Answer to Question #290169 in Computer Networks for raj

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

will assume that string X will be all ones of 64bits while K will be all ones of 56bits.

Having L0 = ones (32), and R0 = ones (32) after the initial permutation of X. Since, there is no effect of the initial permutation.

And after the fixed permutation of X, having round key K1 as ones (48).

The value of R1 through first round will be obtained using Feistel function represented by f:

R1 = L0 XOR f (R0, K1)

• As per the fixed permutation expansion rule, expand R0 to 48-bit string.
• Since, all the bits in the string R0 are 1, therefore, after expansion also the string R0 will contain all ones.
• This expanded R0 string will be XORed with K1.
• Since, both the key K1, and the string R0 are all 1’s therefore, the XOR operation results in a string of zeroes of 48 bits.
• After this, the resultant string of 48 bits is divided into the 6 bits chunks. The ith bit will be transformed as per the rule of Si box. 000000 is being mapped 8 times with the Si box where i = 1, …, 8.
• And this will produce the sequence of 14, 15, 10, 7, 2, 12, 4, 13.

Thus now, the binary representation of the above sequence will be as:

1110 1111 1010 0111 0010 1100 0100 1101

Hence these bits is being permutated as per the P table:

1101 1000 1101 1000 1101 1011 1011 1100

The result of the function f ((R0), K1) will be:

1101 1000 1101 1000 1101 1011 1011 1100

Now, R1 = L0 XOR f (R0, K1)

L0 = 1111 1111 1111 1111 1111 1111 1111 1111

f ((R0), K1) = 1101 1000 1101 1000 1101 1011 1011 1100

  Hence, R1 = 0010 0111 0010 0111 0010 0100 0100 0011

The final result of the Round 1 = L1 + R1

L1 = R0 = (11 ... 1) (These are 32 ones).

After combining both the bits, the resultant bits are 64 in number and this is the result of round 1.

1111 1111 1111 1111 1111 1111 1111 1111 0010 0111 0010 0111 0010 0100 0100 0011