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Scala | Sieve of Eratosthenes
  • Last Updated : 26 Aug, 2019

Eratosthenes of Cyrene was a Greek mathematician, who discovered an amazing algorithm to find prime numbers.

This article performs this algorithm in Scala.

Step 1 : Creating an Int Stream




def numberStream(n: Int): 
    Stream[Int] = Stream.from(n)
      
println(numberStream(10))

Output of above step is Stream(10, ?).

Step 2 : Sieve of Eratosthenes function






// Defining Sieve of Eratosthenes
def sieve_of_Eratosthenes(stream: Stream[Int]): 
    Stream[Int] = stream.head #:: sieve_of_Eratosthenes(
        (stream.tail) 
        filter (x => x % stream.head != 0)
        )
          
println(sieve_of_Eratosthenes(numberStream(10)))

Output of above step is Stream(10, ?).

Step 3 : Extracting “N” number of primes




val no_of_primes = sieve_of_Eratosthenes(numberStream(2))
  
// Selecting number of primes
println(no_of_primes)
(no_of_primes take 7) foreach { 
    println(_
    }

Output of above step is Stream(2, ?)
2
3
5
7
11
13
17
.

Below is the complete program




def numberStream(n: Int): 
    Stream[Int] = Stream.from(n)
       
println(numberStream(10))
  
// Defining Sieve of Eratosthenes
def sieve_of_Eratosthenes(stream: Stream[Int]): 
    Stream[Int] = stream.head #:: sieve_of_Eratosthenes(
        (stream.tail) 
        filter (x => x % stream.head!= 0)
        )
           
println(sieve_of_Eratosthenes(numberStream(10)))
  
  
val no_of_primes = sieve_of_Eratosthenes(numberStream(2))
  
// Selecting number of primes 
println(no_of_primes)
(no_of_primes take 7) foreach { 
    println(_
    }

Output:

Stream(10, ?)
Stream(10, ?)
Stream(2, ?)
2
3
5
7
11
13
17

Insights from the code

  • Using stream.form(), a stream is created which is generating successive numbers. And this number starts off from the argument.
  • A number stream is given to the “sieve_of_Eratosthenes” method. This method by filtering out the elements, lazily generates the successive elements.

Below is the complete working code with explanation:

Working: abc() method inserts the debug statement in the filter() method. If an element is not evenly divisible by the head, the stream treats it as a good element. The code prints it and return true. Otherwise the filtered out sequence is printed and finally the stream is returned.
Some modification is done in sieve_of_Eratosthenes method so as to use the stream creation – abc() method. Elements are taken out from the recursive stream and is printed.




object Sieve extends App {
    def abc(s: Stream[Int], head: Int) =
        val r = s filter {
            x => {
                if (x % head != 0) {
                    println()
                    println(s"${x} is not evenly divisible by ${head}")
                      
                    true
                    
                else {
                    println()
                    println(s"${x} is evenly divisible by ${head}. So Discard ${x}")
                false
                    }
                }
            }
        r
    }
    def numberStream(g: Int): Stream[Int] = Stream.from(g)
      
    def sieve_of_Eratosthenes(stream: Stream[Int]): 
    Stream[Int] = stream.head #:: sieve_of_Eratosthenes(
        abc(stream.tail, stream.head))
      
    val no_of_primes = sieve_of_Eratosthenes(numberStream(2))
    (no_of_primes take 7) foreach {
        println(_)
    }
}

Output :

2

3 is not evenly divisible by 2
3

4 is evenly divisible by 2. So Discard 4

5 is not evenly divisible by 2

5 is not evenly divisible by 3
5

6 is evenly divisible by 2. So Discard 6

7 is not evenly divisible by 2

7 is not evenly divisible by 3

7 is not evenly divisible by 5
7

8 is evenly divisible by 2. So Discard 8

9 is not evenly divisible by 2

9 is evenly divisible by 3. So Discard 9

10 is evenly divisible by 2. So Discard 10

11 is not evenly divisible by 2

11 is not evenly divisible by 3

11 is not evenly divisible by 5

11 is not evenly divisible by 7
11

12 is evenly divisible by 2. So Discard 12

13 is not evenly divisible by 2

13 is not evenly divisible by 3

13 is not evenly divisible by 5

13 is not evenly divisible by 7

13 is not evenly divisible by 11
13

14 is evenly divisible by 2. So Discard 14

15 is not evenly divisible by 2

15 is evenly divisible by 3. So Discard 15

16 is evenly divisible by 2. So Discard 16

17 is not evenly divisible by 2

17 is not evenly divisible by 3

17 is not evenly divisible by 5

17 is not evenly divisible by 7

17 is not evenly divisible by 11

17 is not evenly divisible by 13
17
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