If a recursive algorithm calls itself twice for each element in a set, does that mean it has n^2 complexity, or is that 2n and therefore n complexity? I am confused.
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Anonymous2012-11-12 16:11
>>1
that would depend on the algorithm (you haven't stated even the base condition
Depends how the size of the set the algorithm works on reduces every time it calls itself.
The simplest case of O(n2) that I can think of is for (i in set) { for (j in set) { do a thing } }, but that's not recursive.
Is there any recursive O(n2) algorithm? On a minute of thinking, I don't see how there could be one that isn't on the logarithmic scale.