Answered: Illustrate the growth of the algorithm…

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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Illustrate the growth of the algorithm by:
i. providing a table that reports the growth of the sorting algorithm in the
implementation above; and

ii. plot a graph to illustrate the growth of the sorting algorithm.
To illustrate the growth of your sorting algorithm, run the program with at least the following n values (n represents the number of order ids that was generated): 1,000; 5,000;32,000; 512,000; 1,000,000. The growth of the algorithm is typically measured in millisecond (ms) or microsecond (µs) or nanosecond (ns).

The growth rate of your sorting algorithm may vary due to your processor speed, memory capacity and etc. Therefore, to illustrate the consistency of the growth rate of your program, you will need to perform and report a minimum of 5 trial runs on each of the n values. You may then take the average of the running time for plotting the graph and for the discussion
in the section below.

#include<bits/stde++.h>
void mergeSort(string a[], string b[], int left, int right) {
cout<<originalorderid[i]<<"t\t"<<originalc-
ost[i]<<endl;
int mid;
if(left < right) {
mid = left + (right-1)/2;
mergeSort(a,b,left,mid);
mergeSort(a,b,mid+1,right);
merge(a,b,left,mid,right);
using namespace std;
break;
for(int i= 0; i< n; it+) {
cout<<orderid[i]<<"t\t"<<cost[i]<<endl;
void merge(string a[], string b[], int left, int mid, int right) {
case 3:
int nL = mid - left + 1;
int nR = right - mid;
string leftA[nL], rightA[nR], leftB[nL], rightB[nR];
for(int i = 0; i<nL; i++) {
leftA[i] = a[nL+i];
leftB[i] = b[nL+i]:
}
for(int i = 0; i< nR; i++) {
rightA[i] = a[mid+i+1];
rightB[i] = b[mid+i+1];
}
int i = 0, j = 0, k = left;
while(i< nL &&j<nR) {
if(leftA[i] <= rightA[j]) {
a[k] = leftA[i];
b[k] = leftB[i];
i++;
}
else {
a[k] = rightA[j]:
b[k] = rightB[j];
j++;
k++;
break;
w = false;
break;
w = false;
case 4:
}
}
default:
int menu) {
int ch;
cout<<"1. Time for sorting\n2. Unsorted List\n3, Sort'n4, Exit'nEnter Choice: ":
cin>>ch;
return ch:
}
return 0;
}
End of document
int main() {
int n;
cout<<"Enter number of order ids: ";
cin>>n;
srand((unsigned) time(0));
string orderid[n], cost[n], originalorderid[n], originalcost[n];:
for(int i= 0; i<n; i++) {
int randNum = (rand()%n)+1;
orderid[i] = "FD" + to_string(randNum);
originalorderid[i] = orderid[i];
float a = 1.0;
float randCostNum = (float(rand())/float(RAND_MAX) * a);
randCostNum = (int)(randCostNum * 100 + 0.5);
randCostNum = (float)randCostNum / 100;
cost[i] = "RM" + to_string(randCostNum);
originalcost[i] = cost[i];
}
time_t start, end;
time(&start);
ios_base:sync_with_stdio(false);
mergeSort(orderid, cost, 0, n-1);
time(&end);
double timeTaken = double(end - start);
bool w = true;
while(w) {
int choice = menu);
switch(choice) {
case 1:
}
}
while(i< nL) {
a[k] = leftA[i];
b[k] = leftB[i];
i++;
k++:
}
while(j < nR) {
a[k] = rightA[j];
b[k] = rightB[j];
j++;
k++;
}
cout << "Time taken by merge sort is : "<< timeTaken << setpreci-
sion(5);
cout <<" sec "<< endl:
break;
for(int i = 0; i<n; i++) {
}
case 2:

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