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Introduction to Enigma
Bruteforce attack
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Bruteforce descryption

Learn how to crack the Enigma machine using a brute force attack

By Kevin McAleer,    3 Minutes

What is a brute force attack?

A brute force attack is a method of cracking a code by trying every possible combination of characters until the correct one is found. This is a very time-consuming process, but it is guaranteed to work eventually.

Bruteforce attack on the Enigma machine

To crack the Enigma machine using a brute force attack, we need to try every possible combination of rotor settings and start positions until we find the correct one. This is a very time-consuming process, as there are 60^3 = 216,000 possible combinations of rotor settings and start positions.

To understand when we have found the correct settings, we can use a small substring of the plaintext message, known as a “crib”. This is a word or phrase that we know is in the plaintext message. If we find the crib in the decrypted message, we know that we have found the correct settings.

import time

# ciphertext = "YJPYITREDSYUPIU"
cribtext = "ROBOT"
message = "SUPSH GPYCV JTPYF TDQWV HIBEW FPDBN TAUEL IQXMS ZBDCT"

rotors = [ "I II III", "I II IV", "I II V", "I III II",
"I III IV", "I III V", "I IV II", "I IV III",
"I IV V", "I V II", "I V III", "I V IV",
"II I III", "II I IV", "II I V", "II III I",
"II III IV", "II III V", "II IV I", "II IV III",
"II IV V", "II V I", "II V III", "II V IV",
"III I II", "III I IV", "III I V", "III II I",
"III II IV", "III II V", "III IV I", "III IV II",
"III IV V", "IV I II", "IV I III", "IV I V",
"IV II I", "IV II III", "IV I V", "IV II I",
"IV II III", "IV II V", "IV III I", "IV III II",
"IV III V", "IV V I", "IV V II", "IV V III",
"V I II", "V I III", "V I IV", "V II I",
"V II III", "V II IV", "V III I", "V III II",
"V III IV", "V IV I", "V IV II", "V IV III" ]

plugboard = "AV BS CG DL FU HZ IN KM OW RX"
slip_ring = "1 1 1"

def decrypt_message(rotor_choice, start_pos, ciphertext):
    from enigma.machine import EnigmaMachine

    machine = EnigmaMachine.from_key_sheet(
    rotors=rotor_choice,
    reflector='B',
    ring_settings=slip_ring,
    plugboard_settings=plugboard)

    machine.set_display(start_pos)

    plaintext = machine.process_text(ciphertext)
    print(plaintext.replace("X"," "))


def find_rotor_start(rotor_choice, cipher_text, crib_text):
    from enigma.machine import EnigmaMachine

    alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

    machine = EnigmaMachine.from_key_sheet(
    rotors=rotor_choice,
    reflector='B',
    ring_settings=slip_ring,
    plugboard_settings=plugboard)

    for rotor1 in alphabet:
        for rotor2 in alphabet:
            for rotor3 in alphabet:
                
                # Generate a possible start position
                start_pos = rotor1 + rotor2 + rotor3
                
                # Set the start position
                machine.set_display(start_pos)

                # Attempt to decrypt the plaintext
                plaintext = machine.process_text(cipher_text)
                print(plaintext)
                if crib_text in plaintext:
                    print("Valid settings found! *************************************************")
                    decrypt_message(rotor_choice, start_pos, message)
                    return rotor_choice, start_pos
         
    # If no valid settings are found, return None
    return rotor_choice, "Cannot find settings"

message = message.replace(" ","")

start_time = time.time()

for rotor_setting in rotors:
    rotor_choice, start_pos = find_rotor_start(rotor_setting, message, cribtext)
    print(rotor_choice, start_pos)
    if start_pos != "Cannot find settings":
        break

end_time = time.time()

duration = end_time - start_time

print(f"Brute force attack completed in {round(duration,1)} seconds on Mac M1")

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