spelling-quiz

Challenge Description

This is, once again, a challenge involving substitution ciphers.

We begin by downloading the given file and seeing its contents.

Contents of the file

After downloading the file using wget <link>, we realise that it is a zip file.

We proceed to unzip this file (public.zip), creating another directory called public, with 3 files in it, namely encrypt.py, study-guide.txt and flag.txt.

The encrypted flag and the script used to encrypt it

Running cat on both the flag.txt and encrypt.py files provides us with more information on how we should continue.

The contents of the flag.txt file is:

brcfxba_vfr_mid_hosbrm_iprc_exa_hoav_vwcrm

For the encrypt.py file, we realise that this script is used to encrypt the contents of all .txtfiles in the current directory as well as its subdirectories.

encrypt.py (with added comments)

import random
import os
 
# Produce a list of all .txt files in the current directory and subdirectories
files = [
   os.path.join(path, file)               # Get the full path to each file
   for path, dirs, files in os.walk('.')  # Walk through the directory tree, starting from the current directory
   for file in files
   if file.split('.')[-1] == 'txt'        # Include only files with a .txt extension
]
 
# Generate a shuffled alphabet for substitution cipher
alphabet = list('abcdefghijklmnopqrstuvwxyz')
random.shuffle(shuffled := alphabet[:])
dictionary = dict(zip(alphabet, shuffled)) # Map each original letter to its shuffled counterpart using a dictionary
 
# Encrypt the contents of each .txt file using the substitution cipher
for filename in files:
   text = open(filename, 'r').read()
   encrypted = ''.join([
       dictionary[c]                      # Substitute character 'c' if it's in the dictionary
       if c in dictionary else c          # Leave the character unchanged if it's not in the dictionary
       for c in text
   ])
   open(filename, 'w').write(encrypted)

Key for this substitution cipher

Since the key(dictionary) for this substitution cipher is produced randomly, we are unable to reverse-engineer to find the key.

Contents of study-guide.txt

Running cat study-guide.txt displays the 272543 encrypted lines in this file.

Number of lines

To display the number of lines in a file, we can run this command:

wc -l < file-name >

This will print the number of lines in a file, followed by the name of the file

Since we are unable to reverse-engineer the key, we can do frequency analysis based on the contents in this file instead. After all, this should be the reason this file was provided.

Getting the Flag

For this challenge, I used this tool from Boxentriq to retrieve the flag. I was able to get the flag by feeding this tool the first 50 lines of the study-guide.txt file, and adding the encrypted flag at the top.

Flag

picoCTF{perhaps_the_dog_jumped_over_was_just_tired}