Question

What is the output of first round of DES algorithm when plain text and key both are all ones

Answer 1

Because all inputs are 1s and IP, PC−1, PC−2 and the subkey rotations are all simple permutations, we can discount them knowing that they will output all 1s.

Because the subkeys consist of all 1s, we can discount them from the f

function calculation by inverting the S-box input (since a⊕1=a$^{-1}$ where a$^{-1}$

denotes a bitwise NOT operation). We can also discount the E expansion since it’s guaranteed to produce all 1s.

$L_0=FFFFFFFF_{16}$

$L_1=R_0=FFFFFFFF_{16}$

$R_1$ is calculated as follows:

$R_1=L_0 \oplus f(R_0)$ $=FFFFFFFF_{16} \oplus f(FFFFFF_{16})=(f(FFFFFFFF_{16}))^{-1}$

The input to the S-boxes is the all-1s E output inverted by the all-1s key, the f function is calculated as shown;

$f(FFFFFFFF_{16})=P(S_1(0),S_2(0),S_3(0),S_4(0),S_5(0),S_6(0),S_7(0),S_8(0)$

We will call the intermediate value before P is applied R′$_1$ and after applying all the S-boxes, it is as follows:

$11101111101001110010110001001101_1$

To finish calculating R$_1$ we need to apply P(R′$_1$ ) and then invert the result:

$P(R^{'}_1)=11011000110110001101101110111100_2$

$(P(R^{'}_1))^{-1}=00100111001001110010010001000011_2$

Therefore: $L_1=FFFFFFFF_{16}$

$R_1=27272443_{16}$