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Lab 2

Solution authors:

Language: Python 3

The main.py script prints all the answers to stdout.

Exercise 1: why we do all of this

Solution:

I,(Petra Vaňková, Martin Hula, Petr Stejskal) , understand that cryptography is easy to mess up, and that I will not carelessly combine pieces of cryptographic ciphers to encrypt my users' data. I will not write crypto code myself, but defer to high-level libraries written by experts who took the right decisions for me, like NaCL.

Exercise 2: encrypt single-block AES

What is the ciphertext of encrypting the plaintext 90 miles an hour with the key CROSSTOWNTRAFFIC? Answer in hex.

Solution:

Ciphertext is 092fb4b0aa77beddb5e55df37b73faaa.

Exercise 3: decrypt single-block AES

What is the decryption of the 16 byte block fad2b9a02d4f9c850f3828751e8d1565 with the key VALLEYSOFNEPTUNE?

Solution:

Original plaintext is I feel the ocean.

Exercise 4: implement PKCS#7 padding

What is the output of pad("hello")?

Solution:

Output of function is b'hello\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b'.

Exercise 5: implement PKCS#7 unpadding

What is the output of unpad("hello\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b")?

Solution:

Output of function is hello.

Exercise 6: implement ECB encryption

What is the encryption of the following plaintext? Use the key vdchldslghtrturn. Answer in hex.

Well, I stand up next to a mountain and I chop it down with the edge of my hand

Solution:

Ciphertext is

883319258b745592ef20db9dda39b076a84f4955a48ba9caecd1583641cf3acac86acd5e5795de7895fab54481e9d8c3afc179c39412282eb8445ea2450e763df7282998a74baf19887c843b658f8891

Exercise 7: implement ECB decryption

What is the decryption of the following ECB encoded ciphertext? Use the key If the mountains.

792c2e2ec4e18e9d3a82f6724cf53848
abb28d529a85790923c94b5c5abc34f5
0929a03550e678949542035cd669d4c6
6da25e59a5519689b3b4e11a870e7cea

Solution:

Original plaintext is If the mountains fell in the sea / Let it be, it ain't me.

Exercise 8: ECB ciphertext manipulation (cut and paste 1)

  1. Have a quick look at the hex file. Can you quickly spot some obvious patterns? What fact can you deduce about the song lyrics?

    Solution:

    Some lines are completely repeated e.g. line 9 and 21 or lines 2,4,6,8,12.... In a lot of lines there is the second block repeated '0a5f54b2886d308f7ffd76cd41c3e1a9' eg. line 9,16,19... Song has repeating verses and refrains.

  2. At the end of the file, you can see a 16 byte block "alone"; however, all lines of the song lyrics are really 32 bytes long. Can you explain the presence of this last 16 byte block? Can you guess the plaintext of this block?

    Solution:

    At first we could think, it is a shorter text, a signature or a motto, but the answer is more obvious. It is an empty line, we often do at the end of files.

  3. Then, decrypt the text with key TLKNGBTMYGNRTION. The first line should start with "People" -- what is the rest of this line?

    Solution:

    Content of line is People try to put us d-down.

Exercise 9: ECB message crafting (cut and paste 2)

  1. What is the ciphertext of welcome("Jim")? Answer in hex.

    Solution: Ciphertext is d4d7730a2d4255c88dead80a2ad924f2b114fddb898d7ef8abdfefef30d552863f62b0605102e0186402df7666edcec7

  2. Use these blocks (and perhaps other calls to welcome) to craft an encrypted message whose plaintext starts by "Your name is " and finishes with " and you are an admin". In your report, write down your crafted ciphertext.

    Solution:
    37a9b711912fb58496097b8f6cc23abe0f07a3c84535a484e76f651644f4c37140b8d021d8079fbc03d6aa56f466423f

  3. Decrypt your encrypted message to make sure it is correct. In your report, write the decrypted message.

    Solution: After calling produce_fake_welcome('Petra') we get hex output above in 5, after decryption we get Your name is Petra and you are an admin

  4. Could you quickly describe a real-world scenario where this could be a security issue?

    Answer: As real-world scenario we can imagine situation, where we send a message via internet. Attacker can catch the message and change the content. Without any manipulation the check mechanism is not possible to find out if message is original or manipulated. It can be a message to internet banking (a transaction request) or in some chat.

Exercise 10: ECB decryption (cracking) byte-at-a-time

  1. Call hide_secret() with an input containing fifteen times (16-1) the character A: "AAAAAAAAAAAAAAA". In the first block to be encrypted, what will be the last plaintext byte?

    Answer: Last character of first block will be first character of secret message, in this situation character 't'.

  • You could continue cracking new bytes, but they will always have the value 1. Why?

    Answer: In situation when we crack the last char of secret (plaintext), we decrease count of A chars at the beginning. After that we try to crack next byte, which is one byte of padding with hex value 0x01 (because plaintext moved to left and padding is filled to 16 byte block). We take this byte and concatenate it to the already known secret string.

    In the next round we decrease number of As and again concatenate the known secret string with As block. So in the last block we have 15 bytes (known secret plus one byte with 0x01 hex value) and the algorithm again adds 1 byte of padding to last position (0x01). After that the value we always obtain is the same padding byte repeatedly generated by our cracking algorithm.

  • The length of the response to your initial call hide_secret("AAA..AA") will be exactly a + s + 1 where a is the number of A's (between 0 and 15) and s is how many secret bytes you know so far. Why is this true only when the secret is complete?

    Answer: When the secret is completely reached in formula a + s + 1, a is equal to number of A's in input to hide_secret, s is equal whole known secret length and +1 is one padding byte. So length of output is corresponding with this formula, when we can say, we have reached whole secret. We name this length as L. Why is this true only when the secret is complete?

    • In situation of complete secret, the padding is 1 byte long. Generally we can see that output string from hide_secret() has shortest length if padding has 1 byte length.
    • There can be more of these situations when padding length is 1. Such situation is when length of the secret is % 16 == 0.
    • We can create formula for this shortest length of output situations: a + s + u + 1, where u is number of uncracked characters.
    • Then it is true that a + s + u + 1 is equal to formula a + s + 1 in situation of complete secret.
    • When secret is not complete, u >= 1 in a + s + u + 1, we can see that a + s + 1 must be lower than length of produced output string.
    • So a + s + u + 1 equals to L only of u is zero and this state happens if we have reached the whole secret.