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// A NahamCon CTF 2022 Challenge...

Hey Hackers! This is the write up for the UNIMOD challenge from NahamCon CTF 2022. I will attach a video walk through of the challenge in case if you to visually see how it was executed.


// Starting Off:

The challenge gives us a program which is inspired off of the Rotation 13 [ROT13] cipher.

import random

flag = open('flag.txt', 'r').read()
ct = ''
k = random.randrange(0,0xFFFD)
for c in flag:
    ct += chr((ord(c) + k) % 0xFFFD)

open('out', 'w').write(ct)

They also give us an output file containing the encrypted flag.


// Analysis:

After analyzing the code, i started to understand what it was doing. It picked a random number from 0 to 65,533 [0xFFFD = 65,533] and used it as the encryption key. Iterated through each character in the flag and encrypted it using the following algorithm.

ct += chr((ord(c) + k) % 0xFFFD)

For those of you who took our Breaking Ciphers course, you should immediately recognize this algorithm. For those who don't, it is the algorithm that powers the caesar cipher/ROT13. Here is an example of that same algorithm, decrypting uppercase text.

c += chr((ord(letter) - key - 65) % 26 + 65)

I have a separate article explaining how this works here if you are interested. Regardless, all we need to do is figure out what key they are using and decrypt.

// Key Discovery:

We can discover the key in two main ways. First, you could attempt to decrypt the flag with each key ranging form 1 to 65,533, and filter any out put that starts with "flag". Alternatively, you can do what we did.

I called up Fyzz for help to tag team the challenge. He was able to discover the key that was being used by developing this script.

# Fyzz's code to discover the key
for i in range(0,65533):
    ct = chr((ord('f') + i) % 0xFFFD)
    if ct == '饇':

The output of this code told us that the key being used was "39137". All we had to do was reformat the code and plug in the key.

// Decryption:

I changed up the code a bit to be able to decrypt our flag using our new found key.

import random

flag = open('flag.txt', 'r').read()
ct = ''

k = 39137
attempt = ''
for c in flag:
    attempt += chr((ord(c) - k) % 39137)
ct += f'{attempt}\n'

open('out',  'w').write(ct)

Upon executing the code we get our flag outputted to our out file.


Thanks for reading, and as always,

Happy Hacking!

// Socials:

© 2022 by Cosmodium CyberSecurity LLC

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