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) {
//     //     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 
//    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) = ...
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
}


}