Suppose you want to send the following 2 16-bits numbers: 11010010 01111010 and 10011010 01011001. What would be the checksum? The spaces are used to make the sequence of bits more readable. Options: A) 01101100 00101011 B) 10010011 00101011 C) 10010011 11010100 D) 01101100 11010100
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- Here is a permutation: location in new byte 1 2 3 4 5 6 7 8 location in source byte 1 3 5 4 2 8 6 7 The source is a byte (8 bits) and the result is a byte (8 bits). Bit numbering starts at one, on the left. If the source byte is 1011 1000 what is the new byte? (Write the answer as 8 characters 0 or 1 with a space in the middle.)You have two 32-bit numbers, N and M, as well as two bit positions, I and j. Create a method for inserting Minto N in such a way that M begins at bit j and ends at bit i. You can assume that bits j through I have enough space to accommodate all of M. That is, if M = 10011, you can presume at least 5 bits exist between j and i. You wouldn't have j = 3 and I = 2, for example, because M couldn't completely fit between bit 3 and bit 2.Input: N=10000000000, M=10011,i=2,j=6Output: N=10001001100Suppose you have a 4-letter alphabet {A, B, C, D} and you wish to devise a binary code (with nospaces) for this alphabet such that no two words would be encoded the same. For example, if A = 1and B = 11, then AA and B would be encoded the same. It takes one second to transmit eachcharacter of the binary code (so, 101 would take 3 seconds and 110101 would take 6 seconds).(a) If each letter occurs with the same frequency, determine the most efficient code for the alphabetand explain why it is the most efficient.(b) Now suppose letter A occurs about 40% of the time, B and C both occur about 25% of thetime each, and C occurs about 10% of the time. Determine the most efficient code and explainwhy it is the most efficient.
- Write a program that reads any file of text and prints the bit values corresponding to each bit in each character. Access the individual bit values using bit shifting and/or masking. For example: a b c 0 1 2 Output should look like: a => 00011101 (97) b => 00011102 (98) ... Please use the C program language with screenshots.“Hamming,” will read in an integer “k” and a bit string “s” from the command line, calculate the “Hamming Distances,” and prints all bit strings that have Hamming distance of “k” from “s.” Note: The Hamming Distance is equal to the number of bits in which the two strings differ. A sample run would be as follows. >java Hamming 2 00111111 1001 1010 0101 0110 0000Assume that D has the value 1010101010 and that the 5-bit generator G=10011 is used. What does R stand for? B. C. Does corruption exist? Publish your work. Repetition of parts a and b with the addition of the following: the numbers 1001010101, 0101101010, and 1010100000
- You receive the following data fragment: 0110 0111 1100 1111 0111 1101. You know that the protocol uses bit stuffing. Show the data after destuffingWrite a program to set in 8060 (bit 7) and clear (bit 0) of a byte wide number stored in address 0200-2460H, Save the modified bytes in address 0200-2700HUse DATA: 6FH, 3DH, 27H, 83H, DEH, 13H, 78H, 09H, F5H, 03H, 84H,7EH, 0BH, 87H, FEH, 11HYou are given two bit locations, I and j, together with the 32-bit values N and M. To insert Minto N, create a method that starts M at bit j and finishes it at bit i. It is safe to presume that all of M can fit in the bits j through i. Therefore, you may infer that there are at least 5 bits between j and I if M = 10011. M could not completely fit between bit 3 and bit 2, thus you would not, for instance, have j = 3 and I = 2.EXAMPLE Key in: NN 10000000000, M 10001001100 as output
- Write a program to transfer 8-bit numbers from 9080H to 9090H if D5 is 1 and D3 is 0. otherwise transfer data by changing bit D2 and D6 from 1 to 0 and 0 to 1. Assume there are ten numbers.How many 88-bit binary strings are there subject to each of the following restrictions? 1. The string starts with 001001. 2. The string starts with 001001 or 1010. 3. The string has exactly 4 zero’s. 4. The string has exactly 4 zeros’s and the first bit is 1.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). (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.