--- a/pre_testing1/collatz.scala	Sat Oct 31 16:47:46 2020 +0000
+++ b/pre_testing1/collatz.scala	Sun Nov 01 01:21:31 2020 +0000
@@ -1,37 +1,51 @@
-object CW6a {
+// Basic Part about the 3n+1 conjecture
+//==================================
+
+// generate jar with
+//   > scala -d collatz.jar  collatz.scala
+
+object CW6a { // for purposes of generating a jar
 
-//(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 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 =
+  if (n == 1) 0 else
+    if (n % 2 == 0) 1 + collatz(n / 2) else 
+      1 + collatz(3 * n + 1)
+
+
+def collatz_max(bnd: Long): (Long, Long) = {
+  val all = for (i <- (1L to bnd)) yield (collatz(i), i)
+  all.maxBy(_._1)
 }
 
-def collatz(n: Long) : Long = n match {
-    case n if(n>0) => stepsCounter(n,0)
-    case n if(n<=0) => stepsCounter(1,0)
+//collatz_max(1000000)
+//collatz_max(10000000)
+//collatz_max(100000000)
+
+
+/* some test cases
+val bnds = List(10, 100, 1000, 10000, 100000, 1000000)
+
+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 ${steps}")
+}
+
+*/
+
+def is_pow(n: Long) : Boolean = (n & (n - 1)) == 0
+
+def is_hard(n: Long) : Boolean = is_pow(3 * n + 1)
+
+def last_odd(n: Long) : Long = 
+  if (is_hard(n)) n else
+    if (n % 2 == 0) last_odd(n / 2) else 
+      last_odd(3 * n + 1)
+
+
+//for (i <- 130 to 10000) println(s"$i: ${last_odd(i)}")
+for (i <- 1 to 100) println(s"$i: ${collatz(i)}")
+
 }
 
 
 
-//(2) Complete the collatz_max function below. It should
-//    calculate how many steps are needed for each number 
-//    from 1 up to a bound and then calculate the maximum number of
-//    steps and the corresponding number that needs that many 
-//    steps. Again, you should expect bounds in the range of 1
-//    up to 1 Million. The first component of the pair is
-//    the maximum number of steps and the second is the 
-//    corresponding number.
-
-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
-}
-
-}