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I just found there is an old program of mine where I implemented the following prime sieving algorithm:

#define MAX=1000001
#define SQRT=1000
// ...
not_prime[1] = 1;
for (int a = 2; a <= SQRT; a++)
  if (!not_prime[a])
     for (int c = MAX / a, b = a * c; c >= a; c--, b -= a)
        if (!not_prime[c])
           not_prime[b] = 1;

It seems like my logic that time is: For any given composite a and all corresponding composite cs such that MAX >= b; b = a * c. But I forgot why I set restriction on c that c >= a? Could anyone help explain this line for me and help me analyze how efficient this algorithm is?

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  • $\begingroup$ I forgot why I … don't write, never present undocumented/uncommented code… (Not to add to post above as already on the verge between off- & on-topic: what comment did you leave out above?) $\endgroup$
    – greybeard
    Dec 13 '20 at 12:40
  • $\begingroup$ (When c has fallen below a, b would be less than a² and have one divisor less than a unless prime. Pretty much standard Eratosthenes?!) $\endgroup$
    – greybeard
    Dec 13 '20 at 12:56
  • $\begingroup$ LoL, I just found that !not_prime[a] means a is prime while I said in my post it was a composite... I think your comment is correct btw. @greybeard $\endgroup$
    – Kindred
    Dec 13 '20 at 13:33
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    $\begingroup$ (Negative names are rarely best choice - consider is_composite. Or try and draw upon Code Review@SE.) $\endgroup$
    – greybeard
    Dec 13 '20 at 13:39
  • $\begingroup$ That’s a weird looking inner loop you’ve got there, and I don’t understand the reason for even having the if condition inside it. You know that b is a multiple of something, so just mark it unconditionally as a composite. As for time complexity, say you’re sieving up to maximum $N$. It takes $O(N)$ time to prepare the sieve and read it back. Marking off multiples of 2 takes $N/2$ time, multiples of 3 take $N/3$ time, and so all up you get $N + N/2 + N/3 + \cdots + N/\sqrt{N} = O(N \log N)$ (by a property of harmonic numbers). A slightly tighter analysis can get $O(N \log \log N)$. $\endgroup$
    – Joppy
    Dec 14 '20 at 11:49
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The loop "int c ... " is much too complicated and difficult to get right, just start with c = a, b = a*a, and iterate upwards.

Most people create a sieve for odd numbers only. Some create a sieve only for the eight out of 30 numbers that are not multiples of 2, 3 or 5. That's a factor of 3.75 in space. And removing multiples of 2, 3 and 5 takes substantial time.

Only checking multiples of 2, 3 or 5 also allows you to check fewer multiples of the prime. Another factor 3.75.

If you want to be fast, you use a bit array, and use SIMD instructions to remove multiples of small primes, say up to 100, which saves a huge amount of time. Checking whether c is still considered a prime is time consuming and likely takes more time than it saves, especially since you will end up with lots of mispredicted branches.

Now for primes up to a million that's all irrelevant, but for larger ranges it makes a huge difference.

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