# Tag Info

1

Perhaps. You indicate this is from an embedded device. This strongly implies everything the device needs to decrypt it is already present in the firmware of the machine, including both the algorithm and the key. Here is a blog post on how a talented engineer accomplished a similar feat through reverse engineering. Note that this was not a cryptographic ...

3

As long as the IV is chosen correctly, every individual block of the encrypted output will be uniformly random over the set of all bit-patterns of the given size. Each block is independent from the clear text, but they are not independent from each other. The first block contains the IV itself, which by construction is uniformly random and independent from ...

7

In the first block, the IV provides the "randomness", and in subsequent blocks you just use the previous block of ciphertext instead. Based on the assumption, that the cipher is not weak and behaves like a pseudorandom permutation, this is basically the same: You XOR something unpredictable on the plaintext, and then encrypt. As long as the IV is chosen ...

1

This is why we use random initialization vectors (IVs) for all such algorithms.

4

All modern block ciphers are supposed to be pseudorandom permutations, meaning that they cannot be efficiently distinguished from a truly random permutation without knowledge of the key. (If a practical distinguisher were to be found for a particular cipher, that cipher would be considered broken by modern standards.) This also implies that no two secure ...

2

Yes there is a significant difference concerning brute-force. ECB suffers from multi target attacks whenever you encrypt the same message block. This is always possible in a chosen-plaintext attack and often possible in practice with a known-plaintext attack. With CBC the IV means that the plaintexts passed to the blockcipher are almost certainly unique ...

2

As Maarten Bodewes already wrote in a comment, if you ignore the computational overhead of XOR, then there is essentially no difference in CBC and ECB for a bruteforce attack. However, the question is actually mixing oranges and apples (and it is not obvious), because the security weakness of modes of operation has nothing to do with the underlying ...

2

In short, yes. The complementation property of DES states that if $DES_K(P) = C$, then $DES_{\overline{K}}(\overline{P}) = \overline{C}$, where $\overline{X}$ is the complement of a string $X$. ECB with DES takes a message $M_1M_2\cdots M_\ell$ and computes $C_1C_2\cdots C_\ell$, where $C_i = DES_K(M_i)$, for $1\le i\le\ell$. Therefore, if you encrypt ...

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