Write and execute a program that can read 16 numbers from keyboard and sort them by using selection sort. Selection sort inner loop To maintain algorithm invariants: • Move the pointer to the right. i++; in final order 1 • Identify index of minimum entry on right. int min = i; for (int j = i+1; j < N; j++) if (less (a[j], a [min])) min - j; in final order ↑ Exchange into position. ** exch (a, i, min); in final order

Write and execute a program that can read 16 numbers from keyboard and sort them by using selection sort. Selection sort inner loop To maintain algorithm invariants: • Move the pointer to the right. i++; in final order 1 • Identify index of minimum entry on right. int min = i; for (int j = i+1; j < N; j++) if (less (a[j], a [min])) min - j; in final order ↑ Exchange into position. ** exch (a, i, min); in final order

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Exercise 1:In this problem, we would like to implement a variation of the Bubble Sort algorithm. The algorithm differs from a bubble sort in that it sorts in both directions on each pass through the list. The algorithm is illustrated as in the following figure: For the first step, we perform bubble sort from the index 1 to n (n is thenumber of elements in the array). The next step, we perform a reserved bubble sort from the index n to 1. The process is repeated until all the array is sorted. Propose a pseudo-code to complete the Bubble Sort algorithm. Implement and test this algorithm in C/C++. Analyze and compute the complexity of this algorithm in the best, average and worst scenarios.Exercise 2:Re-implement Exercise 1 using a linear data structure: List, Stack, Queue. Justify your choice of data structure.
: In searching an element in an array, linear search can be used, even though simple to implement, but not efficient, with only O(n) time complexity. Assuming the array is already in sorted order, modify the search function below, using a better algorithm, so the average time complexity for the search function is O(log n). include <iostream> using namespace std; int search(int al), int s, int v) { 1/ Modify below codes. for (int i = 0; i <s; i++) { if (a[i] = v) return i; return -1; int main() { int intArray:10] = { 5, 7, 8, 9, 10, 12, 13, 15, 20, 34); // Search for element '12' in 10-elements integer array. cout << search(intArray, 10, 12); // '5' will be printed out. // Search for element '35' in 10-elements integer array. cout << search(intArray, 10, 35); // '-1' will be printed out. // Index '-l' means that the element is not found. return 0;
Sorting 4. Consider the following array of int values. [17, 11, 6, 22, -3, 8, 4, 22, 2, 35, -4, 10] a. Write the contents of the array after 4 passes of the outermost loop of selection sort. Assume selection sort is sorting slots from low-to-high b. Write the contents of the array during each of the the recursive calls of merge sort.
Language: Python 3 Autocomplete Ready O 1 v import ast 3. Hybrid Sort input() lst %3D 3 lst = ast.literal_eval(lst) 4 Insertion sort is a simple sorting algorithm that builds the final sorted array one item at a time. In each iteration, insertion sort inserts an element into an already sorted list (on left). The position where the item will be inserted is found through linear search. You decided to improve insertion sort by using binary search to find the position p where the new insertion should take place. 6 print(BinaryInsertionSort(lst)) Algorithm BinarylnsertionSort Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n begin BinarylnsertionSort for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p a[j + 1]= a[i] j= j-1 end for a[p] = val i i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part…
Quicksort SPLIT (the 2-pointer algorithm covered in class) is applied to the integer array [4,3,5,7,9,2,1], using the first entry as the pivot. Show the series of item swaps that are performed in the split process, and the array after each of these swaps, up to and including the step of placing the pivot at its correct location. You only need to show the split of the original array, you are not required to continue working on the subarrays after the split.
Complete the following function definition to recursively print the index of a unique value in an array or -1 if the value is not found:1 int getIndex(int *a, int s, int v) {2 if( ) { // if no values are found3 4 }5 if(a[s-1] == v) { // if a value is found6 7 }8 return getIndex( ); // recurse to check next value9 }Hint:Recurse through the array by counting down from size. 1. Complete the one line of code for line 2:2. Write one line of code for line 3:3. Write one line of code for line 6:4. Complete the one line of code for line 8: please send an atom code. not other software
Select the appropriate code that performs selection sort. a) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length-1; k++) { if(arr[k] < arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } b) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length; k++) { if(arr[k] < arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } c) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length-1; k++) { if(arr[k] > arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; } d) int min; for(int j=0; j<arr.length-1; j++) { min = j; for(int k=j+1; k<=arr.length; k++) { if(arr[k] > arr[min]) min = k; } int temp = arr[min]; arr[min] = arr[j]; arr[j] = temp; }
Your task is sorting the given list by dictionary order, sortingoperation must be realized using the bubble sort algorithm. Additionally,this list implementation should be written using a two-dimensional char ar-ray.Bubble Sort Algorithm1 function swap ( a , b)2 // F i l l own your own !3 end function45 function compare ( a , b)6 // Compare f unc t i on should be implemented7 // to r e a l i z e the s o r t i n g c r i t e r i a .8 // For example , e l ement s in a are s t r i n g s and9 // the y are to be s o r t e d in d i c t i o n a r y order .10 // strcmp f unc t i on can be used in p l a c e11 // of compare f unc t i on .12 end function1314 function bubbl e s o r t ( a , n)15 while n != 016 high = 017 for i=0 to n=2 incremented by 118 i f compare ( a [ i ] , a [ i +1]) > 0 then19 swap ( a [ i ] , a [ i +1])20 high = i+121 end i f22 end for23 n = high24 end while25 end function1
Create a MIPS program that gets a set of numbers from the user, sorts them using selection sort, and displays the sorted numbers on the screen. You should create a subprogram to do the selection sort, sending it the length and address of the array. Pseudocode for selection sort: for (int i = 0; i < aLength-1; i++) { /* find the min element in the unsorted a[i .. aLength-1] */ /* assume the min is the first element */ int jMin = i; /* test against elements after i to find the smallest */ for (int j = i+1; j < aLength; j++) { /* if this element is less, then it is the new minimum */ if (a[j] < a[jMin]) { /* found new minimum; remember its index */ jMin = j; } } if (jMin != i) { swap(a[i], a[jMin]); }}
function Sum(A,left,right) if left > right: return 0else if left = right: return A[left] mid = floor(N/2) lsum = Sum(A,left,mid) rsum = Sum(A,mid+1,right) return lsum + rsum function CreateB(A,N)B = new Array of length 1 B[0] = Sum(A,0,N-1) return B Building on the above, in a new scenario, given an array A of non-negative integers of length N, additionally a second array B is created; each element B[j] stores the value A[2*j]+A[2*j+1]. This works straightforwardly if N is even. If N is odd then the final element of B just stores A[N-1] as we can see in the figure below: (added in image) The second array B is now introducing redundancy, which allows us to detect if there has been a hardware failure: in our setup, such a failure will mean the values in the arrays are altered unintentionally. The hope is that if there is an error in A which changes the integer values then the sums in B are no longer correct and the algorithm says there has been an error; if there were an error in B…
See the pseudo-code of the Binary Search using recursion below. Fill in the XXXX and YYYY in the code Algorithm BinarySearch (A,v,low,hi) Input: array A indexed from low to hi with items sorted from smallest to largest. We are searching for the item v Output: returns a location index of item v in array A; if v is not found, -1 is returned. if (low > hi) then return (-1); mid = (lo + hi)/2; if (A[mid] = v) then == return(mid); if (A[mid] < v) then return(BinarySearch(A, XXXX ,hi, v)); else return(BinarySearch(a, lo, YYYY, v)); О а. ХX: lo+1, YҮҮ: hi O b. XXXX: mid, YYYY: mid О с. ХXXX: lo, YYYY: hi O d. XXXX: lo+1, YYYY: hi-1 O e. XXXX: mid+1. YYYY: mid-1
Language: Python 3 • Autocomplete Ready O 1 v import ast lst = input(O lst = ast.literal_eval(lst) def binarysearch(lst,x,low,high): if low - high x: 10 11 Algorithm BinarylnsertionSort 12 Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n 13 begin BinarylnsertionSort return binarysearch (lst, x, mid, high) 14 for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p alj + 1]= a[j] j= j-1 end for 15 else: 16 return mid 17 18 def BinaryInsertionSort(lst): 19 print (BinaryInsertionSort(lst)) 20 a[p] = val j=i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part returns the position p where the val will be inserted. After finding p, the data values in the subsequent positions j = i- 1, ..., p are sequentially moved one position up to i, ..., p+1 so that the value val can be inserted into the proper…