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320
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object CW6a {
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167
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320
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//(1) Complete the collatz function below. It should
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// recursively calculate the number of steps needed
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// until the collatz series reaches the number 1.
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// If needed, you can use an auxiliary function that
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// performs the recursion. The function should expect
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// arguments in the range of 1 to 1 Million.
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281
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126
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320
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// def collatz(n: Long) : Long = {
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// if (n == 1) 1 //else
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// // if (n % 2 == 0) {
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// // collatz(n/2)
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// // steps + 1
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// // } //else
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// // if (n % 2 != 0) {
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// // collatz((3 * n) + 1)
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// // steps + 1
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// // }
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// }
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// val steps: Long = 1
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// val lst = List()
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// def collatz(n: Long) : Long = {
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// if (n == 1) { steps + 1 }
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// else if (n % 2 == 0) {
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// collatz(n/2);
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// }
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// else {
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// collatz((3 * n) + 1);
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// }
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// steps + 1
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// }
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// collatz(6)
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126
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320
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def collatz(n: Long, list: List[Long] = List()): Long = {
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if (n == 1) {
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n :: list
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list.size.toLong
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}
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else if (n % 2 == 0) {
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collatz(n / 2, n :: list)
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}
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else {
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collatz((3 * n) + 1, n :: list)
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}
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}
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val test = collatz(6)
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//(2) Complete the collatz_max function below. It should
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// calculate how many steps are needed for each number
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// from 1 up to a bound and then calculate the maximum number of
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// steps and the corresponding number that needs that many
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// steps. Again, you should expect bounds in the range of 1
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// up to 1 Million. The first component of the pair is
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// the maximum number of steps and the second is the
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// corresponding number.
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//def collatz_max(bnd: Long) : (Long, Long) = ...
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def collatz_max(bnd: Long) : (Long, Long) = {
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val stepsTable = for (n <- (1 to bnd.toInt).toList) yield (collatz(n), n.toLong)
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//println(stepsTable)
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stepsTable.max
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126
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}
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281
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127
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}
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126
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