Comparing CBC mode to OFB mode, only in OFB mode... A. each block is encrypted separately B. an initialization vector is utilized C. each ciphertext block depends on the previous ciphertext block D. a key stream is utilized

Alice encrypts a message using AES, and sends the ciphertext to Bob. Unfortunately, during the calculation of the third block, a lightning struck Alice’s house, corrupting the 2nd bit of the AES output (for the third block). Fortunately, nothing else of the calculation was affected. How much of the plaintext can Bob still recover if the mode of encryption is one of the followings: CBC, CFB, OFB, or CTR?

Suppose an error occurs in a block of ciphertext on transmission using CBC. What effect is produced on the recovered plaintext blocks?

2. You have written a terminal emulation program which sends and receives individual 8-bit keystrokes. Now you want to add encryption, to provide confidentiality. You decide to make use of AES-256-CFB - that is AES with a 256-bit key, operating in cipher feedback mode. Once a connection is set up, assuming no padding, how many blocks will AES encrypt in order to transmit a line of 32 keystrokes?

Modes of operation such as CFB and CBC require the use of an initial variable (IV), which must be known by both the sender and receiver. Discuss the pros and cons of the following suggestions for selection of IVs Randomly generate each IV; Use a counter that is incremented after each message as the IV; Use the last ciphertext block of the previous message as the IV

Let E'(T,X) be a tweakable block cipher that has already been keyed. Given plaintext P = P1 || P2 || ... || Pn (ie, P is an n-block plaintext). Ci = E'(i, Pi) for i=1..nsum = P1 xor P2 xor ... xor Pntag = E'(0,sum) For simplicity let's say that E'(T,X) = ROTL(X,T+1)  (ie, X rotated left T+1 bits. If the block cipher block size is 8 bits and you are encrypting the two byte plaintext 67 89, what ciphertext and tag would be created? Fill in each box as a two-digit hex value. C1____? C2____? tag____?

Problem with constructing MAC. The messages X to be authenticated consists of z independent blocks, so that: X = x1||x2|| . . . ||xz, where every xi consists of |xi| = 8 bits. The input blocks are consecutively put into the compression function  ci = h(ci−1, xi) = ci−1⊕xi At the end, the MAC value: MACk(X) = cz+k mod 28 is calculated, where k is a 64-bit long shared key. Describe how exactly the (effective part of the) key k can be calculated with only one known message X.

What is the biggest advantage of CBC mode of operation   a. It does not need Initialization Vector (IV) b. A cipher text block depends on all blocks before it c. It does not propagate errors in the cipher text d. It can encrypt in parallel different parts of the messages

Please explain the reasoning behind the blocks being affected thank you.