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Why is Sieve of Eratosthenes more efficient than the simple "dumb" algorithm?

Asked 2011-04-13T12:18:45.363
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If you need to generate primes from 1 to N, the "dumb" way to do it would be to iterate through all the numbers from 2 to N and check if the numbers are divisable by any prime number found so far which is less than the square root of the number in question.

As I see it, sieve of Eratosthenes does the same, except other way round - when it finds a prime N, it marks off all the numbers that are multiples of N.

But whether you mark off X when you find N, or you check if X is divisable by N, the fundamental complexity, the big-O stays the same. You still do one constant-time operation per a number-prime pair. In fact, the dumb algorithm breaks off as soon as it finds a prime, but sieve of Eratosthenes marks each number several times - once for every prime it is divisable by. That's a minimum of twice as many operations for every number except primes.

Am I misunderstanding something here?

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In the naive method, you do O(sqrt(num)) operations for each number num you check for primality. Ths is O(n*sqrt(n)) total.

In the sieve method, for each unmarked number from 1 to n you do n / 2 operations when marking multiples of 2, n / 3 when marking those of 3, n / 5 when marking those of 5 etc. This is n*(1/2 + 1/3 + 1/5 + 1/7 + ...), which is O(n log log n). See here for that result.

So the asymptotic complexity is not the same, like you said. Even a naive sieve will beat the naive prime-generation method pretty fast. Optimized versions of the sieve can get much faster, but the big-oh remains unchanged.

The two are not equivalent like you say. For each number, you will check divisibility by the same primes 2, 3, 5, 7, ... in the naive prime-generation algorithm. As you progress, you check divisibility by the same series of numbers (and you keep checking against more and more as you approach your n). For the sieve, you keep checking less and less as you approach n. First you check in increments of 2, then of 3, then 5 and so on. This will hit n and stop much faster.

answered 2011-04-13T12:35:09.583

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