Question

Below is an algorithm using C++ language

Question : Analyze the algorithm and provide in terms of exact number of steps A(n) or the algorithm.
A(n) =

Given your analysis provide an upper bound in terms of Big-O for Algorithm .
O(n) =

#include <iostream>

#include <cstdlib>
#include <cmath>
using namespace std;

int random_value(int range, long seed) {
   srand(seed);
   float generator = (float) rand() / (float) RAND_MAX;
   float random = floor(range * generator);
   return random;
}

int main(){
   int n = 16;
   int m = 512;
   long seed = 36028797153181696;
   int initial[n];
   int sorted[n];

   for(int i = 0; i < n; i++) {
       initial[i] = random_value(m, seed/(i+1)) + 1;
   }
   for(int i = 0; i < n; i++){
       sorted[i] = 0;
   }
  
   cout << "L: (";
   for(int i = 0; i < n; i++){
       if(i > 0){
           cout << ", ";
       }
       cout << initial[i];
   }
   cout << ")\n";
   for(int i = 0; i < n; i++) {
       int v = m + 1;
       int s = 0;
      
       for(int j = 0; j < n; j++) {
           if(initial[j] > 0 && initial[j] < v) {
               v = initial[j];
               s = j;
           }
       }
       sorted[i] = initial[s];
       initial[s] = 0;
   }
   cout << "S: (";
   for(int i = 0; i < n; i++){
       if(i > 0){
           cout << ", ";
       }
       cout << sorted[i];
   }
   cout << ")\n\n";
   return 0;
}

Answer 1

Solution:

Explanation:

We can see in main() function ,

=>for(int i =0;i<n;i++) will run n times from i=0 to i=n-1 independenty.

=>After that there is a loop for(int i =0 ;i<n;i++) which will run n times from i = 0 to i = n-1 indpendently.

=>After that there is a loop for(int i =0 ;i<n;i++) which will run n times from i = 0 to i = n-1 indpendently.

=>After loop there is a loop for(int i =0 ;i<n;i++) which will run n times from i = 0 to i = n-1 indpendently.

=> After that there is a loop for(int j =0 ;j<n;j++) from j = 0 to j = n-1 which will run n times for every value of i=0 to i=n-1

=>After loop there is a loop for(int i =0 ;i<n;i++) which will run n times from i = 0 to i = n-1 indpendently.

From the above analysis we can calculate the number of times above loops will be executed :

=> n + n + n + n*n + n => 4*n + n*n

So A(n) = 4n + n2 (Exact number of steps)

Calculating upper bound:

Explanation:

=> From the above observations we found that 4n + n2 number of times loop will be executed.

=>Converting into upper bound:4n + n2 = O(n2)          (Asymototically)

So the upper bound of the algorithm is O(n2)

I have explained each and every part of the algorithm along with exact number of steps as well as an upper bound.