Prime Sieve of Eratosthenes

The Sieve of Eratosthenes is an algorithm for generating a list of all prime numbers up to a given integer N. It does this using $$O(N)$$ space and $$O\left(\frac{N \log \log N}{\log N}\right)$$ time. (Note that $$N$$ is the number. The size of the input would be the number of bits representing the number, which can be stored in $$\log N$$ bits for $$N$$.)

Algorithm
The Prime Sieve of Eratosthenes algorithm precalculates from the bottom up - given a table of $$N$$ values, we "knock off" numbers that can't possibly be a prime - by counting up from hasn't been knocked off.

Example
Let $$N = 30$$:

1 2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

We know that 1 isn't a prime, so it's safe to cross it off. So we'll start at 2:

2 3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

The smallest number that hasn't been processed is two, now let's knock out multiples of twos - by first going to $$4, 6, 8\ldots$$ - all you need is addition!

*2 3     5     7     9   11    13    15    17    19   21    23    25    27    29

Now that two is done, we can cross out multiples of threes - 6 is already crossed out, but 9 isn't, so we're going to cross that out, continuing..

*2 *3    5     7         11    13          17    19         23    25          29

And continuing on.

Pseudocode
In its simplest form, without optimizations:

func sieve( var N ) var PrimeArray as array of size N  initialize PrimeArray to all true

for i from 2 to N     for each j where i divides j, j from i + 1 to N         set PrimeArray( j ) = false

At the end, $$PrimeArray( x )$$ will be true if $$x$$ is a prime number.

Optimizations

 * 1) If $$m$$ is a multiple of $$i$$ and $$i$$ is a multiple of $$p$$, then $$m$$ is also a multiple of $$p$$. Thus, if $$i$$ has already been crossed out when the algorithm reaches it, we need not go crossing out its multiples, because they would have already been crossed out.
 * 2) If a number $$m$$ is not prime, then it has a factor that is $$\le \sqrt m$$. (This is because if $$m=pq$$, then at least one of $$p$$ and $$q$$ must be $$\le \sqrt m$$.) Thus, any number that is ever crossed out will be crossed out first by a number less than or equal to its square root. Putting this another way, we need not cross out multiples $$m$$ of $$i$$ that are less than $$i^2$$, because if $$m<i^2$$ is composite, it would have already been crossed out by some number smaller than $$i$$.
 * 3) By the above, as we will do nothing for $$i$$ such that $$i^2$$ exceeds $$N$$, we can stop when that happens.

With these observations, we can write the following algorithm: Start with an array is_prime[$$2..N$$], all initialized to true. For each $$i$$ such that $$2\le i$$ and $$i^2 \le N$$, do  If isprime[$$i$$] is true, For each multiple $$m$$ of $$i$$ such that $$i^2 \le m\le n$$, "cross out $$m$$" &mdash; set is_prime[$$m$$] to false

This algorithm is usually what is meant when "the sieve of Eratosthenes" is mentioned.

Other optimizations

 * There is no need to store even numbers, as there is only one even prime number. (2, of course.)
 * You can use bitmaps to pack the memory.
 * If you are looking for only prime numbers you can generate them and keep adding them to a list.

Implementations

 * Implementation in C
 * Implementation in C++
 * Sieve of Eratosthenes illustrated explanation. C++ and Java implementations.