Welcome to JustAnswer.
How soon do yo need it?
Ok cool. I'ill work on it over the weekend.
Just checking. How is the assignment going so far?
I am still on it but seeing that its taking longer than I anticipated I'll keep it open for other experts by opting out while doing it.
More Google nonsense. Could you possible paste the questions into the Messages stream? Or is everyone in such a rush to have Google pirate the entirety of their content?
1- Compare the security and performance advantages and disadvantages of each variant of CBC mode: a fixed IV, a counter IV, a random IV, and a nonce-generated IV.
2- Suppose you, as an attacker, observe the following 32-byte ciphertext C1 (in hex)
4D 11 30 28 13 C7 8C EB 31 21 20 B0 70 B4 C3 B4 FA 3A 20 12 02 13 1C FD C5 44 07 25 D3 96 1F 74
and the following 32-byte ciphertext C2 (also in hex)
50 15 30 30 56 E6 91 BF 22 6E 22 B0 7D E0 CB B5 F0 6E 3A 22 15 0D 19 E5 8A 40 1A 31 D1 8D 03 74
Suppose you know these ciphertexts were generated using CTR mode with the same nonce. You also know that the plaintext P1 corresponding to C1 is
49 6C 69 6B 65 49 6E 74 72 6F 64 75 63 74 69 6F 6E 74 6F 43 72 79 70 74 6F 67 72 61 70 68 79 21
What information, if any, can you infer about the plaintext P2 corresponding to C2?
3- Let P1,P2 be a message that is two blocks long, and let P'1 be a message that is one block long. Let C0, C1, C2 be the encryption of P1, P2 using CBC mode with a random IV and a random key, and let C'0, C'1 be the encryption of P'1 using CBC mode with a random IV and the same key. Suppose an attacker knows P1, P2 and suppose the attacker intercepted and thus know C0, C1, C2 and C'0, C'1. Further suppose that, by random chance, C'1=C2. Show that the attacker can compute P'1