Module 2 - Case

Data Encryption Standards

Assignment Overview

Data Encryption Standards: DES and Triple DES (Part I)

DES Encryption

The overall scheme for DES encryption is illustrated in Figure 1. As with any encryption scheme, there are two inputs to the encryption function: the plaintext to be encrypted and the key. In this case, the plaintext must be 64 bits in length and the key is 56 bits in length.

Figure 1. General Depiction of DES Encryption Algorithm

Looking at the left-hand side of the figure, we can see that the processing of the plaintext proceeds in three phases. First, the 64-bit plaintext passes through an initial permutation (IP) that rearranges the bits to produce the permuted input. This is followed by a phase consisting of sixteen rounds of the same function, which involves both permutation and substitution functions. The output of the last (sixteenth) round consists of 64 bits that are a function of the input plaintext and the key. The left and right halves of the output are swapped to produce the preoutput. Finally, the preoutput is passed through a permutation [IP -1] that is the inverse of the initial permutation function, to produce the 64-bit ciphertext.

The right-hand portion of Figure 1 shows the way in which the 56-bit key is used. Initially, the key is passed through a permutation function. Then, for each of the sixteen rounds, a subkey (Ki) is produced by the combination of a left circular shift and a permutation. The permutation function is the same for each round, but a different subkey is produced because of the repeated shifts of the key bits.

DES works on bits, or binary numbers -- the 0s and 1s common to digital computers. Each group of four bits makes up a hexadecimal, or base 16, number. Binary "0001" is equal to the hexadecimal number "1", binary "1000" is equal to the hexadecimal number "8", "1001" is equal to the hexadecimal number "9", "1010" is equal to the hexadecimal number "A", and "1111" is equal to the hexadecimal number "F".

DES works by encrypting groups of 64 message bits, which is the same as 16 hexadecimal numbers. To do the encryption, DES uses "keys" which are also apparently 16 hexadecimal numbers long, or apparently 64 bits long. However, every 8th key bit is ignored in the DES algorithm, so that the effective key size is 56 bits. But, in any case, 64 bits (16 hexadecimal digits) is the round number upon which DES is organized.

For example, if we take the plaintext message "8787878787878787", and encrypt it with the DES key "0E329232EA6D0D73", we end up with the ciphertext "0000000000000000". If the ciphertext is decrypted with the same secret DES key "0E329232EA6D0D73", the result is the original plaintext "8787878787878787".

A DES Example

We now work through an example and consider some of its implications. Although you are not expected to duplicate the example by hand, you will find it informative to study the hex patterns that occur from one step to the next.

DES is a block cipher -- meaning it operates on plaintext blocks of a given size (64-bits) and returns ciphertext blocks of the same size. Thus DES results in a permutation among the 2^64 (read this as: "2 to the 64th power") possible arrangements of 64 bits, each of which may be either 0 or 1. Each block of 64 bits is divided into two blocks of 32 bits each, a left half block L and a right half R. (This division is only used in certain operations.)

Example: Let M be the plain text message M = 0123456789ABCDEF, where M is in hexadecimal (base 16) format. Rewriting M in binary format, we get the 64-bit block of text:

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

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.

Example: Let K be the hexadecimal 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

The DES algorithm uses the following steps:

Step 1: Create 16 subkeys, each of which is 48-bits long.

The 64-bit key is permuted according to the following table, PC-1. Since the first entry in the table is "57", this means that the 57th bit of the original key K becomes the first bit of the permuted key K+. The 49th bit of the original key becomes the second bit of the permuted key. The 4th bit of the original key is the last bit of the permuted key. Note only 56 bits of the original key appear in the permuted key.

                            PC-1

 

              57   49    41   33    25    17    9

               1   58    50   42    34    26   18

              10    2    59   51    43    35   27

              19   11     3   60    52    44   36

              63   55    47   39    31    23   15

               7   62    54   46    38    30   22

              14    6    61   53    45    37   29

              21   13     5   28    20    12    4

Example: From the original 64-bit key

K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001

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.

Example: 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 C_n and D_n, 1<=n<=16. Each pair of blocks C_n and D_n is formed from the previous pair C_n-1 and D_n-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.

Example: From original 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 K_n, for 1<=n<=16, by applying the following permutation table to each of the concatenated pairs C_nD_n. Each pair has 56 bits, but PC-2 only uses 48 of these.

                              PC-2

 

                 14    17   11    24     1    5

                  3    28   15     6    21   10

                 23    19   12     4    26    8

                 16     7   27    20    13    2

                 41    52   31    37    47   55

                 30    40   51    45    33   48

                 44    49   39    56    34   53

                 46    42   50    36    29   32

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

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

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

K₁ = 000110 110000 001011 101111 111111 000111 000001 110010

For the other keys we have

K₂ = 011110 011010 111011 011001 110110 111100 100111 100101
K₃ = 010101 011111 110010 001010 010000 101100 111110 011001
K₄ = 011100 101010 110111 010110 110110 110011 010100 011101
K₅ = 011111 001110 110000 000111 111010 110101 001110 101000
K₆ = 011000 111010 010100 111110 010100 000111 101100 101111
K₇ = 111011 001000 010010 110111 111101 100001 100010 111100
K₈ = 111101 111000 101000 111010 110000 010011 101111 111011
K₉ = 111000 001101 101111 101011 111011 011110 011110 000001
K₁₀ = 101100 011111 001101 000111 101110 100100 011001 001111
K₁₁ = 001000 010101 111111 010011 110111 101101 001110 000110
K₁₂ = 011101 010111 000111 110101 100101 000110 011111 101001
K₁₃ = 100101 111100 010111 010001 111110 101011 101001 000001
K₁₄ = 010111 110100 001110 110111 111100 101110 011100 111010
K₁₅ = 101111 111001 000110 001101 001111 010011 111100 001010
K₁₆ = 110010 110011 110110 001011 000011 100001 011111 110101

So much for the subkeys. We will continue look at the message itself in Module 2 SLP.

To learn more about DES, check the following sites:

1. DES – the algorithm
   https://www.youtube.com/watch?v=3BZRBfhpIb0
2. Simplified DES examples
   https://www.youtube.com/watch?v=qHZKze24kVo
3. Data Encryption Standard
   https://www.youtube.com/watch?v=kPBJIhpcZgE
4. Data Encryption Standard
   https://www.youtube.com/watch?v=kg4pWTqVfeU
5. Data Encryption Standard (DES)
   https://www.lri.fr/~fmartignon/documenti/systemesecurite/4-DES.pdf
6. DES
   https://www.youtube.com/watch?v=eCAHcfA-2c8
7. Block Ciphers and Data Encryption Standard (DES)
   https://engineering.purdue.edu/kak/compsec/NewLectures/Lecture3.pdf

Case Assignment

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 (the key and the plaintext are the same), namely:

The plaintext and key is:

     0000 0001 0010 0011 0100 0101 0110 0111

     1000 1001 1010 1011 1100 1101 1110 1111

1. Derive K₁, the first-round subkey. Please show step-by-step how you derive K₁ (HINT: First, pass the 64-bit input through PC-1 to produce a 56-bit result. Then perform a left circular shift separately on the two 28-bit halves, C₀ and D₀. Finally, pass the 56-bit result through PC-2 to produce the 48-bit K₁).
2. Derive L₀ and R₀ (HINT: follow the Step 2: Encode each 64-bit block of data).

Assignment Expectations

Use information from the modular background readings as well as the given resources. Also, you could use any good quality resource you can find. Please cite all sources and provide a reference list at the end of your paper.

The following items will be assessed in particular:

1. Your ability to consolidate ideas from reading materials and your understanding of the materials.
2. Your ability to write a report with strong argument.
3. Some in-text references to modular background readings.