john-users - Cracking 1Password8 iOS password. How to use JTR if 1password2john isn't suited for my case?

Follow @Openwall on Twitter for new release announcements and other news
[<prev] [next>] [thread-next>] [day] [month] [year] [list]
Message-ID:
 <BN8PR20MB2529615AE8AAB2D5809AE1CDDE829@BN8PR20MB2529.namprd20.prod.outlook.com>
Date: Sat, 18 Mar 2023 16:09:09 +0000
From: Gonn Zerg <gonnzerg@...look.com>
To: "john-users@...ts.openwall.com" <john-users@...ts.openwall.com>
Subject: Cracking 1Password8 iOS password. How to use JTR if 1password2john
 isn't suited for my case?

Hi all,

I’m trying to format this as best as I can for the mailing list, I
apologize in advance if it doesn’t meet standards.

I’m looking for advice on how to achieve this using JTR. I have
tried passing my .sqlite file to 1password2john but it doesn’t
appear to be compatible. The account only exists in iOS so I can’t
obtain a macOS or Windows database.

So, I’ve written a script in Python that goes through the same
process 1Password goes through to generate the AUK. I do this
for each password in a dictionary.
I would like to know how the following process could be achieved
using JTR. Or maybe this method is the wrong way to go about this.

I relied heavily on this repository https://github.com/dschuetz/1password.

So these things I know or have from a backup of the device:
- secret Key (version, account id, secret)
- email
- algorithm
- iterations
- salt
- encrypted sqlite database (keysets, account, vaults, etc.)

1. With this I then get the HKDF salt:

hkdf_salt = HKDF(ikm=p2s, len=32, salt=email, hash=SHA256, count=1, info=algorithm)

2. Then the derived password key using the HKDF salt:

password_key = PBKDF2(sha256, password, salt=hkdf_salt, iterations=p2c, 32 bytes)

3. Then the HKDF secret key:

hkdf_key = HKDF(ikm=secret, len=32, salt=AcctID, hash=SHA256, count=1, info=version)

4. Then XOR the password key and the HKDF secret key:

auk = bytes(a ^ b for a, b in zip(password_key, hkdf_key))

5. Then I check if the resulting auk is valid by trying to decrypt and verify the data in the symmetric key:

C = AES.new(auk, AES.MODE_GCM, enc_sym_key_iv, mac_len=16)
    try:
        PT_enc_sym_key = C.decrypt_and_verify(enc_sym_key_data[:-16], enc_sym_key_data[-16:])
        if PT_enc_sym_key is not None:
                jwk_loaded = json.loads(jwk_json)
                decrypted_kid = jwk_loaded['kid']
                if(decrypted_kid == keyset_uuid):
                    # Found password!
    except ValueError:
        continue

I did manage to divide the dictionary into chunks and divide it into 12
processes which was better than my first copy/paste Apple Shortcut.

After attempting this with a new 1Password account on another iOS
device, getting the data needed from the backup and going through a
wordlist that contained the correct password I managed to confirm that
this works and it's very likely that it will work on the main device.

So, how can I do better?

I tried to optimize the code bit by bit and I understand there's
plenty more to improve in it. However, I'd like to move to doing
something like this on an environment that would make this a lot
faster or more efficient. I know I won't be able to code more
efficient solutions than the ones already out there by experts, even
if I wanted to reinvent the wheel.

I started to feel the lack of optimization in my solution when John
The Ripper output a wordlist in the almost hundreds of millions of
lines from a set of rules I thought could contain the right password
but no luck yet. I know I have to grow the list even more but optimize
my approach even more. I would appreciate any help pointing me in the
right direction.

Thank you for reading me!
Confused about mailing lists and their use? Read about mailing lists on Wikipedia and check out these guidelines on proper formatting of your messages.