lactf

nine solves

Category: Reverse Engineering

Given a Binary file named nine-solves let's open it using ida.

int __cdecl main(int argc, const char **argv, const char **envp)
{
  __int64 i; // rsi
  unsigned int v4; // eax
  int v5; // ecx
  int v6; // edx
  char v8[6]; // [rsp+0h] [rbp-18h] BYREF
  char v9; // [rsp+6h] [rbp-12h]

  puts("Welcome to the Tianhuo Research Center.");
  printf("Please enter your access code: ");
  fflush(stdout);
  fgets(v8, 16, stdin);
  for ( i = 0LL; i != 6; ++i )
  {
    v4 = v8[i];
    if ( (unsigned __int8)(v8[i] - 32) > 0x5Eu )
      goto LABEL_14;
    v5 = yi[i];
    if ( !v5 )
      goto LABEL_14;
    v6 = 0;
    while ( (v4 & 1) == 0 )
    {
      ++v6;
      v4 >>= 1;
      if ( v5 == v6 )
        goto LABEL_9;
LABEL_6:
      if ( v4 == 1 )
        goto LABEL_14;
    }
    ++v6;
    v4 = 3 * v4 + 1;
    if ( v5 != v6 )
      goto LABEL_6;
LABEL_9:
    if ( v4 != 1 )
      goto LABEL_14;
  }
  if ( !v9 || v9 == 10 )
  {
    eigong(yi);
    return 0;
  }
LABEL_14:
  puts("ACCESS DENIED");
  return 1;
}

Simply this is a Collatz transformation code just by looking at v4 = 3 * v4 + 1;. So the first step is we need to reverse the method and then we need to find out what the yi value is.

image

From the binary, the yi is an array of value {0x1B, 0x26, 0x57, 0x5F, 0x76, 0x9} that can be converted to {27, 38, 87, 95, 118, 9}

From that, we can make the script to generate the access code to get the flag

Solve.py

yi = [27, 38, 87, 95, 118, 9]

def collatz_steps(n, steps):
    for _ in range(steps):
        if n == 1:
            return False
        if n % 2 == 0:
            n = n // 2
        else:
            n = 3 * n + 1
    return n == 1

access_code = []
for steps in yi:
    for c in range(32, 127):  
        if collatz_steps(c, steps):
            access_code.append(chr(c))
            break

access_code = ''.join(access_code)
print(f"Access Code: {access_code}")

Output Access Code: AigyaP

Flag

lactf{REDACTED}