Scala performance - Sieve

Question

Right now, I am trying to learn Scala . I've started small, writing some simple algorithms . I've encountered some problems when I wanted to implement the Sieve algorithm from finding all all prime numbers lower than a certain threshold .

My implementation is:

import scala.math

object Sieve {

    // Returns all prime numbers until maxNum
    def getPrimes(maxNum : Int) = {
        def sieve(list: List[Int], stop : Int) : List[Int] = {
            list match {
                case Nil => Nil
                case h :: list if h <= stop => h :: sieve(list.filterNot(_ % h == 0), stop)
                case _ => list
            }
        }
        val stop : Int = math.sqrt(maxNum).toInt
        sieve((2 to maxNum).toList, stop)
    }

    def main(args: Array[String]) = {
        val ap = printf("%d ", (_:Int)); 

        // works
        getPrimes(1000).foreach(ap(_))

        // works 
        getPrimes(100000).foreach(ap(_))

        // out of memory
        getPrimes(1000000).foreach(ap(_))

    }
}

Unfortunately it fails when I want to computer all the prime numbers smaller than 1000000 (1 million) . I am receiving OutOfMemory .

Do you have any idea on how to optimize the code, or how can I implement this algorithm in a more elegant fashion .

PS: I've done something very similar in Haskell, and there I didn't encountered any issues .

Answer 1

The code in question is not tail recursive, so Scala cannot optimize the recursion away. Also, Haskell is non-strict by default, so you can't hardly compare it to Scala. For instance, whereas Haskell benefits from foldRight, Scala benefits from foldLeft.

There are many Scala implementations of Sieve of Eratosthenes, including some in Stack Overflow. For instance:

(n: Int) => (2 to n) |> (r => r.foldLeft(r.toSet)((ps, x) => if (ps(x)) ps -- (x * x to n by x) else ps))

Answer 2

I would go with an infinite Stream. Using a lazy data structure allows to code pretty much like in Haskell. It reads automatically more "declarative" than the code you wrote.

import Stream._

val primes = 2 #:: sieve(3)

def sieve(n: Int) : Stream[Int] =
      if (primes.takeWhile(p => p*p <= n).exists(n % _ == 0)) sieve(n + 2)
      else n #:: sieve(n + 2)

def getPrimes(maxNum : Int) = primes.takeWhile(_ < maxNum)

Obviously, this isn't the most performant approach. Read The Genuine Sieve of Eratosthenes for a good explanation (it's Haskell, but not too difficult). For real big ranges you should consider the Sieve of Atkin.

Answer 3

The following answer is about a 100 times faster than the "one-liner" answer using a Set (and the results don't need sorting to ascending order) and is more of a functional form than the other answer using an array although it uses a mutable BitSet as a sieving array:

object SoE {
  def makeSoE_Primes(top: Int): Iterator[Int] = {
    val topndx = (top - 3) / 2
    val nonprms = new scala.collection.mutable.BitSet(topndx + 1)
    def cullp(i: Int) = {
      import scala.annotation.tailrec; val p = i + i + 3
      @tailrec def cull(c: Int): Unit = if (c <= topndx) { nonprms += c; cull(c + p) }
      cull((p * p - 3) >>> 1)
    }
    (0 to (Math.sqrt(top).toInt - 3) >>> 1).filterNot { nonprms }.foreach { cullp }
    Iterator.single(2) ++ (0 to topndx).filterNot { nonprms }.map { i: Int => i + i + 3 }
  }
}

It can be tested by the following code:

object Main extends App {
  import SoE._
  val top_num = 10000000
  val strt = System.nanoTime()
  val count = makeSoE_Primes(top_num).size
  val end = System.nanoTime()
  println(s"Successfully completed without errors. [total ${(end - strt) / 1000000} ms]")
  println(f"Found $count primes up to $top_num" + ".")
  println("Using one large mutable1 BitSet and functional code.")
}

With the results from the the above as follows:

Successfully completed without errors. [total 294 ms]
Found 664579 primes up to 10000000.
Using one large mutable BitSet and functional code.

There is an overhead of about 40 milliseconds for even small sieve ranges, and there are various non-linear responses with increasing range as the size of the BitSet grows beyond the different CPU caches.

Answer 4

It looks like List isn't very effecient space wise. You can get an out of memory exception by doing something like this

1 to 2000000 toList

Answer 5

I "cheated" and used a mutable array. Didn't feel dirty at all.

  def primesSmallerThan(n: Int): List[Int] = {
    val nonprimes = Array.tabulate(n + 1)(i => i == 0 || i == 1)
    val primes = new collection.mutable.ListBuffer[Int]
    for (x <- nonprimes.indices if !nonprimes(x)) {
      primes += x
      for (y <- x * x until nonprimes.length by x if (x * x) > 0) {
        nonprimes(y) = true
      }
    }
    primes.toList
  }