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0 pointscjfd6y ago0 comments

clearly O(n^2) for time. Well, perhaps O(n^2/log n) or something, but that is not much different. The thing is, every prime p gets passed through a filter that filters multiples of primes p' for all p' < p.

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merijnv6y ago· 1 in thread

> The thing is, every prime p gets passed through a filter that filters multiples of primes p' for all p' < p

But the filter for multiples of 3 only sees the values that weren't already filtered by 2. So if we have N filters we don't evaluate all N filters for each value. The 2 is applied to everything, the 3 filter is only applied to things that are not multiples of 2, the 5 filter is only applied to things that aren't multiple of 2 and 3, etc. That doesn't sound very quadratic to me.

cjfdOP6y ago

The thing is that the prime numbers have to go through all of the filters. Although the prime numbers are a minority among the numbers they are not a very small minority. In fact, a random number N has about a probability of 1/log(N) to be prime. So that could reduce N^2 to maybe N^2/log(N) but not any further. So I suppose the complexity actually is somewhere between N^2 and N^2/log(N), but that is not much less than N^2. The trick of only checking primes smaller than sqrt(N) reduces the complexity to N^1.5, also maybe involving some division by log(N), but that is actually an improvement in the exponent.

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