3.6 SENDING BOB A MESSAGE

EXERCISE 3.6: SENDING BOB A MESSAGE

Using either a modification of the preceding program or your AES encryptor from the beginning of the chapter, create a couple of meetup messages from Alice to Bob. Also create a few from Bob to Alice. Make sure that you can correctly encrypt and decrypt the messages.

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# Suppose the following code is in a file named: ex3_6.py 

from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes
from cryptography.hazmat.backends import default_backend

class Padding:
    @staticmethod    
    def pad_it_up(msg: bytes) -> bytes: 
        '''
        This function is used to pad the message.

        Technique: Determine the number of padding bytes required.
        This is a number n which satisfies 1 <= n <= 16
        and n + len(msg) is a multiple of 16. Pad the plaintext by appending
        n bytes, each with value n. (Read more in Chapter 4 of the book  
        "Cryptography Engineering by Niels Ferguson, Bruce Schneier, Tadayoshi Kohno".)
        ''' 
        n = 16 - len(msg) % 16
        return msg + bytes([n] * n)
    
    @staticmethod
    def unpad_it_up(msg: bytes) -> bytes: 
        '''this function is used to remove the pads from the message''' 
        n = msg[-1]
        # then ignore the last n bytes
        return msg[:-1*n]

# Alice and Bob's Shared Key 
test_key = bytes.fromhex('00112233445566778899AABBCCDDEEFF')

def encrypt_using_aes_ecb(plaintext: str, key: bytes) -> str: 
    proper_plaintext = Padding.pad_it_up(plaintext.encode('utf-8'))
    aesCipher = Cipher(algorithms.AES(key), modes.ECB(), backend=default_backend())
    aesEncryptor = aesCipher.encryptor()
    ciphertext = aesEncryptor.update(proper_plaintext)
    return ciphertext.hex(chr(10),-16) # this will insert '\n' between every 16-byte block. 

def decrypt_using_aes_ecb(ciphertext: str, key: bytes) -> str: 
    proper_ciphertext = bytes.fromhex(ciphertext.replace('\n',''))
    aesCipher = Cipher(algorithms.AES(key), modes.ECB(), backend=default_backend())
    aesDecryptor = aesCipher.decryptor()
    plaintext = aesDecryptor.update(proper_ciphertext)
    plaintext = Padding.unpad_it_up(plaintext)
    return plaintext.decode()

def EATSAform(from_codename: str, to_codename: str, date: str, location: str, time: str): 
    # Note that EATSA stands for East Antarctica Truth Spying Agency 😉
    return f"FROM: FIELD AGENT {from_codename}\n" \
    f"TO: FIELD AGENT {to_codename}\n" \
    f"RE: Meeting\n" \
    f"DATE: {date}\n" \
    f"\nMeet me today at the {location} at {time}."

First generate the messages:

Then encrypt both of the messages:

Look at the two ciphertext outputs of these messages side-by-side.

No	Ciphertext 1 Blocks	Ciphertext 2 Blocks
1	e5423e662cb98ca3ecf2c66b31cf1c8b	e5423e662cb98ca3ecf2c66b31cf1c8b
2	f7c240c61a9c1453aae2edb028fa8459	f7c240c61a9c1453aae2edb028fa8459
3	3d6ec4443f76ebc5a8abb94879113e58	3d6ec4443f76ebc5a8abb94879113e58
4	88191d257c5970af6774a9f696681766	88191d257c5970af6774a9f696681766
5	9473f734c616d05dae998037a11f67ab	a9f7e4995ec96761b811c24953c19907
6	ee0c3ea2b40c09d80e9558cacf1a60ef	f70cc1a151a388f34c59b7b83ae0fb08
7	66651c2dc32f4c25f276968d8efda22d	4d91ea64099f6cb3dea4d1c0edcab02f
8	[empty]	5a31ed03bf733b150ca46118d8fd8e95

Note that the first 4 ciphertext blocks are the same.

Remember that AES in its raw mode is like a code book. For every input and key, there is exactly one output, independent of any other inputs. Thus, because much of the message header is shared between messages, much of the output is also the same.

Last step, check if the ciphertexts decrypt correctly:

Or: