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- Need #3 onlyNext time you snap a picture, you realize you are capturing millions of pixels into a buffer. The buffer data is read and converted into JPEG in real time. Each pixel in that buffer is an unsigned int (four bytes ): Alpha, Blue, Green, and Red. Let us ignore alpha for now. As you know, a unsigned byte can have a value 0 to 255. In remote sensing jargons, it is called blue channel, green channel, and red channel. Each channel provides valuable information such as, say farm lands, forest fire, drought, landscape, diseases , If a pixel has a value (say in hex) = 0x00a1b1c1 , then 0xc1 is the red pixel, 0xb1 is the green pixel, 0xa1 is the blue and 00 is the alpha. #define RED 1 #define GREEN 2 #define BLUE 3 then, develop a function void calculateSum ( unsigned int *ptr , int count , unsigned char channel , unsigned int *sum, float *average ) { *sum = 0; if (channel == RED ) calculate sum and average for red channel else if (channel == GREEN )…Checksum solvation. Below.
- Suppose 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.Internet Checksum. Consider the six sixteen bit numbers: 10110101 01000110 01001001 01101011 10110100 01000110 01001001 01101101 10110101 01010110 01011010 01101111 Compute the Internet Checksum field value of these six words (each word has sixteen bits) Enter the 2 bytes each as an 8-bit number with only O's and 1's, and make a single blank space between the two 8-bit numbers (e.g., 01010101 00101000). [Note: you must use bitwise XOR logic function and don't use 2's or 1's complement] Not quite. This answer is incorrect.3. How many bit strings of length eight either begin with two Os or end with three 1s* 26 +25 - 23 25 +24 - 23 P (8,5) 26 + 24 - 2³ 23
- Given are the following codewords for an error control coding scheme data 000 011 100 codeword 00011000 00000111 10101000 . a.What is the minimum Hamming distance? b. What is the coding rate for this scheme? c. what raw data rate would be required if we need a throughput of 6 Mpbs? d. in a different situation, the number of data bits to transfer is 5, and the minimum hamming distance is dmin=5. find the maximum coding rate for this situationGiven rax = 0x0000000200000100, rbx = 0x0000000000000100, and rcx = 0x0000000000000001,and the following values in memoryaddress -> byte at that address0x0000000000000100 -> 0x010x0000000000000101 -> 0x000x0000000000000102 -> 0x000x0000000000000103 -> 0x000x0000000000000104 -> 0x020x0000000000000105 -> 0x000x0000000000000106 -> 0x000x0000000000000107 -> 0x00what is the new value in %rax after the following operation?subq -0x04(%rbx, %rcx, 4), %raxTopic: Binary Fill in the boxes Will give you high rating thank you!!
- Internet Checksum. Consider the two sixteen bit numbers: 10110100 01000110 01001000 01101111 Compute the Internet Checksum of these two values Enter the 2 bytes each as an 8-bit number with only O's and 1's, and make a single blank space between the two 8-bit numbers (e.g., 01010101 00101000).Binary Arithmetic Coding: Consider integer implementation of binary arithmetic coding: Use 4 bit integers to store LOW and HIGH. At certain stage, suppose the counters for 0 and 1 are 30 and 10 respectively LOW =2, HIGH = 13, and E3 scaling counter Scale3 = 2; Describe the detains of the encoding process when the next two inputs bits are both 1;Consider a 32-bit number stored in a byte-addressable memory as shown in the table below (note that the value is written in hex format): a. If the machine is little endian and uses 2's complement representation for integers, give the decimal representation of the 32-bit integer that starts at address 100. b. If the machine is big endian and the number is an IEEE single precision floating point value, give the decimal equivalent. Note the following: • When converting the binary significand to its base 10 equivalent, you only need to use the first 8 bits (not all 23) The final answer should be written as a simple real number (i.e., not in exponential format), with the fractional component rounded to 2 decimal places (e.g., 126.46 or -9768.83) Value Address 100 101 102 103 40 92 9C FB