object CW6a {

//(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 collatz(n: Long) : Long =
    if ( n == 1) 1;
    else if (n % 2 == 0) 1 + collatz( n / 2);
    else 1 + collatz( n * 3 + 1);


//(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) =
     ((1.toLong to bnd).toList.map
        (n => collatz(n)).max ,
            (1.toLong to bnd).toList.map
                (n => collatz(n)).indexOf((1.toLong to bnd).toList.map
                    (n => collatz(n)).max) + 1);

//(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.
//idk
 def is_pow_of_two(n: Long) : Boolean =
    if ( n & ( n - 1) == 0) true;
    else false;

def is_hard(n: Long) : Boolean =
    if ( (3*n + 1) & 3*n == 0) true;
    else false;


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



}