# Tag Info

0

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 ...

2

Let's consider what you know: you have the last block, the IV, and the entire ciphertext. What you don't know is the key or the plaintext. Since CBC-mode encryption uses the IV directly only with the first block you must find the first block, before the IV even becomes useful for your attack. Since the CBC mode uses AES(plaintext[i] XOR ciphertext[i-1], key) ...

8

This is a problem created by the library. Either the HMAC algorithm is skipped entirely, or - more likely - the HMAC authentication tag is generated and then forgotten. The reason is likely the spotty to non-existent handing of authenticated ciphers in the OpenSSL command line and (likely) higher-level functions. The string is recognized by the cipher ...

2

How can I construct an attack to show that this mode is not secure? I assume you're asking about CBC mode (you didn't specify), and while this smells like a homework question, it's been long enough for the assignment to have been due, and so I'll just give you the answer. What you do is ask to encrypt three consecutive messages, and the first block of the ...

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