Tag Info

New answers tagged


You can do this slightly better with an additional $\mathcal{0}(2^{56})$ memory and with $\mathcal{0}(2^{56})$ time. You can notice that the relation $c \leftarrow E_{k_1}(E_{k_2}(m))$ can be rewritten as $D_{k_1}(c) = E_{k_2}(m)$ (just apply the decrypt function on both sides. First step consists in the generation of every pair $(k_2, E_{k_2}(m))$ and ...


No. That would amount to cracking AES, which is not feasible.


The simple answer is no, even if one can choose "the original unencrypted (cleartext) file".

Top 50 recent answers are included