使用trace进行算法还原
本帖最后由 neilwu 于 2021-11-2 11:51 编辑## 一、题目
### 1、android 逆向题
小伙伴发来的题，说是某比赛线下题目
### 2、涉及算法
#### AES
#### 魔改RC5
## 二、解题过程
### 1、java层
### 2、native层
sub_F08
```
int __fastcall sub_F08(JNIEnv *a1, int a2, int a3)
{
const char *v3; // r9
size_t v4; // r6
char *v5; // r4
void *v6; // r8
_DWORD *v7; // r5
int v8; // r9
unsigned __int8 *v9; // r6
int v10; // r0
char *v11; // r1
int v12; // r2
int v13; // t1
int v14; // t1
v3 = (*a1)->GetStringUTFChars(a1, a3, 0);
v4 = strlen(v3);
v5 = malloc(v4);
v6 = malloc(4 * v4 + 64);
v7 = malloc(0x20u);
*v7 = 0x1010101;
v7 = 0x1010101;
v7 = 0x1010101;
v7 = 0x1010101;
qmemcpy(v5, v3, v4);
v8 = sub_D68(v6, v5, v4, v7);
free(v5);
memset(v7, 2, 0x20u);
v9 = malloc(4 * v8 + 32);
v10 = sub_E80(v9, v6, v8, v7);
v11 = &c_cipher;
while ( v10 )
{
v13 = *v11++;
v12 = v13;
--v10;
v14 = *v9++;
if ( v14 != v12 )
return 0;
}
return 1;
}
```
#### （1）对伪码进行分析，流程如下
```
第一加密 v8 = sub_D68(v6, v5, v4, v7);
第二次加密 v10 = sub_E80(v9, v6, v8, v7);
解密结果与 &c_cipher进行比较
```
#### （2）动态调试
##### c_cipher
```
unsigned char c_cipher = {
0xCA, 0x60, 0x55, 0x30, 0xB5, 0xDB, 0xD4, 0xA6, 0x01, 0x15, 0x3F, 0xB8, 0xBC, 0x4C, 0x9C, 0x88,
0xEA, 0xF4, 0x76, 0xDD, 0x8D, 0x7B, 0x1A, 0x26, 0xDA, 0x74, 0x2C, 0x1D, 0x28, 0x63, 0x4B, 0x88,
0x44, 0x22, 0x7E, 0x21, 0x0E, 0x6C, 0xF4, 0xAE, 0xE4, 0x21, 0xC7, 0x67, 0x21, 0x40, 0xC5, 0x3B,
0xB2, 0x55, 0x92, 0x21, 0x9B, 0x29, 0xFA, 0x33
};
```
##### 第一加密 v8 = sub_D68(v6, v5, v4, v7);
AES的S盒
```
输入 flag{0123456789}
key 01010101010101010101010101010101
加密结果
7E D8 C7 01 11 B4 88 270E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E03 B7 5B B1 5B C9 4B 6C
```
验证一下
##### 第二次加密 v10 = sub_E80(v9, v6, v8, v7);
```
输入
7E D8 C7 01 11 B4 88 270E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E03 B7 5B B1 5B C9 4B 6C
key
02020202020202020202020202020202
02020202020202020202020202020202
加密结果
8B A8 1D 1F DF E9 09 9878 B0 0A CA F7 8F DE 1D
56 6F B4 65 E0 32 DC BABE 7B 8A 81 A1 41 59 BA
B2 55 92 21 9B 29 FA 33
```
###### sub_E80
```
unsigned int __fastcall sub_E80(int a1, int a2, int a3, int a4)
{
unsigned int result; // r0
int i; // r10
int v9; // r9
int v10; // r1
int v11; // r4
int j; // r5
__int64 v13; // kr00_8
sub_DA8(a4);
result = sub_994(a1, a2, a3, 8);
for ( i = 0; i != result >> 3; ++i )
{
v9 = 2 * i + 1;
v10 = *(a1 + 8 * i) + dword_5E50;
v11 = *(a1 + 4 * v9) + unk_5E54;
for ( j = 0; j != 12; ++j )
{
v13 = *&dword_5E50;
v10 = __ROR4__(v10 ^ v11, 32 - v11) + v13;
v11 = __ROR4__(v10 ^ v11, 32 - v10) + HIDWORD(v13);
}
*(a1 + 8 * i) = v10;
*(a1 + 4 * v9) = v11;
}
return result;
}
```
我使用trace进行的算法还原
###### 对trace进行分析，算法如下
```
输入
7E D8 C7 01 11 B4 88 270E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E03 B7 5B B1 5B C9 4B 6C
加密结果
8B A8 1D 1F DF E9 09 9878 B0 0A CA F7 8F DE 1D
56 6F B4 65 E0 32 DC BABE 7B 8A 81 A1 41 59 BA
B2 55 92 21 9B 29 FA 33
```
输入由32补全40位，08补全
```
输入
7E D8 C7 01 11 B4 88 270E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E03 B7 5B B1 5B C9 4B 6C
08 08 08 08 08 08 08 08
```
输入长度 40 >> 3 = 5
进行了5次大循环，每次循环进行了12+1次
每次小循环使用的数组：
```
uint32_t dword_C30DDE50[] = {
0x751EF375, 0x9A777FE8,
0xCF327F57, 0x378B39DC,
0xB0CF04C8, 0xA0AA7A08,
0xCE01CEA5, 0xF400183,
0x58FBD0BB, 0x8D44D42A,
0x59F0BB54, 0xB632AA3A,
0x40030399, 0xA9C58CAC,
0xB1F1F560, 0x29796A99,
0x513FE4A6, 0x8204DA32,
0xC435BFCA, 0x465706D,
0xECB351E9, 0x9A83C02C,
0x4615414F, 0x8728979D,
0x8CED0CFA, 0x15647C9C};
```
逆推
还原算法
```
uint32_t dword_C30DDE50[] = {
0x751EF375, 0x9A777FE8,
0xCF327F57, 0x378B39DC,
0xB0CF04C8, 0xA0AA7A08,
0xCE01CEA5, 0xF400183,
0x58FBD0BB, 0x8D44D42A,
0x59F0BB54, 0xB632AA3A,
0x40030399, 0xA9C58CAC,
0xB1F1F560, 0x29796A99,
0x513FE4A6, 0x8204DA32,
0xC435BFCA, 0x465706D,
0xECB351E9, 0x9A83C02C,
0x4615414F, 0x8728979D,
0x8CED0CFA, 0x15647C9C};
uint32_t plaintext[] = {0x01C7D87E, 0x2788B411, 0x310B420E, 0x6342CB59, 0x374D4368, 0x0E9FF6D7, 0xB15BB703, 0x6C4BC95B, 0x08080808, 0x08080808};
int len = sizeof(plaintext) / sizeof(uint32_t);
for(int i = 0; i < len/2; i++){
int R1 = plaintext;
int R4 = plaintext;
int R2,R5,R6,R9;
int R3=0xC30DDE50;
for(int j = 0; j < 13; j++){
if(j == 0){
R1 = R1 + dword_C30DDE50;
R4 = R4 + dword_C30DDE50;
}else{
R1 = R1 ^ R4;
R6 = 0x20 - R4;
R1 = ror(R1, R6);
R2 = dword_C30DDE50;
R6 = dword_C30DDE50;
R1 = R1 + R2;
R2 = R1 ^ R4;
R4 = 0x20 - R1;
R2 = ror(R2, R4);
R4 = R2 + R6;
}
}
printf("0x%x,0x%x, \n", R1, R4);
}
```
对加密算法进行逆运算
```
uint32_t dword_C30DDE50[] = {
0x751EF375, 0x9A777FE8,
0xCF327F57, 0x378B39DC,
0xB0CF04C8, 0xA0AA7A08,
0xCE01CEA5, 0xF400183,
0x58FBD0BB, 0x8D44D42A,
0x59F0BB54, 0xB632AA3A,
0x40030399, 0xA9C58CAC,
0xB1F1F560, 0x29796A99,
0x513FE4A6, 0x8204DA32,
0xC435BFCA, 0x465706D,
0xECB351E9, 0x9A83C02C,
0x4615414F, 0x8728979D,
0x8CED0CFA, 0x15647C9C};
uint32_t cipher[] = {0x305560CA, 0xA6D4DBB5, 0xB83F1501, 0x889C4CBC, 0xDD76F4EA, 0x261A7B8D,
0x1D2C74DA, 0x884B6328, 0x217E2244, 0xAEF46C0E, 0x67C721E4, 0x3BC54021,
0x219255B2, 0x33FA299B};
// 解密
int len2 = sizeof(cipher) / sizeof(uint32_t);
for(int i = 0; i< len2/2; i++){
int R1_cipher = cipher;
int R4_cipher = cipher;
int R1, R4;
for(int j = 12; j >=0; j--){
if(j == 0){
R1_cipher = R1_cipher - dword_C30DDE50;
R4_cipher = R4_cipher - dword_C30DDE50;
}else{
R4_cipher = rol(R4_cipher - dword_C30DDE50, 0x20 - R1_cipher) ^ R1_cipher;
R1_cipher = rol(R1_cipher - dword_C30DDE50, 0x20 - R4_cipher) ^ R4_cipher;
}
}
printf("%x %x \n", R1_cipher, R4_cipher);
}
```
使用AES Decrypt
vcdemon 发表于 2021-11-15 11:58
1 请问是怎么看出来并确定是aes加密的呢？是否有某种技巧可以判断出来是用的什么加密方法？
2 trace是个什 ...
顶这个帖子 我也没看到作者对于itrace有很清晰的定义。我认为这个分析不是特别solid 1 请问是怎么看出来并确定是aes加密的呢？是否有某种技巧可以判断出来是用的什么加密方法？
2 trace是个什么东西呢？是某个工具还是什么？能否简单介绍下呢？
感谢大佬解惑。 谢谢分享学习了~~~~~~~~~~~~~~ 感谢分享，学习去了 太厉害了，完全不懂 目前还在加密加壳方面研究， 算法还没开始接触 感谢分享 学习了
谢谢分享学习了~~~ 不错，看看大佬的思路 感谢分享 学习了