目录
1 根据输入的秘钥得到16个子秘钥
1.1 大致流程
graph TB;a[输入一个64位长度的字符串,即为秘钥K_0]b[根据PC-1表挑出56位作为新秘钥K_1]c[K_1拆分,循环左移得到16个子秘钥k0...k15]d[使用PC-2表从各子秘钥56位中挑出48位]a-->b-->c-->d
1.2 利用PC-1从K_0中挑出K_1
接下来我们以输入的秘钥为"0x123456789abcdeff"为例, 及二进制编码为:
接下来我们利用PC-1表生成新秘钥K_1
1.3 利用PC-2从K_1中挑出16个子秘钥
对于从1.2得来的一串56位二进制码, 我们将其拆分成两串28位二进制码, 即为:
接下来, 我们利用下表, 进行循环左移, 规则如下:
之后我们按照上面的规则, 展示子秘钥k_1是如何生成的:
按照同样的逻辑, 在C_0和D_0已被循环左移的基础上, 重复上述步骤, 我们得到如下16个子秘钥:
PS: 上述输出为自己写的代码, 在文末会进行展示.
得到了这16个子秘钥后, 我们就可以对明文进行加密了, 我们继续!
2 利用16个子秘钥对明文进行加密
2.1 大致流程
graph TB;a[对64位的明文进行初始变化,简称IP-Initial Permutation]b[将IP后的明文平分成32位的L0-左半边和R0-右半边]c[接下来使用如下公式对L_n和R_n进行变换]d["1.L_n=R_n-1<br>2.R_n=L_n-1 + f(R_n-1, k_n)"]e["f的具体过程如下<br>1.将R_n-1扩展成48位并与k_n-1相加"得到S盒<br>2.使用8张S表将48位的S盒运算成32位输出P<br>3.使用P表对输出P进行位置变换]f["得到输出L_16和R_16, 拼接成R_16L_16的64位输出"]g["对这64位输出使用IP-1表进行位置变换, 即得到最终输出"]a-->b-->c-->d-.-e-.->f-->gd--重复16次-->d
2.2 将R_0扩展成48位的二进制串, 并与子秘钥k模二加
我们以输入的明文为"0x0123456789abcdef"为例, 即二进制编码为:
2.3 使用S1-S8表将48位变回32位
为什么上面S盒要备注8个6位呢? 原因就是要以8个6位为输入, 分别经过S1-S8盒的运算,得到8个4位输出, 也就变成了32位.
2.4 使用P表对32位输出进行位置变换, 并和L0相加
我们已经得到经过"扩位"和"缩位"的变换, 现在要按照P表进行简单的位置变换:
2.5 往复这样的步骤16次, 将得到的R_16和L_16拼接并经IP^-1变换
如题, 得到最终结果:
3 代码实现
/*
* @Description:
* @Autor: Shmily
* @Date: 2021-03-01 23:40:00
* @LastEditTime: 2021-03-02 22:35:49
*/
#include <bits/stdc++.h>
using namespace std;
const int8_t PC_1[56] = {
57, 49, 41, 33, 25, 17, 9, 1,
58, 50, 42, 34, 26, 18, 10, 2,
59, 51, 43, 35, 28, 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 int8_t PC_2[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 int8_t left_move[16] = {
1, 1, 2, 2, 2, 2, 2, 2,
1, 2, 2, 2, 2, 2, 2, 1
};
const int8_t S8[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 int8_t PTable[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, 4,
22, 11, 4, 25
};
const int8_t IP[2][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, 16, 11, 3,
61, 53, 45, 37, 29, 21, 13, 5,
63, 55, 47, 39, 31, 23, 15, 7
},
{
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
}
};
void out_int8_t(int8_t* out, int len) {
for (int i = 0; i < len; i++) {
printf("%d", out[i]);
if ((i + 1) % 4 == 0) cout << " ";
}
cout << endl;
}
void fun_left_move(int8_t* C_0, int8_t* D_0, int times_left) {
for (int i = 0; i < times_left; i++) {
int8_t temp1 = C_0[0];
int8_t temp2 = D_0[0];
for (int j = 1; j < 28; j++) {
C_0[j - 1] = C_0[j];
D_0[j - 1] = D_0[j];
}
C_0[27] = temp1;
D_0[27] = temp2;
}
}
void Extend32To48(int8_t (*R) [32], int8_t (*RExtend) [48], int times) {
RExtend[times][0] = R[times][31];
RExtend[times][47] = R[times][0];
for (int i = 1; i < 47; i++) {
// 6个位一组, 即为一行
int LineNumber = (int)(i / 6);
RExtend[times][i] = R[times][i - 1 - LineNumber*2];
}
}
void XOR(int8_t (*RExtend) [48], int8_t (*k) [48], int8_t (*S) [48], int times) {
for (int i = 0; i < 48; i++) {
// 进行模二加运算
S[times][i] = (RExtend[times][i] + k[times][i]) % 2;
}
}
void XOR(int8_t (*L) [32], int8_t (*Pchanged) [32], int8_t (*R) [32], int times) {
for (int i = 0; i < 32; i++) {
R[times + 1][i] = (L[times][i] + Pchanged[times][i]) % 2;
}
}
void UseS48To32(int8_t (*S) [48], int8_t (*P) [32], int times) {
for (int i = 0; i < 8; i++) {
int line = 0, row = 0;
// 计算出行号及列号
line = line + S[times][i*6]*2 + S[times][5 + i*6];
row = row + S[times][1 + i*6]*8 + S[times][2 + i*6]*4 + S[times][3 + i*6]*2 + S[times][4 + i*6];
// 从S表中挑出答案
int8_t ans = S8[i][line][row];
for (int j = 0; j < 4; j++) {
int subtrahend = pow(2, (3 - j));
if (ans - subtrahend >= 0) {
P[times][j + i*4] = 1;
ans -= subtrahend;
}
else P[times][j + i*4] = 0;
}
}
}
void UsePChangeP(int8_t (*P) [32], int8_t (*PChanged) [32], int times) {
for (int i = 0; i < 32; i++) {
PChanged[times][i] = P[times][PTable[i] - 1];
}
}
void UseIPChange(int8_t* in, int8_t* out, int8_t choice) {
for (int i = 0; i < 64; i++) {
out[i] = in[IP[choice][i] - 1];
}
}
int main(void) {
// 对秘钥进行处理
// 输入为64位秘钥， 输出位16个32位子秘钥
int8_t K_0[64], K_1[56], k[16][48];
// 秘钥为"0x123456789abcdeff"
char in[] = "0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 1111";
int len = strlen(in);
// 将输入的秘钥存储到K_0中
for (int i = 0, j = 0; i < len; i++) {
if (in[i] == '0') K_0[j++] = 0;
else if (in[i] == '1') K_0[j++] = 1;
}
// 利用PC-0表, 从K_0中取出K_1
for (int i = 0; i < 56; i++) {
// 注意-1
K_1[i] = K_0[PC_1[i]-1];
}
// 上述完成了第一步利用PC-1表
// 接下来利用PC-2表
int8_t C_0[28], D_0[28];
// 将K_1拆分
for (int i = 0; i < 56; i++) {
if (i < 28) C_0[i] = K_1[i];
else D_0[i - 28] = K_1[i];
}
// 循环16次生成16个子秘钥
for (int i = 0; i < 16; i++) {
int8_t times_left = left_move[i];
fun_left_move(C_0, D_0, times_left);
for (int j = 0; j < 48; j++) {
int8_t sub = PC_2[j] - 1;
if (sub < 28) k[i][j] = C_0[sub];
else k[i][j] = D_0[sub - 28];
}
// printf("子秘钥k[%.2d]: ", i + 1);
// out_int8_t(k[i], 48);
}
// 对明文进行处理
// 明文为"0x0123456789abcdef"
char in_text[] = "0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111";
int8_t PlainText[64], IP_PlainText[64], EnchiperTextInitial[64], EnchiperText[64];
int8_t L[17][32], R[17][32], RExtend[16][48], S[16][48], P[16][32], PChanged[16][32];
int len_text = strlen(in_text);
// 64位明文存储到PlainText中
for (int i = 0, j = 0; i < len; i++) {
if (in_text[i] == '0') PlainText[j++] = 0;
else if (in_text[i] == '1') PlainText[j++] = 1;
}
// 使用IP对PlainText进行变换
UseIPChange(PlainText, IP_PlainText, 0);
// 将PlainText分左右存储到L_0及R_0中
for (int i = 0; i < 64; i++) {
if (i < 32) L[0][i] = IP_PlainText[i];
else R[0][i - 32] = IP_PlainText[i];
}
// 循环16次
for (int times = 0; times < 16; times++) {
// L_n = R_n-1
for (int i = 0; i < 32; i++) {
L[times + 1][i] = R[times][i];
}
// L_n = L_n-1 + f(R_n-1, k_n)
// 对R进行扩展
Extend32To48(R, RExtend, times);
// 将RExtend与k_n-1模二加
XOR(RExtend, k, S, times);
// 使用S表将S盒48位压缩成32位
UseS48To32(S, P, times);
// 使用P表对P进行位置变换
UsePChangeP(P, PChanged, times);
// 生成的PChanged就是经过f之后的
// 模二加后放入R_n
XOR(L, PChanged, R, times);
}
for (int i = 0; i < 64; i++) {
if (i < 32) EnchiperTextInitial[i] = R[16][i];
else EnchiperTextInitial[i] = L[16][i - 32];
}
// 使用IP^-1对EnchiperTextInitial的进行变换得到Enchiper
UseIPChange(EnchiperTextInitial, EnchiperText, 1);
//最终结果输出
out_int8_t(EnchiperText, 64);
return 0;
}