Rating:
Inspecting [chall.py](https://ctf.nullcon.net/files/c4ceafb8a9a0fc03840e413b2d6b69d1/chall.py?token=eyJ1c2VyX2lkIjoyNDAzLCJ0ZWFtX2lkIjo5MjIsImZpbGVfaWQiOjQ2fQ.Z6BsoQ.gJHm7oJzi66UJVjoMinh73Pq4qU) and running it with a custom `text.txt` and `flag.txt`, we see that every token (substring of lowercase characters separated by spaces) is Caesar shifted by an amount to produce [output.txt](https://ctf.nullcon.net/files/4fedc0914e89798b8233bb383fe996c9/output.txt?token=eyJ1c2VyX2lkIjoyNDAzLCJ0ZWFtX2lkIjo5MjIsImZpbGVfaWQiOjQ3fQ.Z6BsoQ.e2SN4nDX8iDO-_KrF5bgAceRp54). Specifically, the `count`th token is shifted by `chars.index(flag[count % len(flag)])` for all tokens. We annotate the code below.
```py
def caesar(msg, shift):
return ''.join(chars[(chars.index(c) + shift) % len(chars)] for c in msg)
i = 0
count = 0
while i < len(text):
if text[i] not in string.ascii_lowercase:
print(text[i], end = '')
i += 1
else:
# Get the next token
j = i
while text[j] in string.ascii_lowercase: j += 1
# Shift the token text[i:j]
print(caesar(text[i:j], chars.index(flag[count % len(flag)])), end = '')
count += 1
i = j
```
Since each token is Caesar ciphered, we can recover each token by checking all possible `len(chars) = 65` possible keys for each individual token and seeing which keys yields alphabetic English words.
For example, trying all shifts for the first token `AtvDxK` in [output.txt](https://ctf.nullcon.net/files/4fedc0914e89798b8233bb383fe996c9/output.txt?token=eyJ1c2VyX2lkIjoyNDAzLCJ0ZWFtX2lkIjo5MjIsImZpbGVfaWQiOjQ3fQ.Z6BsoQ.e2SN4nDX8iDO-_KrF5bgAceRp54) yields the alphabetic plaintext : key pairs
```
piksmz : l
ohjrly : m
ngiqkx : n
mfhpjw : o
legoiv : p
kdfnhu : q
jcemgt : r
ibdlfs : s
hacker : t
```
Clearly, the first word of the decrypted [output.txt](https://ctf.nullcon.net/files/4fedc0914e89798b8233bb383fe996c9/output.txt?token=eyJ1c2VyX2lkIjoyNDAzLCJ0ZWFtX2lkIjo5MjIsImZpbGVfaWQiOjQ3fQ.Z6BsoQ.e2SN4nDX8iDO-_KrF5bgAceRp54) is `hacker`, meaning the first character of the flag is `t`.
Running the following script to list the possible plaintext : key pairs and manually choosing the English words as well as corresponding keys yields the flag when concatenating all the keys together.
```py
import string
text = "AtvDxK lAopjz /i + vhw c6 uwnshnuqjx ymfy kymhi Kyv 47+3l/eh Bs kpfkxkfwcnu Als 9phdgj9 +ka ymzuBGxmFq 6fdglk8i CICDowC, sjxir bjme+pfwfkd 6li=fj=kp, nCplEtGtEJ, lyo qeb INKLNBM vm ademb7697. ollqba lq DitCmA xzhm fx ef7dd7ii, wIvv eggiww GB kphqtocvkqp, 3d6 MAx ilsplm /d rpfkd vnloov hc nruwtAj xDxyjrx vexliv KyrE +3hc Gurz, jcemgt ixlmgw 9f7gmj5/9k obpmlkpf/ib mzp 8k/=64c ECo sj qb=eklildv. =k loGznlEpD qzC qo+kpm+obk=v, vHEEtuHKtMBHG, huk h7if75j/d9 mofs+=v, zkloh lqAkwCzioqvo rfqnhntzx fhynAnynjx b/a7 JKvrCzEx hexe BE ecwukpi 63c397. MAxLx wypujpwslz 3/c ql irvwhu 9bbcj1h9cb fsi f tswmxmzi zDGrtK ed FBpvrGL vjtqwij ixlmgep 5f8 =lkpqor=qfsb tmowuzs."
chars = string.ascii_letters + string.digits + '+/='
def caesar(msg, shift):
return ''.join(chars[(chars.index(c) - shift) % len(chars)] for c in msg)
i = 0
count = 0
while i < len(text):
if text[i] not in chars:
i += 1
else:
j = i
while j < len(text) and text[j] in chars: j += 1
for k in range(len(chars)):
word = caesar(text[i:j], k)
if word.isalpha() and word.islower():
print(word, ":", chars[k])
print()
count += 1
i = j
```
Flag: `ENO{th3_d1ffer3nce5_m4ke_4ll_th3_diff3renc3}`
