fs(n) = 1 + fs([n/2]) when n > 1, f;(1) = 1 %3D
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Find its time complexity using Big-O Notation.
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- #include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…Code: #include <bits/stdc++.h> using namespace std; void BUBBLE(int A[],int N){ for(int k=0;k<N-1;++k){ for(int ptr=0;ptr<N-k-1;++ptr){ if(A[ptr]>A[ptr+1]){ int temp = A[ptr]; A[ptr] = A[ptr+1]; A[ptr+1]=temp; } } }} //function to print the arrayvoid printArray(int arr[],int n){ int i; for(i=0;i<n;i++) cout<<arr[i]<<" "; cout<<endl;} //driver function to test the modulesint main(){ int arr[] ={15,17,5,3,25,66,14,7,59,100}; int n=sizeof(arr)/sizeof(arr[0]); cout<<"\nOriginal array: "; printArray(arr,n); cout<<"\n\nOutput of Bubble sort are shown below:\n"; BUBBLE(arr,n); printArray(arr, n); return 0;} Q: Remove the Function from the above codeSolution 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…
- In questions 4-10 estimate the Big O value by analyzing the code. Note the algorithms are written in English. Hint: you are interested in the number of operations for each algorithm.11 - The code segment below has time complexity? for (int i=0; iA = [1 1; 1 -1];b = [80; 20]; A_inverse = inv(A);x = A\b; disp('A inverse is');disp(A_inverse);disp(['x1 is ', num2str(x(1)), ' and x2 is ', num2str(x(2))]); if isequal(round(x), [50; 30]) disp('Success');else disp('Incorrect: Please try again');end. Output matlab. .Debug Program3: int element(int arr[], int num) { in i, ele; ele = arr[0]; for (i = 1; i < num; i++) if (arr[i] > ele) ele = arr[i]; return ele; } main() { arr[] = {1, 24, 145, 20, 8, -101, 300}; int n = sizeof(arr)/sizeof(arr[0]); printf("element of array is %d", element(arr[], n)); } 24 145 -101 300 Program4: main() { int m,n; for(i 0; i << 5; i++) { printf("\t\t\t\t"); for(j 0; j << 5; j+++) printf("*"); printf("/n"); } Program 5: int func(int) main() { int num = 7, c; int f= func(num); printf("\n\n%d of %d\n\n", num, f); } int func(aj); { if(aj=1 aj=0) return 1; else return (aj*fun(aj-1); } 49 5040 7117 none of…typedef struct { short data[4];} MatrixElement; void copy_matrix(MatrixElement m1[], MatrixElement m2[], int ROWS, int COLS) { int i, j, k; for (i = 0; i < ROWS; i++) { for (j = 0; j < COLS; j++) { for (k = 0; k < 4; k++) { m1[i*COLS+j].data[k] = m2[i*COLS+j].data[k]; } } }} void copy_matrix_transpose(MatrixElement m1[], MatrixElement m2[], int ROWS, int COLS) { int i, j, k; for (i = 0; i < ROWS; i++) { for (j = 0; j < COLS; j++) { for (k = 0; k < 4; k++) { m1[i*COLS+j].data[k] = m2[j*ROWS+i].data[k]; } } }} You can assume the following conditions: The matrix m1 is allocated at memory address 0, and matrix m2 immediately follows it. Indices i, j, and k are kept in registers. ROWS and COLS are constants. The cache is initially empty before the function call. The cache is write-back (i.e., only writes back to memory when a line is…void funi (int iist , int s) { int sum=0; for (int i = 0; iSEE MORE QUESTIONSRecommended textbooks for youSystems ArchitectureComputer ScienceISBN:9781305080195Author:Stephen D. BurdPublisher:Cengage Learning
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