Answer to Question #249616 in Algorithms for john

Question

What is the output of the first round of the DES algorithm when the plaintext

and the key are both all ones?

Answer 1

The 64-bit input is x₀=00...0 (64-zeroes). The initial permutation has no effect.

Hence L₀=00...0 (32-zeroes) and R₀=00...0 (32-zeroes). Applying the key schedule which is a fixed permutation on the input bits of the key yields the round key K₁= (00..0) (48-zeroes).

The round computes R₁=L₀XOR f(R₀, K₁)

The f-function

First expands R₀ into 48-bit long bitstring using a fixed permuted expansion rule.

Since only permutations and repetitions are used this will yield a 48-bit 0 string.

The result is XORed with K₁, which produces a 48-bit zero string.

The 48-bit 0 string is divided into eight 6-bit chunks and the ith chunk is transformed under the rule specified in the S_i box. 000000 is mapped 8 times with boxes S_i i=1,..,8 and produces the following sequence of 4-bit values: 14, 15, 10, 7, 2, 12, 4, 13. In binary we obtain the following sequence:

1110 1111 1010 0111 0010 1100 0100 1101

Finally the bit-string is permuted according to the P table to the following:

1101 1000 1101 1000 1101 1011 1011 1100

This is the result of f(R₀),K₁)

The right half R₁=L₀XOR f(R₀),K₁) is simply the output of f(R₀),K₁) since the L₀ is zero.

L₁= R₀= (00..0) (32-zeroes). Concatenating both yields the following 64-bit string, which is the output of round 1.

0000 0000 0000 0000 0000 0000 0000 0000 1101 1000 1101 1000 1101 1011 1011 1100