picoCTF - 13 Challenge Write-Up

Challenge Information

Challenge Name: 13
Category: Cryptography
Difficulty: Easy
Source: https://sukepaja18.medium.com/picoctf-13-1ca39bcb98c5

Challenge Description

The challenge presents an encrypted string and mentions ROT13:

cvpbPGS{abg_gbb_onq_bs_n_ceboyrz}

Like the MOD 26 challenge, this is another ROT13 encryption problem.

Understanding ROT13

ROT13 is a simple substitution cipher that rotates each letter by 13 positions in the alphabet. It's a form of the Caesar Cipher and has the following characteristics:

• Encryption Formula: (c + 13) % 26
• Decryption Formula: (c - 13) % 26

Where c represents the ASCII/position value of the character to be transformed.

Key points: - It works on letters (both uppercase and lowercase) - Numbers and special characters typically remain unchanged - Applying ROT13 twice returns the original text (since 13 + 13 = 26)

Solution

To decrypt the flag, we need to apply ROT13 decryption to the given ciphertext.

Method 1: Online Tools (Recommended)

Use CyberChef:

1. Navigate to CyberChef
2. Paste the encrypted string: cvpbPGS{abg_gbb_onq_bs_n_ceboyrz}
3. Search for and select the "ROT13" recipe
4. The decrypted output will appear instantly

Method 2: Manual Decryption with Python

Using the codecs module:

import codecs

ciphertext = "cvpbPGS{abg_gbb_onq_bs_n_ceboyrz}"
plaintext = codecs.encode(ciphertext, 'rot_13')
print(plaintext)

Or implementing the algorithm manually:

def rot13_decrypt(text):
    result = ""
    for char in text:
        if 'a' <= char <= 'z':
            result += chr((ord(char) - ord('a') + 13) % 26 + ord('a'))
        elif 'A' <= char <= 'Z':
            result += chr((ord(char) - ord('A') + 13) % 26 + ord('A'))
        else:
            result += char
    return result

ciphertext = "cvpbPGS{abg_gbb_onq_bs_n_ceboyrz}"
print(rot13_decrypt(ciphertext))

Flag

After applying ROT13 decryption:

picoCTF{not_too_bad_of_a_problem}

Key Takeaways

1. ROT13 is symmetric - applying it twice returns the original text
2. Simple substitution ciphers are weak - only 25 possible shifts to brute force
3. Hints are crucial - the challenge description explicitly mentions ROT13
4. Reuse your tools - once you understand one ROT13 problem, others become trivial
5. Challenge naming - the challenge number "13" is a hint to the shift value used in ROT13