Suppose you are collecting performance data for the selection sort shown below, and you collect elapsed times as shown and graph those times. What trend would you expect the graph to show? void selection_sort(int a[], int size) { } int next; for (next = 0; next < size - 1; next++) { } int min_pos= min_position (a, next, size - 1); if (min_pos != next) { } int before = time(0); swap(a [min_pos], a[next]); int after cout << (after before) << endl; Constant time Linear Log Quadratic time(0);
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- void ExtractMinimumWeightPath(int path[MAX], int weights[MAX][MAX], int pathWeights[MAX][MAX], int rowc, int colc){int rowsCount = rowc;int colsCount = colc;path[rowsCount];int col = 0;int i = 0;for(i=1;i<colsCount;i++){if(pathWeights[rowsCount-1][i]<pathWeights[rowsCount-1][col]){col = i;}} int row = rowsCount-1;do{path[row] = col + 1;if (col>0 && pathWeights[row-1][col-1] + weights[row][col] == pathWeights[row][col]){col = col - 1;} else if (col < colsCount-1 && pathWeights[row-1][col+1] + weights[row][col] == pathWeights[row][col]){col = col + 1;}row--;} while(row>0);path[0] = col + 1;} This is a code in C, I need to change the code as much as possible, but on condition to perform the same work as this. Changing the var names and the loops is enoughint binsearch (int X , int V [] , int n ) { int low , high , mid , i ; low = 0; high = n - 1; for ( i = 0; i < high ; i ++) { if( V[ i ] > V [ i +1]) return -2; } while ( low <= high ) { mid = ( low + high )/2; if ( X < V [ mid ]) high = mid - 1; else if ( X > V [ mid ]) low = mid + 1; else return mid ; } return -1; } This code takes as input a sorted array V of size n, and an integer X, if X exists in the array it will return the index of X, else it will return -1. 1. Draw a CFG for binsearch(). 2. From the CFG, identify a set of entry–exit paths to satisfy the complete statement coverage criterion. 3. Identify additional paths, if necessary, to satisfy the complete branch coverage criterion. 4. For each path identified above, derive their path predicate…Use loops to create a matrix in which the value of each element is two times its rownumber minus three times its column number. For example, the value of element (2,5) 2*2 - 3*5 = -11
- public class testRun{public static int find(int[] data, int target,int start, int end) {int[] stored = sort(data);if (start <= end) {int midPoint = (start + end / 2);if (stored[midPoint] == target) {return midPoint;}if (stored[midPoint] > target) {return find(data, target, start, midPoint - 1);//return first half,exclusive itself}if (stored[midPoint] < target) {return find(data, target, start, midPoint + 1);//return last half,exclusive itself.}}return -1;}public static int[] sort(int[] data){// Find the smallest element in the listfor (int i = 0; i < data.length -1 ; i++){ //generate the index position,not =< b/c goes outer boundint min = i; //assign index position to min in each iterationfor (int j = i + 1; j < data.length ; j++){if (data[j] < data[min]){ //if the element in the second index number smaller than// the element in previous index numbermin = j; // assign j to the smallest number}}swap (data,i,min);// data[i] = min the smallest number assign to…Function PrintArray(integer array(?) dataList) returns nothing integer i for i = 0; i < dataList.size; i = i + 1 dataList[i] = Get next input Put dataList to output Put "_" to output // Your solution goes here. Modify as needed i = 0 Complete the PrintArray function to iterate over each element in dataList. Each iteration should put the element to output. Then, put "_" to output. Ex: If dataList's elements are 2 4 7, then output is: 2_4_7_ Function Main() returns nothing integer array(3) userNums integer i for i = 0; i < userNums.size; i = i + 1 userNums[i] = Get next input PrintArray(userNums)Design and implement a service that simulates PHP loops. Each of the three loop variants should be encapsulated in an object. The service can be controlled via a parameter and execute three different simulations. The result is returned as JSON. The input is an array consisting of the letters $characters = [A-Z]. -The For loop should store all even letters in an array.-The Foreach loop simulation is to create a backward sorted array by For loop, i.e. [Z-A] .-The While loop should write all characters into an array until the desired character is found. Interface:-GET Parameter String: loopType ( possible values: REVERSE, EVEN, UNTIL )-GET parameter String: until (up to which character) Output:JSON Object: {loopName: <string>, result: <array> }
- Use loops to create a matrix in which the value of each element is two times its rownumber minus three times its column number. For example, the value of element (2,5) 2*2 - 3*5 = -11 BY USING MATLATCreate an ABM function that takes the following parameters: n := number of paths to be simulated m := number of discretization points per path S0 := initial starting point dS=μdt+σdW Program the function by using two nested "for loops" def ABM(n,m,S0,mu,sigma,dt): np.random.seed(999) arr = # create 2D zeros array with the correct dimensions arr[,] = #initialize column 0 # fill in array entries for i in : for j in : arr[i,j] = return arrUse nested for...loops to create a (6x6) matrix in which the value of an element is three times its row number minus four times its column number provided the row number is greater than the column by 1. Otherwise, the element of the matrix is equal to 0.
- make flowchart for this one plz #include<stdio.h>int sortingbook(int array[], int N){ int count = 0, i , j , k , minimum , index , flag = 0; for(i=0;i<N;i++) { minimum = array[i]; //ith element as the minimum index = i; //setting index value as i for( j = i; j < N; j++) { if(array[j] < minimum) { minimum = array[j]; //calculating minimum index = j; //assigning the minimum position into index } } for(k = index; k>i ; k--) //shifting one position to the right from index to i array[k] = array[k-1]; if(array[i]!=minimum) count++; //if the value of min chganged from previous assumption array[i] = minimum; } return count;}int main(){ int array[20],i,N,c; printf("\nEnter number of books(N) : "); scanf("%d",&N); //number of books printf("\nEntercurrent sequence of %d books : \n=> ",N); for(i=0;i<N;i++) scanf("%d",&array[i]); //reading sqeuence c = sortingbook(array,N); printf("\nSorted sequence is :\n=>…Example: This sample loop creates a matrix M with consecutively numbered elements M:= for ie 0. 1 for je 0.2 M.+3-i + j 1,j M 0 1 2 M = 3 45Implement the function below. void swap(int pos1, int pos2){} Initial code to be completed: class ArrayList : public List { int* array; int index; int capacity; void dyn_all_add(){ int cap = ceil(capacity * 1.5); array = (int*)realloc(array,cap * sizeof(int)); capacity = cap; } void dyn_all_rem(){ int cap = capacity - (capacity/3); array = (int*)realloc(array,cap * sizeof(int)); capacity = cap; } public: // CONSTRUCTOR ArrayList() { capacity = 4; array = (int*)malloc(capacity); index = 0; } int add(int num) { if (index == capacity){ dyn_all_add(); } *(array + index) = num; index++; return index; } int get(int pos){ if (pos-1 < index){ return *(array + pos-1); } return -1; } int size(){ return index; }…