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to try again, I guess the attacker need try to modify the last byte No. Some other (symmetric) padding oracle attacks do that, but here the attacker can't tweak the prior block and try another decryption because the MAC error already caused the session to be terminated. Instead the attacker causes the client (browser) to make a new request with the same ...


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This article is quite sloppily written... We suppose that the padding is valid, this means that upon decryption of $C_1' + C_2$, the resulting plaintext will be correctly padded. And we want to show that then $P2'[16] = \mathtt{01}$. We can show this by contraposition: with our construction of $C_1'$ as 15 random bytes followed by $\mathtt{00}$, if $P_2'[16] ...


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Actually, we don't care about the amount of padding the original message had; we care whether the modified plaintext (that is, the result of the decryption of the modified ciphertext) has good padding or not. The padding will be valid if the last block had one of these patterns: XX XX XX XX XX XX XX 01 XX XX XX XX XX XX 02 02 XX XX XX XX XX 03 ...


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Which MAC algorithm is faster - CBC based MAC's or HMAC - depends completely on which ciphers and hashes are used. Furthermore, it depends on the runtime environment that contains the hash and cipher implementations. With regard to the leading CPU architecture for PC's, there are the Intel whitepapers. Both AES and SHA-2 performance can be enhanced using ...



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