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JDye's solution

to Prime Factors in the Ruby Track

Published at Dec 08 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!

For installation and learning resources, refer to the Ruby resources page.

For running the tests provided, you will need the Minitest gem. Open a terminal window and run the following command to install minitest:

gem install minitest

If you would like color output, you can require 'minitest/pride' in the test file, or note the alternative instruction, below, for running the test file.

Run the tests from the exercise directory using the following command:

ruby prime_factors_test.rb

To include color from the command line:

ruby -r minitest/pride prime_factors_test.rb

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.

prime_factors_test.rb

require 'minitest/autorun'
require_relative 'prime_factors'

class PrimeFactorsTest < Minitest::Test
  def test_1
    assert_equal [], PrimeFactors.of(1)
  end

  def test_2
    skip
    assert_equal [2], PrimeFactors.of(2)
  end

  def test_3
    skip
    assert_equal [3], PrimeFactors.of(3)
  end

  def test_4
    skip
    assert_equal [2, 2], PrimeFactors.of(4)
  end

  def test_6
    skip
    assert_equal [2, 3], PrimeFactors.of(6)
  end

  def test_8
    skip
    assert_equal [2, 2, 2], PrimeFactors.of(8)
  end

  def test_9
    skip
    assert_equal [3, 3], PrimeFactors.of(9)
  end

  def test_27
    skip
    assert_equal [3, 3, 3], PrimeFactors.of(27)
  end

  def test_625
    skip
    assert_equal [5, 5, 5, 5], PrimeFactors.of(625)
  end

  def test_901255
    skip
    assert_equal [5, 17, 23, 461], PrimeFactors.of(901_255)
  end

  def test_93819012551
    skip
    assert_equal [11, 9539, 894_119], PrimeFactors.of(93_819_012_551)
  end
end
require 'benchmark'

module PrimeFactors
  def self.of(number)
    factors = []
    i = 1
    n_plus = (6 * i) - 1
    n_minus = (6 * i) + 1

    while number % 2 == 0
      factors.append(2)
      number /= 2
    end

    while number % 3 == 0
      factors.append(3)
      number /= 3
    end


    while n_plus ** 2 <= number || n_minus ** 2 <= number
      if number % n_plus == 0
        factors.append(n_plus)
        number /= n_plus

      elsif number % n_minus == 0
        factors.append(n_minus)
        number /= n_minus

      else
        i += 1
        n_plus = (6 * i) + 1
        n_minus = (6 * i) - 1
      end
    end

    if number > 1
      factors.append(number)
    end
    factors
  end
end

# Benchmark.measure { 1000.times do PrimeFactors.of(600_851_475_143) end }
#       user        system       total         real
#      0.063000    0.000000     0.063000     (0.062008)

Community comments

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Avatar of JDye
Solution Author
commented 44 days ago

My 5th revision of this program. This decreased running time by 80%. The code isn't as clean as it could be and some further optimisations could be made but I'm just gonna say I'm done with it.

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