perceived knowledge of $IV$ all but those bits are affected by $e$ can be recovered (but you need to know which bits of the IV are vulernable, otherwise, the plaintext is lost).

For all the subsequent blocks, it will not be possible anymore. Even if $e$ is Hamming weight $1$, assuming the cipher is strong enough, it will make the first block of ciphertext look totally random. When this is fed to the next block, it will produce completely off ciphertext blocks.

| improve this answer | |
$\endgroup$

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