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问题是他们究竟是喜欢他们是使用软件时,速度快还是速度慢好.(当然在3秒以内).无论是他们输入资料时,查找资料时,分析资料时.算法 这是摘自清华BBS的一篇文章,洋文的,小弟把它翻成中文请列位高手指导。
分号(;)后的话是小弟的翻译,井号(#)后的是小弟的一点感触。
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>
;下面是引见,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.)
;从用户处失掉一个64位的密钥。(每8位一组,每组的第8位是校验位。假如校验
准确,每一个字节应当有一个为1的
1.2 Calculate the key schedule.
;盘算密钥表
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位的密钥停止以下的置换。(去失落校验位,密钥的实践长度是56位。置换后的
;第一名是原密钥的第57位,第二位是原第49位,第五十六位就是本来密钥的第4位。)
# 乖僻的置换,哪位年老能写出算式?
# 好象是分红两部
# 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,不晓得对不合错误?请斧正。
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]两局部,各28位。
1.2.3 Calculate the 16 subkeys. Start with i = 1.
;盘算16个子密钥,从i=1入手下手。
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.
;分离对C[i-1]和D[i-1]停止左移一到两位的位移操作,失掉C[i]和D[i]。每次
;位移数量以下:
# 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.
;以下表,改动C[i]和D[i]的分列,失掉48位长的k[i]。
# 不懂 :(
# 是否是丢失落了某些位?
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.
;反复 1.2.3.1 入手下手的进程,算出16个字密钥。
2 Process a 64-bit data block.
;处置一个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.
;获得一个64位的数据块。假如数据块不到64位,就补足64位。
# 多是用0补吧。
2.2 Perform the following permutation on the data block.
;对数据块停止以下置换。
# 又是分红两局部停止,先是偶数位。
# 对照复杂,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].
;将数据块等分为L[0]和R[0]两局部。
2.4 Apply the 16 subkeys to the data block. Start with i = 1.
;从i=1入手下手,用16个子密钥对数据块停止加密。
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]以上面划定规矩停止扩大。
# 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
楼上说交互性不好,太牵强了吧。在微软提供的一套框架中,利用asp做网站,开发效率高,使用人数少,减少不必要的开销。交互性是互动方式,是有开发人员决定的。 |
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