示例#1
0
def decrypt_aes_cbc_2(text, key, iv):
    # Create blocks of the key's length each
    blocks = [text[i:i + len(key)] for i in range(0, len(text), len(key))]
    results = []
    for _, block in enumerate(blocks):
        # Decrypt the current block with the given key
        deciphered_block = set1.decrypt_aes_ecb(block, key)
        # XOR the deciphered block with the IV
        xored_deciphered_block = set1.XOR(deciphered_block, iv)
        # Decode the result as a string, and add it to the result set
        results.extend(xored_deciphered_block)
        # Set the IV of the next block as the current block's ciphertext
        iv = block
    return set2.pkcs7_remove_padding(bytes(results))
示例#2
0
def aes_cbc_mode_decrypt(cipher, key, iv):
    BLOCK_SIZE = len(key)
    num_blocks = math.ceil(len(cipher) / BLOCK_SIZE)
    blocks = []
    for i in range(num_blocks):
        blocks.append(cipher[(i*BLOCK_SIZE):(i+1)*BLOCK_SIZE])

    output = b''
    previous_block = iv

    for b in blocks[0:]:
        decrypted_block = set1.fixed_xor(set1.decrypt_aes_ecb(b, key), previous_block)
        output += decrypted_block
        previous_block = b

    # TODO remove padding length :)

    return output
示例#3
0
def challenge_25():
    key = set2.random_bytes(16)
    nonce = random.randrange(1, 10**10)
    # Get plaintext
    with open('inputs/25.txt') as file:
        contents = base64.b64decode(file.read())
    plaintext = set2.pkcs7_remove_padding(
        set1.decrypt_aes_ecb(contents, 'YELLOW SUBMARINE'))
    # Generate ciphertext
    ciphertext = set3.encrypt_aes_ctr(plaintext, key, nonce)

    # Create vulnerable function
    def vulnerable_ctr_stream_edit_function(ciphertext, offset, newtext):
        return edit_ctr_stream(ciphertext, key, nonce, offset, newtext)

    # Find the plaintext using the above function
    obtained_plaintext = find_ctr_plaintext(
        ciphertext, vulnerable_ctr_stream_edit_function)
    # Verify found plaintext matches with the original plaintext
    assert_true(obtained_plaintext == plaintext)
示例#4
0
def get_oracles():
    key = random_bytes(16)
    return lambda x: encrypt_aes_ecb(x, key), lambda x: set1.unpad(
        set1.decrypt_aes_ecb(x, key))
示例#5
0
def encrypt_aes_ecb(text, key, padding=True):
    from Crypto.Cipher import AES
    # Create new AES object with the given key in ECB mode
    obj = AES.new(key, AES.MODE_ECB)
    if padding:
        # Compute the padding necessary to make the length of the input text a multiple of len(key)
        padding = len(key) - len(text) % len(key)
        # Apply padding
        text = pkcs7_add_padding(text, len(text) + padding)
    # Encrypt given text
    return obj.encrypt(text)


assert set1.unpad(
    set1.decrypt_aes_ecb(
        encrypt_aes_ecb(bytes("O Brave New World", 'utf-8'),
                        "YELLOW SUBMARINE"),
        "YELLOW SUBMARINE")) == bytes("O Brave New World", 'utf-8')


def decrypt_aes_cbc(text, key, iv, unpad=True):
    # Create blocks of the key's length each
    blocks = [text[i:i + len(key)] for i in range(0, len(text), len(key))]
    results = []
    for _, block in enumerate(blocks):
        # Decrypt the current block with the given key
        deciphered_block = set1.decrypt_aes_ecb(block, key)
        # XOR the deciphered block with the IV
        xored_deciphered_block = set1.XOR(deciphered_block, iv)
        # Decode the result as a string, and add it to the result set
        results.extend(xored_deciphered_block)
        # Set the IV of the next block as the current block's ciphertext
示例#6
0
    def test_aes_ecb_mode(self):
        with open('7.txt') as f:
            contents = bytes(base64.b64decode(f.read()))
            deciphered = set1.decrypt_aes_ecb(contents, 'YELLOW SUBMARINE')

            self.assertIn(b"I'm back and I'm ringin' the bell \nA rockin'", deciphered)