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R74's solution

to Collatz Conjecture in the Lua Track

Published at Apr 30 2020 · 0 comments
Instructions
Test suite
Solution

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.

Examples

Starting with n = 12, the steps would be as follows:

  1. 12
  2. 6
  3. 3
  4. 10
  5. 5
  6. 16
  7. 8
  8. 4
  9. 2
  10. 1

Resulting in 9 steps. So for input n = 12, the return value would be 9.

Running the tests

To run the tests, run the command busted from within the exercise directory.

Further information

For more detailed information about the Lua track, including how to get help if you're having trouble, please visit the exercism.io Lua language page.

Source

An unsolved problem in mathematics named after mathematician Lothar Collatz https://en.wikipedia.org/wiki/3x_%2B_1_problem

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

collatz-conjecture_spec.lua

local conjecture = require('collatz-conjecture')

describe('collatz-conjecture', function()
  it('zero steps for one', function()
    assert.are.equal(0, conjecture(1))
  end)

  it('divide if even', function()
    assert.are.equal(4, conjecture(16))
  end)

  it('even and odd steps', function()
    assert.are.equal(9, conjecture(12))
  end)

  it('large number of even and odd steps', function()
    assert.are.equal(152, conjecture(1000000))
  end)

  it('zero is an error', function()
    assert.has_error(
      function() conjecture(0) end,
      'Only positive numbers are allowed'
    )
  end)

  it('negative value is an error', function()
    assert.has_error(
      function() conjecture(-15) end,
      'Only positive numbers are allowed'
    )
  end)
end)
return function (n)
    local steps_num = 0
    while n ~= 1 do
        if n % 2 == 0 then
            n = n / 2
        else
            n = 3 * n + 1
        end
    steps_num = steps_num + 1
    end
    return steps_num
end

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