In this problem we will work through a round of DES.

For notational simplicity, assume it is the first round.

Please enter all answers as strings of 0's and 1's.

The input is the 64 bit block

00000000000000000010000000000000000000000000000101001010000000000000000000000000001000000000000000000000000000010100101000000000

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

000000001000000000000000000000000000000000000000000000001000000000000000000000000000000000000000

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

Now find L0L0 and R0R0, the left and right halves of the state. L0=L0=

R0=R0=

What is the result of applying the expansion box to R0R0?

E(R0)=E(R0)=

What is the result of XORing the subkey with E(R0)E(R0)?

k1E(R0)=k1E(R0)=

We now apply the S-box transformation.

S(k1E(R0))=S(k1E(R0))=

Finally we apply the permutation box to complete the function ff.

f(R0)=P(S(k1E(R0)))=f(R0)=P(S(k1E(R0)))=

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

L1=L1=

R1=R1=

Entered Answer Preview 00000000001001 00000000000000000000010000 0001000000010010 0000010000 0000000010000 OOOOOO00001000000010000000 0000000010000000000 00000000000000 Atletne of the answers above is NOT correct (1 port this question concem the DES IP box Please enter an answers as strings of O's and 1 Consider the 64 DE DES block below 000000000000000000000000000000011010010000000000000000000000000 Suppose that these the insulinput 10 DES What will the block look like after the di permutation (Pyss applied? 000100000001 000000000100 it we apply the Pransformation to the above what will the resut be? 00000000 10000010000000 1000004