def reverse_crc(target_crc, length): """Return the list of bytes that must have been fed to the CRC to get target_crc.""" # Walk backwards length steps, assuming the *last* processed byte is unknown. # We'll treat each step as "what byte could we have processed last?" # Because CRC is linear, we can just brute‑force each step (256 possibilities) # and keep the one that leads to a feasible state. With 9 steps it is trivial. bytes_rev = [] crc = target_crc for _ in range(length): # Find a byte b such that there exists a previous CRC value. # Because the CRC algorithm is bijective for a fixed length, any byte works; # we simply pick the one that yields a CRC that is a multiple of 2**8. # The easiest way: try all 256 possibilities and keep the first that makes # the high‑byte of the previous CRC zero (which will be the case for the # correct sequence). for b in range(256): # Reverse the step prev = ((crc ^ TABLE[(crc ^ b) & 0xFF]) << 8) | ((crc ^ b) & 0xFF) prev &= 0xFFFFFFFF # After reversing one byte, the CRC must be divisible by 2**8 for the # next reverse step (since we are moving leftwards). This property holds # for the true sequence. if (prev & 0xFF) == 0: bytes_rev.append(b) crc = prev >> 8 break else: raise RuntimeError("No suitable byte found – something went wrong") return list(reversed(bytes_rev))
def crc32_step_rev(crc, b): """Reverse one CRC‑32 step (process byte b at the *end* of the stream).""" # The forward step is: crc = (crc >> 8) ^ TABLE[(crc ^ b) & 0xFF] # Reversing: idx = (crc ^ b) & 0xFF prev_crc = (crc ^ TABLE[idx]) << 8 prev_crc |= idx return prev_crc & 0xFFFFFFFF Adeko 9 Crack 56
# Pre‑compute forward CRC table (standard) def crc32_table(): tbl = [] for i in range(256): c = i for _ in range(8): c = (c >> 1) ^ POLY if (c & 1) else c >> 1 tbl.append(c & 0xFFFFFFFF) return tbl bytes_rev = [] crc = target_crc for _
# ------------------------------------------------------------ if __name__ == "__main__": TARGET = 0x56C9A4F2 for b in range(256): # Reverse the step