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I will assume you generally understand the mechanics of the padding oracle attack. Say we have some ciphertext block $c_i = c_{i-1} \oplus E(k,m_i)$ and we want to find the last byte of $m_i$. Let $b$ denote the last byte of $m_i$, which we are trying to find. Consider ciphertexts of the form $(c_{i-1} \oplus \texttt{00} \cdots \texttt{00} g,c_i)$, where $g$ ...


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If I know the set of possible characters in the plaintext will only be lowercase English letters, then is there any way in which I can speed up the deciphering process? If so, then specifically what would this look like? Yes, it helps. First, remember that the padding oracle works on the fact that the server sends back only one information at all; the ...


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It is a block cipher with block size 3. Therefore for a given known-plaintext, we can build the current key's action table that has $2^3=8$ values. We will use the decryption part $$P_i = D(C_i) \oplus C_{i-1} \oplus P_{i-1}$$ with $C_{0} \oplus P_{0} = IV$ with conversion $$D(C_i) = P_i \oplus C_{i-1} \oplus P_{i-1}$$ P_i 000 001 010 011 100 101 ...


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Am I thinking correctly and will an malicious user (which has access to encryption keys) not (or very hard) be able to generate a collision in my IV-ciphertext combination so that the server is grouping unequal messages together? If the server checks all the bytes in the IV-ciphertext (and declares a collision only every byte for two encrypted messages are ...


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