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Naive Parallel Prime Numbers Sieve
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Naive Parallel Prime Numbers Sieve

This program calculates prime numbers using a new* method for building a sieve.

Usage: ˘

%% in this example the sieve of primes numbers will run from 1 to 1001 max.

The idea behind this program is that each number will be running on its own Erlang process and it will know how to eliminate itself from the sieve.

All Erlang process will cooperate in creating the sieve and together they will arrive to the solution.

I got inspired by this paper: Chemical Computing, tho I'm not necessarily claiming that I'm implementing that.

Yes, I know, one probably shouldn't be using Erlang for this anyway.

The sieve method

The sieve method uses the following property:

If 2n + 1 is Prime, then n can't be congruent to (k-1)/2 modulo k, for some k > n.

More specifically:

for n < k < sqrt(2n + 1), n % k != (k-1) / 2 where % stands for the modulus operation.

because if:

n = kj + (k-1) / 2 => k divides 2n + 1 for some j > 0.

Replacing n for kj + (k-1) / 2 we get:

2(kj + ((k-1)/2)) + 1 =
2kj + 2((k - 1)/2) + 1 =
2kj + k - 1 + 1 =
2kj + k =

The numbers that pass this test are kept in the sieve, and then they are returned when collect is called by multiplying them by 2 and then adding 1.

Since I'm not a professional mathematician we wary of this method, since so far I haven't seen a sieve produced using these properties of numbers.


Algorithm to check if n would be a candidate for making n*2+1=p where p stands for some prime number. This uses the algorithm notation introduced by Knuth in TAOCP.

S1. [Initialize] Set k <- 3, res <- (k-1)/2, t <- n.
S2. [Is t lesser than k?] if t < k, terminate. n is candidate.
S3. [Is k greater than sqrt(2t+1)?] if k > sqrt(2*t+1), terminate, n is candidate.
S4. [Is t congruent to res?] if t % k == res, terminate. n is not a candidate.
S5. [Recycle] Set k <- k + 2, res <- res + 1, and go back to S2.

Or in another notation presented by Knuth in TAOCP:

Here n is the number we want to test, n > 0, k >= 3, r > 0. The return value will be n when n is a candidate or -1 when n is not.

f((n)) = (n, 3, 1, 1);
f((n, k, r, 1)) = (n)  if n < k, (n, k, r, 2) otherwise;
f((n, k, r, 2)) = (n)  if k > sqrt(2*k+1), (n, k, r, 3) otherwise;
f((n, k, r, 3)) = (-1) if n % k == r, (n, k, r, 4) otherwise;
f((n, k, r, 4)) = (n, k+2, r+1, 1).


Write the numbers from 2 to 100 and then scratch of those n that satisfy: n % 3 == 1 and are greater than 3. So [4, 7, 10, 13 ... 100] will be removed.

Then add 2 to 3 getting 5 and then add 1 to 1 getting 2. This means now you will scratch those n that satisfy: n % 5 == 2.

Do the same with those n that satisfy: n % 7 == 3 and so on.

The remaining numbers:

[1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18 ... 99] when multiplied by 2 and then adding 1 to the result, will give you a prime number.

This sequence is A005097.

  • I call this method new because so far I haven't seen it in the various math books I've been consulting. By new I mean the mathematical properties outlined here, not the one process per number part. I'm not claiming this is some advanced Number Theory result or anything of the like. If you have seen this method before, please let me know by opening an issue.
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