Question: Given a message m2-101 110, a key k1 = 10 for encryption and a key k2-00 for message authentication, assume each block has 3 bits

Given a message m2-101 110, a key k1 = 10 for encryption and a key k2-00 for message authentication, assume each block has 3 bits and a random initialization vector IV- 101, - compute a ciphertext and its tag using the Encrypt-then-Authenticate approach also explain why other approaches, such as Encrypt-and-Authenticate and Authenticate-then-Encrypt, are not suitable to protect both data privacy and message authentication. The encryption algorithm uses CBC mode and the Mac generation algorithm also uses CBC mode. When we compute a tag for a ciphertext, we assume that the initialization vector IV is a part of a ciphertext The PRF/PRP used in this block cipher is described as below 000 001 010011 100 101 110 111 k = 00. Fh(2)| 10010101011101|1 1110001001|110 k 01, Fk () 010011101 111000 001 110100 k 10. Fe(2)| 101|1 1110001001|1101001010|011 k 11, Fk(a) 111101 000001 100 110011 010 Given a message m2-101 110, a key k1 = 10 for encryption and a key k2-00 for message authentication, assume each block has 3 bits and a random initialization vector IV- 101, - compute a ciphertext and its tag using the Encrypt-then-Authenticate approach also explain why other approaches, such as Encrypt-and-Authenticate and Authenticate-then-Encrypt, are not suitable to protect both data privacy and message authentication. The encryption algorithm uses CBC mode and the Mac generation algorithm also uses CBC mode. When we compute a tag for a ciphertext, we assume that the initialization vector IV is a part of a ciphertext The PRF/PRP used in this block cipher is described as below 000 001 010011 100 101 110 111 k = 00. Fh(2)| 10010101011101|1 1110001001|110 k 01, Fk () 010011101 111000 001 110100 k 10. Fe(2)| 101|1 1110001001|1101001010|011 k 11, Fk(a) 111101 000001 100 110011 010
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