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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 theData Encryption Standard (DES)A step by step tutorialVersion 1.2The Data Encryption Standard (DES) algorithm, adopted by the U.S.government in 1977, is a block cipher that transforms 64-bit data blocksunder a 56-bit secret key, by means of permutation and substitution. Itis officially described in FIPS PUB 46. The DES algorithm is used formany applications within the government and in the private sector.This is a tutorial designed to be clear and compact, and to provide anewcomer to the DES with all the necessary information to implement ithimself, without having to track down printed works or wade through Csource 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 aparity bit. For a key to have correct parity, each byte should containan 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 paritybits are discarded, reducing the key to 56 bits. Bit 1 of the permutedblock is bit 57 of the original key, bit 2 is bit 49, and so on with bit56 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 17 91 58 50 42 34 26 1810 2 59 51 43 35 2719 11 3 60 52 44 3663 55 47 39 31 23 157 62 54 46 38 30 2214 6 61 53 45 37 2921 13 5 28 20 12 41.2.2 Split the permuted key into two halves. The first 28 bits arecalled 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] andD[i-1] to get C[i] and D[i], respectively. The number of shifts periteration 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 16Left Shifts 1 1 2 2 2 2 2 2 1 2 2 2 2 2 2 11.2.3.2 Permute the concatenation C[i]D[i] as indicated below. Thiswill 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 53 28 15 6 21 1023 19 12 4 26 816 7 27 20 13 241 52 31 37 47 5530 40 51 45 33 4844 49 39 56 34 5346 42 50 36 29 321.2.3.3 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, itshould 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 10 260 52 44 36 28 20 12 462 54 46 38 30 22 14 664 56 48 40 32 24 16 857 49 41 33 25 17 9 159 51 43 35 27 19 11 361 53 45 37 29 21 13 563 55 47 39 31 23 15 72.3 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 thebit-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 54 5 6 7 8 98 9 10 11 12 1312 13 14 15 16 1716 17 18 19 20 2120 21 22 23 24 2524 25 26 27 28 2928 29 30 31 32 12.4.2 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 areB[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]. Startwith 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-bitvalue (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 70 15 7 4 14 2 13 1 10 6 12 11 9 5 3 84 1 14 8 13 6 2 11 15 12 9 7 3 10 5 015 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13S[2]15 1 8 14 6 11 3 4 9 7 2 13 12 0 5 103 13 4 7 15 2 8 14 12 0 1 10 6 9 11 50 14 7 11 10 4 13 1 5 8 12 6 9 3 2 1513 8 10 1 3 15 4 2 11 6 7 12 0 5 14 9S[3]10 0 9 14 6 3 15 5 1 13 12 7 11 4 2 813 7 0 9 3 4 6 10 2 8 5 14 12 11 15 113 6 4 9 8 15 3 0 11 1 2 12 5 10 14 71 10 13 0 6 9 8 7 4 15 14 3 11 5 2 12S[4]7 13 14 3 0 6 9 10 1 2 8 5 11 12 4 1513 8 11 5 6 15 0 3 4 7 2 12 1 10 14 910 6 9 0 12 11 7 13 15 1 3 14 5 2 8 43 15 0 6 10 1 13 8 9 4 5 11 12 7 2 14S[5]2 12 4 1 7 10 11 6 8 5 3 15 13 0 14 914 11 2 12 4 7 13 1 5 0 15 10 3 9 8 64 2 1 11 10 13 7 8 15 9 12 5 6 3 0 1411 8 12 7 1 14 2 13 6 15 0 9 10 4 5 3S[6]12 1 10 15 9 2 6 8 0 13 3 4 14 7 5 1110 15 4 2 7 12 9 5 6 1 13 14 0 11 3 89 14 15 5 2 8 12 3 7 0 4 10 1 13 11 64 3 2 12 9 5 15 10 11 14 1 7 6 0 8 13S[7]4 11 2 14 15 0 8 13 3 12 9 7 5 10 6 113 0 11 7 4 9 1 10 14 3 5 12 2 15 8 61 4 11 13 12 3 7 14 10 15 6 8 0 5 9 26 11 13 8 1 4 10 7 9 5 0 15 14 2 3 12S[8]13 2 8 4 6 15 11 1 10 9 3 14 5 0 12 71 15 13 8 10 3 7 4 12 5 6 11 0 14 9 27 11 4 1 9 12 14 2 0 6 10 13 15 3 5 82 1 14 7 4 10 8 13 15 12 9 0 3 5 6 112.4.4.4 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 P16 7 20 2129 12 28 171 15 23 265 18 31 102 8 24 1432 27 3 919 13 30 622 11 4 252.4.6 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-bitblock 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 3239 7 47 15 55 23 63 3138 6 46 14 54 22 62 3037 5 45 13 53 21 61 2936 4 44 12 52 20 60 2835 3 43 11 51 19 59 2734 2 42 10 50 18 58 2633 1 41 9 49 17 57 25This has been a description of how to use the DES algorithm to encryptone 64-bit block. To decrypt, use the same process, but just use the keysK[i] in reverse order. That is, instead of applying K[1] for the firstiteration, 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 <= 16C[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 <= 16L[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 <= 16R[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-describedprocess for each block in succession. This is called Electronic Codebook(ECB) mode. A stronger method is to exclusive-or each plaintext blockwith the preceding ciphertext block prior to encryption. (The firstblock is exclusive-or'ed with a secret 64-bit initialization vector(IV).) This is called Cipher Block Chaining (CBC) mode. The other twomodes 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. Oneis to simply append zeros. Two suggested by FIPS PUB 81 are, if the datais binary data, fill up the block with bits that are the opposite of thelast bit of data, or, if the data is ASCII data, fill up the block withrandom bytes and put the ASCII character for the number of pad bytes inthe last byte of the block. Another technique is to pad the block withrandom 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） 关闭本页

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