1 // Preliminary Part about the 3n+1 conjecture

     2 //============================================

     3 

     4 object CW6a {

     5 

     6 //(1) Complete the collatz function below. It should

     7 //    recursively calculate the number of steps needed 

     8 //    until the collatz series reaches the number 1.

     9 //    If needed, you can use an auxiliary function that

    10 //    performs the recursion. The function should expect

    11 //    arguments in the range of 1 to 1 Million.

    12 

    13 def collatz(n: Long) : Long = ???

    14 

    15 

    16 //(2) Complete the collatz_max function below. It should

    17 //    calculate how many steps are needed for each number 

    18 //    from 1 up to a bound and then calculate the maximum number of

    19 //    steps and the corresponding number that needs that many 

    20 //    steps. Again, you should expect bounds in the range of 1

    21 //    up to 1 Million. The first component of the pair is

    22 //    the maximum number of steps and the second is the 

    23 //    corresponding number.

    24 

    25 def collatz_max(bnd: Long) : (Long, Long) = ???

    26 

    27 //(3) Implement a function that calculates the last_odd

    28 //    number in a collatz series.  For this implement an

    29 //    is_pow_of_two function which tests whether a number 

    30 //    is a power of two. The function is_hard calculates 

    31 //    whether 3n + 1 is a power of two. Again you can

    32 //    assume the input ranges between 1 and 1 Million,

    33 //    and also assume that the input of last_odd will not 

    34 //    be a power of 2.

    35 

    36 def is_pow_of_two(n: Long) : Boolean = ???

    37 

    38 def is_hard(n: Long) : Boolean = ???

    39 

    40 def last_odd(n: Long) : Long = ???

    41 

    42 }

    43