Time complexity of Sieve of Eratosthenes algorithm

Posted by eSKay on Stack Overflow See other posts from Stack Overflow or by eSKay
Published on 2010-04-06T05:06:23Z Indexed on 2010/04/06 5:13 UTC
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time-complexity

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From Wikipedia:

The complexity of the algorithm is O(n(logn)(loglogn)) bit operations.

How do you arrive at that?

That the complexity includes the loglogn term tells me that there is a sqrt(n) somewhere.

Suppose I am running the sieve on the first 100 numbers (n = 100), assuming that marking the numbers as composite takes constant time (array implementation), the number of times we use mark_composite() would be something like

n/2 + n/3 + n/5 + n/7 + ... + n/97        =      O(n)

And to find the next prime number (for example to jump to 7 after crossing out all the numbers that are multiples of 5), the number of operations would be O(n).

So, the complexity would be O(n^2). Do you agree?

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