diff -r 6e93040e3378 -r cdfa6a293453 core_templates1/collatz.scala
--- a/core_templates1/collatz.scala	Sat Oct 08 00:30:51 2022 +0100
+++ b/core_templates1/collatz.scala	Tue Nov 01 15:03:48 2022 +0000
@@ -3,36 +3,18 @@
 
 object C1 {
 
-//(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.
+// ADD YOUR CODE BELOW
+//======================
 
+
+//(1) 
 def collatz(n: Long) : Long = ???
 
 
-//(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.
-
+//(2) 
 def collatz_max(bnd: Long) : (Long, Long) = ???
 
-//(3) Implement a function that calculates the last_odd
-//    number in a collatz series.  For this implement an
-//    is_pow_of_two function which tests whether a number 
-//    is a power of two. The function is_hard calculates 
-//    whether 3n + 1 is a power of two. Again you can
-//    assume the input ranges between 1 and 1 Million,
-//    and also assume that the input of last_odd will not 
-//    be a power of 2.
-
+//(3)
 def is_pow_of_two(n: Long) : Boolean = ???
 
 def is_hard(n: Long) : Boolean = ???