实验1 安全数据传输综合实验Word文档下载推荐.docx
- 文档编号:17661060
- 上传时间:2022-12-07
- 格式:DOCX
- 页数:15
- 大小:86.01KB
实验1 安全数据传输综合实验Word文档下载推荐.docx
《实验1 安全数据传输综合实验Word文档下载推荐.docx》由会员分享,可在线阅读,更多相关《实验1 安全数据传输综合实验Word文档下载推荐.docx(15页珍藏版)》请在冰豆网上搜索。
3.编程实现实验内容1,请给出每个中间过程的关键数据及源代码。
对称密码算法采用AES(密钥长度在128位、192位、256位中进行3选1即可),公钥密码算法采用RSA。
六、实验要求
1.明文为:
00112233445566778899aabbccddeeff,AES算法的密钥为:
000102030405060708090a0b0c0d0e0f1011121314151617,RSA算法的公私钥自定。
2.完成实验内容1和实验内容2。
3.对于实验内容3的实现部分,可以2人1小组,也可以1人1小组。
2人小组需要实现AES和RSA两个加密算法,而1人小组只需实现RSA加密算法即可,有余力者也可以实现两个算法。
不论是2人小组还是1人小组,实现全部两个加密算法都可以获得适当的实验加分。
AES算法要求实现效果范例:
RSA算法要求实现效果范例:
4.完成实验报告,要求有实验过程、代码实现和说明、效果截图和心得体会。
5.保存并提交这次实验的代码,命名为aes.cpp和rsa.cpp,然后放在一个独立的文件夹下。
七、实现提示和部分代码
1.关于AES算法的实现
1)如何存储明文和密钥
//数组temp保存key.
//数组temp2保存明文.
unsignedchartemp[32]=
{0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,0x10,0x11,0x12,0x13,0x14,0x15,0x16,0x17};
unsignedchartemp2[32]=
{0x00,0x11,0x22,0x33,0x44,0x55,0x66,0x77,0x88,0x99,0xaa,0xbb,0xcc,0xdd,0xee,0xff};
2)密钥扩展KeyExpansion
//Keyexpansion函数生成Nb(Nr+1)个轮密钥roundkey用来加密每轮的state
voidKeyExpansion()
{
inti,j;
unsignedchartemp[4];
for(i=0;
i<
Nk;
i++)//将密钥Key的初始值赋给轮密钥RoundKey
{
RoundKey[4*i]=Key[4*i];
RoundKey[4*i+1]=Key[4*i+1];
RoundKey[4*i+2]=Key[4*i+2];
RoundKey[4*i+3]=Key[4*i+3];
}
for(i=Nk;
Nb*(Nr+1);
i++)//生成Nb(Nr+1)个轮密钥roundkey
temp[0]=RoundKey[4*(i-1)];
temp[1]=RoundKey[4*(i-1)+1];
temp[2]=RoundKey[4*(i-1)+2];
temp[3]=RoundKey[4*(i-1)+3];
if(i%Nk==0)
{
//循环左移RotWord
chartemp1=temp[0];
temp[0]=temp[1];
temp[1]=temp[2];
temp[2]=temp[3];
temp[3]=temp1;
//S-Box操作SubWord
for(j=0;
j<
4;
j++)
{
intnum=(int)temp[j];
temp[j]=(char)getSBoxValue(num);
}
//temp=temp^Rcon
temp[0]=temp[0]^(char)Rcon[i/Nk];
}
elseif(Nk==8&
&
(i%Nk==4))//此处用于256位密钥长度时的操作
RoundKey[4*i]=RoundKey[4*(i-Nk)]^temp[0];
RoundKey[4*i+1]=RoundKey[4*(i-Nk)+1]^temp[1];
RoundKey[4*i+2]=RoundKey[4*(i-Nk)+2]^temp[2];
RoundKey[4*i+3]=RoundKey[4*(i-Nk)+3]^temp[3];
}
3)S-Box操作、逆S-Box操作和轮常数数组Rcon
//S-box操作
intgetSBoxValue(intnum)
intsbox[256]={
//0123456789ABC
DEF
0x63,0x7c,0x77,0x7b,0xf2,0x6b,0x6f,0xc5,0x30,0x01,0x67,0x2b,0xfe,0xd7,0xab,0x76,//0
0xca,0x82,0xc9,0x7d,0xfa,0x59,0x47,0xf0,0xad,0xd4,0xa2,0xaf,0x9c,0xa4,0x72,0xc0,//1
0xb7,0xfd,0x93,0x26,0x36,0x3f,0xf7,0xcc,0x34,0xa5,0xe5,0xf1,0x71,0xd8,0x31,0x15,//2
0x04,0xc7,0x23,0xc3,0x18,0x96,0x05,0x9a,0x07,0x12,0x80,0xe2,0xeb,0x27,0xb2,0x75,//3
0x09,0x83,0x2c,0x1a,0x1b,0x6e,0x5a,0xa0,0x52,0x3b,0xd6,0xb3,0x29,0xe3,0x2f,0x84,//4
0x53,0xd1,0x00,0xed,0x20,0xfc,0xb1,0x5b,0x6a,0xcb,0xbe,0x39,0x4a,0x4c,0x58,0xcf,//5
0xd0,0xef,0xaa,0xfb,0x43,0x4d,0x33,0x85,0x45,0xf9,0x02,0x7f,0x50,0x3c,0x9f,0xa8,//6
0x51,0xa3,0x40,0x8f,0x92,0x9d,0x38,0xf5,0xbc,0xb6,0xda,0x21,0x10,0xff,0xf3,0xd2,//7
0xcd,0x0c,0x13,0xec,0x5f,0x97,0x44,0x17,0xc4,0xa7,0x7e,0x3d,0x64,0x5d,0x19,0x73,//8
0x60,0x81,0x4f,0xdc,0x22,0x2a,0x90,0x88,0x46,0xee,0xb8,0x14,0xde,0x5e,0x0b,0xdb,//9
0xe0,0x32,0x3a,0x0a,0x49,0x06,0x24,0x5c,0xc2,0xd3,0xac,0x62,0x91,0x95,0xe4,0x79,//A
0xe7,0xc8,0x37,0x6d,0x8d,0xd5,0x4e,0xa9,0x6c,0x56,0xf4,0xea,0x65,0x7a,0xae,0x08,//B
0xba,0x78,0x25,0x2e,0x1c,0xa6,0xb4,0xc6,0xe8,0xdd,0x74,0x1f,0x4b,0xbd,0x8b,0x8a,//C
0x70,0x3e,0xb5,0x66,0x48,0x03,0xf6,0x0e,0x61,0x35,0x57,0xb9,0x86,0xc1,0x1d,0x9e,//D
0xe1,0xf8,0x98,0x11,0x69,0xd9,0x8e,0x94,0x9b,0x1e,0x87,0xe9,0xce,0x55,0x28,0xdf,//E
0x8c,0xa1,0x89,0x0d,0xbf,0xe6,0x42,0x68,0x41,0x99,0x2d,0x0f,0xb0,0x54,0xbb,0x16};
//F
returnsbox[num];
//逆S-box操作
intgetSBoxInvert(intnum)
intrsbox[256]=
{0x52,0x09,0x6a,0xd5,0x30,0x36,0xa5,0x38,0xbf,0x40,0xa3,0x9e,0x81,0xf3,0xd7,0xfb
0x7c,0xe3,0x39,0x82,0x9b,0x2f,0xff,0x87,0x34,0x8e,0x43,0x44,0xc4,0xde,0xe9,0xcb
0x54,0x7b,0x94,0x32,0xa6,0xc2,0x23,0x3d,0xee,0x4c,0x95,0x0b,0x42,0xfa,0xc3,0x4e
0x08,0x2e,0xa1,0x66,0x28,0xd9,0x24,0xb2,0x76,0x5b,0xa2,0x49,0x6d,0x8b,0xd1,0x25
0x72,0xf8,0xf6,0x64,0x86,0x68,0x98,0x16,0xd4,0xa4,0x5c,0xcc,0x5d,0x65,0xb6,0x92
0x6c,0x70,0x48,0x50,0xfd,0xed,0xb9,0xda,0x5e,0x15,0x46,0x57,0xa7,0x8d,0x9d,0x84
0x90,0xd8,0xab,0x00,0x8c,0xbc,0xd3,0x0a,0xf7,0xe4,0x58,0x05,0xb8,0xb3,0x45,0x06
0xd0,0x2c,0x1e,0x8f,0xca,0x3f,0x0f,0x02,0xc1,0xaf,0xbd,0x03,0x01,0x13,0x8a,0x6b
0x3a,0x91,0x11,0x41,0x4f,0x67,0xdc,0xea,0x97,0xf2,0xcf,0xce,0xf0,0xb4,0xe6,0x73
0x96,0xac,0x74,0x22,0xe7,0xad,0x35,0x85,0xe2,0xf9,0x37,0xe8,0x1c,0x75,0xdf,0x6e
0x47,0xf1,0x1a,0x71,0x1d,0x29,0xc5,0x89,0x6f,0xb7,0x62,0x0e,0xaa,0x18,0xbe,0x1b
0xfc,0x56,0x3e,0x4b,0xc6,0xd2,0x79,0x20,0x9a,0xdb,0xc0,0xfe,0x78,0xcd,0x5a,0xf4
0x1f,0xdd,0xa8,0x33,0x88,0x07,0xc7,0x31,0xb1,0x12,0x10,0x59,0x27,0x80,0xec,0x5f
0x60,0x51,0x7f,0xa9,0x19,0xb5,0x4a,0x0d,0x2d,0xe5,0x7a,0x9f,0x93,0xc9,0x9c,0xef
0xa0,0xe0,0x3b,0x4d,0xae,0x2a,0xf5,0xb0,0xc8,0xeb,0xbb,0x3c,0x83,0x53,0x99,0x61
0x17,0x2b,0x04,0x7e,0xba,0x77,0xd6,0x26,0xe1,0x69,0x14,0x63,0x55,0x21,0x0c,0x7d};
returnrsbox[num];
//轮常数数组,Rcon[i]
intRcon[255]={
0x8d,0x01,0x02,0x04,0x08,0x10,0x20,0x40,0x80,0x1b,0x36,0x6c,0xd8,0xab,0x4d,0x9a,
0x2f,0x5e,0xbc,0x63,0xc6,0x97,0x35,0x6a,0xd4,0xb3,0x7d,0xfa,0xef,0xc5,0x91,0x39,
0x72,0xe4,0xd3,0xbd,0x61,0xc2,0x9f,0x25,0x4a,0x94,0x33,0x66,0xcc,0x83,0x1d,0x3a,
0x74,0xe8,0xcb,0x8d,0x01,0x02,0x04,0x08,0x10,0x20,0x40,0x80,0x1b,0x36,0x6c,0xd8,
0xab,0x4d,0x9a,0x2f,0x5e,0xbc,0x63,0xc6,0x97,0x35,0x6a,0xd4,0xb3,0x7d,0xfa,0xef,
0xc5,0x91,0x39,0x72,0xe4,0xd3,0xbd,0x61,0xc2,0x9f,0x25,0x4a,0x94,0x33,0x66,0xcc,
0x83,0x1d,0x3a,0x74,0xe8,0xcb,0x8d,0x01,0x02,0x04,0x08,0x10,0x20,0x40,0x80,0x1b,
0x36,0x6c,0xd8,0xab,0x4d,0x9a,0x2f,0x5e,0xbc,0x63,0xc6,0x97,0x35,0x6a,0xd4,0xb3,
0x7d,0xfa,0xef,0xc5,0x91,0x39,0x72,0xe4,0xd3,0xbd,0x61,0xc2,0x9f,0x25,0x4a,0x94,
0x33,0x66,0xcc,0x83,0x1d,0x3a,0x74,0xe8,0xcb,0x8d,0x01,0x02,0x04,0x08,0x10,0x20,
0x40,0x80,0x1b,0x36,0x6c,0xd8,0xab,0x4d,0x9a,0x2f,0x5e,0xbc,0x63,0xc6,0x97,0x35,
0x6a,0xd4,0xb3,0x7d,0xfa,0xef,0xc5,0x91,0x39,0x72,0xe4,0xd3,0xbd,0x61,0xc2,0x9f,
0x25,0x4a,0x94,0x33,0x66,0xcc,0x83,0x1d,0x3a,0x74,0xe8,0xcb,0x8d,0x01,0x02,0x04,
0x08,0x10,0x20,0x40,0x80,0x1b,0x36,0x6c,0xd8,0xab,0x4d,0x9a,0x2f,0x5e,0xbc,0x63,
0xc6,0x97,0x35,0x6a,0xd4,0xb3,0x7d,0xfa,0xef,0xc5,0x91,0x39,0x72,0xe4,0xd3,0xbd,
0x61,0xc2,0x9f,0x25,0x4a,0x94,0x33,0x66,0xcc,0x83,0x1d,0x3a,0x74,0xe8,0xcb};
4)列混合变换和逆列混合变换
//xtime是计算16进制乘以{02}的宏
#definextime(x)((x<
<
1)^(((x>
>
7)&
1)*0x1b))
//Multiplty是计算在域GF(2^8)中的数字乘法的宏
#defineMultiply(x,y)(((y&
1)*x)^((y>
1&
1)*xtime(x))^((y>
2&
1)*xtime(xtime(x)))^((y>
3&
1)*xtime(xtime(xtime(x))))^((y>
4&
1)*xtime(xtime(xtime(xtime(x))))))
//Matrix是列混合变换所需的矩阵
unsignedcharMatrix[16]=
{0x02,0x03,0x01,0x01,0x01,0x02,0x03,0x01,0x01,0x01,0x02,0x03,0x03,0x01,0x01,0x02};
//InvMatrix是逆列混合变换所需的矩阵
unsignedcharInvMatrix[16]=
{0x0e,0x0b,0x0d,0x09,0x09,0x0e,0x0b,0x0d,0x0d,0x09,0x0e,0x0b,0x0b,0x0d,0x09,0x0e};
//列混合变换
voidMixColumns()
charstate1[4][4];
Nb;
i++)//先将state复制给state1,state是明文变换为密文过程中的中间结果
for(j=0;
state1[i][j]=state[i][j];
i++)
//将state1和矩阵Matrix的乘积赋给state,state是明文变换为密文过程中的中间结果
state[j][i]=Multiply(state1[0][i],Matrix[4*j])^Multiply(state1[1][i],Matrix[4*j+1])^Multiply(state1[2][i],Matrix[4*j+2])^Multiply(state1[3][i],Matrix[4*j+3]);
//逆列混合变换
voidInvMixColumns()
//将state1和矩阵InvMatrix的乘积赋给state,state是明文变换为密文过程中的中间结果
state[j][i]=
Multiply(state1[0][i],InvMatrix[4*j])^Multiply(state1[1][i],InvMatrix[4*j+1])^
Multiply(state1[2][i],InvMatrix[4*j+2])^Multiply(state1[3][i],InvMatrix[4*j+3]);
5)加密函数
//加密函数.
voidCipher()
inti=0;
intj=0;
intk=0;
intround=0;
//明文in赋值给state,state是明文变换为密文过程中的中间结果
state[j][i]=in[k];
//state(i,j)->
state(j,i)
k++;
AddRoundKey(round);
//前Nr-1轮加密
for(round=1;
round<
Nr;
round++)
SubBytes();
ShiftRows();
MixColumns();
AddRoundKey(round);
//最后一轮加密
SubBytes();
ShiftRows();
AddRoundKey(Nr);
k=0;
//密文state赋值给o
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 实验1 安全数据传输综合实验 实验 安全 数据传输 综合