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 <mfischer@heinous.isca.uiowa.edu>
；上面是介绍，我就不翻了。 ;)
Here's how to do it, step by step:
1Process the key.
；生成密钥
1.1Get 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
1.2Calculate the key schedule.
；计算密钥表
1.2.1Perform 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<64;j++)
# {
# for(i=j;i<0;i-=8)
# {
# 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 25 179
1 58 50 42 34 26 18
102 59 51 43 35 27
19 113 60 52 44 36
63 55 47 39 31 23 15
7 62 54 46 38 30 22
146 61 53 45 37 29
21 135 28 20 124
1.2.2Split 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。
1.2.3Calculate the 16 subkeys. Start with i = 1.
；计算16个子密钥，从i=1开始。
1.2.3.1Perform 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 # 123456789 10 11 12 13 14 15 16
Left Shifts 1122222212222221
1.2.3.2Permute 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 2415
3 28 156 21 10
23 19 124 268
167 27 20 132
41 52 31 37 47 55
30 40 51 45 33 48
44 49 39 56 34 53
46 42 50 36 29 32
1.2.3.3Loop 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。
2Process a 64-bit data block.
；处理一个64位的数据块。
2.1Get 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。
2.2Perform 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 18 102
60 52 44 36 28 20 124
62 54 46 38 30 22 146
64 56 48 40 32 24 168
57 49 41 33 25 1791
59 51 43 35 27 19 113
61 53 45 37 29 21 135
63 55 47 39 31 23 157
2.3Split 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。
2.4Apply 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。
2.4.1Expand 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)
3212345
456789
89 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 31 321
2.4.2Exclusive-or E(R[i-1]) with K[i].
；用K[i]对E(R[i-1])进行异或操作。
2.4.3Break 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]。
2.4.4Substitute 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！！！ :(
2.4.4.1Take 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。
2.4.4.2Take 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。
2.4.4.3Replace B[j] with S[j][m][n].
；用S[j][m][n]置换B[j]。
Substitution Box 1 (S[1])
144 1312 15 1183 106 125907
0 1574 142 131 106 12 119538
41 148 1362 11 15 12973 1050
15 128249175 113 14 1006 13
S[2]
1518 146 1134972 13 1205 10
3 1347 1528 14 1201 1069 115
0 147 11 104 13158 126932 15
138 1013 1542 1167 1205 149
S[3]
1009 1463 1551 13 127 11428
13709346 10285 14 12 11 151
136498 1530 1112 125 10 147
1 10 13069874 15 143 1152 12
S[4]
7 13 143069 101285 11 124 15
138 1156 1503472 121 10 149
10690 12 117 13 1513 145284
3 1506 101 138945 11 1272 14
S[5]
2 12417 10 116853 15 130 149
14 112 1247 13150 15 103986
421 11 10 1378 159 125630 14
118 1271 142 136 1509 10453
S[6]
121 10 1592680 1334 1475 11
10 15427 129561 13 140 1138
9 14 15528 123704 101 13 116
432 1295 15 10 11 1417608 13
S[7]
4 112 14 1508 133 12975 1061
130 117491 10 1435 122 1586
14 11 13 1237 14 10 15680592
6 11 13814 107950 15 1423 12
S[8]
132846 15 111 1093 1450 127
1 15 138 10374 1256 110 1492
7 11419 12 14206 10 13 15358
21 1474 108 13 15 1290356 11
2.4.4.4Loop 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。
2.4.5Permute 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
167 20 21
29 12 28 17
1 15 23 26
5 18 31 10
28 24 14
32 2739
19 13 306
22 114 25
2.4.6Exclusive-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])。）
2.4.7L[i] = R[i-1].
；L[i] = R[i-1].
2.4.8Loop 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。
2.5Perform 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)
408 48 16 56 24 64 32
397 47 15 55 23 63 31
386 46 14 54 22 62 30
375 45 13 53 21 61 29
364 44 12 52 20 60 28
353 43 11 51 19 59 27
342 42 10 50 18 58 26
331 419 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 <= i <= 16
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 <= i <= 16
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 <= 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])
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）
……