Check if number is prime in O(sqrt(n)) in Scala - scala

Question

When checking if n is a prime number in Scala, the most common solutions is concise one-liner which can be seen in almost all similar questions on SO
def isPrime1(n: Int) = (n > 1) && ((2 until n) forall (n % _ != 0))
Moving on, it's simple to rewrite it to check only odd numbers
def isPrime2(n: Int): Boolean = {
if (n < 2) return false
if (n == 2) return true
if (n % 2 == 0) false
else (3 until n by 2) forall (n % _ != 0)
}
However, to be more efficient I would like to combine checking only odds with counting up to sqrt(n), but without using Math.sqrt. So, as i < sqrt(n) <==> i * i < n, I would write C-like loop:
def isPrime3(n: Int): Boolean = {
if (n < 2) return false
if (n == 2) return true
if (n % 2 == 0) return false
var i = 3
while (i * i <= n) {
if (n % i == 0) return false
i += 2
}
true
}
The questions are:
1) How to achieve the last version in the first version nice Scala functional style?
2) How can I use Scala for to this? I thought of something similar to below, but don't know how.
for {
i <- 3 until n by 2
if i * i <= n
} { ??? }

Answer 1

Here is a method to verify if n is prime until sqrt(n) without using sqrt:
def isPrime3(n: Int): Boolean = {
if (n == 2) {
true
} else if (n < 2 || n % 2 == 0) {
false
} else {
Stream.from(3, 2).takeWhile(i => i * i < n + 1).forall(i => n % i != 0)
}
}
If you want to do it until n/2, which is also a possible optimization (worse than sqrt(n)), you can replace the last line with:
(3 to n/2 + 1 by 2).forall(i => n % i != 0)
If you prefer, you could also make a tail recursive version, something along these lines:
import scala.annotation.tailrec
def isPrime3(n: Int): Boolean = {
if (n == 2 || n == 3) {
true
} else if (n < 2 || n % 2 == 0) {
false
} else {
isPrime3Rec(n, 3)
}
}
#tailrec
def isPrime3Rec(n:Int, i: Int): Boolean = {
(n % i != 0) && ((i * i > n) || isPrime3Rec(n, i + 2))
}