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Daily Alpacahack - XOR (Crypto / Easy) Writeup

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crypto xor ctf

Daily Alpacahack Writeup: XOR (Crypto / Easy)

Category: Crypto
Difficulty: Easy

Challenge Description

We’re given a Python encryption script and a ciphertext hex string.

import os
import secrets
import string
from itertools import cycle

flag = os.getenv("FLAG", "Alpaca{FAKEFAKEFAKEFAKE}").encode()
assert flag.startswith(b"Alpaca{")

# key = b"???????", e.g, abcdefg
key = b"".join(secrets.choice(string.ascii_letters).encode() for _ in range(7))
assert len(key) == 7

c = bytes([c1 ^ c2 for c1, c2 in zip(flag, cycle(key))])
print(c.hex())

Ciphertext:

031b13072d280a2c1816392f3b041d07020d2f1619232817153b24141d000c3925281a3704161b

Analysis

The encryption scheme is a repeating-key XOR cipher:

  • A random 7-byte key is generated using secrets.choice(string.ascii_letters)
  • The flag is XOR’d against the key, repeating the key cyclically with itertools.cycle

XOR has the useful property:

A ^ B = C  →  C ^ A = B

So if we know both the ciphertext and the original plaintext, we can recover the key:

key = ciphertext ^ plaintext

Exploit: Known-Plaintext Attack

We know the flag format is Alpaca{...}, meaning the first 7 bytes of the plaintext are always Alpaca{.

Crucially, 7 bytes = exactly the key length. This means the first 7 bytes of ciphertext were encrypted with exactly one full cycle of the key — giving us a direct, clean recovery.

Recovered key: BwcfNIq

Solve Script

from itertools import cycle

ciphertext = bytes.fromhex("031b13072d280a2c1816392f3b041d07020d2f1619232817153b24141d000c3925281a3704161b")
known = b"Alpaca{"

# XOR known plaintext prefix against ciphertext to recover key
key = bytes([c ^ p for c, p in zip(ciphertext, known)])

# Decrypt full ciphertext with recovered key
flag = bytes([c ^ k for c, k in zip(ciphertext, cycle(key))])

print(f"Key:  {key}")
print(f"Flag: {flag.decode()}")

Output:

Key:  b'BwcfNIq'
Flag: Alpaca{nou_aru_paka_ha_tsume_wo_kakusu}

Flag

Alpaca{nou_aru_paka_ha_tsume_wo_kakusu}

Takeaway

This challenge demonstrates a known-plaintext attack against a repeating-key XOR cipher.

The attack works because:

  1. We know the flag prefix Alpaca{ — 7 bytes of plaintext
  2. The key is exactly 7 bytes long
  3. XOR-ing the known plaintext with the ciphertext directly reveals the full key
  4. With the full key, decrypting the rest of the flag is trivial

The flag itself is a play on the Japanese proverb 能ある鷹は爪を隠す (“A capable hawk hides its talons”), with “taka” (hawk) replaced by “paka” (alpaca) — a nod to the CTF’s theme.

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