Exercise 4 Solution

　

1. This problem provides a numerical example of encryption using a one-round version of DES. We start with the same bit pattern for the key and the plaintext, namely,

In hexadecimal notation: 0 1 2 3 4 5 6 7 8 9 A B C D E F

In binary notation: 0000 0001 0010 0011 0100 0101 0110 0111

                               1000 1001 1010 1011 0100 1101 1110 1111

　

a. Derive K₁, the first-round subkey.

64-bit key è PC1 (64 à 56) è shift registers C&Dè PC2(56à 48)

10110000

11001100

10101010

00001010

10101100

11001111

00000000

 

è         L = 1011 0000 1100 1100 1010 1010 0000

             R = 1010 1010 1100 1100 1111 0000 0000

 

 

C₁(L) = 0110 0001 1001 1001 0101 0100 0001

D₁(R) = 0101 0101 1001 1001 1110 0000 0001

　

1-8            0110 0001

9-16          1001 1001

17-24        0101 0100

25-32        0001 0101

33-40        0101 1001

41-48        1110 0000

49-56        0000 0001

K₁ = 00000011  00000010  01100111   10010011   00000001   10010101

 

b. Derive L₀, R₀.

(Initial permutation)

L₀ = 11001100 00000000 11001100 11111111

R₀ = 10110000 10101010 11110000 10101010

 

c. Expand R₀ to get E[R₀].

 

 

E[R₀] = 010110100001 010101010101 011110100001 010101010101

 

d. Calculate A = E[R₀]ÅK₁.

 

E[R₀] = 010110100001 010101010101 011110100001 010101010101

K₁    = 000000110000 001001100111 100100110000 000110010101

           010110010001 011100110010 111010010001 010011000000

 

 

e. Group the 48-bit result of (d) into sets of 6 bits and evaluate the corresponding S-box substitutions.

S₁(010110)= 1100

S₂(010001)= 1100

S₃(011100)=0100

S₄(110010)= 0001

S₅(111010)= 0011

S₆(010001) = 0110

S₇(010011) = 0011

S₈(000000) = 1101

f. Concatenate the results of (e) to get a 32-bit result, B.

 

B = 1100 1100 0100 0001 0011 0110 0011 1101

 

g. Apply the permutation to get P(B).

P(B) = 10101010 10101001 10001100 10111000

 

h. Calculate R₁ = P(B)ÅL₀

 

R₁ = P(B)Å L₀

 

P(B) = 1010 1010 1010 1001 1000 1100 1011 1000

L₀    = 1100 1100 0000 0000 1100 1100 1111 1111

        0110 0110 1010 1001 0100 0000 0100 0111

 

R₁ =0110 0110 1010 1001 0100 0000 0100 0111

 

i. Write down the ciphertext.

L₁ = R₀ = 10110000 10101010 11110000 10101010

 

Ciphertext = IP^-1(R₁L₁) = 0001 0001 0110 0011 0100 0001 0011 0010

                                        1000 1000 1111 1010 0100 1101 1011 1010

　

　

2. Show that in DES the first 24 bits of each subkey come from the same subset of 28 bits of the initial key and that the second 24 bits of each subkey come from a disjoint subset of 28 bits of the initial key.

 

 

For PC1:

 

C₁ comes from the red half of the bits, and D₁ comes from the blue half of the bits. For all subkeys K_i, the first 24 bits are come from C_i, and second 24 bits are come from D_i, since PC2 takes C_i to generate the first 24 bits and D_i to generate the second 24 bits. There for, first 24 bits of each subkey come from the same subset of 28 bits of the initial key (in red) and the second 24 bits of each subkey come from a disjoint subset of 28 bits of the initial key (in blue).