Going through the wiki for modes of operation I see that the section error propagation says that an error in one block in the ciphertext in CBC mode only impacts two blocks. I do not quite get that. If the second plaintext block is decrypted wrong, then why would it not impact the third block?
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In CBC mode the decryption equation is $P_i = D_k(C_i) \oplus C_{i-1}$. If you received a corrupted $C_i$, $P_i$ and $P_{i+1}$ will be decrypted wrong, but $P_{i+2}$ no longer depends on $C_i$ and will be correct.
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$\begingroup$ Oops.. the xor is done with the plaintext and NOT the cipher text. One of those dumb moments. Thanks. $\endgroup$ May 26, 2014 at 19:20