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Thursday Code Puzzler: Sieve of Eratosthenes

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It's been a long time since I posted a Puzzler, but as I was perusing Khan's CS courses this morning (which look really cool!) I came across this fascinating discourse on prime numbers: Sieve of Eratosthenes.

Your challenge today is simple - watch the video (via the link above or embed below) - and once the theory is discussed, do not look at the full solution. Write up a solution in your language of choice and post your answer below. You can also use a Gist or Pastebin link.

Published at DZone with permission of Raymond Camden, author and DZone MVB. (source)

(Note: Opinions expressed in this article and its replies are the opinions of their respective authors and not those of DZone, Inc.)


Vijay Nathani replied on Thu, 2013/01/17 - 1:17pm


def allPrimesTill (maximum) {
	def (primes, number, allNumbers) =[ [], 0, new LinkedList(2..maximum) ]
	while (allNumbers) {
		if (allNumbers.first() * allNumbers.first() >= maximum) { primes.addAll(allNumbers); break }
		primes.add(number = allNumbers.remove())
		for (int i=number * number; i <= maximum; i+=number)
assert allPrimesTill(10) == [2,3,5,7]

sun east replied on Fri, 2013/01/18 - 4:26am

scala> def primesToN(n: Int): Seq[Int] = {
     |   def sieve(xs: Seq[Int]): Seq[Int] = xs match {
     |     case Seq()   => Seq()
     |     case p +: rs => p +: sieve(rs diff (p*p to n by p))
     |   }
     |   sieve(2 to n)
     | }
primesToN: (n: Int)Seq[Int]

scala> primesToN(20)
res0: Seq[Int] = List(2, 3, 5, 7, 11, 13, 17, 19)


Jim Passmore replied on Wed, 2013/01/23 - 6:56pm


[2, 3, 5, 7]

from math import sqrt
def sieve(max):
  # create array
  primes = []
  for num in range(2,max+1):
  while primes[index] <= sqrt(max):
    for j in range(primes[index]*primes[index], max+1, primes[index]):
    index += 1
  return primes
if __name__ == '__main__':
  primes = sieve(10)
  print primes

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