本题防守方采用了多层VM的设计思路，比赛中VM常见，多层VM罕见，让攻击者欲罢不能。
当攻击者们在寻找快速方案如手解等方式拨开第一层VM后，发现后面还是VM，套娃感凸现，一望无际，定力稍有不慎，可能就会被吓住。这时候需要攻击者使用各种看家本领透视VM业务逻辑，层层击破，得到可阅读的代码。
当VM面纱都揭开时，高层VM对底层VM的运行时修改以及暗藏逻辑的触发联动机制也需要一一解开，才能联合起来得到完整的验证逻辑进行解题。高层VM对支撑其运作的低层VM修改和联动，除了套娃感，也略带虚拟逃逸感。总体设计非常优秀，相信各位攻击者都有所得。
设计思路由作者 ccfer 提供
LightsOut求解直接用sage也是迅速得出结果。
crc暴力穷举会慢点，不过4字节得相当于可逆，有快速破解方法。
设计思路由作者 k1ee 提供
struct vm_sub{int param1; //6, 6, 6, 3int param2; //ins1, ins2, ins3, input_hexchar* vm_ins;int size;int idk_0;int id;int idk7;int idk8;};struct vm_fin{unsigned char* input_hex;int* len_buf;vm_sub* vmsub;};struct vm_context{vm_sub subs[4];vm_fin fin;};
...void disassembly_vm1(vm_sub* ctx){char* eip = ctx->vm_ins;char* esp = eip + 2 * ctx->size;ks_engine* ks;ks_err err;err = ks_open(KS_ARCH_X86, KS_MODE_32, &ks);if (err != KS_ERR_OK){cout << "Keystone open error." << endl;return;}ostringstream dasm = ostringstream();dasm << "push 6;" << endl;dasm << "push 0x20000;" << endl;dasm << "call vmins_0;" << endl;dasm << "jmp vmins_ret;" << endl;while (eip < ctx->vm_ins + 0x792){int vm_offset = eip - ctx->vm_ins;dasm << "vmins_" << vm_offset << ":" << endl;int ins = *eip++;switch (ins){case 17:{dasm << "push ebx;" << endl;break;}......完整源码在此处查看：https://bbs.pediy.com/thread-263829.htm
int vm3_memcpy(char* dst, char* src, int len){vm_fin* v18 = &ctx.fin;unsigned char * v17 = v18->input_hex;int* v16 = new int[10];v16[2] = 0xDEC0CCAE;*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);*(v16 + 2) = crc32(0xFFFFFFFF, v16 + 2);if (*(v16 + 2) == 0xDE05629C)return 1;return 0;}
result = d54b1112 target = de05629cresult = 5ba49ea3 target = d54b1112result = f16f3846 target = 5ba49ea3result = 84b4f299 target = f16f3846result = 3731ce56 target = 84b4f299result = 74f3e321 target = 3731ce56result = 20558f1 target = 74f3e321result = dec0ccae target = 20558f1result = f812fce7 target = dec0ccae
int vm3_memset_1(char* dst, char val, int len){vm_fin* v18 = &ctx.fin;char* hex = (char *)v18->input_hex;int* v16 = v18->len_buf;int* v6 = v16 + 4;*v6 = sub_E21(hex); // equals D540v6 = v16 + 2;*v6 = sub_109A(hex + 4, (char*)v16 + 256);memset(v16 + 1024, 1, 100);sub_1517((char*)v16 + 256, (char*)v16 + 4096);return 0;}
unsigned int __cdecl sub_109A(char* a1, char* a2){int v3; // [esp+3ECh] [ebp-14h]char* v4; // [esp+3F0h] [ebp-10h]unsigned __int8 v5; // [esp+3F4h] [ebp-Ch]unsigned int v6; // [esp+3F8h] [ebp-8h]unsigned int v7; // [esp+3FCh] [ebp-4h]v4 = a2;v7 = 0;v3 = 0;while (v7 < 15){v6 = 0;v5 = a1[v7];v3 <<= 1;v3 |= (unsigned int)v5 >> 7;while (v6 < 7){*v4 = v5 & 1;v5 >>= 1;++v4;++v6;}++v7;}return (((v3 << 8) + ((unsigned int)(unsigned __int8)a1[14] >> 2)) << 8) + (unsigned __int8)a1[15];}
AE CC C0 DE 0C 32 56 F7 5E 37 A6 BF A2 27 A2 ED 3D 54 AC 96 4B 43 54 46 32 30 32 30 46 6C 61 67int vm3_memset_2(char* dst, char val, int len){vm_fin* v18 = &ctx.fin;char* v17 = (char *)v18->input_hex;int* v16 = v18->len_buf;int* v6 = v16 + 2;*v6 = sub_179A(v17 + 32, (char *)v16 + 64*4);if (*v6 == 2){memset(v16 + 1024, 1, 1600);sub_2FB9((char*)v16 + 4*64, (char*)v16 + 4*1024);}return 0;}
uint32_t limit = pow(2, 20) - 1;for (uint32_t val = 0; val <= limit; ++val){for (int i = 0; i < 20; ++i){uint8_t bit = (val >> i) & 0b1;mat[0][i] = bit;mat[0][39 - i] = bit;}memset(table, 1, sizeof(table));for (int i = 0; i < 40; ++i){for (int k = 0; k < 40; ++k){uint8_t bit = mat[i][k];table[i][k] ^= bit;if (i > 0)table[i - 1][k] ^= bit;if (i < 39)table[i + 1][k] ^= bit;if (k > 0)table[i][k - 1] ^= bit;if (k < 39)table[i][k + 1] ^= bit;}if (i != 39){for (int k = 0; k < 40; ++k){mat[i + 1][k] = table[i][k];}}}if (memcmp(table, truth, sizeof(truth)) == 0){printf("Resultn");printf("arr = []n");for (int i = 0; i < 40; ++i){printf("arr.append([");for (int k = 0; k < 40; ++k){printf("%d%s", mat[i][k], k == 39 ? "" : ", ");}printf("]n");}}}
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本文始发于微信公众号（看雪学院）：2020 KCTF秋季赛 | 第五题设计思路及解题思路
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