Writeup of the reverse challenge Opacity in DEFCON2023 Quals: A DRM-protected circuit emulation.

I found this challenge to be highly intriguing, as it took us a considerable amount of time to unravel its true purpose. The implementation of a fundamental gate in this challenge was particularly fascinating, which is why I felt compelled to create a detailed writeup for it.

DRM Protected Environment (init_drm)⌗

The DRM is actually a patched qemu-aarch64 which will use user provided license as PAC key.

This is how PAC key is set:

  • license is a 16-bytes sequence, which will applied to qemu_base+0x48D390 @ .rodata
  • qemu_base+0x22BCCC(do_prctl_reset_keys) is replaced with code that perform memcpy from qemu_base+0x48D390 to PAC_KEY_B (&env->keys.apib)
  • A runtime (init_drm) perform patch and exec itself to a patched qemu and run circuit emulator (run_prog)

run_prog (Circuit Emulator)⌗

PAC as NAND⌗

Using PAC operation as a way to inhibit the nature of bitwise operation is such a cool idea.

Here’s how PAC as NAND is made:

Every gate have two output state (0 or 1):

  • Use a random number to represent the state 0 or 1 Building NAND gate:
  1. Suppose GateA and GateB both output 1, therefore we have two random that represent 1 here (we called it gateA_rand_1 and gateB_rand_1)
  2. Use gateA_rand_1 as Pointer and gateB_rand_1 as modifier, thus we’ll get result_pac which is signed gateA_rand_1
  3. Check other cases, where GateA and GateB will output (00, 01, 10), AUT should fail in those cases
    • If the check failed, the modifier got incremented (modifier++)
    • Repeat until all check success
  4. Save information for NAND
    • Random values for 0 and 1 from both input
    • Modifier increment

Using(Evaluate) NAND Gate:

  1. Acquire output state numbers from both input
  2. Perform AUT
    • If success, then both output is 1
    • If failed, output might be (00, 01, 10)
  3. Left shift the AUT result, leave the high byte there
    • If the AUT success the number will be original value where LSB should be 000000
    • If the AUT failed the number will be invalidate where LSB leave non-zero number
  4. Use the shifted result as output of NAND gate
    • The PAC perform AND operation, and shift perform NOT operation, which result in NAND gate

Solution⌗

Figure Out Opcodes⌗

After our team get a consensus about the VM and the circuit, we found out it’s actually a CPU run a tiny program on it. We attempt to guess all the opcode one by one by observing the state of the machine before/after an operation is done.

After we built a opcode table, we can start disassemble target program.

Bytecode Disassembly⌗

The program makes a checksum based on user-input. We can just brute-force this checksum function to find a valid input and send it to remote server.

0: 00000100
1: 11000111 cmp r0, 0x00
2: 00100110 jmp 4
3: 11000100   invalid
4: 00001000 read
5: 00000111 cmp r0, 0x3f
6: 01000110 jz 8
7: 01000100   hlt

8: 00111011 add r3, 0x3f
9: 00001000  read
10: 11000111  cmp r0, 0x00
11: 10111110  jz TLE
12: 00011011  mul r1, 0x3f
13: 10100001  ROT3 r2, 2
14: 11110001  ROT3 r3, 3
15: 00101001  XOR r2, r0
16: 00111001  XOR r3, r0
17: 01010001  ROT3 r1, 1
18: 10100110  jz 20; break
19: 01001010 jmp 9

20: 00001000 read
21: 00000111 cmp r0, 0x3f
22: 11000110 jz 24
23: 10111010   TLE ; jmp 23
24: 01100111 cmp r2, 0xd5
25: 11011110 jz 27
26: 10111010   TLE
27: 10110111 cmp r3, 0x54
28: 11110110 jz 30; FLG
29: 10111010   TLE
30: 01100000 FLG
31: 00000000