不知道大家对DES有没有兴趣，今天在整理的时候，看到我在一年半前翻译的一篇文章。

如何实现 DES 算法（全）。

这是摘自清华BBS的一篇文章，洋文的，小弟把它翻成中文请各位高手指点。
分号（；）后的话是小弟的翻译，井号（#）后的是小弟的一点感想。


                          How to implement the
                     Data Encryption Standard (DES)

                        A step by step tutorial
                              Version 1.2


The Data Encryption Standard (DES) algorithm, adopted by the U.S.  
government in 1977, is a block cipher that transforms 64-bit data blocks  
under a 56-bit secret key, by means of permutation and substitution. It  
is officially described in FIPS PUB 46. The DES 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 DES 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.
Matthew Fischer

；上面是介绍，我就不翻了。 ;)


Here's how to do it, step by step:

1  Process the key.
；生成密钥

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.)
；从用户处得到一个64位的密钥。（每8位一组，每组的第8位是校验位。如果校验
正确，每个字节应该有一个为1的

1.2  Calculate the key schedule.
；计算密钥表

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.)
；对64位的密钥进行如下的置换。（去掉校验位，密钥的实际长度是56位。置换后的
；第一位是原密钥的第57位，第二位是原第49位，第五十六位就是原来密钥的第4位。）
# 古怪的置换，哪位大哥能写出算式？
# 好象是分成两部
#       for(j=57;j#       {
#               for(i=j;i#               {
#                       if(k=28)
#                               break;
#                       c[k]=i;
#                       k++;
#               }
# 这是前28位，不知道对不对？请指正。


                        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].
；把置换后的密钥分为C[0] 和D[0]两部分，各28位。

1.2.3  Calculate the 16 subkeys. Start with i = 1.
；计算16个子密钥，从i=1开始。

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.
；分别对C[i-1]和D[i-1]进行左移一到两位的位移操作，得到C[i]和D[i]。每次
；位移数目如下：
# 共16次

    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.
；如下表，改变C[i]和D[i]的排列，得到48位长的k[i]。
# 不懂 :(
# 是不是丢掉了某些位？

                        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.
；重复 1.2.3.1 开始的过程，算出16个字密钥。

2  Process a 64-bit data block.
；处理一个64位的数据块。

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.
；获取一个64位的数据块。如果数据块不到64位，就补足64位。
# 可能是用0补吧。

2.2  Perform the following permutation on the data block.
；对数据块进行如下置换。
# 又是分成两部分进行，先是偶数位。
# 比较简单，算式就不写了。

                        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].
；将数据块平分为L[0]和R[0]两部分。

2.4  Apply the 16 subkeys to the data block. Start with i = 1.
；从i=1开始，用16个子密钥对数据块进行加密。

2.4.1  Expand the 32-bit R[i-1] into 48 bits according to the  
bit-selection function below.
；将数据块的后32位R[i-1]以下面规则进行扩展。
# 不会写算式。:(
                             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].
；用K[i]对E(R[i-1])进行异或操作。

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].
；将上一步的操作结果分成8块，每块6位，命名为B[1]到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.
；把所有的B[j]在S框中进行置换，S框中所有的值的宽（长）度应是4位。
# 不懂！！！ :(

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.
；把B[j]中的第一位和第六位命名为m，表示S[j]在置换时的行。

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.
；把B[j]二到五位命名为n，表示S[j]在置换时的列。

2.4.4.3  Replace B[j] with S[j][m][n].
；用S[j][m][n]置换B[j]。

                       Substitution Box 1 (S[1])

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

                                  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.
；重复2.4.4.1开始的步骤，直至8个数据块都被置换。

2.4.5  Permute the concatenation of B[1] through B[8] as indicated below.
；以下面的方法改变B[1]到B[8]的顺序 。

                             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]).)
；用L[i-1]对上一步的结果进行异或操作。如此就有以下结果：R[i] = L[i-1] xor ；

P(S[1](B[1])...S[8](B[8]))。这里，B[j]是六位的数据块，它是E(R[i-1]) xor
；K[i]的结果。（R[i]的函数可以写成R[i] = L[i-1] xor f(R[i-1], K[i])。）

2.4.7  L[i] = R[i-1].
；L[i] = R[i-1].

2.4.8  Loop back to 2.4.1 until K[16] has been applied.
；重复2.4.1开始的步骤，直至所有的子密钥都被使用过。
# 就是再重复15次，每次使用不同的子密钥。

2.5  Perform the following permutation on the block R[16]L[16].
；对R[16]L[16]进行如下的置换。

                       Final Permutation (IP**-1)

                        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 DES 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].
；以上就是怎样用DES算法对一个64位的数据块进行加密的过程。至于解密，只需要
；在以上过程中把子密钥的顺序倒过来用就可以了。也就是说，在加密时用子密钥
；K[1]，在解密过程中就用K[16]；在加密时用子密钥K[2]，在解密过程中就用K[12]。


Summaries:
；摘要
# 以下是生成子密钥，加密和解密的公式化叙述。

Key schedule:
  C[0]D[0] = PC1(key)
  for 1    C[i] = LS[i](C[i-1])
   D[i] = LS[i](D[i-1])
   K[i] = PC2(C[i]D[i])

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

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


To encrypt or decrypt more than 64 bits there are four official modes  
(defined in FIPS PUB 81). One is to go through the above-described  
process for each block in succession. This is called Electronic Codebook  
(ECB) mode. A stronger method is to exclusive-or each plaintext block  
with the preceding ciphertext block prior to encryption. (The first  
block is exclusive-or'ed with a secret 64-bit initialization vector  
(IV).) This is called Cipher Block Chaining (CBC) mode. The other two  
modes are Output Feedback (OFB) and Cipher Feedback (CFB).
；对超过64位的加密和解密，(美国)联邦信息处理标准 PUB 81 中定有四种方法。
；一种是连续的对每个数据块进行上述操作。这种方法被称 ECB mode。另一种更
；高强度的方法是在加密前，用前述的密文块对明文块进行异或操作。
# 括号里那句话不懂 :(
；这种方法被称为 CBC mode。还有两种方法是 OFB mode 和 CFB mode。


When it comes to padding the data block, there are several options. One  
is to simply append zeros. Two suggested by FIPS PUB 81 are, if the data  
is binary data, fill up the block with bits that are the opposite of the  
last bit of data, or, if the data is ASCII data, fill up the block with  
random bytes and put the ASCII character for the number of pad bytes in  
the last byte of the block. Another technique is to pad the block with  
random bytes and in the last 3 bits store the original number of data bytes.
；在填充数据块时（还记不记得，当数据块不足64位时要进行填充），有以下几种
；选择：一种就是填0。第二种是被(美国)联邦信息处理标准 PUB 81所建议的，如
；果数据是二进制的，就填入和数据位最后一位相反的数；如果数据块是ASCII码，
；就填入随机字节，并且将填充数目写入最后一个字节。另一种技术就是填入随机
；字节，并且将最后原数据字节数写入最后的三位。（注意：是位，bit）