// Part 1 about the 3n+1 conceture

//=================================

//(1) Complete the collatz function below. It should

//    recursively calculate the number of steps needed 

//    until the collatz series reaches the number 1.

//    If needed you can use an auxilary function that

//    performs the recursion. The function should expect

//    arguments in the range of 1 to 1 Million.

def collatz(n: Long): List[Long] =

  if (n == 1) List(1) else

    if (n % 2 == 0) (n::collatz(n / 2)) else 

      (n::collatz(3 * n + 1))

// an alternative that calculates the steps directly

def collatz1(n: Long): Long =

  if (n == 1) 1 else

    if (n % 2 == 0) (1 + collatz1(n / 2)) else 

      (1 + collatz1(3 * n + 1))

import scala.annotation.tailrec

@tailrec

def collatz2(n: Long, acc: Long): Long =

  if (n == 1) acc else

    if (n % 2 == 0) collatz2(n / 2, acc + 1) else 

      collatz2(3 * n + 1, acc + 1)

//(2)  Complete the collatz bound function below. It should

//     calculuate how many steps are needed for each number 

//     from 1 upto a bound and return the maximum number of

//     steps and the corresponding number that needs that many 

//     steps. You should expect bounds in the range of 1

//     upto 1 million. 

def collatz_max(bnd: Long): (Long, Long) = {

  (1L to bnd).view.map((i) => (collatz1(i), i)).maxBy(_._1)

}

// some testing harness

//val bnds = List(10, 100, 1000, 10000, 100000, 1000000)

val bnds = List(10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 2000000000)

for (bnd <- bnds) {

  val (steps, max) = collatz_max(bnd)

  println(s"In the range of 1 - ${bnd} the number ${max} needs the maximum steps of ${steps}")

}

//val all = for (i <- (1 to 100000).toList) yield collatz1(i)

//println(all.sorted.reverse.take(10))