2. Say you have an n block of bits message. You split it to 8 bits each (assume that 8 divides n). For each of the 8 bit we choose a permutation of the bits. How many different codes could come out of this?
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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…Msg3. B→A : B, N1, Ks, Message2, h(B, N1, Message2) Whereas: h(m) represents the digest of the message (m). {m}{K} represents the encryption of message (m) using the key (K) PK is public key N1 is a random number. 1. What are the main problems in Msg3?**Please do C****C IS different from B** 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…
- **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…Assume we have a code composed of 5 ternary codewords of varying lengths (1, 1, 2, 2, 3). How many unique messages comprised of two message symbols result in a series of four code symbols after encoding?a. Suppose you have the following packet with the bits: 01111001 11001111 and 11110011 10011000. What is the Internet checksum of this packet? b. Suppose you have the following packet with the bits: 11000011 10101010, 10010001 01000101 and 11000011 10011000. What is the Internet checksum of this packet? c. For the packet in a), give an example where one bit is flipped in each of the two 16-bit words and yet the checksum doesn't change.
- Assume you are using any correct plaintext padding method, such as those described in lecture, with a 128-bit (16-byte) block cipher. If you are sending a message that is 184 bytes long, how many padding bytes would you need to add? Justify your answer. For the remaining questions, consider a 4-bit block cipher, described in hexadecimal by the following table: Plaintext Ciphertext Plaintext Ciphertext 0 a 8 e 1 c 9 d 2 f a 0 3 6 b 7 4 3 c 5 5 8 d b 6 4 e 9 7 2 f 1 You can think of this as a simple substitution cipher for hexadecimal digits. There is no “key” other than the table itself.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).2) Suppose someone suggests the following way to confirm that the two of you are both in possession of the same secret key. You create a random bit string the length of the key, XOR it with the key, and send the result over the channel. Your partner XORs the incoming block with the key (which should be the same as your key) and sends it back. You check, and if what you receive is your original random string, you have verified that your partner has the same secret key, yet neither of you has ever transmitted the key. Is there a flaw in this scheme? please give right solution and do not use chatgpt and chegg or any AI tool please
- Suppose Alice and Bob are going to communicate using AES in CBC mode. Unfortunately Alice's message length (in bytes) is not a multiple of 16. Suppose the last block of her message is just a single zero byte. How can she pad out the last block so that she can use CBC mode? Since this needs to be a reversible operation, how does Bob recognize the padding and remove it?4. Suppose we have to transmit a list of five 4 bit numbers that we need to send to a destination. Show the calculation using checksum method at the sender and receiver side if the set of numbers is ( 3, 7, 9, 11, 13).The very last block of a message encrypted with AES-256 in CBC mode is 15 bytes long and ends with the bytes: 0102030405060708090a0b0c0d0e0f When padded in accordance with PKCS #7 with 16-byte blocks, what is the complete value of the last block? *Please explain how you have solved this, thank you!!