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I'll answer in order: Output size = input size That's correct, GCM uses CTR internally. It encrypts a counter value for each block, but it only uses as many bits as required from the last block. CTR turns the block cipher into a stream cipher. IV of any size For GCM a 12 byte IV is strongly suggested as other IV lengths will require additional ...


3

Yes, it appears that it can be solved in practical time in $GF(2^n)$, if the attacker gets $n+\epsilon$ random $a_i$ values, even if he gets a single bit of the $a_i \times k$ values. The chief observation is that the mapping from $a_i$ to bit $j$ of $a_i \times k$ (which I'll refer to as $bit_j(a_i \times k)$) is bitwise linear (for constant $j$, $k$). ...


1

Thanks @poncho for providing a correct answer. I investigated it deeply, viewing it as a linear algebra problem. Here's what I obtained: in $GF(2^n)$, the series of equations $r_i=a_i\times k$ can be written as $R = K \cdot A$ where: $A$ is a known $n \times n$ matrix, where each column is a bit representation of $n$, linearly independent, $a_i$ $R$ is a ...


1

GCM is sometimes called a 1.5 pass AEAD cipher, where the CTR encryption counts for 1 and the GMAC counts for 0.5. So you would indeed expect it to be faster than encryption + CMAC and HMAC with regards to the amount of CPU instructions. That is: as long as the encryption is using AES for both solutions. GCM requires a 128 bit block cipher while CMAC and ...


1

Reusing an IV once opens you up to someone finding the XOR of those two plaintext, seriously compromising their confidentiality. Moreover, with GCM, a single IV reuse leaks significant information about the key used for authentication; if there are even a few pairs of reused IVs (not even one IV used many times; a few IVs each of which are used twice is ...



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