At the end of the Cybersecurity: Encryption Intro tutorial, there’s a Your Turn in the Substitution Ciphers page where you created a scrambled_alphabet_cipher script. This kind of function is a much better defense against brute force attacks. Instead of testing 93 possible shifts, a brute force attack would have to try decrypting with 26! permutations of the alphabet. The term 26! is pronounced 26 factorial.
3! = 3 * 2 * 1 = 6
4! = 4 * 3 * 2 * 1 = 24
5! = 5 * 4 * 3 * 2 * 1 = 120
26! = 403,291,461,126,605,635,584,000,000
Wow! That would be a lot of rearrangements of the alphabet for a brute force algorithm to crack. Also, who would try to look at all those combinations to find the intelligible text?
Example script: scrambled_alphabet_cipher
- Enter, save, and flash scrambled_alphabet_cipher into a micro:bit.
# scrambled_alphabet_cipher from microbit import * # Scrambled alphabet cipher. def scramble(text, encrypt): alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" crypta = "PTQGMKCRSVEXADZBJFOWYNIHLU" result = "" if encrypt is False: temp = alpha alpha = crypta crypta = temp for letter in text: letter = letter.upper() index = alpha.find(letter) result = result + crypta[index] return result # The script starts executing statements from here. sleep(1000) print("Set your keyboard to CAPS LOCK.") print() while True: text = input("Enter a CAPS LOCK string: ") result = scramble(text, True) print("scrambled result =", result) result = scramble(result, False) print("unscrambled result =", result)
- Click the Open Serial button.
- Type HAPPY into the terminal.
- Verify that the scrambled result is RPBBL. That’s the encrypted result.
- Verify that the unscrambled, decrypted result is HAPPY.