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3 of 4 Fix to the key search in both lists.

Decrypt the ciphertext with every possible key and store the result: $$2^{56}$$ decryptions. Now encrypt the (known) plaintext of the ciphertext with every possible key: $$2^{56}$$ encryptions. Now you have to check every entry, which is in both lists and try it with another plaintext-ciphertext pair. If you can successfully decrypt that, you are very likely to have found the correct key. All in all $$2^{56} + 2^{56} = 2^{57}$$ operations, not $$2^{112}$$. Just the normal Meet-in-the-middle attack.

 answered Apr 18 '15 at 14:19 Nova 3,74011 gold badge1212 silver badges2222 bronze badges