1.5 KiB
1.5 KiB
Mein Kampf
The main idea finding the flag is knowing Enigma Machine library.
Step-1:
After reading the given message:
M4 UKW $ Gamma 2 4 $ 5 9 $ 14 3 $ 5 20 fv cd hu ik es op yl wq jm
Google searches gave some sense of Enigma Machine.
Step-2:
So, I quickly searched for such libraries in python at got it at: https://pypi.org/project/py-enigma/
Step-3:
So, I wrote a exploit.py
script with help from official documentation.
from enigma.machine import EnigmaMachine
ROTORS = ['I', 'II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'Beta', 'Gamma']
REFLECTORS = ['B', 'C', 'B-Thin', 'C-Thin']
state = 'M4 UKW $ Gamma 2 4 $ 5 9 $ 14 3 $ 5 20 fv cd hu ik es op yl wq jm'
enc = 'zkrtwvvvnrkulxhoywoj'
rings = '4 9 3 20'
plug = 'fv cd hu ik es op yl wq jm'.upper()
pos = '2 5 14 5'
pos = ''.join(chr(int(x) - 1 + ord('A')) for x in pos.split())
for rf in REFLECTORS:
for r2 in ROTORS:
for r3 in ROTORS:
for r4 in ROTORS:
rotors = ['Gamma', r2, r3, r4]
e = EnigmaMachine.from_key_sheet(rotors=rotors, ring_settings=rings,
reflector=rf, plugboard_settings=plug)
e.set_display(pos)
txt = e.process_text(enc).lower()
if 'csictf' in txt:
print(txt)
Step-4:
When I ran the script as python3 exploit.py
, I got the flag:
csictfnoshitsherlock
Step-5:
Finally the flag becomes:
csictf{no_shit_sherlock}