> It would be fun to work out mathematically how much they're limiting the password space with these rules. Alright, you nerd sniped me, but I'm lazy so I just

> It would be fun to work out mathematically how much they're limiting the password space with these rules. Alright, you nerd sniped me, but I'm lazy so I just simulated it, this cuts down the password space by ~35%. I didn't take into account passwords with leading 0s in my sampling, but this shouldn't change the result by much. from collections import Counter from random import randint def is_valid(password: str) -> bool: # cannot have the same number 3 times in a row if any(password[i] == password[i + 1] == password[i + 2] for i in range(len(password) - 2)): return False # cannot have 4 ascending numbers if any(all(password[i] == str(int(password[i + j]) - j) for j in range(4)) for i in range(len(password) - 3)): return False # cannot have 4 descending numbers if any(all(password[i] == str(int(password[i + j]) + j) for j in range(4)) for i in range(len(password) - 3)): return False # cannot have the same number appear more than 5 times if max(Counter(password).values()) > 5: return False # cannot have pairs next to each other if the second pair is one number higher if any(int(password[i:i+2]) == int(password[i+2:i+4]) - 1 for i in range(len(password) - 3)): return False return True total_valid = 0 samples = 10**5 for i in range(samples): password = str(randint(10**6, 10**20 - 1)) if is_valid(password): total_valid += 1 print(f"valid passwords: {total_valid}") print(f"valid percentage: {total_valid / samples * 100:.2f}%") valid passwords: 65045 valid percentage: 65.05%