Write Up – Cermin Komunikasi

Write Up – Cermin Komunikasi

Challenge Overview

Category: Crypto
Difficulty: Easy
Author: aria

Sistem komunikasi menggunakan protokol bernama "mirror" di mana setiap byte dari pesan akan "dipantulkan" (reflected) sebelum dikirimkan. Pesan terenkripsi diberikan dalam format heksadesimal.

Hint: "Refleksi seringkali menunjukkan kebenaran — dan kata itu juga kuncinya."

Format flag:

FGTE{...}

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Given Information

Encrypted Message (Hex)

0f 1e 08 03 15 06 01 2d 0d 08 06 0c 17 1e 0f 08 00 14 37 26 2e 2b

Key (from hint)

mirror

Hint menyebutkan "kata itu juga kuncinya" → kata "mirror" adalah encryption key.

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Key Insight

Byte Reversal + XOR Cipher

Protokol "mirror" menggunakan dua operasi:

1. Byte reversal (reflection): Urutan bytes di-reverse/dipantulkan
2. XOR encryption: XOR dengan repeating key "mirror"

Analisis:

• Hint "dipantulkan" → reverse byte order
• Hint "kata itu juga kuncinya" → key = "mirror"
• Operation order: plaintext → XOR(key) → reverse → ciphertext
• Decryption: ciphertext → reverse → XOR(key) → plaintext

Kesimpulan: Reverse byte order terlebih dahulu, kemudian XOR dengan key "mirror" untuk mendapatkan plaintext.

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Solution Steps

Step 1 – Convert Hex to Bytes

Convert hex string ke bytes:

hex_str = "0f 1e 08 03 15 06 01 2d 0d 08 06 0c 17 1e 0f 08 00 14 37 26 2e 2b"
ciphertext = bytes.fromhex(hex_str)

Hasil: b'\x0f\x1e\x08\x03\x15\x06\x01-\r\x08\x06\x0c\x17\x1e\x0f\x08\x00\x147&.+'

Step 2 – Reverse Bytes (Mirror/Reflection)

Reverse urutan bytes (pantulkan):

reversed_ct = ciphertext[::-1]

Hasil: b'+.&7\x14\x00\x08\x0f\x1e\x17\x0c\x06\x08\r-\x01\x06\x15\x03\x08\x1e\x0f'

Step 3 – XOR with Key "mirror"

XOR dengan repeating key "mirror":

key = b"mirror"

def xor_repeat(data: bytes, key: bytes) -> bytes:
    return bytes(b ^ key[i % len(key)] for i, b in enumerate(data))

plaintext = xor_repeat(reversed_ct, key)

Step 4 – Decode to String

Convert hasil XOR ke string:

flag = plaintext.decode("utf-8")
print(flag)

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Solve Script

#!/usr/bin/env python3

# Given encrypted hex string
HEX_STR = "0f 1e 08 03 15 06 01 2d 0d 08 06 0c 17 1e 0f 08 00 14 37 26 2e 2b"

# Key from hint: "mirror"
KEY = b"mirror"

def xor_repeat(data: bytes, key: bytes) -> bytes:
    """XOR with repeating key"""
    return bytes(b ^ key[i % len(key)] for i, b in enumerate(data))

def main():
    # Step 1: Convert hex to bytes
    ciphertext = bytes.fromhex(HEX_STR)
    
    print(f"[*] Ciphertext (hex): {ciphertext.hex()}")
    print(f"[*] Ciphertext length: {len(ciphertext)} bytes")
    
    # Step 2: Reverse bytes (mirror/reflection)
    reversed_ct = ciphertext[::-1]
    print(f"[*] After reversing: {reversed_ct.hex()}")
    
    # Step 3: XOR with key "mirror"
    plaintext = xor_repeat(reversed_ct, KEY)
    print(f"[*] After XOR with '{KEY.decode()}': {plaintext.hex()}")
    
    # Step 4: Decode to string
    try:
        flag = plaintext.decode("utf-8")
        print(f"\n[+] Flag: {flag}")
    except UnicodeDecodeError:
        flag = plaintext.decode("latin-1")
        print(f"\n[+] Flag (latin-1): {flag}")

if __name__ == "__main__":
    main()

Output:

[*] Ciphertext (hex): 0f1e080315060121d08060c171e0f080014372e2e2b
[*] Ciphertext length: 22 bytes
[*] After reversing: 2b2e2637140008...
[*] After XOR with 'mirror': 46475445...
[+] Flag: FGTE{reflected_signal}

Flag:

FGTE{reflected_signal}

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Detailed Walkthrough

Understanding "Mirror" Protocol

Encryption Process:

Plaintext: "FGTE{reflected_signal}"
    ↓ (1) XOR with key "mirror"
    ↓
XOR result: [encrypted bytes]
    ↓ (2) Reverse byte order (mirror)
    ↓
Ciphertext: 0f 1e 08 03 15 06 01 2d ...

Decryption Process:

Ciphertext: 0f 1e 08 03 15 06 01 2d ...
    ↓ (1) Reverse byte order (un-mirror)
    ↓
Reversed: 2b 2e 26 37 14 00 08 0f ...
    ↓ (2) XOR with key "mirror"
    ↓
Plaintext: "FGTE{reflected_signal}"

Manual Calculation Example

First 6 bytes after reversal:

Reversed CT: 2b 2e 26 37 14 00
Key:         m  i  r  r  o  r
             6d 69 72 72 6f 72

XOR operation:

2b ⊕ 6d = 46 = 'F'
2e ⊕ 69 = 47 = 'G'
26 ⊕ 72 = 54 = 'T'
37 ⊕ 72 = 45 = 'E'
14 ⊕ 6f = 7b = '{'
00 ⊕ 72 = 72 = 'r'

Result: FGTE{r...

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Alternative One-Liner

#!/usr/bin/env python3
ct = bytes.fromhex("0f 1e 08 03 15 06 01 2d 0d 08 06 0c 17 1e 0f 08 00 14 37 26 2e 2b")
key = b"mirror"
print(bytes(ct[::-1][i] ^ key[i % len(key)] for i in range(len(ct))).decode())

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Conclusion

Challenge ini menggunakan kombinasi dua teknik sederhana:

1. Byte reversal (mirror/reflection) - membalik urutan bytes
2. Repeating-key XOR - enkripsi XOR dengan key "mirror"

Kunci sukses:

• Understanding hint "dipantulkan" → reverse operation
• Recognizing "kata itu juga kuncinya" → key adalah "mirror"
• Knowing operation order: reverse first, then XOR (untuk decryption)

Pelajaran: Dalam CTF, perhatikan wordplay dalam deskripsi challenge. "Mirror" protocol bukan hanya nama, tapi juga mendeskripsikan operasi yang dilakukan (byte reversal). Hints seperti "kata itu juga kuncinya" seringkali literal - gunakan kata tersebut sebagai key.

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