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.
For installation and learning resources, refer to the exercism help 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 nth_prime_test.rb
To include color from the command line:
ruby -r minitest/pride nth_prime_test.rb
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.
require 'minitest/autorun'
require_relative 'nth_prime'
# Common test data version: 1.0.0 016d65b
class NthPrimeTest < Minitest::Test
def test_first_prime
# skip
assert_equal 2, Prime.nth(1)
end
def test_second_prime
skip
assert_equal 3, Prime.nth(2)
end
def test_sixth_prime
skip
assert_equal 13, Prime.nth(6)
end
def test_big_prime
skip
assert_equal 104743, Prime.nth(10001)
end
def test_there_is_no_zeroth_prime
skip
assert_raises(ArgumentError) { Prime.nth(0) }
end
# Problems in exercism evolve over time, as we find better ways to ask
# questions.
# The version number refers to the version of the problem you solved,
# not your solution.
#
# Define a constant named VERSION inside of the top level BookKeeping
# module, which may be placed near the end of your file.
#
# In your file, it will look like this:
#
# module BookKeeping
# VERSION = 1 # Where the version number matches the one in the test.
# end
#
# If you are curious, read more about constants on RubyDoc:
# http://ruby-doc.org/docs/ruby-doc-bundle/UsersGuide/rg/constants.html
def test_bookkeeping
skip
assert_equal 1, BookKeeping::VERSION
end
endclass Prime
def self.nth(n)
raise ArgumentError if n < 1
prime = 2
until n == 0
n -= 1 unless (2..Math.sqrt(prime)).any? { |i| prime%i == 0 }
prime += 1
end
return prime - 1
end
end
module BookKeeping
VERSION = 1
endA 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.
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Community comments
Nice solution!!
A couple of suggestions:
Ruby is an expressive language so there is no need to have a return at the end of a method. Instead of loop consider using an enumerator that would eliminate mutating the accumulator (prime) outside the block. The disadvantages of using loop is that you have to manage the initial states of the local variable (prime); and it decreases the readability of the code since the reader is left guessing whether the mutation was intentional or not. Have a look at reduce or inject enumerators.
@chrisZingel commented:
Nice solution!!
Thank you!!
A couple of suggestions:
Ruby is an expressive language so there is no need to have a return at the end of a method.
I know, but I use it sometimes, because I think that with the return statement the function becomes more expressive.
Instead of loop consider using an enumerator that would eliminate mutating the accumulator (prime) outside the block. The disadvantages of using loop is that you have to manage the initial states of the local variable (prime); and it decreases the readability of the code since the reader is left guessing whether the mutation was intentional or not. Have a look at reduce or inject enumerators.
Can you show me how would you do that with my code?
Hmmm eating some humble pie. It looked like you could easy replace it with reduce or inject enumerators. My apologies, it’s not easy at all. In fact, your solution only has one loop where I have resorted to having two loops. :-)
@chrisZingel commented:
Hmmm eating some humble pie. It looked like you could easy replace it with reduce or inject enumerators. My apologies, it’s not easy at all. In fact, your solution only has one loop where I have resorted to having two loops. :-)
It's okay :)