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For CBC mode, the IV must be Never used twice with the same key Unpredictable So, in your example (filename$\oplus$key), if you ever encrypt two files that have the same filename with the same key, you will violate #1. Now, you may be tempted to say "but I always generate a new key for every file that I encrypt, so that example doesn't apply". Fine, ...


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Remark: in One Time Pad the pad is used once, thus this is not OTP, since here $k$ is reused. Hint for part 1: Write the relations between $k$, the message blocks $m_i$, the ciphertext blocks $\small C_i$ with the convention $\small\text{IV}=\small C_0$. Then, find equations that allow computing the desired $m_3⊕m_4$ from known quantities. Hint for part 2, ...


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First of all, I suggest you to try use as IV first 16 bytes of encrypted file. Because in general IV is the first block of ciphertext. But if that doesn't work, then – of course – you can decrypt all message except first block. Just use first block as IV, and start to decrypt from second block. That will work because CBC does not provide integrity, and ...


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SHORT: The key remains the same. CBC mode is one of the ways which allow block cipher to securely process multi block input. Block cipher with single key produces same output for same input, as it is relatively common for input to contain duplicate blocks, instead of using block cipher as is, a more complex construct is needed. The idea of CBC mode is ...


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Just for completeness sake, CBC is defined as follows: The error you have made is that: $$M;N = (M_1, ..., M_n, N_1 ⊕ \mathbf{T_m}, N_2, ..., N_n)$$ (I've changed notation from M||N to M;N to reflect this isn't just concatenation) You need to cancel the tag from the message $M$, not the tag from the $N$ message. In that case, $T_{M;N}=T_N$ as required.



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