本帖最后由 Ericky 于 2017-12-1 20:49 编辑
前言
从第一题题目来看，感觉这次的题目都出得很灵活，主要考的是密码学这一块的东西。这道题目也不例外，由于不是科班出身，根本没有接触过密码学，那只能现学现卖，其中踩了不少坑，但是不要怕，我们可以慢慢爬出来再继续踩。初一看这道题后面的“双重保镖”很是霸气，很容易被吓跑，但是仔细认真做了之后会发现，Oh，hash碰撞也不是那么难的。好了，废话扯多了估计要被拉黑，我选择的依旧是安卓的题目，下面进入正题，我将此题分成两大部分，主要阐述这两大部分。
0x1 拉开帷幕
官方具体表演如下：
[Asm] 纯文本查看 复制代码 v17 = 0;
v15 = "aLSG";
v16 = "71b";
v19 = 0;
v20 = 0;
v21 = 0;
v22 = 0;
v18 = getpid();
v0 = dlopen("/data/local/tmp/libZapus.so", 0);
v1 = v0;
if ( !v0 )
goto LABEL_18;
v2 = (void (__fastcall *)(int *))dlsym(v0, "zapus_get");
if ( !v2 )
goto LABEL_18;
v2(&v19);
dlclose(v1);
((void (__fastcall *)(int *))sub_F90)(&v19);
v3 = memcmp(&v19, &v15, 0x10u);
if ( v3 )
goto LABEL_18;
v4 = fopen("/data/local/tmp/libZapus.so", "rb");
v5 = v4;
if ( !v4 )
goto LABEL_18;
fseek(v4, 0, 2);
v6 = ftell(v5);
fseek(v5, 0, 0);
v7 = malloc(v6);
v8 = v7;
if ( !v7 )
1.判断指定路径有没有so，没有或者是打开失败就跪。
2.找zapus_get符号，符号表里没有就跪。
3.有了的话，把该函数初始化的指针带进到图中的sub_F90函数，然后开始计算。
4.小插曲，这里说一下这个题目第一次发的有一个bug，解密后的代码与固定两个字符串的指针进行了比较，导致每次对比的值是变的，我做到这里的时候怀疑了一下人生。what？比较的值还是变的？还不能改程序内存？在心中无限崩溃循环中还好官网出来及时修补了bug。
计算过程如下:
[Asm] 纯文本查看 复制代码 unsigned char input[] = "1234567890abcdef";
unsigned char tmp;
for ( j = 0; (unsigned int) j < 0x80; j++ )
{
tmp = 0;
for ( k = 0; (unsigned int) k < 0x80; ++k )
{
tmp = ((((signed int)(unsigned char)buffer[16 * j + k / 8] >> k % 8) & ((signed int)*( (unsigned char *) input + k / 8) >> (7 - k % 8) ) ) ^ tmp) & 1;
}
s[j/8] |= tmp << (7 - j % 8);
}
实际上第一步要找出这段算法的逆运算。程序中有一个buffer，简算过后如下：
[Asm] 纯文本查看 复制代码 unsigned char buffer[] 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程序将输入的16bytes(也就是128bit)，记为input，buffer的大小是2048bytes，分成了128个128bit，记为x[0]到x[127]，根据之前的运算，input将和x数组进行128次运算，每一次运算生成1bit，一共生成128个bit，即16字节，这16字节就是最后算出来的结果，也是需要比较的结果。
其实是一个128元1次的异或方程组，采用高数中的高斯消元法来求解，这个就算没有学过密码学也会，因为这就是高数中的内容啊，看到这个名字是不是很熟悉？对，就是矩阵化简！对于高数课逃课的同学，关于高斯消元法，给个链接：
https://en.wikipedia.org/wiki/Gaussian_elimination
这里已经非常详细
然后根据本题目的特征google一下就能搜到不少解该类方程组的模板，选取一个，代码如下：
[Asm] 纯文本查看 复制代码 int Gauss(){
int max_r,col,k;
free_num=0;
for(k=0,col=0;k<equ&&col<var;k++,col++){
max_r=k;
for(int i=k+1;i<equ;i++){
if(abs(a[i][col])>abs(a[max_r][col]))
max_r=i;
}
if(a[max_r][col]==0){
k--;
free_x[free_num++]=col;
continue;
}
if(max_r!=k){
for(int j=col;j<var+1;j++)
swap(a[k][j],a[max_r][j]);
}
for(int i=k+1;i<equ;i++){
if(a[i][col]!=0){
for(int j=col;j<var+1;j++)
a[i][j]^=a[k][j];
}
}
}
for(int i=k;i<equ;i++)
if(a[i][col]!=0)
return -1; //无解
printf("you jie k:%d var :%d \n",k,var);
if(k<var) return var-k; //解不唯一，返回解个数
//解唯一，生成解集
for(int i=var-1;i>=0;i--){
x[i]=a[i][var];
for(int j=i+1;j<var;j++)
x[i]^=(a[i][j]&&x[j]);
printf("%d",x[i] );
}
return 0;
}
再根据IDA中看到的信息，确认正确答案的前面7字节是"GSLab17"，最后四字节是pid，编写代码如下：
[Asm] 纯文本查看 复制代码 void * zapus_get(void *abc)
{ int j,k,i;
int pid = getpid();
LOGI("pid:%d\n",pid );
unsigned char key[16]= "GSLab17";
unsigned char s[16] = {0};
key[7] = 0;
*((int *)key+2) = 0;
*((int *)key+3) = pid;
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
//***initial zeng guang ju zhen
for ( j = 0; (unsigned int) j < 0x80; j++ )
{
for ( k = 0; (unsigned int) k < 0x80; ++k )
{
a[j][k] = (((signed int)(unsigned char)buffer[16 * j + k / 8] >> k % 8) & 1);
}
}
for ( k = 0; (unsigned int) k < 0x80; ++k )
{
a[k][128] =((signed int)*( (unsigned char *) key + k / 8) >> (7 - k % 8) ) & 1;
}
k = Gauss();
for(int i=0;i<128;i++){
s[i/8] |= x[i] << (7 - i % 8);
}
LOGI("HELLO I'm here.");
memcpy(abc, s, 16);
}
再调试看看结果，自己挖了什么坑自己填填坑，即可过第一关，进入下一关。
0x2 孪生兄弟
接下来来到孪生兄弟把守的最后大关，只要越过这道防线，就可以飞向光明顶。
之所以说它是孪生兄弟，是因为它对程序进行了hash值的校验，而且是两种校验，两个校验值都必须值为0x614C5347，其实就是“GSLa”。
什么是hash函数？
Hash函数H将可变长度的数据M作为输入，产生固定长度的Hash值h。
Hash函数，哈希函数，散列函数，杂凑函数它们说的都是同一个含义，后续我们都称之为Hash函数。
定义
Hash函数H将可变长度的数据M作为输入，产生固定长度的Hash值h。
Hash函数，哈希函数，散列函数，杂凑函数它们说的都是同一个含义，后续我们都称之为Hash函数。
h=H(M)
单向性
给定输入M，通过函数H可以很容易计算出输出h；但如果给定h，则找到M在计算上不可行。
数据完整性
输入数据M中任何1个bit发生变化，都将导致输出M发生很大的变化。
Hash冲突（突破本题的关键）
在Hash函数中，M称之为h原像,，因为H函数是一个多对一的映射,，对于任意给定的Hash数值h,，可能会有多个原像,，如果满足如下条件, 则称之为发生了哈希碰撞，也就是哈希冲突。
x!=yandH(x)==H(y)
一个优良的Hash函数必须满足如下几个性质:
任意y，找x，使得H(x) = y，非常困难
给定x1, 找x2, 使得H(x1) == H(x2), 非常困难
找任意的x1, x2, 使得H(x1) == H(x2), 非常困难
这里感觉本题是第一条或者第二条，如果是第三条的话（可惜本题并不是），可以参考生日定理来解题，这里放个链接：
https://en.wikipedia.org/wiki/Birthday_problem
介绍完基本概念，言归正传，本题用的两个校验是crc32校验和fnvhash函数校验，这里再给两个链接：
crc32：http://blog.csdn.net/zhaodm/article/details/3711034
fnvhash：http://blog.csdn.net/taochenchang/article/details/7319739
通过分析代码可以得到：
1.crc32 校验
标准算法没改动过：
[Asm] 纯文本查看 复制代码 unsigned long calc_crc(unsigned char *p, unsigned int len)
{
register unsigned long crc;
unsigned int count=0;
crc = 0xFFFFFFFF;
while (count < len){
crc = (crc>>8) ^ crc_table[ (crc^p[count++]) & 0xFF ];
}
return( crc^0xFFFFFFFF );
}
单独改crc32是没问题的.
2.fnv_hash
标准算法没改动过
[Asm] 纯文本查看 复制代码 unsigned long calc_fnvhash(unsigned char *p, unsigned int len)
{
register unsigned long hash;
unsigned int count=0;
hash = 0x811C9DC5;
while (count < len){
hash = 0x1000193 * (p[count++] ^ hash);
}
return hash;
}
因为crc32的值是可以求出来的，再给个链接，其实这篇文章最早在看雪：
http://www.cnblogs.com/thinksea/articles/2024199.html
所以思路就变成了暴力碰撞fnvhash的值即可。
分析完了，写代码进行碰撞，关键代码：
[Asm] 纯文本查看 复制代码 //8字节碰撞
unsigned char byte[4] ;
unsigned char byte2[4] ;
srand( (unsigned) time(NULL) * getpid());
long i =0;
while(1)
{
byte[0]=rand()%256 + byte[1] ;
byte[1]=rand()%256 + byte[2];
byte[2]=rand()%256 + byte[3]; ;
byte[3]=rand()%256 + byte2[0]; ;
byte2[0]=rand()%256 + byte2[1];
byte2[1]=rand()%256 + byte2[2];
byte2[2]=rand()%256 + byte2[3] ;
byte2[3]=rand()%256 + byte[0] ;
*(unsigned int*)(filemap+offset -4) = *(unsigned int*)byte;
*(unsigned int*)(filemap+offset -8) = *(unsigned int*)byte2;
tweakcrc(filemap, filestat.st_size, crc, offset); //修改crc32 值为固定值
fnvhash=calc_fnvhash_op(filemap+filestat.st_size -12, 12);
if (fnvhash >= 0x614C5340 && fnvhash <= 0x614C5349)
{
printf("$$$$$$$$$$$$$$ near success $$$$$$$$$$$$$\n");
printf("%x %x %x %x\n", byte[0],byte[1],byte[2],byte[3]);
printf("%x %x %x %x\n", byte2[0],byte2[1],byte2[2],byte2[3]);
printf("calc_fnvhash_op: 0x%X\n", fnvhash);
printf("$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\n");
if (fnvhash == 0x614C5347)
{
printf("********************************\n");
printf("**************success***********\n");
printf("********************************\n");
printf("%x %x %x %x\n", byte[0],byte[1],byte[2],byte[3]);
printf("%x %x %x %x\n", byte2[0],byte2[1],byte2[2],byte2[3]);
printf("calc_fnvhash_op: 0x%X\n", fnvhash);
break;
}
}
i++;
if (i % 100000000 == 0)
{
printf("已经碰撞%ld次 继续努力\n", i);
}
}
优化后的crc32和fnvhash函数的计算，代码对比如下，这里的值只针对我自己的so进行的优化：
[Asm] 纯文本查看 复制代码 //crc
unsigned long calc_crc(char *p, unsigned int len)
{
register unsigned long crc;
unsigned int count=0;
crc = 0xFFFFFFFF;
while (count < len){
crc = (crc>>8) ^ crc_table[ (crc^p[count++]) & 0xFF ];
}
return( crc^0xFFFFFFFF );
}
unsigned long calc_crc_op(char *p, unsigned int len)
{
register unsigned long crc;
unsigned int count=0;
crc = 0x81820a5b;
while (count < len){
crc = (crc>>8) ^ crc_table[ (crc^p[count++]) & 0xFF ];
}
return( crc^0xFFFFFFFF );
}
//fnvhash
unsigned long calc_fnvhash(unsigned char *p, unsigned int len)
{
register unsigned long hash;
unsigned int count=0;
hash = 0x811C9DC5;
while (count < len){
hash = 0x1000193 * (p[count++] ^ hash);
}
return hash;
}
unsigned long calc_fnvhash_op(unsigned char *p, unsigned int len)
{
register unsigned long hash;
unsigned int count=0;
hash = 0x799362BC;
while (count < len){
hash = 0x1000193 * (p[count++] ^ hash);
}
return hash;
}
这里再说明一下，一开始没有优化算法，导致计算缓慢（被卡了一波，标记，其实也不是没做出来的主要原因，自己挖了个坑，周末一直在挂机，没注意到错误，以为hash碰撞很难，但是过了很久没跑出结果就感觉不太对了，于是我就跟了一下，解决掉大坑，到达胜利光明顶），优化算法后效率提升了大概2100倍，大概1秒钟碰撞1E次，瞬间得出解。跑了大概半分钟，得出4个解（感觉任意指定hash都可以算，然后再感觉一天给任意hash1-2W个解也是没有压力的），如下给出四组附加bytes：
[Asm] 纯文本查看 复制代码 1.
9a 14 7 99
56 8c 8 1f
calc_fnvhash_op: 0x614C5347
2.
84 1e 16 23
92 18 3c 34
calc_fnvhash_op: 0x614C5347
3.
5a f5 69 0
a0 d1 39 29
calc_fnvhash_op: 0x614C5347
4.
28 f0 98 98
a1 82 b0 e7
calc_fnvhash_op: 0x614C5347
经过此役，发现hash碰撞很有趣，做题使人快速学习新东西，复习老东西。
谢谢观看，不足之处请看官斧正！
2017.7.15
By Ericky |
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发表于 2017-12-1 20:43
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发表于 2017-12-4 17:33
大牛