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The PC-2 substitution transforms the 56-bit concatenation of the 28-bit $C$ and $D$ after they have been appropriately rotated, into the 48-bit subkey $K_i$ for round $i$, before that is combined using XOR with the 48-bit output of expansion $E$ and divided into 6-bit entries for each of the 8 S-tables. The table defining PC-2 is correspondingly organized as ...


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Remember that the initial value is split into two halves, and each half is shifted independently. If all the bits in each half are either 0 or 1, then the key used for any cycle of the algorithm is the same for all the cycles of the algorithm. This can occur if the key is entirely 1s, entirely 0s, or if one half of the key is entirely 1s or the other half is ...


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I cannot think of how to attack this using exactly the same resources as the classic meet-in-the-middle attack against double-DES, but there is a way to solve it with similar computational and memory resources (i.e. with about $2^{57}$ time and memory), but using $2^{56}$ chosen plaintexts and $2^{56}$ (adaptive) chosen ciphertexts. First, notice that if we ...


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Let $m$ and $m'$ are two distinct plaint text. We have: $E_k(m)=DES_{k1}(DES_{k2}(m)) \oplus k3$ $E_k(m')=DES_{k1}(DES_{k2}(m')) \oplus k3$ So $E_k(m) \oplus E_k(m')=DES_{k1}(DES_{k2}(m))\oplus DES_{k1}(DES_{k2}(m'))$ and this means that $k_3$ have not affect the security of this system. So security of this system is same as double $DES$. Also for ...


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Permuted Choice 1 selects 28 bits for C and 28 bits for D from the input 64 bits. The remaining 8 bits are odd byte parity. From FIPS 46-3, Data Encryption Standard (DES) (withdrawn May 19, 2005) We see Permuted Choice 1 expressed (PDF page 24): Reading the text associated with the table you'll find there are two groups of 28 input values each ...


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The parity bits act as error checking. In effect there is only 56 bits being used for entropy. Also This question has already been answered. See similar question here or here


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Each 56-bit key has a unique 8-bit parity value. For this reason there are only $2^{56}$ keys.


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If you're talking about the S-DES developed by Professor Edward Schaefer of Santa Clara University, I can try to explain you the reason. The algorithm takes in input a 10-bit key (K) but, using a key generation algorithm, the plaintext is encrypted using two different subkeys (K1 and K2) that are generated by the algorithm. First, you should pass the ...



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