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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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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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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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