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Sieve of Eratosthenes

Math / Number Theory O(n log log n) O(n)

Finds every prime up to n by crossing out the multiples of each prime, starting from its square.

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1/67

Assume every number from 2 to 60 is prime.

InvariantEvery number still uncrossed below p² has no factor ≤ p, so it is prime.

sieve-of-eratosthenes.c
void sieve(int n) {
    int prime[n + 1];
    for (int i = 0; i <= n; i++) prime[i] = 1;
    prime[0] = prime[1] = 0;
 
    for (int p = 2; p * p <= n; p++) {
        if (!prime[p]) continue;
        for (int m = p * p; m <= n; m += p)
            prime[m] = 0;      /* cross out multiple */
    }
 
    for (int i = 2; i <= n; i++)
        if (prime[i]) printf("%d ", i);
}

How Sieve of Eratosthenes works

List 2..n; repeatedly take the next surviving number (a prime) and cross out its multiples starting at p². Anything still standing at the end was never a multiple of a smaller prime.

When to use it

Generating all primes up to n, smallest-prime-factor tables for fast factorization.