Question

Very similar complexity examples. I am trying to understand as to how these questions vary. Exam coming up tomorrow 🙁 Any shortcuts for find the complexities here.

CASE 1:

void doit(int N) { 
   while (N) {
      for (int j = 0; j < N; j += 1) {}
   N = N / 2;   
   }
}

CASE 2:

void doit(int N) { 
   while (N) {
      for (int j = 0; j < N; j *= 4) {}
   N = N / 2;   
   }
}

CASE 3:

void doit(int N) { 
   while (N) {
      for (int j = 0; j < N; j *= 2) {}
   N = N / 2;   
   }
}

Thank you so much!

Answer 1

void doit(int N) { 
   while (N) {
     for (int j = 0; j < N; j += 1) {}
   N = N / 2;   
   }
}

To find the O() of this, notice that we are dividing N by 2 each iteration. So, (not to insult your intelligence, but for completeness) the final non-zero iteration through the loop we will have N=1. The time before that we will have N=a(2), then before that N=a(4)… where 0< a < N (note those are non-inclusive bounds). So, this loop will execute a total of log(N) times, meaning the first iteration we see that N=a2^(floor(log(N))).

Why do we care about that? Well, it’s a geometric series which has a nice closed form:

Sum = \sum_{k=0}^{\log(N)} a2^k = a*\frac{1-2^{\log N +1}}{1-2} = 2aN-a = O(N). 

If someone can figure out how to get that latexy notation to display correctly for me I would really appreciate it.