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Perform DES encryption on the first 64-bit plaintext only, what are these 0's and 1's? Do they mean that I need to use this value 111101 to derive $K$?

I want to know how to use initial permutation to derive $K$ using permuted choice 1 and permute choice 2.

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    $\begingroup$ It is absolutely unclear what you're asking. $\endgroup$ – Maeher Mar 24 at 7:03
  • $\begingroup$ Welcome to crypto.SE. Exactly what part fo the specification of DES do you have a problem with? $\endgroup$ – fgrieu Mar 24 at 7:41
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I want to know how to use initial permutation to derive $K$ using permuted choice 1 and permute choice 2

You start with the DES key $K$, usually expressed as 8 bytes. That's 64 bits. DES number these bits from 1 to 64, starting with high-order bit on the first byte in reading order, moving to lower-order bit, then after the 8th bit of a byte moving to the next byte in reading order, so that bit 42 is the second-highest-order bit of to the 6th byte in reading order, testable by binary AND of that byte with 40h.

From these bits 64 bits of $K$ it is extracted two 28-bit quantities $C_0$ and $D_0$ by picking bits per table $PC1$. The top (resp. bottom) of this table gives the bit numbers in $K$ of the bits forming $C_0$ (resp. $D_0$). You'll notice that $PC1$ contains no multiple of 8, meaning that the low-order bit of each byte of $K$ is ignored. This deliberately weakens DES.

Bits of $C_0$ are numbered 1 to 28, and bits of $D_0$ are numbered 29 to 56, in reading order of $PC1$.

For the first round of encryption, from $C_0\mathbin\|D_0$ it is extracted eight 6-bit quantities (the subkeys 1 to 8 for the first round) by picking bits per table $PC2$, which lines gives the bit numbers in $C_0\mathbin\|D_0$ of the bits forming each 6-bit quantities. Eight bits of $C_0\mathbin\|D_0$ are left aside, but will be used in the next round, after some rotation of $C_0$ (resp. $D_0$) yielding $C_1$ (resp. $D_1$).

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