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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}