# NetOn CTF 2021 - SecretMessage

Cryptography – 247 pts (8 solves) – Chall author: X4v1l0k

Simple reversing of a custom encryption function.

## Challenge

So we already have our flag

̜͍͑͋˻͒̽͊˼˻͇̽̓˻͍͉̼͍̀͌˻͑͂̇˻͉͓͍͂˻͇̽̓̀˻͉͉̀͌̕ͅ˻̢̜̭̝̜͖̼̺͍̺͕͇̺͎̼̌͋̎͋̋͌̀̌̎͌͘


but it looks kind of weird… Luckily, they tell us how they encrypted it

def encrypt(pwd, s):
n = 0
for c in pwd: n += ord(c)
lc = string.ascii_lowercase
uc = string.ascii_uppercase
tcyph = str.translate(s, str.maketrans(lc + uc, lc[13:] + lc[:13] + uc[13:] + uc[:13]))
fcyph = ''
for c in tcyph: fcyph += chr(ord(c) + n)
return fcyph


## Solution

Seems we have a simple translation mapping, by swapping the halves of the lower- and uppercase alphabet, and an addition factor. This factor is made by adding the ord values of all charaters in a password. Knowing this factor is zero at the beginning we can find it by guessing the first letter to be ‘N’. Using a simple Python script

# Imports
import string

lc = string.ascii_lowercase
uc = string.ascii_uppercase
# Alphabet in, alphabet out
dic_in = lc + uc
dic_out = lc[13:] + lc[:13] + uc[13:] + uc[:13]
# Offset guess
n = 731
# Flag
flag = [chr(ord(i)-n) for i in list('̜͍͑͋˻͒̽͊˼˻͇̽̓˻͍͉̼͍̀͌˻͑͂̇˻͉͓͍͂˻͇̽̓̀˻͉͉̀͌̕ͅ˻̢̜̭̝̜͖̼̺͍̺͕͇̺͎̼̌͋̎͋̋͌̀̌̎͌͘')]
# Loop over flag
for i,char in enumerate(flag):
if char in lc+uc:
flag[i] = dic_in[dic_out.index(char)]

print(''.join(flag))


we find

Nice job! you earned it, take your award: NETON{n1c3_c0de_my_fr13nd}


Note that we got lucky. I guessed ‘N’ to be the first letter because of the flag format (NETON{}), however it did not start with the flag. Second option would have been to guess the final character to be ‘}’, so we would have found it either way :).