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Well, there's a fairly obvious CPA attack that uses circa $2^{34}$ or so chosen plaintexts, and about $2^{120}$ trial DES operations, and so cryptographical strength is not greatly better than standard 3DES. As for code size and memory size, it'd certainly be larger than standard 3DES (you do more, and have more keying material lying around), but it's hard ...


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and so $E_k((IV << 8) | C_1 | C_2')$ gives an incorrect encryption That's your issue; CFB isn't designed to correct for errors in the plaintext - after all, if the ciphertext is not modified, then the decryptor will get precisely what the encryptor encrypted and so if the original plaintext had an error, and then was correct, what the decryptor will ...


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In MitM for 2DES, in a tabulation phase we compute and keep $2^{56}$ 64-bit values and their associated key, then in a search phase we compute up to $2^{56}$ 64-bit values and search these in the table. There's a hit about once in $2^{64-56}=2^8=256$ searches, that is about $2^{56-8}=2^{48}$ hits, and all except one are false hits. We need to eliminate false ...


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The DES round keys are just certain secret key bits (48 of them for each round). Say $K_1,\ldots,K_{16}$ are the 48 bit round keys with $K_i$ the $i^{th}$ round subkey. Then, the EDE triple DES uses round keys in the following order $$K^{(1)}_1,K^{(1)}_2,\ldots,K^{(1)}_{16}$$ which are the subkeys of the first (E) followed by (D) $$K^{(2)}_{16},K^{(2)}_{...


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DES key schedule takes the 64-bit key and PC-1 (Permuted Choice 1) discards the parity check bits and applies a permutation to the remaining. Then for the rest they key bits are divided into 28-bit halves and for the each round They are rotated left one or two specified for each round. Then 48-bit bits are selected by PC-2 (Permuted Choice 2) which is ...


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