Question

for i := 1 to n do
  j := 2;
  while j < i do
    j := j^4;

I’m really confused when it comes to Big-O notation, so I’d like to know if it’s O(n log n). That’s my gut, but I can’t prove it. I know the while loop is probably faster than log n, but I don’t know by how much!

Edit: the caret denotes exponent.

Answer 1

The problem is the number of iterations the while loop is executed for a given i.

On every iteration j:= j^4 and at the beginning j := 2, so after x iterations j = 24^x

j < i is equivalent to x < log_4(log_2(i))

I’d risk a statement, that the complexity is O(n * log_4(log_2(n)))

You can get rid of constant factors in Big O notation. log_4(x) = log(x) / log(4) and log(4) is a constant. Similarly you can change log_2(x) to log(x). The complexity can be expressed as O(n*log(log(n)))