int alloddBits(int x) { return 2; } * negate return -x Example: negate (1) = -1. Legal ops: ! ~ & ^ Max ops: 5 Rating: 2 */ int negate(int x) { return 2; } | + << >>
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- Solution Floating point representation: It is defined as the representation of floating numbers. It includes sign bit, exponent, and mantissa bits. Based on precision it has 2 types. 1. For IEEE 754 single-precision floating-point numbers, what is the exponent of a denormalized floating-point number in decimal? Solution: In IEEE 754 single-precision, exponent bits are 8. Therefore exponent = 2 ^(n-1) -1 = 2^(8-1) -1 = 127 OPTION D 2. For IEEE 754 single-precision floating-point numbers, how many bits for mantissa? Solution: In IEEE 754 single-precision, there are 23 bits for mantissa. sign = 1 bit exponent = 8 bits mantissa = 23 bits 3. For IEEE 754 single-precision floating-point numbers, which of the following is an example of NAN? Solution: In IEEE 754 single-precision, NAN is a special value where all exponents bits are 1's and the mantissa is non zero. a. 1 111 1 111 0000 0000 1101 0000 0000 0000 Exponent is not all 1's. Not a NAN b. 0 111 1 111 1000 0000…Dessssssssssinclude <bits/stdc++.h> using namespace std; int MSD(int n){ if(n == 0) return 0; int k = log10(n); int x = pow(10,k); int ans = n/x; return ans; } int main() { int n; cin >> n; cout << MSD(n).Assignment for Computer Architecture: N Factual by Recusion *please have comments in the code* You are to write a program in MIPS that computes N! using recursion. Remember N! is the product of all the numbers from 1 to N inclusive, that is 1 x 2 x 3 x (N – 1) x N. It is defined as 1 for N = 0 and is undefined for values less than 0. The programs first requests the user to input the value of N (display a prompt first so the user knows what to do). If the input value is less than 0, the program is to display “N! undefined for values less than 0” and then requests the user to input the value of N again. If the value input is non-negative, it is to compute N! using a recursive function, that is one that calls itself. You are to have your name, the assignment number, and a brief description of the program in comments at the top of your program. Since this is an assembly language program, I expect to see comments on almost every line of code in the program. Also make the…
- Soundex System Soundex is a system that encodes a word into a letter followed by three digtis that roughly describe how the word sounds. That is, similar sounding words have similar four-character codes. For instance, the words carrot and caret are both coded as C123. A slight variation of the Soundex coding algorithm is as follows: 1. Retain the first letter. 2. For the remaining letters, delete all occurrences of a, e, i, o, u, h, y, and w. 3. Replace the letters that remain with numbers so that (a) b, f, p, and v become 1 (b) c, g, j, k, q, s, x, and z become 2 (c) d and t both become 3 (d) l (that is, el) becomes 4 (e) m and n become 5 (f) r becomes 6 4. If the result contains two adjacent identical digits, eliminate the second of them. 5. Keep only the first four characters of what you have left. If you have fewer than four, then add zeros on the end to make the string have length four. Write a program that carries out the algorithm. See Fig. 6.86.int X[900]; int Y[600]; int sum, sum1, sum2, sum3; //parallelism : dividing outer loop in three parts //i = 1 to 300 for(i=1;i<=300;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 301 to 600 for(i=301;i<=600;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 601 to 900 for(i=601;i<=900;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } sum = sum1 + sum2 + sum3;} another way to solve the question that send in the picSoundex System Soundex is a system that encodes a word into a letter followed by three digtis that roughly describe how the word sounds. That is, similar sounding words have similar four-character codes. For instance, the words carrot and caret are both coded as C123. A slight variation of the Soundex coding algorithm is as follows: 1. Retain the first letter. 2. For the remaining letters, delete all occurrences of a, e, i, o, u, h, y, and w. 3. Replace the letters that remain with numbers so that (a) b, f, p, and v become 1 (b) c, g, j, k, q, s, x, and z become 2 (c) d and t both become 3 (d) l (that is, el) becomes 4 (e) m and n become 5 (f) r becomes 6 4. If the result contains two adjacent identical digits, eliminate the second of them. 5. Keep only the first four characters of what you have left. If you have fewer than four, then add zeros on the end to make the string have length four. Write a program that carries out the algorithm. See Fig. 6.86. THIS IS DONE IN VISUAL BASIC
- Q/Complete the following code in Python language (biometrics) for voice recognition and apply the code, mentioning the approved source if it exists import osimport numpy as npfrom pyAudioAnalysis import audioBasicIO, audioFeatureExtraction, audioTrainTestfrom pydub import AudioSegment# Function to capture and save voice samplesdef capture_voice_samples(num_samples, speaker_name):os.makedirs("speakers", exist_ok=True)os.makedirs(f"speakers/{speaker_name}", exist_ok=True)for i in range(num_samples):input(f"Press Enter and start speaking for sample {i + 1}...")# Recording audio using pyAudioAnalysisaudio = audioBasicIO.record_audio(4, 44100)filepath = f"speakers/{speaker_name}/sample_{i + 1}.wav"audioBasicIO.write_audio_file(filepath, audio, 44100)print(f"Sample {i + 1} saved for {speaker_name}")# Function to extract features from voice samplesdef extract_features():speakers = [d for d in os.listdir("speakers") if os.path.isdir(os.path.join("speakers", d))]all_features = []all_labels =…/* segvhunt.cFind and eliminate all code that generates Segmentation Fault*/#include <stdio.h>int main() {char **s;char foo[] = "Hello World";*s = foo;printf("s is %s\n",s);s[0] = foo;printf("s[0] is %s\n",s[0]);return(0);}Multiplication [1 2 3;4 5 6] and [78 9;10 11 12] [7 16 27; 40 55 72] True False O O O
- wave.h: #include <stdint.h> typedef struct { uint16_t nchannels; uint16_t bits_per_sample; uint32_t sample_rate; uint32_t datasize; int16_t* data;} wavdata_t; int read_wav(char *wavfile, wavdata_t *);int write_wav(char *wavfile, wavdata_t *); typedef struct { double r; double i;} complex_t; void dft(int16_t *x, complex_t *X, int N); wave.c: #include <stdio.h>#include <fcntl.h>#include <unistd.h>#include <stdlib.h>#include <math.h>#include "wav.h" int write_wav(char *wavfile, wavdata_t *wav){ int fd = open(wavfile, O_WRONLY | O_CREAT | O_TRUNC, 0644); if (fd == -1) { printf("Could not open file %s\n", wavfile); return -1; } write(fd, "RIFF", 4); uint32_t filesize = wav->datasize + 36; uint8_t data[20]; data[0] = filesize & 0xff; data[1] = (filesize>>8) & 0xff; data[2] = (filesize>>16) & 0xff; data[3] = (filesize>>24) & 0xff; write(fd, data,…Normalisation Function Write a function Float24_t float24_normalise (int32_t oversizeMantissa, int8_t exponent) that is a bit like the float24_init() except that it allows the oversizeMatissa to be much bigger than what is normally allowed for a Float24_t. The function makes the matissa "fit" by scaling the 32-bit oversizeMantissa down to a 16-bit integer (only if necessary) and correspondingly changing the exponent value. For example if dividing the oversizeMantissa by 4 yielded a number whose magnitude could be represented by a int16_t for the mantissa then we would need to add '2' to the exponent (scale factor) of the number. Clearly this dividing down leads to a loss of precision. Note: 1. The maximum value expressible by an int16_t is defined in limits.h as INT16_MAX. All "write a function" questions in this assignment will include this for you. 2. The minimum value, after normalisation, of the mantissa is -INT16_MAX. ie. values must lie in the range +INT16_MAX While there are…* Assignment: Word operations * * Description: * This assignment asks you to implement common word operations that are * not available in the Scala programming language. The intent is to practice * your skills at working with bits. * */ package wordOps /* * Task 1: Population count (number of 1-bits in a word) * * Complete the following function that takes as parameter * a 32-bit word, w, and returns __the number of 1-bits__ in w. * */ def popCount(w: Int): Int = ??? /* * Task 2: Reverse bit positions * * Complete the following function that takes as parameter * a 16-bit word, w, and returns a 16-bit word, r, such that * for every j=0,1,...,15, * the value of the bit at position j in r is equal to * the value of the bit at position 15-j in w. * */ def reverse(w: Short): Short = ??? /* * Task 3: Left rotation * * Complete the following function that takes two parameters * * 1) a 64-bit word, w, and *…