Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa)
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Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa).
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- a. Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV. Details in pages 325-326 Ch.12 of the textbook) b. Suppose Alice computes the Secret prefix MAC (page 322: secret prefix MAC(x) = h(key || x)) for the message ”AM” (in ASCII) with key “G” (in ASCII) that both Alice and Bob know. The hash function that is used is h(x1x2x3)= g(g(x1 XOR x2) XOR x3 ) where each xi is a character represented as 8 bits, and g(x) is a 8-bit string that is equal to the complement of bits in x. For example, g(10110011) = 01001100. The MAC is 8 bits. (8-bit ASCII representation of the characters is given below.) What is the Secret prefix MAC…6. Consider the following simplified version of the CFB mode. The plaintext is broken into 32-bit pieces: P = [P₁, P2, ,], where each P, has 32 bits, rather than the 8 bits used in CFB. Encryption proceeds as follows. An initial 64-bit X₁ is chosen. Then for j = 1,2,..., the following is performed: CjPjL32 (Ek (Xj)) Xj+1 = R32(X₁)||C₁ where L32(X) denotes the 32 leftmost bits of X, R32 (X) denotes the rightmost 32 bits of X, and XY denotes the string obtained by writing X followed by Y. Find the decryption algorithm.Consider a simple system with 8-bit block size. Assume the encryption (and decryption) to be a simple XOR of the key with the input. In other words, to encrypt x with k we simply perform x⊕k giving y. Similarly, to decrypt y, we perform y⊕k giving x. You are given the following 16-bit input “F5C6” in hexadecimal. You are provided IV as “8C” in hexadecimal. The key to be used (where appropriate) is “49” in hexadecimal. Compute the encrypted output with the following methods. Express your final answer, for each of them, as 4 hexadecimal characters so it is easy to read. A. Using ECB B. Using CBC C. Using OFB D. Using CFB
- A symmetric block encryption algorithm is shown below. 16-bit blocks of plaintext P are encrypted using a 32-bit key. Encryption is defined as: C = (P EX_OR Ko) + K1 C is the ciphertext, K is the secret key, Ko is the leftmost 16 bits of K, K1 is the rightmost 16 bits of K, EX_OR is bitwise exclusive OR, and + is binary addition. a. The ciphertext C must be the same size as the plaintext P, that is, it must not be larger than 16 bits. How can this be achieved? b. Show the decryption equation. How will the encrypted message be decrypted?AsapCompute CBC-MAC for a message of 16 bits, “ABCD” (in Hexa). Assume a block size of 8 bits with an IV=C9 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be D8 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV.
- **Please do A or C** a. Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV. Details in pages 325-326 Ch.12 of the textbook) b. Suppose Alice computes the Secret prefix MAC (page 322: secret prefix MAC(x) = h(key || x)) for the message ”AM” (in ASCII) with key “G” (in ASCII) that both Alice and Bob know. The hash function that is used is h(x1x2x3)= g(g(x1 XOR x2) XOR x3 ) where each xi is a character represented as 8 bits, and g(x) is a 8-bit string that is equal to the complement of bits in x. For example, g(10110011) = 01001100. The MAC is 8 bits. (8-bit ASCII representation of the characters is given below.) What is the…To prevent the tapping and use of information on a wireless network which has brought untold mistrust amongst the directors of Okonko Systems, you suggested that every message that is transmitted should go through the following: A systematic two left bit shift; After “a” above, the algorithm f[(6m +2) + k] mode 13 should be applied to the cipher text, where m is the message and k =9 is a key; After “b” above, perform a systematic right bit shift to the new cipher; The message should then be transmitted. If on a trial basis, the message is “genomics” Use the mono – alphabetic letters where a = 0, b = 1, ….z = 25 to find the cipher that would be transmittedIn an RSA system, the public key (n,e) of a given user is (323, 11). 1. What is the value of the exponent in the private key (n, d), of this user? 2. Suppose you want to send this user the message m = 45, write down the expression to generate the ciphertext for this message. 3. Suppose that your ciphertext, c, is 5, write down the expression to generate the plaintext matching this ciphertext. 4. Can you encrypt the message m=322 with this public key? O a. Yes O b. No
- Let's encrypt a message using RSA: Choose p = 7 and q = 11, and then select e=13. a.Compute d b.Select plaintext message x=7. Produce the ciphertext y using the fast exponentiation algorithm. c. Decrypt the ciphertext (y) to verify that the initial plaintext (x) is produced. Again, please use the fast exponentiation algorithm.Compute CBC-MAC for a message of 16 bits, “ABCD” (in Hexa). Assume a block size of 8 bits with an IV=C9 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be D8 (in Hexa).b) Find the inverse of the encrypting function f (x) = (5 – x) mod 26,0 < x < 25, and use it to decrypt the message "ZRQUBFNB". -