0
$\begingroup$

I have the following repetition structure:

sum = 0
for (i = 0; i < n; i++) {
   for (j = 0; j < i*i; j++) {
      for (k = 0; k < j; k++) {
         sum++
      }
   }
}

By the following, I have reasoned the runtime of this structure to be in $O(n^7)$ :

$\sum_{i=1}^{n} (\sum_{j=1}^{i^2}(\sum_{k=1}^{j} k))$

$= \sum_{i=1}^{n}(\sum_{j=1}^{i^2}(\frac{j^2+j}{2}))$

$= \frac{1}{2} (\sum_{i=1}^{n}(\sum_{j=1}^{i^2}j^2 + \sum_{j=1}^{i^2}j))$

$= \frac{1}{2} (\sum_{i=1}^{n}(\frac{i^2(i^2+1)(2i^2+1)}{6} + \frac{i^2(i^2+1)}{2}))$

$= \frac{1}{2} (\sum_{i=1}^{n} \frac{2i^6+3i^4+i^2}{6} + \sum_{i=1}^{n} \frac{i^4+i^2}{2})$

$= \frac{1}{2} (\frac{1}{3} \sum_{i=1}^{n} i^6 + \frac{1}{2} \sum_{i=1}^{n} i^4 + \frac{1}{6} \sum_{i=1}^{n} i^2 + \frac{1}{2} \sum_{i=1}^{n} i^4 + \frac{1}{2} \sum_{i=1}^{n} i^2)$

$=\frac{1}{6} \sum_{i=1}^{n} i^6 + \frac{1}{2} \sum_{i=1}^{n} i^4 +\frac{1}{3} \sum_{i=1}^{n} i^2$

$\approx \frac{n^7}{42} + \frac{n^5}{10} + \frac{n^3}{9}$

$\therefore \text{runtime} \in O(n^7)$

The approximation at the end comes from the rule:

$\sum_{i=1}^{n} i^k \approx \frac{n^{k+1}}{k+1}$

Is this correct? I'm fairly confident that it is but some of the results that I've gotten from running the structure in JMH (Java Micro-Benchmarking Harness) don't really reflect this result so I just want to confirm.

$\endgroup$
4
  • 2
    $\begingroup$ We typically don't verify proofs. We can help you if you are unsure about a particular step. $\endgroup$ Mar 26 '20 at 12:31
  • 1
    $\begingroup$ I’ve got a different power. Second line is wrong. $\endgroup$
    – gnasher729
    Mar 26 '20 at 13:49
  • $\begingroup$ Can you clarify what's wrong in the Second line? $\endgroup$ Mar 26 '20 at 19:25
  • $\begingroup$ Actually the first line is already wrong. $\endgroup$ Mar 26 '20 at 20:44
0
$\begingroup$

As Yuval Filmus and gnasher729 have pointed out my original equation for the repetition structure is indeed wrong. I believe this to be the true correct answer:

$\sum_{i=1}^{n} (\sum_{j=1}^{i^2} j)$

$= \sum_{i=1}^{n} (\frac{i^4+i^2}{2})$

$= \frac{1}{2} \sum_{i=1}^{n}i^4 + \frac{1}{2} \sum_{i=1}^{n}i^2$

$\approx \frac{n^5}{10} + \frac{n^3}{6}$

$\therefore runtime \in O(n^5)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.