Given a number n, determine what the nth prime is.
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
If your language provides methods in the standard library to deal with prime numbers, pretend they don't exist and implement them yourself.
Execute the tests with:
$ elixir nth_prime_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
# @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.
A variation on Problem 7 at Project Euler http://projecteuler.net/problem=7
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("nth_prime.exs", __DIR__) end ExUnit.start() ExUnit.configure(exclude: :pending, trace: true) defmodule NthPrimeTest do use ExUnit.Case # @tag :pending test "first prime" do assert Prime.nth(1) == 2 end @tag :pending test "second prime" do assert Prime.nth(2) == 3 end @tag :pending test "sixth prime" do assert Prime.nth(6) == 13 end @tag :pending test "100th prime" do assert Prime.nth(100) == 541 end @tag :pending test "weird case" do catch_error(Prime.nth(0)) end end
defmodule Prime do @doc """ Generates the nth prime. """ @spec nth(non_neg_integer) :: non_neg_integer # we COULD handle 0 with this implementation, but test says raise error def nth(count) when count <= 0, do: raise ArgumentError def nth(count) do List.last(primes_list(count, , 2)) end # Tradeoffs: # # - Could have not bothered caching the primes we've found so far, and just # counted them... but then we'd have to check ALL numbers up to the limit to # see if the current candidate is a multiple, not just the primes, which are # a small subset. # # - Could have PREpended to the "primes so far" list, and taken the FIRST one # instead of the last in nth, and reversed before the take_while... but even # though appending and taking the last are far less efficient, the former # happens only on primes and the latter happens only once, but they let us # skip the reversal at *every* step. # # - Could have gotten the numbers as a Stream. That was in fact what I first # tried. But it made other things more complex, assuming I wanted to still # cache the "primes so far" list. def primes_list(how_many, primes_so_far, candidate) do cond do Enum.count(primes_so_far) == how_many -> primes_so_far any_factors(primes_so_far, candidate) -> primes_list(how_many, primes_so_far, candidate + 1) true -> primes_list(how_many, primes_so_far ++ [candidate], candidate + 1) end end defp any_factors(primes_so_far, candidate) do primes_so_far |> Enum.take_while(&(&1 <= :math.sqrt(candidate))) |> Enum.any?(&(is_multiple?(&1, candidate))) end defp is_multiple?(factor, multiple) do rem(multiple, factor) == 0 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.