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Ric0chet's solution

to Prime Factors in the Scala Track

Published at Oct 03 2019 · 1 comment
Instructions
Test suite
Solution

Compute the prime factors of a given natural number.

A prime number is only evenly divisible by itself and 1.

Note that 1 is not a prime number.

Example

What are the prime factors of 60?

  • Our first divisor is 2. 2 goes into 60, leaving 30.
  • 2 goes into 30, leaving 15.
    • 2 doesn't go cleanly into 15. So let's move on to our next divisor, 3.
  • 3 goes cleanly into 15, leaving 5.
    • 3 does not go cleanly into 5. The next possible factor is 4.
    • 4 does not go cleanly into 5. The next possible factor is 5.
  • 5 does go cleanly into 5.
  • We're left only with 1, so now, we're done.

Our successful divisors in that computation represent the list of prime factors of 60: 2, 2, 3, and 5.

You can check this yourself:

  • 2 * 2 * 3 * 5
  • = 4 * 15
  • = 60
  • Success!

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.

Please see the learning and installation pages if you need any help.

Source

The Prime Factors Kata by Uncle Bob http://butunclebob.com/ArticleS.UncleBob.ThePrimeFactorsKata

Submitting Incomplete Solutions

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

PrimefactorsTest.scala

import org.scalatest.{Matchers, FunSuite}

/** @version 1.1.0 */
class PrimeFactorsTest extends FunSuite with Matchers {

  test("no factors") {
    PrimeFactors.factors(1) should be(List())
  }

  test("prime number") {
    pending
    PrimeFactors.factors(2) should be(List(2))
  }

  test("square of a prime") {
    pending
    PrimeFactors.factors(9) should be(List(3, 3))
  }

  test("cube of a prime") {
    pending
    PrimeFactors.factors(8) should be(List(2, 2, 2))
  }

  test("product of primes and non-primes") {
    pending
    PrimeFactors.factors(12) should be(List(2, 2, 3))
  }

  test("product of primes") {
    pending
    PrimeFactors.factors(901255) should be(List(5, 17, 23, 461))
  }

  test("factors include a large prime") {
    pending
    PrimeFactors.factors(93819012551l) should be(List(11, 9539, 894119))
  }
}
import scala.collection.mutable.ListBuffer
//  10-03-19

object PrimeFactors {
  def factors(num: Long): List[Long] = findPrimes(num)
  
  // 41 ms -- FASTEST recursion matching divisor
  //          (div :: acc & reverse  is faster than:  acc += div)
  @annotation.tailrec
  private def findPrimes(num: Long,
                         div: Long = 2,
                         acc: List[Long] = List()): List[Long] =
    div match {
      case d if num % d == 0L =>
        findPrimes(num / div, div, div :: acc)
      case d if d >= num => acc.reverse
      case _ => findPrimes(num, div + 1, acc)
    }

  // 57 ms -- SLOWER recursion matching number
  @annotation.tailrec
  private def findPrimes2(num: Long,
                          div: Long = 2,
                          acc: List[Long] = List()): List[Long] =
    num match {
      case 1 => acc.reverse
      case n if n % div == 0L =>
        findPrimes2(num / div, div, div :: acc)
      case _ => findPrimes2(num, div + 1, acc)
    }

  // 43 ms -- FAST brute-force using match case
  private def findPrimes2(num: Long): List[Long] = {
    if (num < 2) return List()

    var facts = ListBuffer[Long]()
    var n = num
    var fact = 2L

    while (fact * fact <= n) fact match {  // magic shortcut
      case f if n % f == 0L =>
        facts += fact
        n /= fact
      case f if f == 2L => fact += 1
      case _ => fact += 2       // odds, only
    }
    if (n != 1L) facts += n     // prime remainder

    facts.toList
  }

  // 43 ms -- FAST brute-force using nested while loops
  private def findPrimes3(num: Long): List[Long] = {
    if (num < 2) return List()

    var facts = ListBuffer[Long]()
    var n = num
    var fact = 2L

    while (fact * fact <= n) {  // magic shortcut
      while (n % fact == 0L) {
        facts += fact
        n /= fact
      }
      fact += 1                 // odds & evens
    }
    if (n != 1L) facts += n     // prime remainder

    facts.toList
  }
}

Community comments

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Avatar of Ric0chet

I also proved to my own satisfaction that using: div :: acc (in L19, L32) plus acc.reverse (in L20, L30) has less overhead than: acc += div & plain acc

Ric0chet's Reflection

Includes four solutions for benchmark comparison (two recursive, two brute-force). The first recursive solution (increasing divisor) has the most efficient algorithm, slightly better than those of the (tied) brute-force methods. But none of the gaps between their benchmarks are wide.