Published at Jul 13 2018
·
2 comments

Instructions

Test suite

Solution

The Collatz Conjecture or 3x+1 problem can be summarized as follows:

Take any positive integer n. If n is even, divide n by 2 to get n / 2. If n is odd, multiply n by 3 and add 1 to get 3n + 1. Repeat the process indefinitely. The conjecture states that no matter which number you start with, you will always reach 1 eventually.

Given a number n, return the number of steps required to reach 1.

Starting with n = 12, the steps would be as follows:

- 12
- 6
- 3
- 10
- 5
- 16
- 8
- 4
- 2
- 1

Resulting in 9 steps. So for input n = 12, the return value would be 9.

The Scala exercises assume an SBT project scheme. The exercise solution source should be placed within the exercise directory/src/main/scala. The exercise unit tests can be found within the exercise directory/src/test/scala.

To run the tests simply run the command `sbt test`

in the exercise directory.

For more detailed info about the Scala track see the help page.

An unsolved problem in mathematics named after mathematician Lothar Collatz https://en.wikipedia.org/wiki/3x_%2B_1_problem

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
import org.scalatest.{Matchers, FunSuite}
/** @version 1.2.0 */
class CollatzConjectureTest extends FunSuite with Matchers {
test("zero steps for one") {
CollatzConjecture.steps(1) should be (Some(0))
}
test("divide if even") {
pending
CollatzConjecture.steps(16) should be (Some(4))
}
test("even and odd steps") {
pending
CollatzConjecture.steps(12) should be (Some(9))
}
test("Large number of even and odd steps") {
pending
CollatzConjecture.steps(1000000) should be (Some(152))
}
test("zero is an error") {
pending
CollatzConjecture.steps(0) should be (None)
}
test("negative value is an error") {
pending
CollatzConjecture.steps(-15) should be (None)
}
}
```

```
import scala.annotation.tailrec
object CollatzConjecture {
def steps(n : Int) : Option[Int] = {
@tailrec
def accumulator(next:Int, tally : Int) : Option[Int] = {
next match {
case 1 => Some(tally)
case x if x < 1 => None
case x if x % 2 == 0 => accumulator(x/2, tally + 1)
case x if x % 2 != 0 => accumulator((x * 3) + 1, tally + 1)
}
}
accumulator(n, 0)
}
}
```

A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

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## Community comments

Like the nested function. I wrote an overloaded one instead. This is nicer.

@steffenalbrecht commented:

Like the nested function. I wrote an overloaded one instead. This is nicer.

It's all good Steffen! Scala is very flexible - a lot of different ways to do the same thing. Kev