# Tag Info

6

the key and plaintext is the same. The attacker knows this and the IVs used but doesn't know the plaintext. Is there anything to learn about the plaintext when multiple ciphertexts are available instead of only one? No. Giving the attacker multiple encryptions of a single plaintext using a randomly chosen IV and a fixed key with AES-CBC does not leak ...

2

If you use a random IV each time you encrypt a file, the result will be different. $$\text{AES-CBC}(IV_1, Key, M) = C_1$$ $$\text{AES-CBC}(IV_2, Key, M) = C_2$$ you do not gain any pieces of informations by having $C_1$ and $C_2$. Even knowing that they share the same key and should they share the same length, an attacker can not even know whether or not ...

3

If you have a nonrepeating (but possibly predictable) value, you can convert that into an unpredictable CBC-mode IV at fairly minimal cost. Here's how: Prepend the 128 bit nonrepeating value to the message CBC mode encrypt the (value, message), using any IV that's not correlated to the nonrepeating value (all 0's work) Use the first 16 bytes of the ...

0

No, it is not possible. The encryption step of CBC is $c_i = E(c_{i-1} \oplus p_i)$. When you double encrypt, the previous block ciphertext gets XORed with an independent block cipher output. The only place where something weird could happen is the beginning. If you assume that the (identical) IV $I$ is prepended to the ciphertext, as is common, you get ...

2

Scheme is not IND-CPA for any message longer than one block. I'll include a image of CBC mode below for reference (Source: Wikipedia). Suppose instead of block cipher encryption we have plaintext xor-ed with the key as you propose. You'll note that for message block 1, $M_1$, the ciphertext block $C_1 = M_1 \oplus IV \oplus Key$. Similarly \$C_2 = M_2 ...

Top 50 recent answers are included