a) The generator polynomial x8 +x2 +x + 1 is used to create a CRC code for the transmission of the message 1011001001001011. Determine the code that will be transmitted. (b) What will be the error produced if the MSB of the transmitted message obtained in (a) became corrupted during transmission.
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- The code containing of 8 code words which encrypt the communications zero to 1 1 1 as follows: State the succeeding questions with explanation on the question above: Is it linear? what is its block length, message length, charge and distance? how many mistakes can it detect, and how many situations is it incorrect? ( to reply this question grant the best range of errors that the code can assurance to end up conscious of or correct. for occasion a convenient parity bit can word two errors so we would say this can end up conscious of completely a single error. No calculations of risk are required.A code must have certain properties in order that any errors can be corrected. Let CC Z be the code C = {23410, 34102, 44113, 21110, 01103}. Which of these scenarios shows that C is not 1-error correcting? O If 44113 is sent and an error changes the second digit 4 to 1, the receiver cannot know what error has occurred. O If the receiver gets the word 00000, they cannot know what message was sent. O If 23410 is sent and an error change the 3rd digit to 1, the receiver cannot know what error has occurred. O If 23410 is sent and an error moves the first digit 2 to the last place, making 34102, the receiver cannot know that any error has occurredTo 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 transmitted
- The parity check matrix is given as shown:Determine the format of the codeword. c = [message | parity bits] c = [parity bits | message] c = [message | parity bits | message] none of the answersAlice and Bob decided to use HMAC to authenticate the communication between them. They already share a key k for the MAC scheme. To sign a message M (which is always of an even bit length), the sender parses the message as M=M_L||M_R, i.e. the message is parsed as a left half and a right half (both halves have equal length), then the sender computes X-M₁MR (where denotes XOR). The sender then sends M and T=HMACK(X) to the other party. Does this construction provide existential unforgeability? Yes NoUsing the matrix A= 1 L-11 -1] 0 decode the received message: -1 4 2 -12 -1 -12 13 26 -265 -14 -23 100 -6 7 -121. (Hint: Assign a number to each letter of the alphabet: A is 1, B is 2, and so on. A space between words is 27). Transmission Message A |Decoded Message Encoding Matrix Decoding Matrix
- A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data. Propose your own Roll as data bits (e.g roll 1 to 100) and show how the sender adds the parity bits and write the final code word that will be sent. Value of Key is 1011 (derived from polynomial function). Also show in the receiver side that the correct data is received.Consider the following message M=1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x² + 1 is?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?
- prove that: There does not exist any protocol for two processes to reach common knowledge about a binary value in an asynchronous message-passing system with unreliable communication.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 channel7. In the proof of the claim that for any n-bit messages, any resulting code C with parity bits satisfying properties (0) and (1) will have minimum distance 3, we use 3 cases: A(r, y) = 1,2 and 3. Select all that are true: a. We don't need the case of A(r, y) = 0 because the code has minimum distance 3. b. We don't need the case of A(r, y) = 0 because we assumed r # y. c. We don't need the case of A(r, y) = 0 because it is taken care of by the other cases. d. We don't need the case of A(r, y) = 0 because the claim doesn't need to be proven for that case. e. We don't need the case of A(r, y) = 4 because the code has minimum distance 3. f. We don't need the case of A(r, y) = 4 because we assumed a # y. g. We don't need the case of A(r, y) = 4 because it is taken care of by the other cases. h. We don't need the case of A(r, y) = 4 because the claim doesn't need to be proven for that case.