Cryptohack - ECB Oracle writeup

writeup.md

Working Method:

from binascii import hexlify
import requests
import json
from string import printable

def encrypt(pt):
    p = hexlify(pt).decode()
    url = "http://aes.cryptohack.org/ecb_oracle/encrypt/"+p
    r = requests.get(url)
    ct = (json.loads(r.text))['ciphertext']
    return ct


flag = ""
n = 31
while 1:
    payload = b'1'*(n-len(flag))
    comp = encrypt(payload)
    for i in printable:
        enc = encrypt(payload + flag.encode() + i.encode())
        if comp[32:64] == enc[32:64]:
            flag += i
            print("flag letter: ", i)
            break
    if flag[-1] == '}':
        break
print(flag)

flag = crypto{p3n6u1n5_h473_3cb}

reference = https://yidaotus.medium.com/breaking-ecb-by-prepending-your-own-message-b7b376d5efbb

Alternate Method that doesn't work:

from binascii import hexlify
import requests
import json
from string import printable

def encrypt(pt):
    p = hexlify(pt).decode()
    url = "http://aes.cryptohack.org/ecb_oracle/encrypt/"+p
    r = requests.get(url)
    ct = (json.loads(r.text))['ciphertext']
    return ct


flag = ""
n = 31
while 1:
    payload = b'1'*(n-len(flag)) + flag.encode()
    comp = encrypt(payload)
    for i in printable:
        enc = encrypt(payload + i.encode())
        if comp[32:64] == enc[32:64]:
            flag += i
            print("flag letter: ", i)
            break
    if flag[-1] == '}':
        break
print(flag)

Reason why the alternate method doesn't work:

In the payload initialising part, in working method payload = b'1'*(n-len(flag)) and in alternate method payload = b'1'*(n-len(flag)) + flag.encode(). The challenge encrypts user entered plain text using padded = pad(plaintext + FLAG.encode(), 16) , the isuue with the alternate method is that we would be giving the flag twice (pt+flag+flag) so we wont be getting the entire flag with this method, we will only get the first character