Recall that in regular DES, 8 bits of the initial 64-bit key are thrown away, so that the actual encryption only proceeds with the remaining 56 bits and so that an attacker Oscar really only needs those 56 bits to reveal the plaintext. (That is, even though the true keyspace has size 2^64, the keyspace for the keys actually used has size 2^56.) Discuss then whether beginning with a 64-bit key has any merit and if Oscar can simply skip the PC-1 step of DES entirely in carrying out a brute force attack.

Question
Recall that in regular DES, 8 bits of the initial 64-bit key are thrown away, so that
the actual encryption only proceeds with the remaining 56 bits and so that an
attacker Oscar really only needs those 56 bits to reveal the plaintext. (That is,
even though the true keyspace has size 2^64, the keyspace for the keys actually
used has size 2^56.) Discuss then whether beginning with a 64-bit key has any
merit and if Oscar can simply skip the PC-1 step of DES entirely in carrying out a
brute force attack.
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Transcribed Image Text

Recall that in regular DES, 8 bits of the initial 64-bit key are thrown away, so that the actual encryption only proceeds with the remaining 56 bits and so that an attacker Oscar really only needs those 56 bits to reveal the plaintext. (That is, even though the true keyspace has size 2^64, the keyspace for the keys actually used has size 2^56.) Discuss then whether beginning with a 64-bit key has any merit and if Oscar can simply skip the PC-1 step of DES entirely in carrying out a brute force attack.

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