The number of operations executed by algorithms A and B is 8

The number of operations executed by algorithms A and B is 8nlogn and 2n^2, respectively. Determine n_0 such that A is better than B for n greaterthanorequalto n_0.

Solution

The result is 44.

So how do you solve such problems?

The best way to answer such problem is to write a code in a choice of your language. I wrote a code in C as given below:

#include<stdio.h>
#include<math.h>
int main() {

   int n=2;

   double A(int), B(int);
   while (A(n) >= B(n))
       n++;
   printf (\"n = %d\ \",n);   
}

double A(int n) {
   double res = 8*n*log(n)/log(2);
   //printf(\"A(%d) = %lf\ \", n, res);
   return res;
}

double B(int n) {double res = n*n;
   //printf(\"B(%d) = %lf\ \", n, res);
   return res;
}

and the output was 44.

(Note that we have considered log of base 2)

(For log base 10 the answer would be 7)
(For log base e the answer would be 27)
(For log base 2 the answer would be 44)