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?
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- Suppose we use a checksum scheme based on the addition of successive 8-bit fields rather than 16-bit fields as in the Internet Transport layer. Now suppose that the following 3 bytes are received, in which the last bytes is the checksum field. 0 1 0 1 0 0 1 1 1 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 Does the receiver detect an error that might have occurred during the transmission of these three bytes?Suppose the information content of a packet is the bit pattern 1110 0110 1001 1101 and an even parity scheme is being used. What would the value of the field containing the parity bits be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum length checksum field is used.Suppose the information content of a packet is the bit pattern 1110101010101111 and an even parity scheme is being used. What would the value of the checksum field be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum-length checksum field is used.
- Computer Science With the CBC/OFB/CFB/CTR mode: (a) if there is a single bit transmission error in block C3 of the ciphertext, which plaintext blocks are affected? CBC and CTR (b) Suppose that there is a bit error in the source version of P1. Through how many ciphertext blocks is this error propagated? What is the effect at the receiver?ACTIVITY : CODING Suppose the following messages are source encoded given by the table. Message Encoded Bits UP 00 DOWN 01 LEFT 10 RIGHT 11 Complete the given table by writing the source encoded message of the following set of directions. Set of Directions Source Encoding LEFT - UP - RIGHT - DOWN 1.) 10001101 UP - DOWN - UP - RIGHT 2.)00010011 DOWN - DOWN - RIGHT - LEFT 3.) 01011110 UP - DOWN - LEFT - RIGHT 4.)00011011 RIGHT - LEFT - RIGHT - UP 5.)11101100 Decode the set of directions given by the message received. Message Received Decoded Message 10010011 1.) 01010010 2.) 11000110 3.) 01100111 4.) 00101111 5.) III. Complete the given table by writing the channel encoded message by using parity check of adding 1 bit in the message. Message Source Encoding Channel Encoding ALLY 000 1.) ENEMY 011 2.) ATTACK 100 3.) RETREAT…For each of the following sets of codewords, please give the appropriate (n,k,d) designation where n is number of bits in each codeword, k is the number of message bits transmitted by each code word and d is the minimum Hamming distance between codewords. Also give the code rate. {111, 100, 001, 010}
- 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).Suppose the message that should be transmitted from host A to host B contains the following four bytes: 01110111, 11011010, 10011101, and 00110011. Find the checksum of this message computed by host A, assuming the checksum is computed as sum of bytes rather than sum of 16 bits. Show your work.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?
- Suppose the sender and the receiver agree to use the bit pattern 01111110 to mark the beginning and the end of a frame. 1)The sender has the following bits to send. What does the sender actually send? 011110000111111011101111101 2) The receiver receives the following bits. What're the original data bits (note: this question has nothing to do with the previous question) 01111110111110111110001101111101000011111001111110Compute 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.Suppose that a binary message – either 0 or 1 – must be transmitted by wire from location Ato location B. However, the data sent over the wire are subject to a channel noise disturbance,so to reduce the possibility of error, the value 2 is sent over the wire when the message is1, and the value −2 is sent when the message is 0. If X, X = {−2,2}, is the value sent atlocation A, the value received at location B, denoted as R, is given byR = X + N ,where N is the channel noise disturbance, which is independent of X. When the message isreceived at location B, the receiver decodes it according to the following rule:if R ≥.5, then conclude that message 1 was sentif R < .5, then conclude that message 0 was sentAssuming that the channel noise, N, is a unit normal random variable and that the message0 or 1 is sent with equal probability, what is the probability that we conclude that the wrongmessage was sent? This is the probability of error for this communication channel