MD5-based password algorithm in Python

Question

I try to call API that specifically uses MD5 hash in one of the steps. In the documentation they specifically show example reference that generates MD5 in the following way

$ openssl passwd -1 -salt stack overflow
$1$stack$MVcBmQ3RlrBu5Xoj74NBA0

or to be more exact, they just use the part after the third $

$ openssl passwd -1 -salt stack overflow | cut -f 4 -d '$'
MVcBmQ3RlrBu5Xoj74NBA0

At first, I tried to use hashlib and got the hexadecimal output that does not resemble the exampla at all.

salt = b'stack'
input = b'overflow'
output = hashlib.md5(salt + input).hexdigest()
print(output)

73868cb1848a216984dca1b6b0ee37bc

I figured that I just need to decode those hex values to characters, but decode does not work for default utf8 or for latin1

salt = b'stack'
input = b'overflow'
output = hashlib.md5(salt + input).digest().decode()
print(output)

UnicodeDecodeError: 'utf-8' codec can't decode byte 0x86 in position 1: invalid start byte

I found some help here python version of openssl passwd and here MD5 hash in Python

I could reproduce this one with crypt

$ openssl passwd -salt stack overflow
st22n6QiCXNQY

salt = 'stack'
input = 'overflow'
output = crypt.crypt(input, salt)
print(output)

st22n6QiCXNQY

But as soon openssl passwd -1 is added, which stands for

-1                  MD5-based password algorithm

I cannot reproduce it anymore.

How can I recreate MD5-based password algorithm in Python? I would preferably use hashlib if possible.

Rostfrei · Accepted Answer · 2023-06-30 08:44:14Z

2

I found the solution using passlib

from passlib.hash import md5_crypt

salt = 'stack'
input = 'overflow'
output = md5_crypt.using(salt=salt).hash(input)
print(output)

$1$stack$MVcBmQ3RlrBu5Xoj74NBA0

answered Jun 30, 2023 at 8:44

Rostfrei

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Comments

Peter234 · Accepted Answer · 2023-06-30 08:39:29Z

1

On Linux you can do this with the crypt module. However it will be removed in Python 3.13.

import crypt
crypt.crypt("overflow", "$1$stack")

Here $1$ specifies the use of MD5 and "stack" is the salt.

edited Jun 30, 2023 at 8:39

answered Jun 30, 2023 at 8:26

Peter234

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1 Comment

Rostfrei Over a year ago

This works! Thank you so much! I would never figure this one out. I also found the solution with the passlib.

Alex Derbov · Accepted Answer · 2025-04-06 00:56:31Z

https://passlib.readthedocs.io/en/stable/lib/passlib.hash.md5_crypt.html

Python code for copy-pasters

MAGIC = '$1$'           # Magic string
ITOA64 = "./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"

from hashlib import md5

def to64 (v, n):
    ret = ''
    while (n - 1 >= 0):
        n = n - 1
    ret = ret + ITOA64[v & 0x3f]
    v = v >> 6
    return ret


def apache_md5_crypt (pw, salt):
    # change the Magic string to match the one used by Apache
    return unix_md5_crypt(pw, salt, '$apr1$')


def unix_md5_crypt(pw, salt, magic=None):
    
    if magic==None:
        magic = MAGIC

    # Take care of the magic string if present
    if salt[:len(magic)] == magic:
        salt = salt[len(magic):]
        

    # salt can have up to 8 characters:
    import string
    salt = string.split(salt, '$', 1)[0]
    salt = salt[:8]

    ctx = pw + magic + salt

    final = md5.md5(pw + salt + pw).digest()

    for pl in range(len(pw),0,-16):
        if pl > 16:
            ctx = ctx + final[:16]
        else:
            ctx = ctx + final[:pl]


    # Now the 'weird' xform (??)

    i = len(pw)
    while i:
        if i & 1:
            ctx = ctx + chr(0)  #if ($i & 1) { $ctx->add(pack("C", 0)); }
        else:
            ctx = ctx + pw[0]
        i = i >> 1

    final = md5.md5(ctx).digest()
    
    # The following is supposed to make
    # things run slower. 

    # my question: WTF???

    for i in range(1000):
        ctx1 = ''
        if i & 1:
            ctx1 = ctx1 + pw
        else:
            ctx1 = ctx1 + final[:16]

        if i % 3:
            ctx1 = ctx1 + salt

        if i % 7:
            ctx1 = ctx1 + pw

        if i & 1:
            ctx1 = ctx1 + final[:16]
        else:
            ctx1 = ctx1 + pw
            
            
        final = md5.md5(ctx1).digest()


    # Final xform
                                
    passwd = ''

    passwd = passwd + to64((int(ord(final[0])) << 16)
                           |(int(ord(final[6])) << 8)
                           |(int(ord(final[12]))),4)

    passwd = passwd + to64((int(ord(final[1])) << 16)
                           |(int(ord(final[7])) << 8)
                           |(int(ord(final[13]))), 4)

    passwd = passwd + to64((int(ord(final[2])) << 16)
                           |(int(ord(final[8])) << 8)
                           |(int(ord(final[14]))), 4)

    passwd = passwd + to64((int(ord(final[3])) << 16)
                           |(int(ord(final[9])) << 8)
                           |(int(ord(final[15]))), 4)

    passwd = passwd + to64((int(ord(final[4])) << 16)
                           |(int(ord(final[10])) << 8)
                           |(int(ord(final[5]))), 4)

    passwd = passwd + to64((int(ord(final[11]))), 2)


    return magic + salt + '$' + passwd


## assign a wrapper function:
md5crypt = unix_md5_crypt

if __name__ == "__main__":
    print unix_md5_crypt("overflow", "stack", "$1$")

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