$(1$ point$)$ This question concerns the DES IP box.

Please enter all answers as strings of $0\text{'}$s and $1\text{'}$s$.$

Consider the $64$ bit DES block below.

$0000000000000000000000000001000100010000001000000000000001000000$

Suppose that this is the initial input to DES. What will the block look like after the initial permutation $($IP$)$ is applied?

If we anolv the $I{P}^{-1}$ transformation to the above. what will the result be$?$For notational simplicity, assume it is the first round.

Please enter all answers as strings of $0\text{'}$s and $1\text{'}$s$.$

The input is the $64$ bit block

$0001000000000000000000000000010000000000001000000000000000000000$

Suppose that the subkey for the current round is this $48$ bit number:

$000010000000000000000000000000000000000000000000$

What does the $64$ bit state look like after the IP transformation is applied to the input?

Now find $L_0$ and $R_0,$ the left and right halves of the state.

$L_0=$

$R_0=$

What is the result of applying the expansion box to $R_0?$

What is the result of XORing the subkey with $E(R_0)?$

$k_{1}o+E(R_0)=$

We now apply the S$-$box transformation.

$S(k_{1}o+E(R_0))=$

Finally we apply the permutation box to complete the function $f.$

$f(R_0)=P(S(k_{1}o+E(R_0)))=$

We can now compute the state of DES going into the next round.

$L_1=$

$R_1=$