The Collatz Conjecture or 3x+1 problem can be summarized as follows:
Take any positive integer n. If n is even, divide n by 2 to get n / 2. If n is odd, multiply n by 3 and add 1 to get 3n + 1. Repeat the process indefinitely. The conjecture states that no matter which number you start with, you will always reach 1 eventually.
Given a number n, return the number of steps required to reach 1.
Starting with n = 12, the steps would be as follows:
Resulting in 9 steps. So for input n = 12, the return value would be 9.
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:
To include color from the command line:
ruby -r minitest/pride collatz_conjecture_test.rb
An unsolved problem in mathematics named after mathematician Lothar Collatz https://en.wikipedia.org/wiki/3x_%2B_1_problem
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
require 'minitest/autorun' require_relative 'collatz_conjecture' # Common test data version: 1.2.1 d94e348 class CollatzConjectureTest < Minitest::Test def test_zero_steps_for_one # skip assert_equal 0, CollatzConjecture.steps(1) end def test_divide_if_even skip assert_equal 4, CollatzConjecture.steps(16) end def test_even_and_odd_steps skip assert_equal 9, CollatzConjecture.steps(12) end def test_large_number_of_even_and_odd_steps skip assert_equal 152, CollatzConjecture.steps(1_000_000) end def test_zero_is_an_error skip assert_raises(ArgumentError) do CollatzConjecture.steps(0) end end def test_negative_value_is_an_error skip assert_raises(ArgumentError) do CollatzConjecture.steps(-15) end end end
class CollatzConjecture def self.steps(n) raise ArgumentError if n < 1 return 0 if n == 1 return steps(n / 2) + 1 if n.even? return steps(n * 3 + 1) + 1 if n.odd? 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.