--- 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
 }
 
-
 }
-