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If I have a function that evaluates every pair of elements in a list, including pairs of an element with itself and with elements it has previously been evaluated against, the complexity of that function is, obviously, $O(n^2)$:

for (let i=0; i < n; i++) {
  for (let j=0; j < n; j++) {
    // this line will be executed n * n times
    operate(l[i], l[j]);
  }
}

However, if I have a variant of that loop that only considers pairs of elements that have not yet been compared, is the complexity of that function still $O(n^2)$?

for (let i=0; i < n; i++) {
  for (let j=i+1; j < n; j++) {
    // this line will be executed (n * n-1) / 2 times?
    operate(l[i], l[j]);
  }
}

There are a few things about this that I still don't understand:

  • How can we calculate that the furthering reduction in j will divide the operations by 2? (I'm only inferring the $/2$ because I know that matchup tables have half the table redundant; I wouldn't know how to evaluate the reduction if, say, only even indices were compared against odd indices.)
  • Why doesn't that $/2$ matter? How do we determine the asymptote? How does $(n * (n-1)) / 2$ get reduced to $O(n^2)$? Are we just removing terms in reverse order of operations, by picking which one incorporates higher-order operations on variables (so $(n * (n-1)) / 2$ becomes $(n * (n-1))$, which becomes $(n * n)$)?
  • How does this generalize for $x$-tuples? If we're comparing triplets, is the complexity $O(n^3)$? Does it go up as $O(n^x)$? Is the non-asymptotic complexity still divided by 2 for orders of 3 or more? Would the numerator be $(n * (n-1) * (n-2))$? How would that be expressed generally for orders of $x$ (would it look like $(n * (n - 1) * ... * (n - (x - 1) ))$?)?
Raphael
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Stuart P. Bentley
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  • @Raphael Okay, cool. Maybe, since this is a Q & A site, I could get some answers that could actually help me to go at this in such a structured way? Maybe described in simpler terms, more accessible to someone who is struggling with this question because they do not have a deep background in computer science, than a ten-page essay describing the problem as $\qquad\displaystyle \mathbb{E}[C_{\text{swaps}}] = \frac{1}{n!} \sum_{A} \sum_{i=0}^{n-2} \sum_{j=0}^{n-i-2} C_{5,9}(A^{(i,j)})$? – Stuart P. Bentley Aug 05 '16 at 07:59
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    There are plenty of [tag:algorithm-analysis+loops] questions around you can check out for examples. I also think that if you actually tried and read the reference question, you'd have found what you need (namely how to translate loops into sums). Note that even on Q&A sites, some prior knowledge and own effort can be expected. (This is not a deep question.) – Raphael Aug 05 '16 at 09:58