# Kotlin sieve using sequences

While learning about functional programming the simple sieve example was brought up in Haskell. I wanted to try a Kotlin implementation using sequences.

Are there any other quick wins without modifying the structure too much?

import kotlin.coroutines.experimental.buildSequence
import kotlin.system.measureTimeMillis

fun sieve(isPrime: Int, ints: Sequence<Int>): Boolean = with(ints.first()){
return when {
isPrime  < 2 -> false
isPrime == 2 -> true
isPrime == this -> true
isPrime.rem(2) == 0 -> false
//        isPrime.and(1) == 0 -> false // same way to check if number is even
this > isPrime -> false
else -> sieve(isPrime, ints.filter { n -> n.rem(this) != 0 })
}
}

fun main(args: Array<String>) {
val lazySeq = buildSequence { for (i in 3..Int.MAX_VALUE step 2) yield(i) }
println("Duration = ${measureTimeMillis { println("isPrime(4057) =${sieve(4057, lazySeq)}") }}ms")
//    (2..200).forEach { println("isPrime($it) =${sieve(it, lazySeq)}") }
}


Here's my take on this:

import kotlin.system.measureTimeMillis

fun sieve(xs: Sequence<Int>): Sequence<Int> = sequence {
val tail = xs.drop(1).filter { it % head != 0 }
for (i in sieve(tail))
yield(i)
}

val primes = sieve(generateSequence(2) { it + 1 })
fun isPrime(n: Int) = primes.contains(n)

val durationMs = measureTimeMillis {
println("isPrime(4057) = ${isPrime(4057)}") } println("Duration =$durationMs ms")

1. Start with the first element in the sequence - number 2
2. Filter out all numbers divisible by 2
3. Yield 2
4. Yield the rest recursively:
1. Start with the first element in the sequence - number 3
2. Filter out all numbers divisible by 3
3. Yield 3
4. Yield the rest recursively...