ECB Oracle

Description

ECB is the most simple mode, with each plaintext block encrypted entirely independently. In this case, your input is prepended to the secret flag and encrypted and that's it. We don't even provide a decrypt function. Perhaps you don't need a padding oracle when you have an "ECB oracle"?

Help

This page offers a convenient way for you to interact with the challenge functions. You can also use GET requests to send and receive data directly from the listed routes/endpoints if you wish. For more information see the FAQ.

Your aim is to recover the FLAG value. Once you have it, submit it on the CryptoHack Symmetric Ciphers page.

Source

from Crypto.Cipher import AES
from Crypto.Util.Padding import pad, unpad


KEY = ?
FLAG = ?


@chal.route('/ecb_oracle/encrypt/<plaintext>/')
def encrypt(plaintext):
    plaintext = bytes.fromhex(plaintext)

    padded = pad(plaintext + FLAG.encode(), 16)
    cipher = AES.new(KEY, AES.MODE_ECB)
    try:
        encrypted = cipher.encrypt(padded)
    except ValueError as e:
        return {"error": str(e)}

    return {"ciphertext": encrypted.hex()}