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