# aweidner's solution

## to Collatz Conjecture in the Lua Track

Published at Aug 05 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)
if n < 1 then
error("Only positive numbers are allowed")
end

steps = 0
while n ~= 1 do
if n % 2 == 0 then
n = n / 2
else
n = 3 * n + 1
end
steps = steps + 1
end
return steps
end``````

### What can you learn from this solution?

A huge amount can be learned from reading other peopleโ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.

• What compromises have been made?