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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A: The answer is given in the below step

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A: The answer is below;:-

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A: I Have answered this question in step 2.

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A: In step 2, I have provided   answer  of  above  question------------------

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A: Solution:

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Q: tradeoff

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A: the answer is given below:-

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A: As per our honor code, we are answering the first question.

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A: Refer to step 2 for the answer,

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A: Here is the solution which is mentioned below:

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A: Please upvote/like. I am providing the correct answer below.         Unlike symmetric encryption,…

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