Data encryption standard

If you want to see a DES algorithm in arduino language then look here.
This is the link to my other arduino projects.

The Data Encryption Standard (DES) algorithm, adopted by the U.S.  government in 1977, is a block cipher that transforms 64-bit data blocks is officially described in FIPS PUB 46. The data encryption standard  algorithm is used for many applications within the government and in the private sector.

This is a tutorial designed to be clear and compact, and to provide a newcomer to the data encryption standard with all the necessary information to implement it himself, without having to track down printed works or wade through C source code.
I welcome any comments.

Here’s how to do the data encryption standard, step by step:

1  Process the key.

1.1  Get a 64-bit key from the user. (Every 8th bit is considered a parity bit. For a key to have correct parity, each byte should contain an odd number of “1” bits.)

1.2  Calculate the key schedule.

1.2.1  Perform the following permutation on the 64-bit key. (The parity bits are discarded, reducing the key to 56 bits. Bit 1 of the permuted block is bit 57 of the original key, bit 2 is bit 49, and so on with bit 56 being bit 4 of the original key.)

Permuted Choice 1 (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

1.2.2  Split the permuted key into two halves. The first 28 bits are called C[0] and the last 28 bits are called D[0].

1.2.3  Calculate the 16 subkeys. Start with i = 1.

1.2.3.1  Perform one or two circular left shifts on both C[i-1] and D[i-1] to get C[i] and D[i], respectively. The number of shifts per iteration are given in the table below.

Iteration #   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
Left Shifts   1  1  2  2  2  2  2  2  1  2  2  2  2  2  2  1

1.2.3.2  Permute the concatenation C[i]D[i] as indicated below. This will yield K[i], which is 48 bits long.

Permuted Choice 2 (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

1.2.3.3  Loop back to 1.2.3.1 until K[16] has been calculated.

2  Process a 64-bit data block.
2.1  Get a 64-bit data block. If the block is shorter than 64 bits, it should be padded as appropriate for the application.

2.2  Perform the following permutation on the data block.

Initial Permutation (IP)

58 50 42 34 26 18 10  2
60 52 44 36 28 20 12  4
62 54 46 38 30 22 14  6
64 56 48 40 32 24 16  8
57 49 41 33 25 17  9  1
59 51 43 35 27 19 11  3
61 53 45 37 29 21 13  5
63 55 47 39 31 23 15  7

2.3  Split the block into two halves. The first 32 bits are called L[0], and the last 32 bits are called R[0].

2.4  Apply the 16 subkeys to the data block. Start with i = 1.

2.4.1  Expand the 32-bit R[i-1] into 48 bits according to the bit-selection function below.

Expansion (E)

32  1  2  3  4  5
4  5  6  7  8  9
8  9 10 11 12 13
12 13 14 15 16 17
16 17 18 19 20 21
20 21 22 23 24 25
24 25 26 27 28 29
28 29 30 31 32  1

2.4.2  Exclusive-or E(R[i-1]) with K[i].

2.4.3  Break E(R[i-1]) xor K[i] into eight 6-bit blocks. Bits 1-6 are B[1], bits 7-12 are B[2], and so on with bits 43-48 being B[8].

2.4.4  Substitute the values found in the S-boxes for all B[j]. Start with j = 1. All values in the S-boxes should be considered 4 bits wide.

2.4.4.1  Take the 1st and 6th bits of B[j] together as a 2-bit value  (call it m) indicating the row in S[j] to look in for the substitution.

2.4.4.2  Take the 2nd through 5th bits of B[j] together as a 4-bit value (call it n) indicating the column in S[j] to find the substitution.

2.4.4.3  Replace B[j] with S[j][m][n].

Substitution Box 1 (S[1])
15 12  8  2  4  9  1  7  5 11  3 14 10  0  6 13
14  4 13  1  2 15 11  8  3 10  6 12  5  9  0  7
0 15  7  4 14  2 13  1 10  6 12 11  9  5  3  8
4  1 14  8 13  6  2 11 15 12  9  7  3 10  5  0

S[2] 15  1  8 14  6 11  3  4  9  7  2 13 12  0  5 10
3 13  4  7 15  2  8 14 12  0  1 10  6  9 11  5
0 14  7 11 10  4 13  1  5  8 12  6  9  3  2 15
13  8 10  1  3 15  4  2 11  6  7 12  0  5 14  9

S[3] 10  0  9 14  6  3 15  5  1 13 12  7 11  4  2  8
13  7  0  9  3  4  6 10  2  8  5 14 12 11 15  1
13  6  4  9  8 15  3  0 11  1  2 12  5 10 14  7
1 10 13  0  6  9  8  7  4 15 14  3 11  5  2 12

S[4]

7 13 14  3  0  6  9 10  1  2  8  5 11 12  4 15
13  8 11  5  6 15  0  3  4  7  2 12  1 10 14  9
10  6  9  0 12 11  7 13 15  1  3 14  5  2  8  4
3 15  0  6 10  1 13  8  9  4  5 11 12  7  2 14

S[5]

2 12  4  1  7 10 11  6  8  5  3 15 13  0 14  9
14 11  2 12  4  7 13  1  5  0 15 10  3  9  8  6
4  2  1 11 10 13  7  8 15  9 12  5  6  3  0 14
11  8 12  7  1 14  2 13  6 15  0  9 10  4  5  3

S[6]

12  1 10 15  9  2  6  8  0 13  3  4 14  7  5 11
10 15  4  2  7 12  9  5  6  1 13 14  0 11  3  8
9 14 15  5  2  8 12  3  7  0  4 10  1 13 11  6
4  3  2 12  9  5 15 10 11 14  1  7  6  0  8 13

S[7]

4 11  2 14 15  0  8 13  3 12  9  7  5 10  6  1
13  0 11  7  4  9  1 10 14  3  5 12  2 15  8  6
1  4 11 13 12  3  7 14 10 15  6  8  0  5  9  2
6 11 13  8  1  4 10  7  9  5  0 15 14  2  3 12

S[8]

13  2  8  4  6 15 11  1 10  9  3 14  5  0 12  7
1 15 13  8 10  3  7  4 12  5  6 11  0 14  9  2
7 11  4  1  9 12 14  2  0  6 10 13 15  3  5  8
2  1 14  7  4 10  8 13 15 12  9  0  3  5  6 11

2.4.4.4  Loop back to 2.4.4.1 until all 8 blocks have been replaced.

2.4.5  Permute the concatenation of B[1] through B[8] as indicated below.

Permutation P

16  7 20 21
29 12 28 17
1 15 23 26
5 18 31 10
2  8 24 14
32 27  3  9
19 13 30  6
22 11  4 25

2.4.6  Exclusive-or the resulting value with L[i-1]. Thus, all together, your R[i] = L[i-1] xor P(S[1](B[1])…S[8](B[8])), where B[j] is a 6-bit  block of E(R[i-1]) xor K[i]. (The function for R[i] is written as, R[i] = L[i-1] xor f(R[i-1], K[i]).)

2.4.7  L[i] = R[i-1].

2.4.8  Loop back to 2.4.1 until K[16] has been applied.

Final Permutation (IP**-1) 2.5  Perform the following permutation on the block R[16]L[16].

40  8 48 16 56 24 64 32
39  7 47 15 55 23 63 31
38  6 46 14 54 22 62 30
37  5 45 13 53 21 61 29
36  4 44 12 52 20 60 28
35  3 43 11 51 19 59 27
34  2 42 10 50 18 58 26
33  1 41  9 49 17 57 25

This has been a description of how to use the data encryption standard algorithm to encrypt one 64-bit block. To decrypt, use the same process, but just use the keys K[i] in reverse order. That is, instead of applying K[1] for the first iteration, apply K[16], and then K[15] for the second, on down to K[1].

Decipherment:
R[16]L[16] = IP(cipher block)
for 1 <= i <= 16
R[i-1] = L[i] L[i-1] = R[i] xor f(L[i], K[i])
plain block = FP(L[0]R[0])

The description of the Data encryption standard come from Matthew Fischer.
Here you can find my other arduino projects.

Data encryption standard Links:

DES data encryption standard Encryption Package
Huffman Tree Generator Plant your own huffman tree
NerdKits – Halloween Hijinks Huffman-coded Audio
Compressed audio output for Arduino
Arduino Forum – newbie troubles to use avrcryptolib
Huffman Coding A CS2 Assignment
A quick tutorial on generating a huffman tree
Huffman Code Discussion and Implementation
theqlabs-SOSSE · GitHub
The Do’s and Don’ts of Decryption – TopTenREVIEWS