What would be the big o for the algo:

for (i=0; i < n*n; i++)
  for(j=0; j<i*i; j++)

As per my understanding
1st loop will loop upto n2 times.
2nd loop will go around n2 times.

Am I right or is it a O(n6) representation?

Also is there a easy way to say when to use log. I have assumed that whenever the loop variable is multiplied the ln function is applied

e.g.: for (i=1; i<n*n i=i*2) Big o for the loop as per my understanding is O(ln n2) (Am I right again?)

  • What does the log have to do with it? – Brian Carlton Oct 8 '14 at 1:54

These are 2 distinct questions.

1) Yes, this is O(n^6). You can, in fact, directly compute the number of iterations, inner loop makes i^2 iterations, so sum i^2 for i from 0 to n^2. Input this into Wolfram Alpha or Matlab and you get 1/6 * n^2 * (n^2 + 1) * (2 * n^2 + 1)

2) That is a useful mnemonic. Yes, a log naturally arises when you count the number of times you need to multiply something by itself to get something else. That is not the only place it arises though. Also, O(ln(n^2)) = O(2 * ln(n)) = O(ln(n))

  • what other places do log arise? sorry if this question is too naive – new Oct 7 '14 at 18:00
  • Pure log? I don't remember any cases off the top of my head. However sorting takes a minimum of n*log(n) in worst case and the log there is far from trivial (That actually arises from the quotient of factorial and 2^n) – Ordous Oct 7 '14 at 18:04
  • @Ordous: only if you are sorting by comparison. – kevin cline Oct 7 '14 at 18:16
  • @kevincline Yeah, I forgot there are other ways, thank you. – Ordous Oct 7 '14 at 18:20

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