PicoCTF: Mod26 write-up

In this challenge, we aim to “crack” a simple letter substitution cipher. Cryptography can be straightforward if you understand the basics, such as ROT13. Here’s the encoded text we need to decipher:

cvpbPGS{arkg_gvzr_V'yy_gel_2_ebhaqf_bs_ebg13_jdJBFOXJ}

Solution 1

We are presented with an encoded string of text that resembles a flag (formatted like picoCTF{flag_text_here}). Our task is to decipher the characters using an appropriate method.

The challenge description hints at ROT13 (a simple substitution cipher) as the method for decoding the text. ROT13 works by shifting each letter by 13 positions in the alphabet (e.g., A becomes N, B becomes O, C becomes P, and so on). Notably, ROT13 is an involutory function for a 26-character alphabet (like English), meaning that applying ROT13 twice to a text returns the original text.

We can quickly solve the challenge using a single terminal command:

$ echo "cvpbPGS{arkg_gvzr_V'yy_gel_2_ebhaqf_bs_ebg13_jdJBFOXJ}" | tr 'A-Za-z' 'N-ZA-Mn-za-m'
picoCTF{REDACTED}

Solution 2

Alternatively, we can use an online tool like CyberChef to decipher the encoded text. Simply paste the string into the input field and set up the ROT13 recipe as follows:

Upon applying the ROT13 recipe, the deciphered flag will be revealed.