Null entropy

Hey I am s0ulix from s0ulr1pp3r which got 9th place in this you I used D4rkS0ul username though.

So we have given with a enc.py file and a ciphertext and a desc.txt.

desc.txt{

The management staff has constantly insisted Mr. Chris Brown Chandelier to induce maximum randomness into their security algorithms. He always denied it due to the budget problems, but we think he lacks knowledge of standard applications of symmetric ciphers. Can you show him what problem his actions can produce?

https://github.com/bittentech/cyberhackctf/tree/master/Null%20Entropy

NOTE: The '*' character in the string means the correct byte is redacted.

Author: Ansh Bhawnani
}


enc.py{
from Crypto.Cipher import AES
import binascii, sys
from flag import flag

key = "3N7g309d6Y7enT**"
IV = flag

message = 'Security is not a joke, mind it. But complete security is a myth'

def encrypt(message,passphrase):
	aes = AES.new(passphrase, AES.MODE_CBC, IV)
	return aes.encrypt(message)

print("Encrypted data: ", binascii.hexlify(encrypt(message,key)).decode())
}

ciphertext{
TlRRcUtpb3FLaW9xS2lvcUtpb3FLaW9xS2lvcUtpb3FLaW9xS2pNMlpEUXpZU29xS2lvcUtpb3FLaW9xS2lvcUtpb3FLaW9xS2lvcUtpb3FNVFkxT0dVM0tpb3FLaW9xS2lvcUtpb3FLaW9xS2lvcUtpb3FLaW9xS2lwaVlqSmpaR0ZpWXpBeVlUQTFPR1k0WXpNNU16WmhOV1prTm1Sa1pETmpOVEEwT1RFPQ==
}

This seems intersting, some bytes from aes enc are removed, so i searched somthing about it and get some writeup which i will share down.
so just decode the ct first

we got from base64 two times

{54**************************36d43a**************************1658e7**************************bb2cdabc02a058f8c3936a5fd6ddd3c50491
}

Just replace * with 0 i.e. null bytes.

{540000000000000000000000000036d43a000000000000000000000000001658e700000000000000000000000000bb2cdabc02a058f8c3936a5fd6ddd3c50491}

refer to this writeup to get some clearity about aes:-

https://tildeho.me/ritsec-ctf-writeup-recover-aes-cbc-iv/

Now break message and ct in 4 equal parts.
Why? Just read this writeup first

We will now be using there script and modifying it
to get the last two digits of key

script1{
from Crypto.Cipher import AES
from operator import xor
import binascii, sys

KEY_first = "3N7g309d6Y7enT"
cipher1 = "e700000000000000000000000000bb2c" 
cipher2 = "dabc02a058f8c3936a5fd6ddd3c50491"
plain1 = " But complete se"
plain2 = "curity is a myth"

def decrypt(cipher, passphrase):
    aes = AES.new(passphrase, AES.MODE_CBC, binascii.unhexlify(cipher1))
    return aes.decrypt(cipher)

# iterate through relavent ascii range
for i in range(32, 126):
    for j in range(32, 126):
        key = KEY_first + chr(i) + chr(j)
        dec_plain2 = decrypt(binascii.unhexlify(cipher2),  key)
        if  str(dec_plain2).startswith("c") and str(dec_plain2).endswith('th'):
            print "decrypted plain2: " + dec_plain2 + " with key: " + key
}

This give us last 2 characters of key : 3N7g309d6Y7enT1p

Now we have to find the full cipher text so last part will give us 3rd, 3rd will give us 2nd,2nd will give us 1st
And the 1st block will give us iv which is flag as given

so using script

script2{
from Crypto.Cipher import AES
import binascii, sys

KEY = "3N7g309d6Y7enT1p"
plain2 = "curity is a myth"

cipher2= "dabc02a058f8c3936a5fd6ddd3c50491"

def decrypt(cipher,passphrase):
    aes = AES.new(passphrase,AES.MODE_CBC,plain2)
    return aes.decrypt(cipher)

# Output result
print "Decrypted data: " + binascii.hexlify(decrypt(binascii.unhexlify(cipher2), KEY))
}

it will give us 3rd block as : e764f66e54397534f7d049506acabb2c

using this script agin with different parameters

script3{
from Crypto.Cipher import AES
import binascii, sys

KEY = "3N7g309d6Y7enT1p"
plain2 = " But complete se"

cipher2= "e764f66e54397534f7d049506acabb2c"

def decrypt(cipher,passphrase):
    aes = AES.new(passphrase,AES.MODE_CBC,plain2)
    return aes.decrypt(cipher)

# Output result
print "Decrypted data: " + binascii.hexlify(decrypt(binascii.unhexlify(cipher2), KEY))
}

This will give us 2nd block as result :  3a4911ae86c1f5138c2d8e17e7721658

For 1st part we will use this

script4{
from Crypto.Cipher import AES
import binascii, sys

KEY = "3N7g309d6Y7enT1p"
plain2 = "a joke, mind it."

cipher2= "3a4911ae86c1f5138c2d8e17e7721658"

def decrypt(cipher,passphrase):
    aes = AES.new(passphrase,AES.MODE_CBC,plain2)
    return aes.decrypt(cipher)

# Output result
print "Decrypted data: " + binascii.hexlify(decrypt(binascii.unhexlify(cipher2), KEY))
}

This will give output 1st block as : 54453ced1207bb9f67894d2c8ae336d4

Now finally we have got the 1st block which will give us the iv

For this we will use this script

script5{
rom Crypto.Cipher import AES
import binascii, sys

KEY = "3N7g309d6Y7enT1p"
IV="Security is not "
    
cipher1="54453ced1207bb9f67894d2c8ae336d4"

def decrypt(cipher,passphrase):
    aes = AES.new(passphrase,AES.MODE_CBC,IV)
    return aes.decrypt(cipher)

print "decrypted data: " + decrypt(binascii.unhexlify(cipher1), KEY)
}

This will give us the iv that is the flag: {W3ak_IV_5uckS5}

flag : "cyberhack{W3ak_IV_5uckS5}"

Hope you liked it:)