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Indeed $H=E_k(0)$ is used to choose from the family. This is not a problem, and here is some intuition on why. The output of the hash function is not leaked in clear, it is "hidden" by xoring with $E_k(iv,ctr=0)$ which is different per each encrypted message (in contrast to $H=E_k(0)$). Otherwise it would be indeed trivial to recover H as the UHF ...


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I've used Algorithm 3 of the paper for this purpose. $M[128] ← e1||0^{120} $ instead of H $P ← 04||0^{120} $ which is 2 with reverse bits


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The polynomial f would be 0xe1, no? it is $e1 \mathbin \|0^{120}$ i.e. 0xe1000000000000000000000000000000 What would be the value of the element P, if it corresponds to the polynomial α? it is 2 reversed i.e. 0x40000000000000000000000000000000


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