// Part 1 about the 3n+1 conceture+
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//=================================+
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+
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//(1) Complete the collatz function below. It should+
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//    recursively calculate the number of steps needed +
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//    until the collatz series reaches the number 1.+
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//    If needed you can use an auxilary function that+
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//    performs the recursion. The function should expect+
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//    arguments in the range of 1 to 1 Million.+
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+
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def collatz(n: Long): List[Long] =+
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  if (n == 1) List(1) else+
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    if (n % 2 == 0) (n::collatz(n / 2)) else +
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      (n::collatz(3 * n + 1))+
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+
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+
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// an alternative that calculates the steps directly+
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def collatz1(n: Long): Long =+
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  if (n == 1) 1 else+
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    if (n % 2 == 0) (1 + collatz1(n / 2)) else +
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      (1 + collatz1(3 * n + 1))+
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+
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import scala.annotation.tailrec+
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+
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@tailrec+
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def collatz2(n: Long, acc: Long): Long =+
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  if (n == 1) acc else+
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    if (n % 2 == 0) collatz2(n / 2, acc + 1) else +
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      collatz2(3 * n + 1, acc + 1)+
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+
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+
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//(2)  Complete the collatz bound function below. It should+
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//     calculuate how many steps are needed for each number +
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//     from 1 upto a bound and return the maximum number of+
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//     steps and the corresponding number that needs that many +
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//     steps. You should expect bounds in the range of 1+
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//     upto 1 million. +
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+
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def collatz_max(bnd: Long): (Long, Long) = {+
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  (1L to bnd).view.map((i) => (collatz1(i), i)).maxBy(_._1)+
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}+
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+
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+
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// some testing harness+
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//val bnds = List(10, 100, 1000, 10000, 100000, 1000000)+
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val bnds = List(10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 2000000000)+
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+
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+
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+
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for (bnd <- bnds) {+
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  val (steps, max) = collatz_max(bnd)+
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  println(s"In the range of 1 - ${bnd} the number ${max} needs the maximum steps of ${steps}")+
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}+
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+
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+
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//val all = for (i <- (1 to 100000).toList) yield collatz1(i)+
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//println(all.sorted.reverse.take(10))+
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