I was trying to do a simple CPA attack against this scheme, to understand better the concept.

Instead of using a new 𝐼𝑉 each time, we decide to use the last block of the previous ciphertext as an initialization vector. Prove this new scheme is vulnerable to a chosen-plaintext attack.

So in this case,

  • the challenger chooses a "game" and a key.
  • After that, we send $(0\ldots 0,1\ldots 1)$
  • and we receive $(IV, c)$.
  • Now we send $(0\ldots 0, 0\ldots 0)$
  • and we receive $(IV'=c , c')$.
  • So if $c$ is equal to $IV'$ then the challenger is playing the left game, right otherwise.

Am I right? Am I confusing the concepts? When we talk about a block, is all the cipher or only the last bit?

  • $\begingroup$ This game is problematic since the first IV generation is not clear. Anyway, assuming that is random for the first time, you need to send $(1\ldots 1)$ as the first data on the second try. Write CBC equations and see better? ( note that you seem to send two blocks) $\endgroup$
    – kelalaka
    Feb 21, 2022 at 18:38
  • $\begingroup$ I don't really see a connection to guess the game. The first time I send the message, I will recive (IV, F(k,mi xor ci-1)), second time ( F(k,mi xor ci-1), F(k,mi' xor ci-1')). I guess I'm missing something required to solve the attack. $\endgroup$ Feb 21, 2022 at 20:47
  • $\begingroup$ maybe I just get it, if I do the cipher of the second try XOR with the IV of the first try, I will get the message of the second try? $\endgroup$ Feb 21, 2022 at 21:02
  • $\begingroup$ Send $((c \oplus IV), (1\ldots 1)$ on the second time? $\endgroup$
    – kelalaka
    Feb 21, 2022 at 21:02
  • $\begingroup$ If this is not homework can you write an answer to your question? $\endgroup$
    – kelalaka
    Feb 21, 2022 at 21:11

1 Answer 1


The attack goes as follows, first the adversary asks for the encryption $(0…0,0…0)$. Then he gets $c_1=(c_{11},c_{12})=(IV,F_{K}(m_{\gamma_1}\oplus IV))=(IV,c=F_{K}(IV))$, since $m_{\gamma_1}=0…0$ regardless of the value of $\gamma$.  Then he asks for the encryption of $(m_{L_2}=0…0,m_{R_2}=c_{12}\oplus IV)$, and gets  $c_2=(c_{21},c_{22})=(F_{K}(IV),F_{K}(F_{K}(IV)\oplus m_{\gamma_2}))$. If $\gamma=L$, he gets $(F_{K}(IV),F_{K}(F_{K}(IV))$ and if $\gamma=R$ he gets $(F_{K}(IV),F_{K}(F_{K}(IV)\oplus c_{12}\oplus IV)=(F_{K}(IV),F_{K}(IV))$ since $c_{12}=F_{K}(IV)$.  In summary if $c_{21}=c_{22}$ he says $\gamma=R$ and he says $\gamma=L$ otherwise.


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