C 203: PBKDF2 and AES (90 pts)

• Any computer with Python 3. I used Debian 10.

Purpose

Installing pip and pycryptodome

sudo apt update
sudo apt install python3-pip -y
python3 -m pip install pycryptodome

Opening Python

python3

Understanding PBKDF2

Using PBKDF2

import hashlib
help(hashlib)

Press q to close the help page.

Execute this command to derive a 16-byte key from the password "apple" with one round of HMAC-MD5 and no salt:

hashlib.pbkdf2_hmac('md5', b"apple", b"", 1, 16).hex()

C 203.1: Hexadecimal Values (5 pts) Compute a 4-byte hash of this password: today7 With these parameters: Hash name: sha1 Salt: aaaabbbb Iterations: 1000 The flag is the hash value in hexadecimal.

Type Conversions

Python3 Conversions

C 203.2: Crack a PIN (10 pts) The password is a three-digit number, between 000 and 999. The hash value, using the same parameters as the previous flag, is: f68923f5 Find the password. That's the flag. Hint: for loops in Python, see project VP 100.

Using PBKDF2 and AES Together

Execute this code to find a 128-bit key from the password "apple" and use that key to encrypt a sentence with AES:

import hashlib
from Crypto.Cipher import AES

cleartext = b'Any nation that does not honor its heroes will not long endure. '
key = hashlib.pbkdf2_hmac('md5', b"apple", b"", 1, 16)
cipher = AES.new(key, AES.MODE_ECB)
ciphertext = cipher.encrypt(cleartext)
print(ciphertext.hex())

To decrypt the ciphertext, enter this code:

import hashlib
from Crypto.Cipher import AES

ciphertext = bytes.fromhex("68aa9afdbd3f2f0a40dfd961ad631bf2a9f56bc1a5808a90609e36e112e429402046b639085aa24c22b60ef2c7fba971c0862ff2587203236d7e3d6afbb5851d")
key = hashlib.pbkdf2_hmac('md5', b"apple", b"", 1, 16)
cipher = AES.new(key, AES.MODE_ECB)
cleartext = cipher.decrypt(ciphertext)
print(cleartext)

C 203.3: Encrypt a Sentence (5 pts) Calculate ciphertext from these parameters: Password: WSYIWYG Hash name: sha256 Salt: 3xrtras4lt Iterations: 50 Plaintext: Abandon all hope Encryption algorithm: AES-ECB Key size: 128 bits The flag is the ciphertext in hexadecimal.

C 203.4: Decrypt (20 pts) The password contains three lowercase letters, between aaa and zzz. Using the same parameters as the previous challenge, decrypt this ciphertext: 7afa663510b006989976f542d0da66a5 The decrypted text contains only valid ASCII bytes, so no value is larger than 127.

Encrypted Virtual Machines

• Hash name: sha1
• Iterations: 10000
• Encryption algorithm: AES-ECB
• Key size: 256 bits

This is a "keysafe": a data structure which contains an encrypted key, which can only be extracted by entering the correct password.

Making these substitutions:

• %25 %
• %2b +
• %2d -
• %2f /
• %3a :
• %3d =

encryption.keySafe = "vmware:key/list/(pair/(phrase/+IOeb4UyCwg=/
pass2key=PBKDF2-HMAC-SHA-1:
cipher=AES-256:
rounds=10000:
salt=tKU/oavut45txJFpUmcDng==,
HMAC-SHA-1,
L7te7YmmbxRzEhmaHIx/RlsenaF3h+qF4rPBlAeGpAarBASv7XVUfU/5Y/AAKRqArknfLGRB7RO+zw6NcjjNIWzQJY23hHUNLpa84xlVzH3+j0xstLpbJYkiYEIOV2XoZKu5XATB59CyBefiBvinGTonjrw=
))"

import hashlib
from Crypto.Cipher import AES
import base64

data = bytes.fromhex(base64.b64decode("L7te7YmmbxRzEhmaHIx/RlsenaF3h+qF4rPBlAeGpAarBASv7XVUfU/5Y/AAKRqArknfLGRB7RO+zw6NcjjNIWzQJY23hHUNLpa84xlVzH3+j0xstLpbJYkiYEIOV2XoZKu5XATB59CyBefiBvinGTonjrw=").hex())

iv = data[0:16]
ciphertext = data[16:64]
salt = bytes.fromhex(base64.b64decode("tKU/oavut45txJFpUmcDng==").hex())
key = hashlib.pbkdf2_hmac('sha1', b"123123", salt, 10000, 32)
cipher = AES.new(key, AES.MODE_CBC, iv)
cleartext = cipher.decrypt(ciphertext)
print(cleartext)

C 203.5: Crack the Safe (20 pts) The password for the keysafe shown below contains six digits. Each digit is repeated twice, like this: 112233, so there are 1000 possible keys from 000000 through 999999. Find the key to crack the safe. That six-digit key is the flag. encryption.keySafe = "vmware:key/list/(pair/(phrase/76IjGGQzjAA=/ pass2key=PBKDF2-HMAC-SHA-1: cipher=AES-256: rounds=10000: salt=DkjJMNgtJIY/3StMtBQ0LA==, HMAC-SHA-1, 0n1UgXgEpQ8Xun9PbStxYjmV54rkENd0+Ew4xXbhqN9yRXTeX4WEGWt5Nblz8LXtlgWV8uK8yjuIiLWsiBagTiVPJO6l2+XGIQn2fLCXs00HquSK9ap2oSx41s30nCH4f4Foozgt8EExi55moOY2dL3Y4ls= ))" The decrypted text contains only valid ASCII bytes, so no value is larger than 127.

C 203.6: Crack a Harder Safe (30 pts) The password for the keysafe shown below contains eight characters. The first two characters are We and the last two characters are ss The other four characters are lowercase letters. Find the key to crack the safe. That eight-character key is the flag. encryption.keySafe = "vmware:key/list/(pair/(phrase/zROLP2SQhFQ=/ pass2key=PBKDF2-HMAC-SHA-1: cipher=AES-256: rounds=10000: salt=sa+HuVbMVcXjFSqkA6wGkg==, HMAC-SHA-1, gYELJ/hydfSzSi8VM8nBA5RPOMFQBE9qdJmzYD286iHUur/JuBiIgAqKB9iQHaL2KCJiWtf4r0bzrSqocRhDES3YcM1dyzVOTFhNlMdFlVtLraeZ+cJdfzi31WDjaQsn0zWX5d5vr4k4GNf2BbOPHcwLxBU= ))" The decrypted text contains only valid ASCII bytes, so no value is larger than 127.

References

Acknowledgement