Published at Jul 13 2018
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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!

Execute the tests with:

```
$ elixir prime_factors_test.exs
```

In the test suites, all but the first test have been skipped.

Once you get a test passing, you can unskip the next one by
commenting out the relevant `@tag :pending`

with a `#`

symbol.

For example:

```
# @tag :pending
test "shouting" do
assert Bob.hey("WATCH OUT!") == "Whoa, chill out!"
end
```

Or, you can enable all the tests by commenting out the
`ExUnit.configure`

line in the test suite.

```
# ExUnit.configure exclude: :pending, trace: true
```

For more detailed information about the Elixir track, please see the help page.

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.

```
if !System.get_env("EXERCISM_TEST_EXAMPLES") do
Code.load_file("prime_factors.exs", __DIR__)
end
ExUnit.start()
ExUnit.configure(exclude: :pending, trace: true)
defmodule PrimeFactorsTest do
use ExUnit.Case
# @tag :pending
test "1" do
assert PrimeFactors.factors_for(1) == []
end
@tag :pending
test "2" do
assert PrimeFactors.factors_for(2) == [2]
end
@tag :pending
test "3" do
assert PrimeFactors.factors_for(3) == [3]
end
@tag :pending
test "4" do
assert PrimeFactors.factors_for(4) == [2, 2]
end
@tag :pending
test "6" do
assert PrimeFactors.factors_for(6) == [2, 3]
end
@tag :pending
test "8" do
assert PrimeFactors.factors_for(8) == [2, 2, 2]
end
@tag :pending
test "9" do
assert PrimeFactors.factors_for(9) == [3, 3]
end
@tag :pending
test "27" do
assert PrimeFactors.factors_for(27) == [3, 3, 3]
end
@tag :pending
test "625" do
assert PrimeFactors.factors_for(625) == [5, 5, 5, 5]
end
@tag :pending
test "901255" do
assert PrimeFactors.factors_for(901_255) == [5, 17, 23, 461]
end
@tag :pending
test "93819012551" do
assert PrimeFactors.factors_for(93_819_012_551) == [11, 9539, 894_119]
end
@tag :pending
# @tag timeout: 2000
#
# The timeout tag above will set the below test to fail unless it completes
# in under two sconds. Uncomment it if you want to test the efficiency of your
# solution.
test "10000000055" do
assert PrimeFactors.factors_for(10_000_000_055) == [5, 2_000_000_011]
end
end
```

```
defmodule PrimeFactors do
@doc """
Compute the prime factors for 'number'.
The prime factors are prime numbers that when multiplied give the desired
number.
The prime factors of 'number' will be ordered lowest to highest.
"""
@spec factors_for(pos_integer) :: [pos_integer]
def factors_for(number) do
do_factors_for(2, number, trunc(:math.sqrt(number)), [], []) |> Enum.reverse
end
def do_factors_for(_, 1, _, _, acc), do: acc
def do_factors_for(candidate, number, limit, _, acc) when candidate > limit do
[number|acc]
end
def do_factors_for(candidate, number, limit, primes, acc, known_prime \\ false) do
next = if candidate > 2, do: candidate + 2, else: candidate + 1
if known_prime || is_prime?(candidate, trunc(:math.sqrt(candidate)), primes) do
# ++ is slow, but we should only have to do it rarely
primes = if known_prime, do: primes, else: primes ++ [candidate]
if rem(number, candidate) == 0 do
new_number = trunc(number/candidate)
do_factors_for(candidate,
new_number,
trunc(:math.sqrt(new_number)),
primes,
[candidate|acc], true)
else
do_factors_for(next, number, limit, primes, acc)
end
else
do_factors_for(next, number, limit, primes, acc)
end
end
def is_prime?(_, limit, [prime|_]) when prime > limit, do: true
def is_prime?(_, _ , [] ) , do: true
def is_prime?(candidate, limit, [prime|more_primes]) do
case rem(candidate, prime) do
0 -> false
_ -> is_prime?(candidate, limit, more_primes)
end
end
end
```

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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