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1

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 ...


0

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


4

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



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