How can I resolve maximum recursion depth error

Question

I'm generating a random password with a desired length. I want it to have at least 2 uppercase letters, 2 lowercase letters, 2 digits and 2 special characters. I've tried multiple things, but every time I get this recursion depth error. Can anybody tell me what I've done wrong?

list_lower =['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
list_upper = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N', 'O','P','Q','R','S','T','U','V','W','X','Y','Z'] 
list_digit = [1,2,3,4,5,6,7,8,9,0]
def generatePassword(desiredLength: int) -> str:
    x = 0
    password = ""
    for x in range (desiredLength):
        password = password + chr(random.randint(33,126))
        list(password)
        list_password = list(password)
        times_lower = 0
        times_upper = 0
        times_digit = 0
        times_special = 0
        for character in list_password:
            if character in list_lower:
                times_lower += 1
            elif character in list_upper:
                times_upper += 1
            elif character in list_digit:
                times_digit += 1
            else:
                times_special +=1
        if times_lower >= 2 and times_upper >= 2 and times_digit >= 2 and times_special >= 2:
            return password
        else:
            return generatePassword(desiredLength)

generatePassword(7)

I get the error in line 30 which makes the function recursive.

Why do it recursively at all? This doesn't seem like a natural use-case. Also -- please don't use things like "line 30". Your editor's line numbering is not preserved in your code. — John Coleman
– John Coleman, Commented Sep 28, 2018 at 15:52
Your breaking condition is based on randomness. No matter what you do, there will always be a non-zero chance of recurring too many times. — mypetlion
– mypetlion, Commented Sep 28, 2018 at 15:53
First, your function's base case is incorrect: when desiredLength is 0, the function returns None instead of an empty string. Second, the function calls itself with the same value of desiredLength (must be desiredLength-1). — DYZ
– DYZ, Commented Sep 28, 2018 at 15:54
shouldn't the if -else branch that makes the recursive call be outside the inner for-loop? Otherwise, it will never possibly have those properties you are hoping (because len(password) == 1) and thus, it will always make the recursive call. Anyway, just use a while loop. No need for recursion. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Sep 28, 2018 at 15:55
Also, even with a while-loop, there's a better approach. Choose, randomly, 2 characters from every category you require. Then randomly choose a category and a character until you fill up the desired length. Then shuffle. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Sep 28, 2018 at 15:58

Caleth · Accepted Answer · 2018-09-28 15:58:59Z

Calling generatePassword(7) will never generate a password with 2 of each of 4 distinct categories.

You don't need recursion at all.

def generatePassword(desiredLength: int) -> str:
    while True:
        password = ""
        for x in range (desiredLength):
            password = password + chr(random.randint(33,126))
        times_lower = 0
        times_upper = 0
        times_digit = 0
        times_special = 0
        for character in password:
            if character in list_lower:
                times_lower += 1
            elif character in list_upper:
                times_upper += 1
            elif character in list_digit:
                times_digit += 1
            else:
                times_special +=1
        if times_lower >= 2 and times_upper >= 2 and times_digit >= 2 and times_special >= 2:
            return password
        else
            print ("Rejecting ", password)

That will loop forever if asked to generate a password of length 7 or less. We can improve that by checking the desired length first

if desiredLength < 8:
    raise ArgumentError("Cannot generate passwords shorter than 8 characters")

bened · Accepted Answer · 2018-09-28 16:02:09Z

0

times_digit will never be >= 2 because it tests stings (e.g. "2" against the integers in your list, (e.g. 2) change your list_digit to

list_digit = ["1","2","3","4","5","6","7","8","9","0"]

and try again.

By the way this could be done much simpler and doensn't need a recursive function.

answered Sep 28, 2018 at 16:02

bened

113 bronze badges

1 Comment

bened Over a year ago

As others have pointed out, there are many more problems in your code.

Bradley Young · Accepted Answer · 2018-09-28 16:32:36Z

If you are generating passwords, it's important that you generate ones that actually have enough randomness to not be predictable.

Random string generation with upper case letters and digits in Python

Has a great breakdown of how to generate a password that's truly random:

''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))

(adding "special characters" and "lowercase characters" omitted to preserve the existing code)

I know that this is a somewhat oblique answer (i.e. it does not answer the question directly), so here's a potential solution if you still need the "it must contain these types of characters" (even though that would actually reduce security):

import random
import string
from collections import Counter

def gen(N):
    return ''.join(random.SystemRandom().choice(string.ascii_letters + string.digits + string.punctuation) for _ in range(N))

while True:
    pw = gen(8)
    counts = Counter(pw)
    upper = lower = digit = special = 0
    for (letter, count) in counts.items():
        if (letter in string.ascii_lowercase):
            lower += 1
        elif (letter in string.ascii_uppercase):
            upper += 1
        elif (letter in string.digits):
            digit += 1
        else:
            special += 1
            pass
    if (lower > 1 and upper > 1 and digit > 1 and special > 1):
        print("password is {}".format(pw))
        break
    print("failed password: {}".format(pw))

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