This problem provides a numerical example of encryption using a one-round versio

Question

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 K and the plaintext, namely: Hexadecimal notation: 0123456789ABCDEF Binary notation: 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 a. Derive Ki, the first-round subkey b. Derive Lo, Ro c. Expand Ro to get E[Rol, where El is the expansion function of Table S.1. d. Calculate A= El Rol??, e. Group the 48-bit result of (d) into sets of 6 bits and evaluate the corresponding f. g. h. i. S-box substitutions Concatenate the results of (e) to get a 32-bit result, B Apply the permutation to get P(B) Calculate R1=P(B)?4 Write down the ciphertext. (a) Initial Permutation (IP) 20 12 32 33 27 41 17 13 15 31 (b) Inverse Initial Permutation 1P-1) 16 15 7 55 31 ? 37 13 12 21 61 28 51 18 41 (c) Expansion Permutation (E) 12 20 17 14 16121 31 (d) Permutation Function (P 12 18 16 8 23 13 10 14 19

Explanation / Answer

Let M be the plain text message M = 0123456789ABCDEF (HexaDecimal).

Writing M in binary we get,

M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 (64bits)
L = 0000 0001 0010 0011 0100 0101 0110 0111 (32bits)
R = 1000 1001 1010 1011 1100 1101 1110 1111 (32bits)

The first bit of M is "0". The last bit is "1". We read from left to right.

DES operates on the 64-bit blocks using key sizes of 56- bits. The keys are actually stored as being 64 bits long, but every 8th bit in the key is not used (i.e. bits numbered 8, 16, 24, 32, 40, 48, 56, and 64). However, we will nevertheless number the bits from 1 to 64, going left to right, in the following calculations. But, as you will see, the eight bits just mentioned get eliminated when we create subkeys.

Since the key is not given assume the key K = 133457799BBCDFF1. This gives us as the binary key (setting 1 = 0001, 3 = 0011, etc., and grouping together every eight bits, of which the last one in each group will be unused):

K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001 (64bit)

From the original 54 bit key,, we get56 bit permutation

we get the 56-bit permutation

K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111

Next, split this key into left and right halves, C0 and D0, where each half has 28 bits.

From the permuted key K+, we get

C0 = 1111000 0110011 0010101 0101111
D0 = 0101010 1011001 1001111 0001111

With C0 and D0 defined, we now create sixteen blocks Cn and Dn, 1<=n<=16. Each pair of blocks Cn and Dn is formed from the previous pair Cn-1 and Dn-1, respectively, for n = 1, 2, ..., 16, using the following schedule of "left shifts" of the previous block. To do a left shift, move each bit one place to the left, except for the first bit, which is cycled to the end of the block.

                     Iteration     Number of

                      Number      Left Shifts

                          1          1

                          2          1

                          3          2

                          4          2

                          5          2

                          6          2

                          7          2

                          8          2

                          9          1

                         10          2

                         11          2

                         12          2

                         13          2

                         14          2

                         15          2

                         16          1

This means, for example, C3 and D3 are obtained from C2 and D2, respectively, by two left shifts, and C16 and D16 are obtained from C15 and D15, respectively, by one left shift. In all cases, by a single left shift is meant a rotation of the bits one place to the left, so that after one left shift the bits in the 28 positions are the bits that were previously in positions 2, 3,..., 28, 1.

From original pair pair C0 and D0 we obtain:

C0 = 1111000011001100101010101111
D0 = 0101010101100110011110001111

C1 = 1110000110011001010101011111
D1 = 1010101011001100111100011110

C2 = 1100001100110010101010111111
D2 = 0101010110011001111000111101

C3 = 0000110011001010101011111111
D3 = 0101011001100111100011110101

C4 = 0011001100101010101111111100
D4 = 0101100110011110001111010101

C5 = 1100110010101010111111110000
D5 = 0110011001111000111101010101

C6 = 0011001010101011111111000011
D6 = 1001100111100011110101010101

C7 = 1100101010101111111100001100
D7 = 0110011110001111010101010110

C8 = 0010101010111111110000110011
D8 = 1001111000111101010101011001

C9 = 0101010101111111100001100110
D9 = 0011110001111010101010110011

C10 = 0101010111111110000110011001
D10 = 1111000111101010101011001100

C11 = 0101011111111000011001100101
D11 = 1100011110101010101100110011

C12 = 0101111111100001100110010101
D12 = 0001111010101010110011001111

C13 = 0111111110000110011001010101
D13 = 0111101010101011001100111100

C14 = 1111111000011001100101010101
D14 = 1110101010101100110011110001

C15 = 1111100001100110010101010111
D15 = 1010101010110011001111000111

C16 = 1111000011001100101010101111
D16 = 0101010101100110011110001111

We now form the keys Kn, for 1<=n<=16, by applying the following permutation table to each of the concatenated pairs CnDn. Each pair has 56 bits, but PC-2 only uses 48 of these.

Therefore, the first bit of Kn is the 14th bit of CnDn, the second bit the 17th, and so on, ending with the 48th bit of Kn being the 32th bit of CnDn.

For the first key we have C1D1 = 1110000 1100110 0101010 1011111 1010101 0110011 0011110 0011110

which, after we apply the permutation PC-2, becomes

(a)

K1 = 000110 110000 001011 101111 111111 000111 000001 110010

(b)

Applying the initial permutation to the block of text M, given previously, we get

M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
IP = 1100 1100 0000 0000 1100 1100 1111 1111 1111 0000 1010 1010 1111 0000 1010 1010

Here the 58th bit of M is "1", which becomes the first bit of IP. The 50th bit of M is "1", which becomes the second bit of IP. The 7th bit of M is "0", which becomes the last bit of IP.

Next divide the permuted block IP into a left half L0 of 32 bits, and a right half R0 of 32 bits.

From IP, we get L0 and R0

L0 = 1100 1100 0000 0000 1100 1100 1111 1111
R0 = 1111 0000 1010 1010 1111 0000 1010 1010

(c)

We calculate E(R0) from R0 as follows:

R0 = 1111 0000 1010 1010 1111 0000 1010 1010
E(R0) = 011110 100001 010101 010101 011110 100001 010101 010101

(d)

(Note that each block of 4 original bits has been expanded to a block of 6 output bits.)

Next in the f calculation, we XOR the output E(Rn-1) with the key Kn:

Kn + E(Rn-1).

For K1 , E(R0), we have

K1 = 000110 110000 001011 101111 111111 000111 000001 110010
E(R0) = 011110 100001 010101 010101 011110 100001 010101 010101
K1+E(R0) = 011000 010001 011110 111010 100001 100110 010100 100111

(e)

Write the previous result, which is 48 bits, in the form:

Kn + E(Rn-1) =B1B2B3B4B5B6B7B8,

where each Bi is a group of six bits. We now calculate

S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8)

where Si(Bi) referres to the output of the i-th S box.

To repeat, each of the functions S1, S2,..., S8, takes a 6-bit block as input and yields a 4-bit block as output.

(g) and (h)

The final stage in the calculation of f is to do a permutation P of the S-box output to obtain the final value of f:

f = P(S1(B1)S2(B2)...S8(B8))

The permutation P is defined in the following table. P yields a 32-bit output from a 32-bit input by permuting the bits of the input block.

From the output of the eight S boxes:

S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8) = 0101 1100 1000 0010 1011 0101 1001 0111

we get

f = 0010 0011 0100 1010 1010 1001 1011 1011

R1 = L0 + f(R0 , K1 )

= 1100 1100 0000 0000 1100 1100 1111 1111
+ 0010 0011 0100 1010 1010 1001 1011 1011
= 1110 1111 0100 1010 0110 0101 0100 0100

(i)

we will have L2 = R1, which is the block we just calculated, and then we must calculate R2 =L1 + f(R1, K2), and so on for 16 rounds. At the end of the sixteenth round we have the blocks L16 and R16. We then reverse the order of the two blocks into the 64-bit block

R16L16

If we process all 16 blocks using the method defined previously, we get, on the 16th round,

L16 = 0100 0011 0100 0010 0011 0010 0011 0100
R16 = 0000 1010 0100 1100 1101 1001 1001 0101

We reverse the order of these two blocks and apply the final permutation to

R16L16 = 00001010 01001100 11011001 10010101 01000011 01000010 00110010 00110100

IP-1 = 10000101 11101000 00010011 01010100 00001111 00001010 10110100 00000101

which in hexadecimal format is

85E813540F0AB405.

This is the encrypted form of M = 0123456789ABCDEF: namely, C = 85E813540F0AB405.

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