diff -r 755d165633ec -r 1f8005b4cdf6 testing1/collatz.scala --- a/testing1/collatz.scala Tue Nov 19 06:38:20 2019 +0000 +++ b/testing1/collatz.scala Fri Nov 22 16:41:45 2019 +0000 @@ -6,48 +6,18 @@ // If needed, you can use an auxiliary function that // performs the recursion. The function should expect // arguments in the range of 1 to 1 Million. +def stepsCounter(n: Long, s: Long) : Long = n match{ + case 1 => s + case n if(n%2==0) => stepsCounter(n/2,s+1) + case _ => stepsCounter(3*n+1, s+1) +} + +def collatz(n: Long) : Long = n match { + case n if(n>0) => stepsCounter(n,0) + case n if(n<=0) => stepsCounter(1,0) +} -// def collatz(n: Long) : Long = { -// if (n == 1) 1 //else -// // if (n % 2 == 0) { -// // collatz(n/2) -// // steps + 1 -// // } //else -// // if (n % 2 != 0) { -// // collatz((3 * n) + 1) -// // steps + 1 -// // } -// } - -// val steps: Long = 1 -// val lst = List() -// def collatz(n: Long) : Long = { -// if (n == 1) { steps + 1 } -// else if (n % 2 == 0) { -// collatz(n/2); -// } -// else { -// collatz((3 * n) + 1); -// } -// steps + 1 -// } -// collatz(6) - -def collatz(n: Long, list: List[Long] = List()): Long = { - if (n == 1) { - n :: list - list.size.toLong - } - else if (n % 2 == 0) { - collatz(n / 2, n :: list) - } - else { - collatz((3 * n) + 1, n :: list) - } -} - -val test = collatz(6) //(2) Complete the collatz_max function below. It should // calculate how many steps are needed for each number @@ -58,13 +28,10 @@ // the maximum number of steps and the second is the // corresponding number. -//def collatz_max(bnd: Long) : (Long, Long) = ... -def collatz_max(bnd: Long) : (Long, Long) = { - val stepsTable = for (n <- (1 to bnd.toInt).toList) yield (collatz(n), n.toLong) - //println(stepsTable) - stepsTable.max +def collatz_max(bnd: Long) : (Long, Long) = { + val allCollatz = for(i<-1L until bnd) yield collatz(i) + val pair = (allCollatz.max, (allCollatz.indexOf(allCollatz.max) +1).toLong) + pair } - } -