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.

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
```

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

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

```
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)
```

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

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.