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.
Follow the setup instructions for Crystal here:
More help installing can be found here:
Execute the tests with:
$ crystal spec
In each test suite all but the first test have been skipped.
Once you get a test passing, you can unskip the next one by changing
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 "spec" require "../src/*" describe "CollatzConjecture" do it "zero steps for one" do CollatzConjecture.steps(1).should eq(0) end pending "divide if even" do CollatzConjecture.steps(16).should eq(4) end pending "even and odd steps" do CollatzConjecture.steps(12).should eq(9) end pending "large number of even and odd steps" do CollatzConjecture.steps(1000000).should eq(152) end pending "zero is an error" do expect_raises(ArgumentError) do CollatzConjecture.steps(0) end end pending "negative value is an error" do expect_raises(ArgumentError) do CollatzConjecture.steps(-15) end end end
class CollatzConjecture def self.steps(n) raise ArgumentError.new("n has to be positive") if n <= 0 _steps(n, 0) end private def self._steps(n, count) return count if n == 1 if n.even? _steps(n / 2, count + 1) else _steps(n * 3 + 1, count + 1) 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.