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Ric0chet's solution

to Collatz Conjecture in the Scala Track

Published at Sep 20 2019 · 0 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.

Examples

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

  1. 12
  2. 6
  3. 3
  4. 10
  5. 5
  6. 16
  7. 8
  8. 4
  9. 2
  10. 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.

Source

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

Submitting Incomplete Solutions

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

CollatzConjectureTest.scala

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
//  09-19-19

object CollatzConjecture {
  // 28 ms -- FASTEST using tail recursion
  @tailrec
  def steps(start: Int, count: Int = 0): Option[Int] = start match {
    case i if i < 1 => None
    case 1 => Some(count)
    case _ => steps(if (start % 2 == 0) start / 2 else start * 3 + 1,
      count + 1)
  }

  // 38 ms -- FAST using recursion
  def steps2(start: Int, count: Int = 0): Option[Int] = start match {
    case i if i < 1 => None
    case 1 => Some(count)
    case _ => steps2(if (start % 2 == 0) start / 2 else start * 3 + 1,
      count + 1)
  }

  // 69 ms -- SLOWER Lambda using (lazy) stream generator
  def steps3(start: Int): Option[Int] =
    if (start < 1) None
    else Some(Stream.iterate(start) { i =>
      if (i % 2 == 0) i / 2 else i * 3 + 1
    }.indexOf(1))

  // 67 ms -- SLOWER using lazy val to generate stream on-demand
  def steps4(start: Int): Option[Int] = {
    if (start < 1) return None

    lazy val st: Stream[Int] = start #:: st.map { i =>
      if (i % 2 == 0) i / 2 else i * 3 + 1 }

    Some(st.indexOf(1))
    // also works:  Some(st.takeWhile(_ > 1).size)
  }

  // 38 ms -- FAST using brute-force while
  def steps5(start: Int): Option[Int] = {
    if (start < 1) return None
    var num = start
    var count = 0

    while (num > 1) {
      num = if (num % 2 == 0) num / 2 else num * 3 + 1
      count += 1
    }
    Some(count)
  }

}

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Ric0chet's Reflection

Includes five solutions for benchmark comparison: 2 recursive, 2 streams, 1 brute-force. The first two methods show that tail recursion is faster than standard (and also faster than brute-force, at least for this data-set).

The stream solutions began as my attempt at an on-demand sequence generator, which rapidly spiraled down the API rabbit hole!