reverse_gambit

Reverse Gambit Challenge

A reverse engineering challenge that includes some self-modifying code. Difficulty level is medium.

Info	Value
Difficulty	Medium
Flag	ccsc{v3ry_e4sy_ch4ll}

Tweaking and Compiling

You can update the decryption key by changing the FLAG_CIPHERTEXT_KEY macro in includes.h. Make sure that the value is at max 32bit only.

Compile using:

cmake

Note: Ignore error challenge.h not found in your IDE, to generate it run the make process once (it's automatically generated by flag_print.c)

Solution 1

Note: FYI don't use this one, it's time consuming and not really the correct approach Brute-force the executable with values ranging from 0x0 - 0xFFFFFFFF

import os, subprocess
cmd = os.path.join(os.path.realpath('.'), "challenge")
for key in range(0x00, 0xffffffff):
    if "ccsc" in subprocess.run(cmd, input="{:x}".format(key), capture_output=True):
        print(f"Key = {hex(key)}")

Solution 2

TODO

In short:

1. Identify BB

   .text:0000000000000771                 lea     rdi, s          ; "Invalid key"
   

2. Trace back to where the decision is made:

   .text:000000000000072B                 call    sub_8A0
   .text:0000000000000730                 cmp     cs:dword_201020, 0E5894855h
   .text:000000000000073A                 jnz     short loc_771
   

3. Function call looks like a XOR encryption

   .text:00000000000008E0 loc_8E0:                                ; CODE XREF: sub_8A0+49↓j
   .text:00000000000008E0                 xor     [rbx], ebp
   .text:00000000000008E2                 add     rbx, 4
   .text:00000000000008E6                 cmp     rax, rbx
   .text:00000000000008E9                 jnz     short loc_8E0
   

4. Take the value 0E5894855h and the encoded/encrypted ciphertext being compared at step #2 (dword_201020) and find a value where *(unsigned long *)dword_201020 ^ K = 0E5894855h. Now all you have to do is just a XOR operation to get the key K = 0E5894855h ^ *(unsigned long *)dword_201020