Question

DES加密和解密的对称性

DES加密和解密的程序是对称的，只有加密和解密的时候用的密钥的顺序是相反的，我想知道加密的时候经过S盒映射成密文，但是它会丢弃那两位行号值，那它解密的时候是怎么恢复的，怎么证明它的对称性。或者我是想问，由于s盒是非线性映射，那它怎么又从密文映射成明文的呢。想好几天了，百思不得其解。
答对了加分
能说得详细一点吗，我是学计算机的，没学加密，懂程序，可是对密码学一窍不通。我想不通DES为什么会有对称性

Answer 1

#include <iostream>
#include <fstream>
using namespace std;
const static char ip[] = { //IP置换
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 static char fp[] = { //zuizhongzhihuan
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 static char sbox[8][64] = { //s_box
/* 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 static char rar[] = { //ya suo zhi huan
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 static char ei[] = { //kuo zhan zhi huan
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 static char Pzh[]={ //P置换
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
};
const static char Keyrar[]={
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
};
bool key[16][48]={0},/*rekey[16][48],*/
char key_in[8];
void ByteToBit(bool *Out,char *In,int bits) //字节到位的转换
{
int i;
for(i=0;i<bits;i++)
Out[i]=(In[i/8]>>(i%8))&1;
}
void BitToByte(char *Out,bool *In,int bits) //位到字节转换
{
for(int i=0;i<bits/8;i++)
Out[i]=0;
for(i=0;i<bits;i++)
Out[i/8]|=In[i]<<(i%8); //"|="组合了位操作符和赋值操作符的功能
}
void Xor(bool *InA,const bool *InB,int len) //按位异或
{
for(int i=0;i<len;i++)
InA[i]^=InB[i];
}
void keyfc(char *In) //获取密钥函数
{
int i,j=0,mov,k;
bool key0[56],temp,keyin[64];
ByteToBit(keyin,In,64); //字节到位的转换
for(i=0;i<56;i++) //密钥压缩为56位
key0[i]=keyin[Keyrar[i]-1];
for(i=0;i<16;i++) //16轮密钥产生
{
if(i==0||i==1||i==8||i==15)
mov=1;
else
mov=2;
for(k=0;k<mov;k++) //分左右两块循环左移
{
// for(int m=0;m<8;m++)
// {
// temp=key0[m*7];
// for(j=m*7;j<m*7+7;j++)
// key0[j]=key0[j+1];
// key0[m*7+6]=temp;
// }
temp=key0[0];
for(int m=0;m<27;m++)
key0[m]=key0[m+1];
key0[27]=temp;
temp=key0[28];
for(m=28;m<55;m++)
key0[m]=key0[m+1];
key0[55]=temp;
}

for(j=0;j<48;j++) //压缩置换并储存
key[i][j]=key0[rar[j]-1];
}
}

void DES(char Out[8],char In[8],bool MS)//加密核心程序,ms=0时加密，反之解密
{
bool MW[64],tmp[32],PMW[64]; //注意指针
bool kzmw[48],keytem[48],ss[32];
int hang,lie;
ByteToBit(PMW,In,64);
for(int j=0;j<64;j++)
{
MW[j]=PMW[ip[j]-1]; //初始置换
}
bool *Li=&MW[0],*Ri=&MW[32];
for(int i=0;i<48;i++) //右明文扩展置换
kzmw[i]=Ri[ei[i]-1]; //注意指针

if(MS==0) //DES加密过程
{
for(int lun=0;lun<16;lun++)
{
for(i=0;i<32;i++)
ss[i]=Ri[i];
for(i=0;i<48;i++) //右明文扩展置换
kzmw[i]=Ri[ei[i]-1]; //注意指针
for(i=0;i<48;i++)
keytem[i]=key[lun][i]; //轮密钥
Xor(kzmw,keytem,48);
/*S盒置换*/
for(i=0;i<8;i++)
{
hang=kzmw[i*6]*2+kzmw[i*6+5];
lie =kzmw[i*6+1]*8+kzmw[i*6+2]*4+kzmw[i*6+3]*2+kzmw[i*6+4];
tmp[i*4+3]=sbox[i][(hang+1)*16+lie]%2;
tmp[i*4+2]=(sbox[i][(hang+1)*16+lie]/2)%2;
tmp[i*4+1]=(sbox[i][(hang+1)*16+lie]/4)%2;
tmp[i*4]=(sbox[i][(hang+1)*16+lie]/8)%2;
}

for(int i=0;i<32;i++) //P置换
Ri[i]=tmp[Pzh[i]-1];
Xor(Ri,Li,32); //异或
for(i=0;i<32;i++) //交换左右明文
{
Li[i]=ss[i];
}
}
for(i=0;i<32;i++)
{
tmp[i]=Li[i];
Li[i]=Ri[i];
Ri[i]=tmp[i];
}
for(i=0;i<64;i++)
PMW[i]=MW[fp[i]-1];
BitToByte(Out,PMW,64); //位到字节的转换
}
else //DES解密过程
{

for(int lun=15;lun>=0;lun--)
{
for(i=0;i<32;i++)
ss[i]=Ri[i];
for(int i=0;i<48;i++) //右明文扩展置换
kzmw[i]=Ri[ei[i]-1]; //注意指针
for(i=0;i<48;i++)
keytem[i]=key[lun][i]; //轮密钥
Xor(kzmw,keytem,48);
/*S盒置换*/
for(i=0;i<8;i++)
{
hang=kzmw[i*6]*2+kzmw[i*6+5];
lie =kzmw[i*6+1]*8+kzmw[i*6+2]*4+kzmw[i*6+3]*2+kzmw[i*6+4];
tmp[i*4+3]=sbox[i][(hang+1)*16+lie]%2;
tmp[i*4+2]=(sbox[i][(hang+1)*16+lie]/2)%2;
tmp[i*4+1]=(sbox[i][(hang+1)*16+lie]/4)%2;
tmp[i*4]=(sbox[i][(hang+1)*16+lie]/8)%2;
}

for(i=0;i<32;i++) //P置换
Ri[i]=tmp[Pzh[i]-1];
Xor(Ri,Li,32); //异或
for(i=0;i<32;i++) //交换左右明文
{
Li[i]=ss[i];
}

}
for(i=0;i<32;i++)
{
tmp[i]=Li[i];
Li[i]=Ri[i];
Ri[i]=tmp[i];
}
for(i=0;i<64;i++)
PMW[i]=MW[fp[i]-1];
BitToByte(Out,PMW,64); //位到字节的转换
}
}
void main()
{
char Ki[8],jm[8],final[8];
int i0;
cout<<"请输入密钥（8字节）："<<endl;
for(i0=0;i0<8;i0++)
cin>>Ki[i0];
// if(i0<8)
// for(i0=0;i0<8;i0++)
// cin//[i0];
keyfc(Ki);
cout<<"请输入明文："<<endl;
for(i0=0;i0<8;i0++)
cin>>jm[i0];
DES(final,jm,0); //加密
for(i0=0;i0<8;i0++)
cout<<final[i0];
cout<<endl;
DES(jm,final,1); //解密
for(i0=0;i0<8;i0++)
cout<<jm[i0];
cout<<endl;
}
这个事用VC实现
给你算法你自己看

Answer 2

对称性是靠Feistal模型来保证的，与S盒无关。。好好看看书，很简单哈。。