Crate primal

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primal puts raw power into prime numbers.

This crates includes

  • optimised prime sieves
  • checking for primality
  • enumerating primes
  • factorising numbers
  • estimating upper and lower bounds for π(n) (the number of primes below n) and pk (the kth prime)

This uses a state-of-the-art cache-friendly Sieve of Eratosthenes to enumerate the primes up to some fixed bound (in a memory efficient manner), and then allows this cached information to be used for things like enumerating and counting primes.

primal takes around 2.8 seconds and less than 3MB of RAM to count the exact number of primes below 1010 (455052511) on my laptop (i7-3517U).


Using this library

Just add the following to your Cargo.toml:

primal = "0.2"


“Indexing” Primes

Let’s find the 10001st prime. The easiest way is to enumerate the primes, and find the 10001st:

// (.nth is zero indexed.)
let p = primal::Primes::all().nth(10001 - 1).unwrap();
println!("The 10001st prime is {}", p); // 104743

This takes around 400 microseconds on my computer, which seems nice and quick, but, Primes is flexible at the cost of performance: we can make it faster. The StreamingSieve type offers a specialised nth_prime function:

let p = primal::StreamingSieve::nth_prime(10001);
println!("The 10001st prime is {}", p); // 104743

This runs in only 10 microseconds! StreamingSieve is extremely efficient and uses very little memory. It is the best way to solve this task with primal.

Since that was so easy, let’s now make the problem bigger and harder: find the sum of the 100,000th, 200,000th, 300,000th, …, 10,000,000th primes (100 in total).

We could call StreamingSieve::nth_prime repeatedly:

// the primes we want to find
let ns = (1..100 + 1).map(|x| x * 100_000).collect::<Vec<_>>();

// search and sum them up
let sum = ns.iter()
            .map(|n| primal::StreamingSieve::nth_prime(*n))
            .fold(0, |a, b| a + b);
println!("the sum is {}", sum);

This takes around 1.6s seconds to print the sum is 8795091674; not so speedy. Each call to nth_prime is individually fast (400 microseconds for 100,000 to 40 milliseconds for 10,000,000) but they add up to something bad. Every one is starting from the start and redoing work that previous calls have done… wouldn’t it be nice if we could just do the computation for 10,000,000 and reuse that for the smaller ones?

The Sieve type is a wrapper around StreamingSieve that caches information, allowing repeated queries to be answered efficiently.

There’s one hitch: Sieve requires a limit to know how far to sieve: we need some way to find an upper bound to be guaranteed to be at least as large as all our primes. We could guess that, say, 1010 will be large enough and use that, but that’s a huge overestimate (spoilers: the 10,000,000th prime is around 2×108). We could also try filtering with exponentially larger upper bounds until we find one that works (e.g. doubling each time), or, we could just take a shortcut and use deeper mathematics via estimate_nth_prime.

// the primes we want to find
let ns = (1..100 + 1).map(|x| x * 100_000).collect::<Vec<_>>();

// find our upper bound
let (_lo, hi) = primal::estimate_nth_prime(10_000_000);

// find the primes up to this upper bound
let sieve = primal::Sieve::new(hi as usize);

// now we can efficiently sum them up
let sum = ns.iter()
            .map(|n| sieve.nth_prime(*n))
            .fold(0u64, |a, b| a + b as u64);
println!("the sum is {}", sum);

This takes around 40 milliseconds, and gives the same output: much better!

(By the way, the version using 1010 as the bound instead of the more accurate estimate still only takes ~3 seconds.)

Counting Primes

Another problem: count the number of primes below 1 million. This is evaluating the prime-counting function π, i.e. π(106).

As above, there’s a few ways to attack this: the iterator, and the sieves.

const LIMIT: usize = 1_000_000;

// iterator
let count = primal::Primes::all().take_while(|p| *p < LIMIT).count();
println!("there are {} primes below 1 million", count); // 78498

// sieves
let sieve = primal::Sieve::new(LIMIT);
let count = sieve.prime_pi(LIMIT);
println!("there are {} primes below 1 million", count);

let count = primal::StreamingSieve::prime_pi(LIMIT);
println!("there are {} primes below 1 million", count);

StreamingSieve is fastest (380 microseconds) followed by Sieve (400) with Primes bringing up the rear at 1300 microseconds. Of course, repeated queries will be faster with Sieve than with StreamingSieve, but that flexibility comes at the cost of extra memory use.

If an approximation is all that is required, estimate_prime_pi provides close upper and lower bounds:

let (lo, hi) = primal::estimate_prime_pi(1_000_000);
println!("there are between {} and {} primes below 1 million", lo, hi);
// 78380, 78573

Searching Primes

Now for something where Primes might be useful: find the first prime where the binary expansion (not including trailing zeros) ends like 00..001 with at least 27 zeros. This condition is checked by:

fn check(p: usize) -> bool {
    p > 1 && (p / 2).trailing_zeros() >= 27

I have no idea how large the prime might be: I know it’s guaranteed to be at least 227 + 1 + 1, but not an upper limit.

The Primes iterator works perfectly for this:

let p = primal::Primes::all().find(|p| check(*p)).unwrap();
println!("the prime is {}", p);

It takes about 3.1 seconds for my computer to spit out 3,221,225,473.

Using a sieve is a little trickier: one approach is to start with some estimated upper bound (like double the absolute lower bound), look for a valid prime. If one isn’t found, double the upper bound and start again. The primes_from method allows for saving a little bit of work: we can start iterating from an arbitrary point in the sequence, such as the lower bound.

let p;
let mut lower_bound = 1 << (27 + 1);
loop {
    // our upper bound is double the lower bound
    let sieve = primal::Sieve::new(lower_bound * 2);
    if let Some(p_) = sieve.primes_from(lower_bound).find(|p| check(*p)) {
        p = p_;
    lower_bound *= 2;
println!("the prime is {}", p);

This takes around 3.5 seconds to print the same number. Slower than the iterator!

I was just using this silly condition as an example of something that doesn’t have an obvious upper bound, rather than a problem that is hard to do fast. There’s a much faster way to tackle it, by inverting the problem: construct numbers that satisfy check, and check the primality of those.

The numbers that satisfy check are k * (1 << (27 + 1)) + 1 for k >= 1, so the only hard bit is testing primality. Fortunately, primal offers the is_prime function which is an efficient way to do primality tests, even of very large numbers.

let mut p = 0;
for k in 1.. {
    p = k * (1 << (27 + 1)) + 1;
    if primal::is_prime(p) { break }
println!("the prime is {}", p);

This takes 6 microseconds: more than 500,000× faster than the iterator!


An iterator over all primes.
A heavily optimised prime sieve.
An iterator over the primes stored in a Sieve instance.
A heavily optimised prime sieve.


Returns integers (y, k) such that x = y^k with k maximised (other than for x = 0, 1, in which case y = x, k = 1).
Return Some((p, k)) if x = p^k for some prime p and k >= 1 (that is, including when x is itself a prime).
Gives estimated bounds for pn, the nth prime number, 1-indexed (i.e. p1 = 2, p2 = 3).
Returns estimated bounds for π(n), the number of primes less than or equal to n.
Test if n is prime, using the deterministic version of the Miller-Rabin test.