DES全称为Data Encryption Standard，即数据加密标准，是一种使用密钥加密的块算法，
1977年被美国联邦政府的国家标准局确定为联邦资料处理标准（FIPS），并授权在非密级政府通信中使用，随后该算法在国际上广泛流传开来。需要注意的是，在某些文献中，作为算法的DES称为数据加密算法（Data Encryption Algorithm,DEA），已与作为标准的DES区分开来。
算法流程
算法流程可以参考百度百科的算法流程图
算法详解
DES算法在《图解密码技术》这本书上一笔带过了，比较好奇书上说的 f() 函数是什么，于是就找了 DES 的源码来看。
DES算法有三个参数：data，key，mode。
- data 要加密的信息
- key 加密的密钥
- mode 加密还是解密
DES算法还使用到了很多置换，搞清楚置换表的作用是非常重要的。
我们通过 key 置换，循环左移的操作得到 subkey，记为 k1，k2，…，kn。
首先将 data 置换，并且平分成两半，记为 L0 和 R0。
第 N 次操作的时候，我们需要将 Ln = R(n-1)，而 Rn = f(L(n-1), R(n-1), kn) 三个参数去计算，计算过程也相当复杂。
首先我们需要将 R(n-1) 通过拓展表拓展成 48 位的，然后跟 subkey 进行异或，然后根据 s 盒表的规则，将对应每 6 位变换成 4 位，重新变成 32 位，再和 L(n-1) 进行异或操作。
最最重要的是！千万不能搞错谁和谁异或，用谁去操作，不然 bug 就找不到了。
代码实现
最后恬不知耻地贴上自己的代码 诶嘿
#include <bits/stdc++.h>
using namespace std;
class DES {
private:
//const table
const char IP_Table[64] = {
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
};
const char IPR_Table[64] = {
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
};
const char PC1_Table[56] = {
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
};
const char PC2_Table[48] = {
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
};
const char LOOP_Table[16] = {
1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1
};
const char E_Table[48] = {
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
};
const char S_Box[8][4][16] = {
// S1
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,
// S2
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,
// S3
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,
// S4
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,
// S5
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,
// S6
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,
// S7
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,
// S8
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
};
const char P_Table[32] = {
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
};
int l[20][40], r[20][40];
int keyl[40], keyr[40];
int key[20][70];
int ext[50];
public:
DES() {}
~DES() {}
void crypt(int *data, int *key, int *ret, int mode = 0) {
for (int i = 0; i <= 64; i++) ret[i] = 0;
if (!mode) { //encrypt
memset(l, 0, sizeof l);
memset(r, 0, sizeof r);
//initialize the L0,R0
for (int i = 1; i <= 32; i++) {
l[0][i] = data[IP_Table[i - 1]];
}
for (int i = 33; i <= 64; i++) {
r[0][i - 32] = data[IP_Table[i - 1]];
}
//initialize the subkey
for (int i = 1; i <= 28; i++) {
keyl[i] = key[PC1_Table[i - 1]];
}
for (int i = 29; i <= 56; i++) {
keyr[i - 28] = key[PC1_Table[i - 1]];
}
for (int i = 1; i <= 16; i++) {
for (int o = 1; o <= LOOP_Table[i]; o++) {
for (int j = 1; j < 28; j++) {
keyl[j] = keyl[j + 1];
}
keyl[28] = keyl[1];
for (int j = 1; j < 28; j++) {
keyr[j] = keyr[j + 1];
}
keyr[28] = keyr[1];
}
for (int o = 1; o <= 48; o++) {
this->key[i][o] = (PC2_Table[o - 1] <= 28 ? keyl[PC2_Table[o - 1]] : keyr[PC2_Table[o - 1] - 28]);
}
}
// Feistal network
for (int i = 1; i <= 16; i++) {
for (int o = 1; o <= 48; o++) ext[o] = r[i - 1][E_Table[o - 1]]; //extend the R(n-1)
for (int o = 1; o <= 48; o++) ext[o] ^= this->key[i][o];//xor the subkey
for (int o = 0; o < 8; o++) { //core algorithm of f()
int row = ext[6 * o + 1] * 2 + ext[6 * o + 6];
int col = ext[6 * o + 2] * 8 + ext[6 * o + 3] * 4 + ext[6 * o + 4] * 2 + ext[6 * o + 5];
int midresult = S_Box[o][row][col];
for (int j = 1; j <= 4; j++) {
ext[32 - 6 * o - j + 1] = midresult % 2;
midresult /= 2;
}
}
for (int o = 1; o <= 32; o++) {
r[i][o] = ext[P_Table[o - 1]] ^ l[i - 1][o];//xor the L(n-1)
}
for (int o = 1; o <= 32; o++) {
l[i][o] = r[i - 1][o];
}
}
for (int i = 1; i <= 64; i++) {
ret[i] = IPR_Table[i - 1] <= 32 ? r[16][IPR_Table[i - 1]] : l[16][IPR_Table[i - 1] - 32];
}
} else { //decrypt
memset(l, 0, sizeof l);
memset(r, 0, sizeof r);
for (int i = 1; i <= 32; i++) {
l[0][i] = data[IP_Table[i - 1]];
}
for (int i = 33; i <= 64; i++) {
r[0][i - 32] = data[IP_Table[i - 1]];
}
for (int i = 1; i <= 28; i++) {
keyl[i] = key[PC1_Table[i - 1]];
}
for (int i = 29; i <= 56; i++) {
keyr[i - 28] = key[PC1_Table[i - 1]];
}
for (int i = 1; i <= 16; i++) {
for (int o = 1; o <= LOOP_Table[i]; o++) {
for (int j = 1; j < 28; j++) {
keyl[j] = keyl[j + 1];
}
keyl[28] = keyl[1];
for (int j = 1; j < 28; j++) {
keyr[j] = keyr[j + 1];
}
keyr[28] = keyr[1];
}
for (int o = 1; o <= 48; o++) {
this->key[i][o] = (PC2_Table[o - 1] <= 28 ? keyl[PC2_Table[o - 1]] : keyr[PC2_Table[o - 1] - 28]);
}
}
for (int i = 1; i <= 16; i++) {
for (int o = 1; o <= 48; o++) ext[o] = r[i - 1][E_Table[o - 1]];
for (int o = 1; o <= 48; o++) ext[o] ^= this->key[17 - i][o];
for (int o = 0; o < 8; o++) {
int row = ext[6 * o + 1] * 2 + ext[6 * o + 6];
int col = ext[6 * o + 2] * 8 + ext[6 * o + 3] * 4 + ext[6 * o + 4] * 2 + ext[6 * o + 5];
int midresult = S_Box[o][row][col];
for (int j = 1; j <= 4; j++) {
ext[32 - 6 * o - j + 1] = midresult % 2;
midresult /= 2;
}
}
for (int o = 1; o <= 32; o++) {
r[i][o] = ext[P_Table[o - 1]] ^ l[i - 1][o];
}
for (int o = 1; o <= 32; o++) {
l[i][o] = r[i - 1][o];
}
}
for (int i = 1; i <= 64; i++) {
ret[i] = IPR_Table[i - 1] <= 32 ? r[16][IPR_Table[i - 1]] : l[16][IPR_Table[i - 1] - 32];
}
}
return;
}
};
int main() { //take a test
DES des = *(new DES());
int a[100]{};
for (int i = 1; i <= 64; i+=2) a[i] = 1;
int b[100]{};
int c[100]{};
int d[100]{};
des.crypt(a, b, c, 0);
des.crypt(c, b, d, 1);
for (int i = 1; i <= 64; i++) cout << d[i];
}
算法改进
在 1999 年的 DES Challenge III 中，破译 DES 密文只用了 22 小时 15 分钟，所以提出了三重 DES(triple-DES)。
三重 DES 并不是运行三次 DES 加密，而是加密解密加密的形式
虽然听上去非常不可思议，但是这样做可以兼容普通的 DES（三次都使用同一个 key）。
我们称使用两个 key 的 TDEA 为 DES-EDE2，使用三个 key 的为 DES-EDE3。
对应的，解密过程为：解密加密解密。
本博客所有文章除特别声明外，均采用 CC BY-SA 3.0协议 。转载请注明出处！